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Aufgaben:Exercise 3.9: Circular Arc and Parabola

2022-04-09T15:17:09Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1106__Mod_A_3_8.png|right|frame|Locus curves in FM: Arc and Parabola]]
We now consider the frequency modulation of a cosine source signal
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t )$$
with amplitude  $A_{\rm N} = 1 \ \rm V$  and frequency $f_{\rm N} = 5 \ \rm kHz$.
*The modulation index (phase deviation) is  $η = 2.4$.
*The corresponding low-pass signal with normalized carrier amplitude  $(A_{\rm T} = 1)$ is:
:$$ s_{\rm TP}(t) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
*This represents an arc.  Within the period $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$  the following phase angles result:
:$$ \phi(0) = 0, \hspace{0.2cm}\phi(0.25 \cdot T_{\rm N}) = \eta, \hspace{0.2cm}\phi(0.5 \cdot T_{\rm N})= 0,\hspace{0.2cm}
\phi(0.75 \cdot T_{\rm N})= -\eta,\hspace{0.2cm}\phi(T_{\rm N})= 0.$$
*Theoretically, the channel bandwidth required to transmit this signal is infinite.

However, if the bandwidth is limited to  $B_{\rm K} = 25 \ \rm kHz$, for example, the equivalent low-pass signal of the received signal can be described as follows:
:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
In this case, the result is a parabolic locus curve
:$$ y^2 + a \cdot x + b = 0,$$
which will be analyzed in this task.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].
*During calculation, assume the following values of the Bessel function:
:$${\rm J}_0 (2.4) \approx 0, \hspace{0.2cm}{\rm J}_1 (2.4) = -{\rm J}_{-1} (2.4)\approx 0.52, \hspace{0.2cm}{\rm J}_2 (2.4) = {\rm J}_{-2} (2.4)\approx 0.43.$$

===Question===

<quiz display=simple>
{What is the modulator constant  $K_{\rm FM}$?
|type="{}"}
$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$

{Calculate the real part $x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$  of the equivalent low-pass signal and give its maximum and minimum.
|type="{}"}
$x_{\rm max} \ = \ $ { 0.86 3% }
$x_{\rm min} \ = \ $ { -0.89--0.83 }

{What is the maximum and minimum of the imaginary part  $y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?
|type="{}"}
$y_{\rm max} \ = \ $ { 1.04 3% }
$y_{\rm min} \ = \ $ { -1.07--1.01 }

{What are the phase values for all multiples of  $T_{\rm N}/2$?
|type="{}"}
$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm degrees$

{What is the maximum phase angle  $ϕ_{\rm max}$?  Interpret the result.
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm degrees$

{Show that the locus curve can be given in the form  $y^2 + a · x + b = 0$ .  Determine the parabolic parameters  $a$  and  $b$.
|type="{}"}
$a\ = \ $ { 0.629 3% }
$b\ = \ $ { 0.541 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  In the frequency modulation of a cosine signal, the modulation index is:
:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} = \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$

'''(2)'''  The equation given for the equivalent low-pass signal when  $γ = ω_{\rm N} · t$  and considering   ${\rm J}_{–1} = –{\rm J}_1$  and  ${\rm J}_{–2} = {\rm J}_2$, is written out as:
:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
*Thus, for the real part in general, including for   $η = 2.4$, that is,  ${\rm J}_0 = 0$  and  ${\rm J}_2 = 0.43$:
:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}
\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$

'''(3)'''  According to the result of question  '''(2)'''  we get an imginary part   $({\rm J}_1 = 0.52)$ of:
:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$

'''(4)'''  The imaginary part is zero at each of these time points and so is the phase function:
:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$
*This fact can also be seen from the sketch on the exercise page.

'''(5)'''  From the sketch it can already be seen that the phase angle reaches its maximum value at  $t = T_{\rm N}/4$ , for example.
*This can be calculated with  $y_{\rm max} = 1.04$  and  $x_{\rm min} = -0.86$  as follows:
:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$
*Without band-limiting, the phase angle here would be $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$ .
*Thus, the maximum deviation of the sink signal from the source signal occurs at time  $t = T_{\rm N}/4$ , for example.

[[File:P_ID1113__Mod_A_3_8_f.png|right|frame|Parabolic progression]]
'''(6)'''  Using  $γ = ω_{\rm N} · t$  and  $\cos(2γ) = 1 - 2 · \cos^2(γ)$  the real and imaginary parts can be written as:

:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$

*These equations can be rearranged:
:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$

$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$

*Thus, the parabolic parameters for  $ {\rm J}_0 = 0$ are:
:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$
*To double-check, we use   $y = 0$   ⇒   $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$
*Thus, the values at  $x = 0$  are     $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.9: Circular Arc and Parabola

2022-04-09T15:15:53Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1106__Mod_A_3_8.png|right|frame|Locus curves in FM: Arc and Parabola]]
We now consider the frequency modulation of a cosine source signal
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t )$$
with amplitude  $A_{\rm N} = 1 \ \rm V$  and frequency $f_{\rm N} = 5 \ \rm kHz$.
*The modulation index (phase deviation) is  $η = 2.4$.
*The corresponding low-pass signal with normalized carrier amplitude  $(A_{\rm T} = 1)$ is:
:$$ s_{\rm TP}(t) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
*This represents an arc.  Within the period $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$  the following phase angles result:
:$$ \phi(0) = 0, \hspace{0.2cm}\phi(0.25 \cdot T_{\rm N}) = \eta, \hspace{0.2cm}\phi(0.5 \cdot T_{\rm N})= 0,\hspace{0.2cm}
\phi(0.75 \cdot T_{\rm N})= -\eta,\hspace{0.2cm}\phi(T_{\rm N})= 0.$$
*Theoretically, the channel bandwidth required to transmit this signal is infinite.

However, if the bandwidth is limited to  $B_{\rm K} = 25 \ \rm kHz$, for example, the equivalent low-pass signal of the received signal can be described as follows:
:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
In this case, the result is a parabolic locus curve
:$$ y^2 + a \cdot x + b = 0,$$
which will be analyzed in this task.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].
*During calculation, assume the following values of the Bessel function:
:$${\rm J}_0 (2.4) \approx 0, \hspace{0.2cm}{\rm J}_1 (2.4) = -{\rm J}_{-1} (2.4)\approx 0.52, \hspace{0.2cm}{\rm J}_2 (2.4) = {\rm J}_{-2} (2.4)\approx 0.43.$$

===Question===

<quiz display=simple>
{What is the modulator constant  $K_{\rm FM}$?
|type="{}"}
$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$

{Calculate the real part $x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$  of the equivalent low-pass signal and give its maximum and minimum.
|type="{}"}
$x_{\rm max} \ = \ $ { 0.86 3% }
$x_{\rm min} \ = \ $ { -0.89--0.83 }

{What is the maximum and minimum of the imaginary part  $y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?
|type="{}"}
$y_{\rm max} \ = \ $ { 1.04 3% }
$y_{\rm min} \ = \ $ { -1.07--1.01 }

{What are the phase values for all multiples of  $T_{\rm N}/2$?
|type="{}"}
$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm degrees$

{What is the maximum phase angle  $ϕ_{\rm max}$?  Interpret the result.
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm degrees$

{Show that the locus curve can be given in the form  $y^2 + a · x + b = 0$ .  Determine the parabolic parameters  $a$  and  $b$.
|type="{}"}
$a\ = \ $ { 0.629 3% }
$b\ = \ $ { 0.541 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  In the frequency modulation of a cosine signal, the modulation index is:
:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} = \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$

'''(2)'''  The equation given for the equivalent low-pass signal when  $γ = ω_{\rm N} · t$  and considering   ${\rm J}_{–1} = –{\rm J}_1$  and  ${\rm J}_{–2} = {\rm J}_2$, is written out as:
:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
*Thus, for the real part in general, including for   $η = 2.4$, that is,  ${\rm J}_0 = 0$  and  ${\rm J}_2 = 0.43$:
:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}
\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$

'''(3)'''  According to the result of question  '''(2)'''  we get an imginary part   $({\rm J}_1 = 0.52)$ of:
:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$

'''(4)'''  The imaginary part is zero at each of these time points and so is the phase function:
:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$
*This fact can also be seen from the sketch on the exercise page.

'''(5)'''  From the sketch it can already be seen that the phase angle reaches its maximum value at  $t = T_{\rm N}/4$ , for example.
*This can be calculated with  $y_{\rm max} = 1.04$  and  $x_{\rm min} = -0.86$  as follows:
:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$
*Without band-limiting, the phase angle here would be $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$ .
*Thus, the maximum deviation of the sink signal from the source signal occurs at time  $t = T_{\rm N}/4$ , for example.

[[File:P_ID1113__Mod_A_3_8_f.png|right|frame|Parabolic progression]]
'''(6)'''  Using  $γ = ω_{\rm N} · t$  and  $\cos(2γ) = 1 - 2 · \cos^2(γ)$  the real and imaginary parts can be written as:

:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$

*These equations can be rearranegd:
:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$

$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$

*Thus, the parabolic parameters for  $ {\rm J}_0 = 0$ are:
:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$
*To double-check, we use   $y = 0$   ⇒   $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$
*Thus, the values at  $x = 0$  are     $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.7: Angular Modulation of a Harmonic Oscillation

2022-04-09T15:14:00Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1103__Mod_Z_3_6.png|right|frame|Demodulator for FM]]
The signal arriving at a receiver is:
:$$ r(t) = 3\,{\rm V} \cdot \cos \hspace{-0.05cm} \big[2 \pi \cdot 1\,{\rm MHz} \cdot t + 3 \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\big]\hspace{0.05cm}.$$
 $r(t)$  is an angle-modulated signal that was neither distorted nor influenced by noise during transmission.

The signals  $v_{\rm PM}(t)$  and  $v_{\rm FM}(t)$  result after ideal demodulation by means of
* a phase demodulator, given by the equation
:$$ v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) \hspace{0.05cm},\hspace{0.3cm} {K_{\rm PM}} = 2\,{\rm V}^{-1}\hspace{0.05cm},$$
* a frequency demodulator, consisting of a PM demodulator, a differentiator and a constant $K$.

In order for all signals to have equal units, this constant $K$ is dimensionally constrained.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].

===Questions===

<quiz display=simple>
{Which statements are definitely true?
|type="[]"}
+ There could be a PM modulation.
+ There could be a FM modulation.
- The message phase is definitely  $ϕ_{\rm N} = 0$.
+ The message phase is definitely  $f_{\rm N} = 10 \ \rm kHz$.

{Calculate the signal $v_{\rm PM}(t)$  after the phase demodulator.  What is the signal value at time  $t = 0$?
|type="{}"}
$v_{\rm PM}(t = 0) \ = \ $ { 1.5 3% } $\ \rm V$

{Calculate the signal $v_{\rm FM}(t)$. What is the message phase  $ϕ_{\rm N}$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { 90 3% } $\ \rm Grad$

{How should  $K$  be chosen so that the amplitude of  $v_{\rm FM}(t)$  is equal to  $1.5 \ \rm V$ ?
|type="{}"}
$K\ = \ $ { 6.28 3% } $\ \rm \cdot 10^4 \ 1/s$

{Which of the following statements is true for the FM-modulated signal?
|type="[]"}
+ The phase deviation is  $ϕ_{\rm max} = 3$.
+ The frequency deviation is  $Δf_{\rm A} = 30 \ \rm kHz$.
+ The instantaneous frequencies are between  $0.97\ \rm MHz$  and  $1.03 \ \rm MHz$ .
- If  $f_{\rm N} = 5 \ \rm kHz$ , the phase deviation would be unchanged.
+ If  $f_{\rm N} = 5 \ \rm kHz$  the frequency deviation would be unchanged.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answers 1, 2 and 4 are correct:
*From the equation for  $r(t)$  it can only be ascertained that it is an angle modulation,
*but not whether it is a phase modulation (PM) or a frequency modulation (FM).
*Based on the equation, it is clear that the message frequency is  $f_{\rm N} = 10 \ \rm kHz$ .
*The phase  $ϕ_{\rm N} = 0$  of the source signal would then only apply, if phase modulation were present.

'''(2)'''  With the modulator constant  $K_{\rm PM} = 2 \ \rm V^{–1}$  this is given by:
:$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) = \frac{3}{2\,{\rm V}^{-1}} \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
*At time   $t = 0$  it therefore holds that:
:$$v_{\rm PM}(t = 0) = {A_{\rm N}} \hspace{0.15cm}\underline {= 1.5\,{\rm V}}\hspace{0.05cm}.$$

'''(3)'''  The output signal   $v_{\rm FM}(t)$  of the FM demodulator – consisting of a PM–demodulator and differentiator – can be written as:
:$$v_{\rm FM}(t) = \frac{{\rm d}v_{\rm PM}(t)}{{\rm d}t} \cdot K = \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot (- \sin(2 \pi \cdot {f_{\rm N}} \cdot t))= \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot \cos(2 \pi \cdot {f_{\rm N}} \cdot t + 90^\circ)\hspace{0.05cm}.$$
*The message phase is thus   $ϕ_{\rm N} \hspace{0.15cm}\underline {= 90^\circ}$.

'''(4)'''  In this case, it must hold that:  
:$$ K ={2 \pi \cdot f_{\rm N}} \hspace{0.15cm}\underline { = 6.28 \cdot 10^{4} \,\,{1}/{ s}} \hspace{0.05cm}.$$

'''(5)''' Answers 1, 2, 3 and 5 are correct:
*The phase deviation is identical to the modulation index, which can be discerned from the equation given:
:$$\phi_{\rm max} = \eta = 3 = \frac{\Delta f_{\rm A}}{ f_{\rm N}} \hspace{0.05cm}.$$
*This leads to the frequency deviation  $Δf_{\rm A} = 3 · f_{\rm N} = 30 \ \rm kHz$.
*With a carrier frequency of   $f_{\rm T} = 1 \ \rm MHz$ , the instantaneous frequency  $f_{\rm T}(t)$  can only take values between  $1±0.03 \ \rm MHz$ .

'''Thus, the following statement is also valid:''':

At half the message frequency, the phase deviation  $η$ doubles, while the frequency deviation  $Δf_{\rm A}$ is unaffected:
:$$\eta = \frac{K_{\rm PM} \cdot A_{\rm N}}{ f_{\rm N}} = 6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} = \eta \cdot f_{\rm N} = 6 \cdot 5\,{\rm kHz} = 30\,{\rm kHz}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.7: Angular Modulation of a Harmonic Oscillation

2022-04-09T15:13:28Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1103__Mod_Z_3_6.png|right|frame|Demodulator for FM]]
The signal arriving at a receiver is:
:$$ r(t) = 3\,{\rm V} \cdot \cos \hspace{-0.05cm} \big[2 \pi \cdot 1\,{\rm MHz} \cdot t + 3 \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\big]\hspace{0.05cm}.$$
 $r(t)$  is an angle-modulated signal that was neither distorted nor influenced by noise during transmission.

The signals  $v_{\rm PM}(t)$  and  $v_{\rm FM}(t)$  result after ideal demodulation by means of
* a phase demodulator, given by the equation
:$$ v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) \hspace{0.05cm},\hspace{0.3cm} {K_{\rm PM}} = 2\,{\rm V}^{-1}\hspace{0.05cm},$$
* a frequency demodulator, consisting of a PM demodulator, a differentiator and a constant $K$.

In order for all signals to have equal units, this constant $K$ is dimensionally constrained.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].

===Questions===

<quiz display=simple>
{Which statements are definitely true?
|type="[]"}
+ There could be a PM modulation.
+ There could be a FM modulation.
- The message phase is definitely  $ϕ_{\rm N} = 0$.
+ The message phase is definitely  $f_{\rm N} = 10 \ \rm kHz$.

{Calculate the signal $v_{\rm PM}(t)$  after the phase demodulator.  What is the signal value at time  $t = 0$?
|type="{}"}
$v_{\rm PM}(t = 0) \ = \ $ { 1.5 3% } $\ \rm V$

{Calculate the signal $v_{\rm FM}(t)$. What is the message phase  $ϕ_{\rm N}$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { 90 3% } $\ \rm Grad$

{How should  $K$  be chosen so that the amplitude of  $v_{\rm FM}(t)$  is equal to  $1.5 \ \rm V$ ?
|type="{}"}
$K\ = \ $ { 6.28 3% } $\ \rm \cdot 10^4 \ 1/s$

{Which of the following statements is true for the FM-modulated signal?
|type="[]"}
+ The phase deviation is  $ϕ_{\rm max} = 3$.
+ The frequency deviation is  $Δf_{\rm A} = 30 \ \rm kHz$.
+ The instantaneous frequencies are between  $0.97\ \rm MHz$  and  $1.03 \ \rm MHz$ .
- If  $f_{\rm N} = 5 \ \rm kHz$ , the phase deviation would be unchanged.
+ If  $f_{\rm N} = 5 \ \rm kHz$  the frequency deviation would be unchanged.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answers 1, 2 and 4 are correct:
*From the eqution for  $r(t)$  it can only be acertained that it is an angle modulation,
*but not whether it is a phase modulation (PM) or a frequency modulation (FM).
*Based on the equation, it is clear that the message frequency is  $f_{\rm N} = 10 \ \rm kHz$ .
*The phase  $ϕ_{\rm N} = 0$  of the source signal would then only apply, if phase modulation were present.

'''(2)'''  With the modulator constant  $K_{\rm PM} = 2 \ \rm V^{–1}$  this is given by:
:$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) = \frac{3}{2\,{\rm V}^{-1}} \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
*At time   $t = 0$  it therefore holds that:
:$$v_{\rm PM}(t = 0) = {A_{\rm N}} \hspace{0.15cm}\underline {= 1.5\,{\rm V}}\hspace{0.05cm}.$$

'''(3)'''  The output signal   $v_{\rm FM}(t)$  of the FM demodulator – consisting of a PM–demodulator and differentiator – can be written as:
:$$v_{\rm FM}(t) = \frac{{\rm d}v_{\rm PM}(t)}{{\rm d}t} \cdot K = \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot (- \sin(2 \pi \cdot {f_{\rm N}} \cdot t))= \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot \cos(2 \pi \cdot {f_{\rm N}} \cdot t + 90^\circ)\hspace{0.05cm}.$$
*The message phase is thus   $ϕ_{\rm N} \hspace{0.15cm}\underline {= 90^\circ}$.

'''(4)'''  In this case, it must hold that:  
:$$ K ={2 \pi \cdot f_{\rm N}} \hspace{0.15cm}\underline { = 6.28 \cdot 10^{4} \,\,{1}/{ s}} \hspace{0.05cm}.$$

'''(5)''' Answers 1, 2, 3 and 5 are correct:
*The phase deviation is identical to the modulation index, which can be discerned from the equation given:
:$$\phi_{\rm max} = \eta = 3 = \frac{\Delta f_{\rm A}}{ f_{\rm N}} \hspace{0.05cm}.$$
*This leads to the frequency deviation  $Δf_{\rm A} = 3 · f_{\rm N} = 30 \ \rm kHz$.
*With a carrier frequency of   $f_{\rm T} = 1 \ \rm MHz$ , the instantaneous frequency  $f_{\rm T}(t)$  can only take values between  $1±0.03 \ \rm MHz$ .

'''Thus, the following statement is also valid:''':

At half the message frequency, the phase deviation  $η$ doubles, while the frequency deviation  $Δf_{\rm A}$ is unaffected:
:$$\eta = \frac{K_{\rm PM} \cdot A_{\rm N}}{ f_{\rm N}} = 6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} = \eta \cdot f_{\rm N} = 6 \cdot 5\,{\rm kHz} = 30\,{\rm kHz}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.7: Angular Modulation of a Harmonic Oscillation

2022-04-09T15:13:07Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1103__Mod_Z_3_6.png|right|frame|Demodulator for FM]]
The signal arriving at a receiver is:
:$$ r(t) = 3\,{\rm V} \cdot \cos \hspace{-0.05cm} \big[2 \pi \cdot 1\,{\rm MHz} \cdot t + 3 \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\big]\hspace{0.05cm}.$$
 $r(t)$  is an angle-modulated signal that was neither distorted nor influenced by noise during transmission.

The signals  $v_{\rm PM}(t)$  and  $v_{\rm FM}(t)$  result after ideal demodulation by means of
* a phase demodulator, given by the equation
:$$ v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) \hspace{0.05cm},\hspace{0.3cm} {K_{\rm PM}} = 2\,{\rm V}^{-1}\hspace{0.05cm},$$
* a frequency demodulator, consisting of a PM demodulator, a differentiator and a constant $K$.

In order for all signals to have equal units, this constant $K$ is dimensionally constrained.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].

===Questions===

<quiz display=simple>
{Which statements are definitely true?
|type="[]"}
+ There could be a PM modulation.
+ There could be a FM modulation.
- The message phase is definitely  $ϕ_{\rm N} = 0$.
+ The message phase is definitely  $f_{\rm N} = 10 \ \rm kHz$.

{Calculate the signal $v_{\rm PM}(t)$  after the phase demodulator.  What is the signal value at time  $t = 0$?
|type="{}"}
$v_{\rm PM}(t = 0) \ = \ $ { 1.5 3% } $\ \rm V$

{Calculate the signal $v_{\rm FM}(t)$. What is the message phase  $ϕ_{\rm N}$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { 90 3% } $\ \rm Grad$

{How should  $K$  be chosen so that the amplitude of  $v_{\rm FM}(t)$  is equal to  $1.5 \ \rm V$ ?
|type="{}"}
$K\ = \ $ { 6.28 3% } $\ \rm \cdot 10^4 \ 1/s$

{Which of the following statements is true for the FM-modulated signal?
|type="[]"}
+ The phase deviation is  $ϕ_{\rm max} = 3$.
+ The frequency deviation is  $Δf_{\rm A} = 30 \ \rm kHz$.
+ Instantaneous frequencies between  $0.97\ \rm MHz$  and  $1.03 \ \rm MHz$ .
- If  $f_{\rm N} = 5 \ \rm kHz$ , the phase deviation would be unchanged.
+ If  $f_{\rm N} = 5 \ \rm kHz$  the frequency deviation would be unchanged.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answers 1, 2 and 4 are correct:
*From the eqution for  $r(t)$  it can only be acertained that it is an angle modulation,
*but not whether it is a phase modulation (PM) or a frequency modulation (FM).
*Based on the equation, it is clear that the message frequency is  $f_{\rm N} = 10 \ \rm kHz$ .
*The phase  $ϕ_{\rm N} = 0$  of the source signal would then only apply, if phase modulation were present.

'''(2)'''  With the modulator constant  $K_{\rm PM} = 2 \ \rm V^{–1}$  this is given by:
:$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) = \frac{3}{2\,{\rm V}^{-1}} \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
*At time   $t = 0$  it therefore holds that:
:$$v_{\rm PM}(t = 0) = {A_{\rm N}} \hspace{0.15cm}\underline {= 1.5\,{\rm V}}\hspace{0.05cm}.$$

'''(3)'''  The output signal   $v_{\rm FM}(t)$  of the FM demodulator – consisting of a PM–demodulator and differentiator – can be written as:
:$$v_{\rm FM}(t) = \frac{{\rm d}v_{\rm PM}(t)}{{\rm d}t} \cdot K = \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot (- \sin(2 \pi \cdot {f_{\rm N}} \cdot t))= \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot \cos(2 \pi \cdot {f_{\rm N}} \cdot t + 90^\circ)\hspace{0.05cm}.$$
*The message phase is thus   $ϕ_{\rm N} \hspace{0.15cm}\underline {= 90^\circ}$.

'''(4)'''  In this case, it must hold that:  
:$$ K ={2 \pi \cdot f_{\rm N}} \hspace{0.15cm}\underline { = 6.28 \cdot 10^{4} \,\,{1}/{ s}} \hspace{0.05cm}.$$

'''(5)''' Answers 1, 2, 3 and 5 are correct:
*The phase deviation is identical to the modulation index, which can be discerned from the equation given:
:$$\phi_{\rm max} = \eta = 3 = \frac{\Delta f_{\rm A}}{ f_{\rm N}} \hspace{0.05cm}.$$
*This leads to the frequency deviation  $Δf_{\rm A} = 3 · f_{\rm N} = 30 \ \rm kHz$.
*With a carrier frequency of   $f_{\rm T} = 1 \ \rm MHz$ , the instantaneous frequency  $f_{\rm T}(t)$  can only take values between  $1±0.03 \ \rm MHz$ .

'''Thus, the following statement is also valid:''':

At half the message frequency, the phase deviation  $η$ doubles, while the frequency deviation  $Δf_{\rm A}$ is unaffected:
:$$\eta = \frac{K_{\rm PM} \cdot A_{\rm N}}{ f_{\rm N}} = 6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} = \eta \cdot f_{\rm N} = 6 \cdot 5\,{\rm kHz} = 30\,{\rm kHz}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.7: Angular Modulation of a Harmonic Oscillation

2022-04-09T15:12:47Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1103__Mod_Z_3_6.png|right|frame|Demodulator for FM]]
The signal arriving at a receiver is:
:$$ r(t) = 3\,{\rm V} \cdot \cos \hspace{-0.05cm} \big[2 \pi \cdot 1\,{\rm MHz} \cdot t + 3 \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\big]\hspace{0.05cm}.$$
 $r(t)$  is an angle-modulated signal that was neither distorted nor influenced by noise during transmission.

The signals  $v_{\rm PM}(t)$  and  $v_{\rm FM}(t)$  result after ideal demodulation by means of
* a phase demodulator, given by the equation
:$$ v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) \hspace{0.05cm},\hspace{0.3cm} {K_{\rm PM}} = 2\,{\rm V}^{-1}\hspace{0.05cm},$$
* a frequency demodulator, consisting of a PM demodulator, a differentiator and a constant $K$.

In order for all signals to have equal units, this constant $K$ is dimensionally constrained.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].

===Questions===

<quiz display=simple>
{Which statements are definitely true?
|type="[]"}
+ There could be a PM modulation.
+ There could be a FM modulation.
- The message phase is definitely  $ϕ_{\rm N} = 0$.
+ The message phase is definitely  $f_{\rm N} = 10 \ \rm kHz$.

{Calculate the signal $v_{\rm PM}(t)$  after the phase demodulator.  What is the signal value at time  $t = 0$?
|type="{}"}
$v_{\rm PM}(t = 0) \ = \ $ { 1.5 3% } $\ \rm V$

{Calculate the signal $v_{\rm FM}(t)$. What is the message phase  $ϕ_{\rm N}$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { 90 3% } $\ \rm Grad$

{How should  $K$  be chosen so that the amplitude of  $v_{\rm FM}(t)$  is equal to  $1.5 \ \rm V$ ?
|type="{}"}
$K\ = \ $ { 6.28 3% } $\ \rm \cdot 10^4 \ 1/s$

{Which of the following statements is true for the FM-modulated signal?
|type="[]"}
+ The phase deviation is  $ϕ_{\rm max} = 3$.
+ The frequency deviation is  $Δf_{\rm A} = 30 \ \rm kHz$.
+ Instantaneous frequencies between  $0.97\ \rm MHz$  And  $1.03 \ \rm MHz$ .
- If  $f_{\rm N} = 5 \ \rm kHz$ , the phase deviation would be unchanged.
+ If  $f_{\rm N} = 5 \ \rm kHz$  the frequency deviation would be unchanged.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answers 1, 2 and 4 are correct:
*From the eqution for  $r(t)$  it can only be acertained that it is an angle modulation,
*but not whether it is a phase modulation (PM) or a frequency modulation (FM).
*Based on the equation, it is clear that the message frequency is  $f_{\rm N} = 10 \ \rm kHz$ .
*The phase  $ϕ_{\rm N} = 0$  of the source signal would then only apply, if phase modulation were present.

'''(2)'''  With the modulator constant  $K_{\rm PM} = 2 \ \rm V^{–1}$  this is given by:
:$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) = \frac{3}{2\,{\rm V}^{-1}} \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
*At time   $t = 0$  it therefore holds that:
:$$v_{\rm PM}(t = 0) = {A_{\rm N}} \hspace{0.15cm}\underline {= 1.5\,{\rm V}}\hspace{0.05cm}.$$

'''(3)'''  The output signal   $v_{\rm FM}(t)$  of the FM demodulator – consisting of a PM–demodulator and differentiator – can be written as:
:$$v_{\rm FM}(t) = \frac{{\rm d}v_{\rm PM}(t)}{{\rm d}t} \cdot K = \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot (- \sin(2 \pi \cdot {f_{\rm N}} \cdot t))= \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot \cos(2 \pi \cdot {f_{\rm N}} \cdot t + 90^\circ)\hspace{0.05cm}.$$
*The message phase is thus   $ϕ_{\rm N} \hspace{0.15cm}\underline {= 90^\circ}$.

'''(4)'''  In this case, it must hold that:  
:$$ K ={2 \pi \cdot f_{\rm N}} \hspace{0.15cm}\underline { = 6.28 \cdot 10^{4} \,\,{1}/{ s}} \hspace{0.05cm}.$$

'''(5)''' Answers 1, 2, 3 and 5 are correct:
*The phase deviation is identical to the modulation index, which can be discerned from the equation given:
:$$\phi_{\rm max} = \eta = 3 = \frac{\Delta f_{\rm A}}{ f_{\rm N}} \hspace{0.05cm}.$$
*This leads to the frequency deviation  $Δf_{\rm A} = 3 · f_{\rm N} = 30 \ \rm kHz$.
*With a carrier frequency of   $f_{\rm T} = 1 \ \rm MHz$ , the instantaneous frequency  $f_{\rm T}(t)$  can only take values between  $1±0.03 \ \rm MHz$ .

'''Thus, the following statement is also valid:''':

At half the message frequency, the phase deviation  $η$ doubles, while the frequency deviation  $Δf_{\rm A}$ is unaffected:
:$$\eta = \frac{K_{\rm PM} \cdot A_{\rm N}}{ f_{\rm N}} = 6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} = \eta \cdot f_{\rm N} = 6 \cdot 5\,{\rm kHz} = 30\,{\rm kHz}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.5Z: Phase Modulation of a Trapezoidal Signal

2022-04-09T15:11:46Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1100__Mod_Z_3_5.png|right|frame|Trapezoidal and rectangular signals]]
A phase modulator with input signal $q_1(t)$  a modulated signal $s(t)$  at the output are described as follows:
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\psi(t) \big ]=
A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q_1(t) \big ] \hspace{0.05cm}.$$
*The carrier angular frequency is  $ω_{\rm T} = 2π · 10^5 \cdot {1}/{\rm s}$.
*The instantaneous angular frequency  $ω_{\rm A}(t)$  is equal to the derivative of the angle function  $ψ(t)$  with respect to time.
*The instantaneous frequency is thus  $f_{\rm A}(t) = ω_{\rm A}(t)/2π$.

The trapezoidal signal  $q_1(t)$  is applied as a test signal, where the nominated time duration is  $T = 10 \ \rm µ s$ .

The same modulated signal  $s(t)$  would result from a frequency modulator with the angular function
:$$\psi(t) = \omega_{\rm T} \cdot t + K_{\rm FM} \cdot \int q_2(t)\hspace{0.15cm}{\rm d}t$$
if the rectangular source signal  $q_2(t)$  is applied according to the lower plot.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].

===Questions===

<quiz display=simple>
{How should the modulator constant  $K_{\rm PM}$  be chosen so that  $ϕ_{\rm max} = 3 \ \rm rad$ ?
|type="{}"}
$K_{\rm PM} \ = \ $ { 1.5 3% } $\ \rm V^{-1}$

{What range of values does the instantaneous frequency  $f_{\rm A}(t)$  take on in the time interval $0 < t < T$ ?
|type="{}"}
$f_\text{A, min} \ = \ $ { 147.7 3% } $\ \rm kHz$
$f_\text{A, max} \hspace{0.06cm} = \ $ { 147.7 3% } $\ \rm kHz$

{What range of values does the instantaneous frequency  $f_{\rm A}(t)$ take on in the time interval  $T < t < 3T$ ?
|type="{}"}
$f_\text{A, min} \ = \ $ { 100 3% } $\ \rm kHz$
$f_\text{A, max} \hspace{0.06cm} = \ $ { 100 3% } $\ \rm kHz$

{What range of values does the instantaneous frequency $f_{\rm A}(t)$ take on in the time interval  $3T < t < 5T$ ?
|type="{}"}
$f_\text{A, min} \ = \ $ { 52.3 3% } $\ \rm kHz$
$f_\text{A, max} \hspace{0.06cm} = \ $ { 52.3 3% } $\ \rm kHz$

{How must the modulator constant  $K_{\rm FM}$ be chosen so that the signal $q_2(t)$  results in the same RF signal $s(t)$  after frequency modulation?
|type="{}"}
$K_{\rm FM} \ = \ $ { 1.5 3% } $\ \cdot 10^5 \ \rm V^{-1}s^{-1}$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The phase function is calculated as  $ϕ(t) = K_{\rm PM} · q_1(t)$.  The phase deviation  $ϕ_{\rm max}$ is equal to the phase resulting from the maximum value of the source signal:
:$$ \phi_{\rm max} = K_{\rm PM} \cdot 2\,{\rm V} = 3\,{\rm rad}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm PM} \hspace{0.15cm}\underline {= 1.5\,{\rm V^{-1}}} \hspace{0.05cm}.$$

'''(2)'''  In the range  $0 < t < T$ , the angular function can be represented as follows:
:$$ \psi(t) = \omega_{\rm T} \cdot t + K_{\rm PM} \cdot 2\,{\rm V} \cdot {t}/{T}\hspace{0.05cm}.$$
*For the instantaneous angular frequency   $ω_{\rm A}(t)$  or the instantaneous frequency   $f_{\rm A}(t)$ , the following holds:
:$$\omega_{\rm A}(t) = \frac{{\rm d}\hspace{0.03cm}\psi(t)}{{\rm d}t}= \omega_{\rm T} + K_{\rm PM} \cdot \frac{2\,{\rm V}}{10\,{\rm µ s}}\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T} + \frac{1.5\,{\rm V}^{-1}}{2 \pi} \cdot 2 \cdot 10^5 \rm {V}/{ s} = 100\,{\rm kHz}+ 47.7\,{\rm kHz}= 147.7\,{\rm kHz}\hspace{0.05cm}.$$
*The instantaneous frequency is constant, so $f_\text{A, min} = f_\text{A, max}\hspace{0.15cm}\underline{ = 147.7 \ \rm kHz}$  holds.

'''(3)'''  Due to the constant source signal, the derivative is zero throughout the time interval  $T < t < 3T$  under consideration, so the instantaneous frequency is equal to the carrier frequency:
:$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} \hspace{0.15cm}\underline {= 100\,{\rm kHz}}\hspace{0.05cm}.$$

'''(4)'''  The linear decay of  $q_1(t)$  in the time interval  $3T < t < 5T$  with slope as calculated in  '''(2)'''  leads to the result:
:$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} - 47.7\,{\rm kHz} \hspace{0.15cm}\underline {= 52.3\,{\rm kHz}}\hspace{0.05cm}.$$

'''(5)'''  By differentiation, we arrive at the instantaneous angular frequency:
:$$ \omega_{\rm A}(t) = \omega_{\rm T} + K_{\rm FM} \cdot q_2(t) \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T}+\frac{ K_{\rm FM}}{2 \pi} \cdot q_2(t)\hspace{0.05cm}.$$
*Using the result from   '''(2)''' , we get:
:$$\frac{ K_{\rm FM}}{2 \pi} \cdot 2\,{\rm V} = \frac{ 3 \cdot 10^5}{2 \pi} \cdot {\rm s^{-1}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} \hspace{0.15cm}\underline {= 1.5 \cdot 10^5 \hspace{0.15cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.9: Circular Arc and Parabola

2022-04-09T15:09:41Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1106__Mod_A_3_8.png|right|frame|Locus curves in FM: Arc and Parabola]]
We now consider the frequency modulation of a cosine source signal
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t )$$
with amplitude  $A_{\rm N} = 1 \ \rm V$  and frequency $f_{\rm N} = 5 \ \rm kHz$.
*The modulation index (phase deviation) is  $η = 2.4$.
*The corresponding low-pass signal with normalized carrier amplitude  $(A_{\rm T} = 1)$ is:
:$$ s_{\rm TP}(t) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
*This represents an arc.  Within the period $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$  the following phase angles result:
:$$ \phi(0) = 0, \hspace{0.2cm}\phi(0.25 \cdot T_{\rm N}) = \eta, \hspace{0.2cm}\phi(0.5 \cdot T_{\rm N})= 0,\hspace{0.2cm}
\phi(0.75 \cdot T_{\rm N})= -\eta,\hspace{0.2cm}\phi(T_{\rm N})= 0.$$
*Theoretically, the channel bandwidth required to transmit this signal is infinite.

However, if the bandwidth is limited to  $B_{\rm K} = 25 \ \rm kHz$, for example,the equivalent low-pass signal of the received signal can be described as follows:
:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
In this case, the result is a parabolic locus curve
:$$ y^2 + a \cdot x + b = 0,$$
which will be analyzed in this task.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].
*During calculation, assume the following values of the Bessel function:
:$${\rm J}_0 (2.4) \approx 0, \hspace{0.2cm}{\rm J}_1 (2.4) = -{\rm J}_{-1} (2.4)\approx 0.52, \hspace{0.2cm}{\rm J}_2 (2.4) = {\rm J}_{-2} (2.4)\approx 0.43.$$

===Question===

<quiz display=simple>
{What is the modulator constant  $K_{\rm FM}$?
|type="{}"}
$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$

{Calculate the real part $x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$  of the equivalent low-pass signal and give its maximum and minimum.
|type="{}"}
$x_{\rm max} \ = \ $ { 0.86 3% }
$x_{\rm min} \ = \ $ { -0.89--0.83 }

{What is the maximum and minimum of the imaginary part  $y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?
|type="{}"}
$y_{\rm max} \ = \ $ { 1.04 3% }
$y_{\rm min} \ = \ $ { -1.07--1.01 }

{What are the phase values for all multiples of  $T_{\rm N}/2$?
|type="{}"}
$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm degrees$

{What is the maximum phase angle  $ϕ_{\rm max}$?  Interpret the result.
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm degrees$

{Show that the locus curve can be given in the form  $y^2 + a · x + b = 0$ .  Determine the parabolic parameters  $a$  and  $b$.
|type="{}"}
$a\ = \ $ { 0.629 3% }
$b\ = \ $ { 0.541 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  In the frequency modulation of a cosine signal, the modulation index is:
:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} = \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$

'''(2)'''  The equation given for the equivalent low-pass signal when  $γ = ω_{\rm N} · t$  and considering   ${\rm J}_{–1} = –{\rm J}_1$  and  ${\rm J}_{–2} = {\rm J}_2$, is written out as:
:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
*Thus, for the real part in general, including for   $η = 2.4$, that is,  ${\rm J}_0 = 0$  and  ${\rm J}_2 = 0.43$:
:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}
\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$

'''(3)'''  According to the result of question  '''(2)'''  we get an imginary part   $({\rm J}_1 = 0.52)$ of:
:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$

'''(4)'''  The imaginary part is zero at each of these time points and so is the phase function:
:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$
*This fact can also be seen from the sketch on the exercise page.

'''(5)'''  From the sketch it can already be seen that the phase angle reaches its maximum value at  $t = T_{\rm N}/4$ , for example.
*This can be calculated with  $y_{\rm max} = 1.04$  and  $x_{\rm min} = -0.86$  as follows:
:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$
*Without band-limiting, the phase angle here would be $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$ .
*Thus, the maximum deviation of the sink signal from the source signal occurs at time  $t = T_{\rm N}/4$ , for example.

[[File:P_ID1113__Mod_A_3_8_f.png|right|frame|Parabolic progression]]
'''(6)'''  Using  $γ = ω_{\rm N} · t$  and  $\cos(2γ) = 1 - 2 · \cos^2(γ)$  the real and imaginary parts can be written as:

:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$

*These equations can be rearranegd:
:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$

$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$

*Thus, the parabolic parameters for  $ {\rm J}_0 = 0$ are:
:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$
*To double-check, we use   $y = 0$   ⇒   $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$
*Thus, the values at  $x = 0$  are     $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.8: Modulation Index and Bandwidth

2022-04-09T15:09:31Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1105__Mod_A_3_7.png|right|frame|Bessel function values]]
A harmonic oscillation of the form
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N})$$
is angle-modulated and then the one-sided magnitude spectrum  $|S_+(f)|$  is obtained.

*with a message frequency of  $f_{\rm N} = 2 \ \rm kHz$  the following spectral lines can be seen with the following weights:
:$$|S_{\rm +}(98\,{\rm kHz})| = |S_{\rm +}(102\,{\rm kHz})| = 1.560\,{\rm V}\hspace{0.05cm},$$ $$|S_{\rm +}(96\,{\rm kHz})| = |S_{\rm +}(104\,{\rm kHz})| = 1.293\,{\rm V}\hspace{0.05cm},$$
:$$ |S_{\rm +}(94\,{\rm kHz})| = |S_{\rm +}(106\,{\rm kHz})| = 0.594\,{\rm V}\hspace{0.05cm}.$$
:Further spectral lines follow each with frequency spacing  $f_{\rm N} = 2 \ \rm kHz$, but are not given here and can be ignored.

*If one increases the message frequency to  $f_{\rm N} = 4 \ \rm kHz$, there occur dominant lines
:$$|S_{\rm +}(100\,{\rm kHz})| = 2.013\,{\rm V}\hspace{0.05cm},$$
:$$|S_{\rm +}(96\,{\rm kHz})|\hspace{0.2cm} = |S_{\rm +}(104\,{\rm kHz})| = 1.494\,{\rm V}\hspace{0.05cm},$$
:$$ |S_{\rm +}(92\,{\rm kHz})|\hspace{0.2cm} = |S_{\rm +}(108\,{\rm kHz})| = 0.477\,{\rm V},$$
:as well as further, negligible Dirac lines with spacing  $f_{\rm N} = 4 \ \rm kHz$.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].

===Questions===

<quiz display=simple>
{Which modulation method is used here?
|type="()"}
- Phase modulation.
+ Frequency modulation.

{What is the modulation index  $η_2$  at message frequency  $f_{\rm N} = 2 \ \rm kHz$?
|type="{}"}
$η_2 \ = \ $ { 2.4 3% }

{What is the carrier amplitude?
|type="{}"}
$A_{\rm T} \ = \ $ { 3 3% } $\ \rm V$

{Specify the bandwidth  $B_2$ for  $f_{\rm N} = 2 \ \rm kHz$  if a distortion factor  $K < 1\%$  is desired.
|type="{}"}
$B_2 \ = \ $ { 17.6 3% } $\ \rm kHz$

{What is the modulation index $η_4$  at message frequency  $f_{\rm N} = 4 \ \rm kHz$?
|type="{}"}
$η_4\ = \ $ { 1.2 3% }

{What channel bandwidth  $B_4$  is now required to ensure  $K < 1\%$ ?
|type="{}"}
$B_4 \ = \ $ { 25.6 3% } $\ \rm kHz$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)''' We are dealing with a frequency modulation⇒   Answer 2.
*In phase modulation, the weights of the Dirac lines would not change when the frequency is doubled.

'''(2)'''  The spectral function given suggests the carrier frequency  $f_{\rm T} = 100 \ \rm kHz$  due to the symmetry properties.
*Since at   $f_{\rm N} = 2 \ \rm kHz$  the spectral line disappears at  $f_{\rm T} = 100 \ \rm kHz$ , we can assume $η_2 \hspace{0.15cm}\underline { ≈ 2.4}$ .
*A check of the other pulse weights confirms this result:
:$$\frac { |S_{\rm +}(f =102\,{\rm kHz})|}{ |S_{\rm +}(f =104\,{\rm kHz})|} = 1.206,\hspace{0.2cm} \frac { {\rm J}_1(2.4)}{ {\rm J}_2(2.4)}= 1.206 \hspace{0.05cm}.$$

'''(3)'''  The weights of the Dirac lines at  $f_{\rm T} + n · f_{\rm N}$  are generally:
:$$D_n = A_{\rm T} \cdot { {\rm J}_n(\eta)} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_1 = A_{\rm T} \cdot { {\rm J}_1(\eta)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} A_{\rm T} = D_1/{\rm J}_1(η) = 1.560\ \rm V/0.520\hspace{0.15cm}\underline { = 3 \ V}.$$

'''(4)'''  Given the requirement  $K < 1\%$ , one can use the following rule of thumb   (''Carson's rule''):
:$$B_{\rm 2} = 2 \cdot f_{\rm N} \cdot (\eta +2) \hspace{0.15cm}\underline {= 17.6\,{\rm kHz}}\hspace{0.05cm}.$$
*Thus, the Fourier coefficients   $D_{–4}$, ... , $D_4$  are available.

'''(5)'''  For frequency modulation, the general rule is:
:$$\eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.05cm}.$$
*Thus, by doubling the message frequency $f_{\rm N}$, the modulation index is halved:   $η_4 = η_2/2\hspace{0.15cm}\underline { = 1.2}$.

'''(6)'''  Using the same calculation as in question   '''(4)''' , the channel bandwidth necessary for   $K < 1\%$  is obtained using
:$$B_4 = 3.2 · 8\ \rm kHz \hspace{0.15cm}\underline {= 25.6 \ \rm kHz}.$$
*Because the modulation index is only half as large, transmitting the Fourier coefficients   $D_{–3}$, ... , $D_3$ is sufficient for limiting the distortion factor to  $1\%$.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.7: Angular Modulation of a Harmonic Oscillation

2022-04-09T15:09:23Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1103__Mod_Z_3_6.png|right|frame|Demodulator for FM]]
The signal arriving at a receiver is:
:$$ r(t) = 3\,{\rm V} \cdot \cos \hspace{-0.05cm} \big[2 \pi \cdot 1\,{\rm MHz} \cdot t + 3 \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\big]\hspace{0.05cm}.$$
 $r(t)$  is an angle-modulated signal that was neither distorted nor influenced by noise during transmission.

The signals  $v_{\rm PM}(t)$  and  $v_{\rm FM}(t)$  result after ideal demodulation by means of
* a phase demodulator, given by the equation
:$$ v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) \hspace{0.05cm},\hspace{0.3cm} {K_{\rm PM}} = 2\,{\rm V}^{-1}\hspace{0.05cm},$$
* a frequency demodulator, consisting of a PM demodulator, a differentiator and a constant $K$.

In order for all signals to have equal units, this constant $K$ is dimensionally constrained.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation|Signal characteristics with frequency modulation]].

===Questions===

<quiz display=simple>
{Which statements are definitely true?
|type="[]"}
+ There could be a PM modulation.
+ There could be a FM modulation.
- The message phase is definitely  $ϕ_{\rm N} = 0$.
+ The message phase is definitely  $f_{\rm N} = 10 \ \rm kHz$.

{Calculate the signal $v_{\rm PM}(t)$  after the phase demodulator.  What is the signal value at time  $t = 0$?
|type="{}"}
$v_{\rm PM}(t = 0) \ = \ $ { 1.5 3% } $\ \rm V$

{Calculate the signal $v_{\rm FM}(t)$. What is the message phase  $ϕ_{\rm N}$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { 90 3% } $\ \rm Grad$

{How should  $K$  be chosen so that the amplitude of  $v_{\rm FM}(t)$  is equal to  $1.5 \ \rm V$ ?
|type="{}"}
$K\ = \ $ { 6.28 3% } $\ \rm \cdot 10^4 \ 1/s$

{Which of the following statements is true for the FM-modulated signal?
|type="[]"}
+ The phase deviation is  $ϕ_{\rm max} = 3$.
+ The frequenCY deviation is  $Δf_{\rm A} = 30 \ \rm kHz$.
+ Instantaneous frequencies between  $0.97\ \rm MHz$  And  $1.03 \ \rm MHz$ .
- If  $f_{\rm N} = 5 \ \rm kHz$ , the phase deviation would be unchanged.
+ If  $f_{\rm N} = 5 \ \rm kHz$  the frequency deviation would be unchanged.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answers 1, 2 and 4 are correct:
*From the eqution for  $r(t)$  it can only be acertained that it is an angle modulation,
*but not whether it is a phase modulation (PM) or a frequency modulation (FM).
*Based on the equation, it is clear that the message frequency is  $f_{\rm N} = 10 \ \rm kHz$ .
*The phase  $ϕ_{\rm N} = 0$  of the source signal would then only apply, if phase modulation were present.

'''(2)'''  With the modulator constant  $K_{\rm PM} = 2 \ \rm V^{–1}$  this is given by:
:$$v_{\rm PM}(t) = \frac{1}{K_{\rm PM}} \cdot \phi_r(t) = \frac{3}{2\,{\rm V}^{-1}} \cdot \cos(2 \pi \cdot 10\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
*At time   $t = 0$  it therefore holds that:
:$$v_{\rm PM}(t = 0) = {A_{\rm N}} \hspace{0.15cm}\underline {= 1.5\,{\rm V}}\hspace{0.05cm}.$$

'''(3)'''  The output signal   $v_{\rm FM}(t)$  of the FM demodulator – consisting of a PM–demodulator and differentiator – can be written as:
:$$v_{\rm FM}(t) = \frac{{\rm d}v_{\rm PM}(t)}{{\rm d}t} \cdot K = \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot (- \sin(2 \pi \cdot {f_{\rm N}} \cdot t))= \frac{K \cdot A_{\rm N}}{2 \pi \cdot f_{\rm N}} \cdot \cos(2 \pi \cdot {f_{\rm N}} \cdot t + 90^\circ)\hspace{0.05cm}.$$
*The message phase is thus   $ϕ_{\rm N} \hspace{0.15cm}\underline {= 90^\circ}$.

'''(4)'''  In this case, it must hold that:  
:$$ K ={2 \pi \cdot f_{\rm N}} \hspace{0.15cm}\underline { = 6.28 \cdot 10^{4} \,\,{1}/{ s}} \hspace{0.05cm}.$$

'''(5)''' Answers 1, 2, 3 and 5 are correct:
*The phase deviation is identical to the modulation index, which can be discerned from the equation given:
:$$\phi_{\rm max} = \eta = 3 = \frac{\Delta f_{\rm A}}{ f_{\rm N}} \hspace{0.05cm}.$$
*This leads to the frequency deviation  $Δf_{\rm A} = 3 · f_{\rm N} = 30 \ \rm kHz$.
*With a carrier frequency of   $f_{\rm T} = 1 \ \rm MHz$ , the instantaneous frequency  $f_{\rm T}(t)$  can only take values between  $1±0.03 \ \rm MHz$ .

'''Thus, the following statement is also valid:''':

At half the message frequency, the phase deviation  $η$ doubles, while the frequency deviation  $Δf_{\rm A}$ is unaffected:
:$$\eta = \frac{K_{\rm PM} \cdot A_{\rm N}}{ f_{\rm N}} = 6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} = \eta \cdot f_{\rm N} = 6 \cdot 5\,{\rm kHz} = 30\,{\rm kHz}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.5Z: Phase Modulation of a Trapezoidal Signal

2022-04-09T15:09:14Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1100__Mod_Z_3_5.png|right|frame|Trapezoidal and rectangular signals]]
A phase modulator with input signal $q_1(t)$  a modulated signal $s(t)$  at the output are described as follows:
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\psi(t) \big ]=
A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q_1(t) \big ] \hspace{0.05cm}.$$
*The carrier angular frequency is  $ω_{\rm T} = 2π · 10^5 \cdot {1}/{\rm s}$.
*The instantaneous angular frequency  $ω_{\rm A}(t)$  is equal to the derivative of the angle function  $ψ(t)$  with respect to time.
*The instantaneous frequency is thus  $f_{\rm A}(t) = ω_{\rm A}(t)/2π$.

The trapezoidal signal  $q_1(t)$  is applied as a test signal, where the nominated time duration is  $T = 10 \ \rm µ s$ .

The same modulated signal  $s(t)$  would result from a frequency modulator with the angular function
:$$\psi(t) = \omega_{\rm T} \cdot t + K_{\rm FM} \cdot \int q_2(t)\hspace{0.15cm}{\rm d}t$$
if the rectangular source signal  $q_2(t)$  is applied according to the lower plot.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].

===Questions===

<quiz display=simple>
{How should the modulator constant  $K_{\rm PM}$  be chosen so that  $ϕ_{\rm max} = 3 \ \rm rad$ ?
|type="{}"}
$K_{\rm PM} \ = \ $ { 1.5 3% } $\ \rm V^{-1}$

{What range of values does the instantaneous frequency  $f_{\rm A}(t)$  take on in the time interval $0 < t < T$ ?
|type="{}"}
$f_\text{A, min} \ = \ $ { 147.7 3% } $\ \rm kHz$
$f_\text{A, max} \hspace{0.06cm} = \ $ { 147.7 3% } $\ \rm kHz$

{What range of values does the instantaneous frequency  $f_{\rm A}(t)$ take on in the time interval  $T < t < 3T$ ?
|type="{}"}
$f_\text{A, min} \ = \ $ { 100 3% } $\ \rm kHz$
$f_\text{A, max} \hspace{0.06cm} = \ $ { 100 3% } $\ \rm kHz$

{What range of values does the instantaneous frequency $f_{\rm A}(t)$ take on in the time interval  $3T < t < 5T$ ?
|type="{}"}
$f_\text{A, min} \ = \ $ { 52.3 3% } $\ \rm kHz$
$f_\text{A, max} \hspace{0.06cm} = \ $ { 52.3 3% } $\ \rm kHz$

{How must the modulator constant  $K_{\rm FM}$ be chosen so that the signal $q_2(t)$  results in the same RF signal $s(t)$  after frequency modulation?
|type="{}"}
$K_{\rm FM} \ = \ $ { 1.5 3% } $\ \cdot 10^5 \ \rm V^{-1}s^{-1}$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The phase function is calculated as  $ϕ(t) = K_{\rm PM} · q_1(t)$.  The phase deviation  $ϕ_{\rm max}$ is equal to the phase resulting from the maximum value of the source signal:
:$$ \phi_{\rm max} = K_{\rm PM} \cdot 2\,{\rm V} = 3\,{\rm rad}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm PM} \hspace{0.15cm}\underline {= 1.5\,{\rm V^{-1}}} \hspace{0.05cm}.$$

'''(2)'''  In the range  $0 < t < T$ , the angular function can be represented as follows:
:$$ \psi(t) = \omega_{\rm T} \cdot t + K_{\rm PM} \cdot 2\,{\rm V} \cdot {t}/{T}\hspace{0.05cm}.$$
*The instantaneous angular frequency   $ω_{\rm A}(t)$  or the instantaneous frequency   $f_{\rm A}(t)$  the following holds:
:$$\omega_{\rm A}(t) = \frac{{\rm d}\hspace{0.03cm}\psi(t)}{{\rm d}t}= \omega_{\rm T} + K_{\rm PM} \cdot \frac{2\,{\rm V}}{10\,{\rm µ s}}\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T} + \frac{1.5\,{\rm V}^{-1}}{2 \pi} \cdot 2 \cdot 10^5 \rm {V}/{ s} = 100\,{\rm kHz}+ 47.7\,{\rm kHz}= 147.7\,{\rm kHz}\hspace{0.05cm}.$$
*The instantaneous frequency is constant, so $f_\text{A, min} = f_\text{A, max}\hspace{0.15cm}\underline{ = 147.7 \ \rm kHz}$  holds.

'''(3)'''  Due to the constant source signal, the derivative is zero throughout the time interval  $T < t < 3T$  under consideration, so the instantaneous frequency is equal to the carrier frequency:
:$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} \hspace{0.15cm}\underline {= 100\,{\rm kHz}}\hspace{0.05cm}.$$

'''(4)'''  The linear decay of  $q_1(t)$  in the time interval  $3T < t < 5T$  with slope as calculated in  '''(2)'''  leads to the result:
:$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} - 47.7\,{\rm kHz} \hspace{0.15cm}\underline {= 52.3\,{\rm kHz}}\hspace{0.05cm}.$$

'''(5)'''  By differentiation, we arrive at the instantaneous angular frequency:
:$$ \omega_{\rm A}(t) = \omega_{\rm T} + K_{\rm FM} \cdot q_2(t) \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T}+\frac{ K_{\rm FM}}{2 \pi} \cdot q_2(t)\hspace{0.05cm}.$$
*Using the result from   '''(2)''' , we get:
:$$\frac{ K_{\rm FM}}{2 \pi} \cdot 2\,{\rm V} = \frac{ 3 \cdot 10^5}{2 \pi} \cdot {\rm s^{-1}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} \hspace{0.15cm}\underline {= 1.5 \cdot 10^5 \hspace{0.15cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.5: PM and FM for Rectangular Signals

2022-04-09T15:09:04Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
}}

[[File:P_ID1099__Mod_A_3_5.png|right|frame|Two signal waveforms in angle modulation]]
Assume a bipolar and rectangular source signal $q(t)$ , as shown in the upper diagram.  This signal can only take on the two signal values  $±A = ±2 \ \rm V$  and the duration of the positive and negative rectangles are each $T = 1 \ \rm ms$.  The period of  $q(t)$  is therefore  $T_0 = 2 \ \rm ms$.

The signals $s_1(t)$  and  $s_2(t)$  display two transmit signals with angle modulation  $\rm (WM)$, each of which can be represented as
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big [\psi (t) \big ]$$
Here, we distinguish between phase modulation  $\rm (PM)$  with the angular function
:$$\psi(t) = \omega_{\rm T} \cdot t + \phi(t) = \omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t)$$
and frequency modulation  $\rm (FM)$, where the instantaneous freqiency is linearly related to $q(t)$:
:$$f_{\rm A}(t) = \frac{\omega_{\rm A}(t)}{2\pi}, \hspace{0.3cm} \omega_{\rm A}(t) = \frac{{\rm d}\hspace{0.03cm}\psi(t)}{{\rm d}t}= \omega_{\rm T} + K_{\rm FM} \cdot q(t)\hspace{0.05cm}.$$
$K_{\rm PM}$  and  $K_{\rm FM}$  denote the dimensionally constrained constants given by the realizations of the PM and FM modulators, respectively.  The frequency deviation  $Δf_{\rm A}$  indicates the maximum deviation of the instantaneous frequency from the carrier frequency.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].

*In anticipation of the fourth chapter, it should be mentioned that phase modulation with a digital input signal is also called ''Phase Shift Keying''  $\rm (PSK)$  and frequency modulation is analogously called ''Frequency Shift Keying''  $\rm (FSK)$ .

===Questions===

<quiz display=simple>
{Which of the signals is due to phase modulation and which is due to frquency modulation?
|type="()"}
- $s_1(t)$  represents a phase modulation.
+ $s_1(t)$  represents a frequency modulation.

{What is the carrier phase  $ϕ_{\rm T}$ that could be measured without a message signal   ⇒   $q(t) \equiv 0$ ?
|type="{}"}
$ϕ_{\rm T} \ = \ $ { 0. } $\ \rm Grad$

{What carrier frequency  $($with respect to  $1/T)$  was used in the graphs?
|type="{}"}
$f_{\rm T} · T \ = \ $ { 6 3% }

{The phase of the PM signal is  $±90^\circ$.  What is the modulator constant?
|type="{}"}
$K_{\rm PM} \ = \ $ { 0.785 3% } $\ \rm V^{-1}$

{What is the frequency deviation  $Δf_{\rm A}$  of the FM signal with respect to  $1/T$?
|type="{}"}
$Δf_{\rm A} · T \ = \ $ { 2 3% }

{What is the FM modulator constant?
|type="{}"}
$K_{\rm FM} \ = \ $ { 6283 3% } $\ \rm (Vs)^{-1}$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answer 2 is correct:
*For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform  $s_2(t)$.
*Frequency modulation (FM), on the other hand, has diverse instantaneous frequencies at different times, as in  $s_1(t)$.

'''(2)'''  When  $q(t) = 0$ , the equations provided for both PM and FM give
:$$s(t) = A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t ) \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \phi_{\rm T} \hspace{0.15cm}\underline {= 0}\hspace{0.05cm}.$$

'''(3)'''  The carrier frequency  $f_{\rm T}$  can be directly determined only from the PM signal   $s_2(t)$ .
*By counting the oscillations of  $s_2(t)$  in the time interval  $T$ , it can be seen that  $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$  was used.
*When frequency modulating a bipolar source signal,   $f_{\rm T}$  does not occur directly.
*However, the graphs do indicate that   $f_{\rm T} · T = 6$  is also used here.

'''(4)'''  The amplitude value  $A = 2 \ \rm V$  results in the phase  $90^\circ$  or  $π/2$  (minus sine wave).  This gives:
:$$K_{\rm PM} = \frac {\pi /2}{2\,{\rm V}} \hspace{0.15cm}\underline {= 0.785\,{\rm V}^{-1}} \hspace{0.05cm}.$$

'''(5)'''  The graph for  $s_1(t)$  shows that either four or eight oscillations arise within a time interval  $T$ :   $4 \le f_{\rm A}(t) \cdot T \le 8\hspace{0.05cm}.$
*Considering the (normalized) carrier frequency  $f_{\rm T} · T = 6$ , the (normalized) frequency deviation is:
:$$\Delta f_{\rm A} \cdot T \hspace{0.15cm}\underline {=2}\hspace{0.05cm}.$$

'''(6)'''  The frequency deviation can also be represented as follows:
:$$\Delta f_{\rm A} = \frac {K_{\rm FM}}{2\pi}\cdot A \hspace{0.05cm}.$$
*With   $Δf_{\rm A} · {\rm A} = 2$  we thus get:
:$$K_{\rm FM} = \frac {2 \cdot 2\pi}{A \cdot T}= \frac {4\pi}{2\,{\rm V} \cdot 1\,{\rm ms}}\hspace{0.15cm}\underline {= 6283 \,{\rm V}^{-1}{\rm s}^{-1}} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Aufgaben:Exercise 3.4: Simple Phase Modulator

2022-04-09T15:04:47Z

Reed: /* Questions */

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1086__Mod_A_3_4.png|right|frame|"Approximate phase modulator"]]
The adjacent circuit allows the approximate realization of a phase-modulated signal. 

From the cosinusoidal carrier,  $z(t)$ , the  $90^\circ$ phase shifter forms a sinusoidal signal of the same frequency, such that the modulated signal can be written as:
:$$ s(t) = z(t) + q(t) \cdot \frac{z(t- T_0/4)}{A_{\rm T}}
= A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t) + q(t) \cdot \sin (\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
The second term describes a "DSB–AM without carrier".  Additionally, the carrier, phase-shifted by $90^\circ$ , is added.  Thus, with a cosine source signal $q(t) = A_{\rm N} \cdot \cos (\omega_{\rm N} \cdot t)$ , we get:
:$$s(t) = A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t) + A_{\rm N} \cdot \cos (\omega_{\rm N} \cdot t) \cdot \sin (\omega_{\rm T} \cdot t) $$
:$$\Rightarrow \hspace{0.3cm}s(t) = A_{\rm T} \cdot \big[\cos (\omega_{\rm T} \cdot t) + \eta \cdot \cos (\omega_{\rm N} \cdot t) \cdot \sin (\omega_{\rm T} \cdot t) \big] \hspace{0.05cm}.$$
We refer to the ratio  $η = A_{\rm N}/A_{\rm T}$  as the modulation index;  in the following, the carrier amplitude is set to   $A_{\rm T} = 1$  for simplicity.

*In contrast to  [[Modulation_Methods/Phasenmodulation_(PM)#Signalverl.C3.A4ufe_bei_Phasenmodulation|ideal phase modulation]]  the modulation index  $η$  and the phase deviation  $ϕ_{\rm max}$ may differ in this "approximate phase modulation".
*Additionally, we can see that the envelope  $a(t) ≠ 1$ .  This means that an unwanted amplitude modulation is superimposed on the phase modulation.

From the representation of the equivalent low-pass signal  $s_{\rm TP}(t)$  in the complex plane (locus curve), the following are to be calculated in this task:
*the envelope  $a(t)$  and
*the phase function $ϕ(t)$.

Then, you are to analyse the distortions arising when an ideal PM demodulator, which sets the sink signal  $v(t)$  proportional to the phase  $ϕ(t)$ , is used on the receiving side of this nonideal PM modulator.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

*You can use the following equations to approximate the distortion factor:
:$$\arctan(\gamma) \approx \gamma - {\gamma^3}/{3} \hspace{0.05cm}, \hspace{0.3cm} \cos^3(\gamma) ={3}/{4} \cdot \cos(\gamma) +{1}/{4} \cdot \cos(3 \cdot \gamma) \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Calculate the equivalent low-pass signal. Which statement is true?
|type="()"}
- The locus curve  $s_{\rm TP}(t)$  is a circular arc.
- The locus curve  $s_{\rm TP}(t)$  is a horizontal straight line.
+ The locus curve  $s_{\rm TP}(t)$  is a vertical straight line.

{Calculate the (normalized) envelope $a(t)$  for  $A_{\rm T} = 1$.  What are its minimum and maximum values when  $η = 1$?
|type="{}"}
$a_{\rm min} \ = \ $ { 1 3% }
$a_{\rm max} \ = \ $ { 1.414 3% }

{Calculate the maximum value of the phase $ϕ(t)$  for  $η = 1$  and  $η = 0.5$.
|type="{}"}
$η = 1.0\text{:} \ \ \ ϕ_{\rm max} \ = \ $ { 45 3% } $\ \rm Grad$
$η = 0.5\text{:} \ \ \ ϕ_{\rm max} \ = \ $ { 26.6 3% } $\ \rm Grad$

{What distortions result after ideal phase demodulation of  $s(t)$?
|type="()"}
- No distortions occur.
- Linear distortions occur.
+ Nonlinear distortions occur.

{Calculate the distortion factor  $K$  considering the trigonometric relationships given on the exercise page.
|type="{}"}
$η = 1.0\text{:} \ \ \ K \ = \ $ { 11.1 3% } $\ \text{%}$
$η = 0.5\text{:} \ \ \ K \ = \ $ { 2.2 3% } $\ \text{%}$

</quiz>

===Solution===
{{ML-Kopf}}
[[File:P_ID1087__Mod_A_3_4_a.png|right|frame|Construction of the "vertical locus" from the pointers]]
'''(1)'''  Answer 3 is correct:
*The equivalent low-pass signal is::$$s_{\rm TP}(t) = A_{\rm T} \cdot \left ( 1 + {\rm j}\cdot \frac {\eta}{2}\cdot \left ({\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}\right) \right)
= A_{\rm T} \cdot \big ( 1 + {\rm j}\cdot {\eta}\cdot \cos (\omega_{\rm N} \cdot t) \big)\hspace{0.05cm}.$$
*The graph illustrates that the locus curve  $s_{\rm TP}(t)$  is now a a vertical straight line in contrast to the ideal PM (circular arc) and DSB–AM  (horizontal straight line). 
*In the following, we set   $A_{\rm T} = 1$ .

'''(2)'''  The envelope is obtained from the time-dependent pointer length as
:$$a(t) = \sqrt{1 + \eta^2 \cdot \cos^2 (\omega_{\rm N} \cdot t)} \hspace{0.3cm}
\Rightarrow \hspace{0.3cm}a_{\rm min} \hspace{0.15cm}\underline { = 1}, \hspace{0.3cm}a_{\rm max} = \sqrt{1 + \eta^2 }\hspace{0.05cm}.$$
*For  $η = 1$  the maximum value becomes  $a_{\rm max} = \sqrt{2}\hspace{0.15cm}\underline { ≈ 1.414}$.

'''(3)'''  The phase function of this simple phase demodulator is given by:
:$$\phi(t) = \arctan \frac{{\rm Im}[s_{\rm TP}(t)]}{{\rm Re}[s_{\rm TP}(t)]} = \arctan (\eta \cdot \cos (\omega_{\rm N} \cdot t)) \hspace{0.05cm}.$$
*The maximum value occurs at time  $t = 0$ , for example, and is   $ϕ_{\rm max} = \arctan(η)$.
:*When  $η = 1$ , one obtains   $ϕ_{\rm max}\hspace{0.15cm}\underline { = 45^\circ}$  $($to compare:  for ideal PM  $57.3^\circ)$,
:*When  $η = 0.5$  one gets   $ϕ_{\rm max}\hspace{0.15cm}\underline { \approx 26.6^\circ}$  $($for ideal PM  $28.7^\circ)$.

'''(4)''' Answer 3 is correct:
*It is  '''not''' true that:      $\arctan\big [η · \cos(γ)\big ] = η · \cos(γ)$.
*This means that the sink signal is not cosine, in contrast to the source signal.
*This points to nonlinear distortions.

'''(5)'''  Using  $γ = η · \cos(ω_N · t)$  and  $\arctan(γ) ≈ γ – γ^3/3$ , we get:
:$$ \phi(t) = \eta \cdot \cos (\omega_{\rm N} \cdot t) - \frac{\eta^3}{3}\cdot \cos^3 (\omega_{\rm N} \cdot t))=
\eta \cdot \cos (\omega_{\rm N} \cdot t) - \frac{\eta^3}{3}\cdot \left [ {3}/{4}\cdot \cos (\omega_{\rm N} \cdot t) + {1}/{4}\cdot \cos (3 \omega_{\rm N} \cdot t)\right ] $$
:$$\Rightarrow \hspace{0.3cm} \phi(t) = \left(\eta - {\eta^3}/{4} \right) \cdot \cos (\omega_{\rm N} \cdot t) - {\eta^3}/{12}\cdot \cos (3\omega_{\rm N} \cdot t) \hspace{0.05cm}.$$
*This means:  using the given series expansion (where 5th and higher order terms are ignored), only the third-order harmonic distortion is non-zero. Thus:
:$$K = K_3 = \frac{\eta^3/12}{\eta-\eta^3/4}= \frac{1}{12/\eta^2 -3} \hspace{0.05cm}.$$
*When  $η = 1$  the numerical value is  $K = 1/9 \hspace{0.15cm}\underline { ≈ 11.1\%}$.
*When  $η = 0.5$  the distortion factor is  $K = 1/45 \hspace{0.15cm}\underline {≈ 2.2\%}$.

A simulation shows that by stopping the series after the third order term, we have made the error of over-estimating the distortion factor:
*The values obtained by simulation are   $K ≈ 6%$  $($for  $η = 1)$  and  $K ≈ 2%$  $($for  $η = 0.5)$.
*Thus, the error increases more than proportionally with increasing  $η$ .

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3Z: Characteristics Determination

2022-04-09T15:03:39Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1085__Mod_Z_3_3.png|right|frame|Spectrum of the analytical signal]]
Let us consider the phase modulation of the harmonic oscillation
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$
which, given a normalized carrier amplitude  $(A_{\rm T} = 1)$ , leads to the following transmitted signal:
:$$ s(t) = \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \big]\hspace{0.05cm}.$$
The spectrum of the corresponding analytical signal $s_{\rm TP}(t)$  is generally:
:$$S_{\rm TP}(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}(\phi_{\rm N}\hspace{0.05cm}+\hspace{0.05cm} 90^\circ) }\cdot \hspace{0.05cm} \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}$$
Here,  $η = K_{\rm PM} · A_{\rm N}$  is called the modulation index.

In the graph, the real and imaginary parts of the spectrum  $S_+(f)$  of the analytical signal  $s_+(t)$ are shown separately. This should be used to determine the characteristics  $f_{\rm T}$,  $f_{\rm N}$,  $ϕ_{\rm N}$  and  $η$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

*For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:
:$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm J}_{2} (\eta)= {2}/{\eta} \cdot {\rm J}_{1} (\eta) - {\rm J}_{0} (\eta) \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{What are the frequencies  $f_{\rm T}$  and  $f_{\rm N}$?
|type="{}"}
$f_{\rm T} \ = \ $ { 40 3% } $\ \rm kHz$
$f_{\rm N} \ = \ $ { 3 3% } $\ \rm kHz$

{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 3 \ \rm kHz)$.
|type="{}"}
$|S_{\rm TP}(f = 3 \ \rm kHz)| \ = \ $ { 0.558 3% }
${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $ { 60 3% } $\ \rm Grad$

{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 6 \ \rm kHz)$.
|type="{}"}
$|S_{\rm TP}(f = 6 \ \rm kHz)| \ = \ $ { 0.232 3% }
${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $ { 120 3% } $\ \rm Grad$

{What is the phase of the source signal  $q(t)$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { -30.9--29.1 } $\ \rm Grad$

{What is the magnitude of the modulation index  $η$ ?
|type="{}"}
$η \ = \ $ { 1.5 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Regarding   $|S_+(f)|$  there is a symmetry with respect to the carrier frequency  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  The distance between the spectral lines is  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.

'''(2)'''  Considering  $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$ , it holds that:
:$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$
:$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$

'''(3)'''  Analogously to in question  '''(2)''' , at frequency   $f = 6 \ \rm kHz$ we obtain:
:$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$
:$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$

'''(4)'''  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   '''(2)''', the phase is:
:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
*Checking this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  '''(3)'''  yields the same value:
:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$

'''(5)'''  The given equation can be rewritten as follows:
:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
*With  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  and  ${\rm J}_2(η) = 0.232$  we thus get:
:$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3Z: Characteristics Determination

2022-04-09T15:03:02Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1085__Mod_Z_3_3.png|right|frame|Spectrum of the analytical signal]]
Let us consider the phase modulation of the harmonic oscillation
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$
which, given a normalized carrier amplitude  $(A_{\rm T} = 1)$ , leads to the following transmitted signal:
:$$ s(t) = \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \big]\hspace{0.05cm}.$$
The spectrum of the corresponding analytical signal $s_{\rm TP}(t)$  is generally:
:$$S_{\rm TP}(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}(\phi_{\rm N}\hspace{0.05cm}+\hspace{0.05cm} 90^\circ) }\cdot \hspace{0.05cm} \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}$$
Here,  $η = K_{\rm PM} · A_{\rm N}$  is called the modulation index.

In the graph, the real and imaginary parts of the spectrum  $S_+(f)$  of the analytical signal  $s_+(t)$ are shown separately. This should be used to determine the characteristics  $f_{\rm T}$,  $f_{\rm N}$,  $ϕ_{\rm N}$  and  $η$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

*For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:
:$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm J}_{2} (\eta)= {2}/{\eta} \cdot {\rm J}_{1} (\eta) - {\rm J}_{0} (\eta) \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{What are the frequencies  $f_{\rm T}$  and  $f_{\rm N}$?
|type="{}"}
$f_{\rm T} \ = \ $ { 40 3% } $\ \rm kHz$
$f_{\rm N} \ = \ $ { 3 3% } $\ \rm kHz$

{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 3 \ \rm kHz)$.
|type="{}"}
$|S_{\rm TP}(f = 3 \ \rm kHz)| \ = \ $ { 0.558 3% }
${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $ { 60 3% } $\ \rm Grad$

{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 6 \ \rm kHz)$.
|type="{}"}
$|S_{\rm TP}(f = 6 \ \rm kHz)| \ = \ $ { 0.232 3% }
${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $ { 120 3% } $\ \rm Grad$

{What is the phase of the source signal  $q(t)$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { -30.9--29.1 } $\ \rm Grad$

{What is the magnitude of the modulation index  $η$ ?
|type="{}"}
$η \ = \ $ { 1.5 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Regarding   $|S_+(f)|$  there is a symmetry with respect to the carrier frequency  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  The distance between the spectral lines is  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.

'''(2)'''  Considering  $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$ , it holds that:
:$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$
:$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$

'''(3)'''  Analogously to in question  '''(2)''' , at frequency   $f = 6 \ \rm kHz$ we obtain:
:$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$
:$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$

'''(4)'''  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   '''(2)''', the phase is:
:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
*Checking this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  '''(3)'''  yields the same value:
:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$

'''(5)'''  The given equation can be rewritten as follows:
:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
*Mit  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  und  ${\rm J}_2(η) = 0.232$  erhält man somit:
:$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3: Sum of two Oscillations

2022-04-09T15:01:48Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1084__Mod_A_3_3.png|right|frame|Two different Bessel spectra]]
The equivalent low-pass signal with phase modulation, when normalized to the carrier amplitude $(A_{\rm T} = 1)$ is: 
:$$ s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}K_{\rm PM}\hspace{0.05cm}\cdot \hspace{0.05cm}q(t) }\hspace{0.05cm},$$
The modulator constant is assumed to be  $K_{\rm PM} = \rm 1/V$  throughout the task.

The upper graph shows the corresponding spectral function  $B_1(f)$, when the source signal is
:$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$
The weights of the Bessel-Dirac lines when  $η_1 = 0.9$  are obtained as follows:
:$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm}
{\rm J}_1 (0.9) = 0.406 \approx 0.4,$$
:$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm}
{\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$
Use the approximations given in the graph to simplify the calculations.

The Bessel function  $B_2(f)$  is obtained for the source signal
:$$q_2(t) = 0.65\,{\rm V} \cdot \cos(2 \pi \cdot 3\,{\rm kHz} \cdot t)$$
The numerical values of the Dirac lines are obtained from the following:
:$${\rm J}_0 (0.65) = 0.897 \approx 0.9,\hspace{0.3cm}{\rm J}_1 (0.65) = 0.308 \approx 0.3, \hspace{0.3cm}{\rm J}_2 (0.65) = 0.051 \approx 0\hspace{0.05cm}.$$
From the graph, it can be seen that due to the cosine source signal  $q_2(t)$  and the cosine carrier signal $z(t)$ , the spectral lines at $±3 \ \rm kHz$  are both positive and imaginary.

In the context of this task, we will now investigate the case where the source signal
:$$q(t) = q_1(t) + q_2(t)$$
is applied to the input of the phase modulator.
*It is worth mentioning that  $|q(t)| < q_{\rm max} = 1.45 \ \rm V$ .
*This maximum value is slightly smaller than the sum  $A_1 + A_2$ of the individual amplitudes when a sine and a cosine function with the given amplitudes are added up.

In the following questionnaire,
*$S_{\rm TP}(f)$  denotes the spectral function of the equivalent low-pass signal,
*$S_+(f)$  denotes the spectral functions of the analytic signal,

in both cases assuming that  $q(t) = q_1(t) + q_2(t)$  holds and that the carrier frequency is $f_{\rm T} = 100 \ \rm kHz$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
*The values of the Bessel functions can be found in formula collections in table form.
*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kind| Bessel functions of the first kind]]  to solve this task.

===Questions===

<quiz display=simple>
{Let  $q(t) = q_1(t)+q_2(t)$.  Which geometric figure describes the given locus curve $s_{\rm TP}(t)$?
|type="()"}
- The locus curve is an ellipse.
- The locus curve is a circle.
+ The locus curve is approximately a semi-circle.
- The locus curve is an arc, with an approximate opening angle of  $90^\circ$.

{Calculate the spectral function $S_{\rm TP}(f)$.  Between what frequencies  $f_{\rm min}$  and  $f_{\rm max}$  do the spectral lines lie?
|type="{}"}
$f_{\rm min} \ = \ $ { -5.15--4.85 } $\ \rm kHz$
$f_{\rm max} \ = \ ${ 5 3% } $\ \rm kHz$

{Calculate the weight of the Dirac function at  $f = 0$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0.72 3% }
${\rm Im}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0. }

{Calculate the weight of the Dirac function at  $f = 1\ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.36 3% }
${\rm Im}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.03 3% }

{Calculate the weight of the  $S_+(f)$–Dirac function at  $f = 98 \ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.09 3% }
${\rm Im}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.12 3% }
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The third answer is correct:
*In angle modulation, the complex pointer   $s_{\rm TP}(t)$  always moves on a circular arc with the following opening angle:
:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
*Using the (admittedly very rough) approximation  $166^\circ \approx 180^\circ$  we indeed get a semicircle.

'''(2)'''  In general,   $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.
*Since  $B_1(f)$  is limited to the frequencies   $|f| ≤ 2 \ \rm kHz$  and  $B_2(f)$  is limited to the range   $±3 \ \rm kHz$ , the convolution product is limited to  $|f| ≤ 5 \ \rm kHz$ .
*It follows that:
:$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$
:$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$

'''(3)'''  The convolution product for frequency   $f = 0$  results from multiplying  $B_1(f)$  with  $B_2(f)$  and summing.
*Only for   $f = 0$  are both  $B_1(f)$  and  $B_2(f)$  non-zero.
*Thus, we get:
:$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$

'''(4)'''  Now, before multiplication and summation there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$ .  This gives:
:$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
+ B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
= 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$

'''(5)'''  The Dirac line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is
:$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +
B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3: Sum of two Oscillations

2022-04-09T15:01:24Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1084__Mod_A_3_3.png|right|frame|Two different Bessel spectra]]
The equivalent low-pass signal with phase modulation, when normalized to the carrier amplitude $(A_{\rm T} = 1)$ is: 
:$$ s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}K_{\rm PM}\hspace{0.05cm}\cdot \hspace{0.05cm}q(t) }\hspace{0.05cm},$$
The modulator constant is assumed to be  $K_{\rm PM} = \rm 1/V$  throughout the task.

The upper graph shows the corresponding spectral function  $B_1(f)$, when the source signal is
:$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$
The weights of the Bessel-Dirac lines when  $η_1 = 0.9$  are obtained as follows:
:$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm}
{\rm J}_1 (0.9) = 0.406 \approx 0.4,$$
:$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm}
{\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$
Use the approximations given in the graph to simplify the calculations.

The Bessel function  $B_2(f)$  is obtained for the source signal
:$$q_2(t) = 0.65\,{\rm V} \cdot \cos(2 \pi \cdot 3\,{\rm kHz} \cdot t)$$
The numerical values of the Dirac lines are obtained from the following:
:$${\rm J}_0 (0.65) = 0.897 \approx 0.9,\hspace{0.3cm}{\rm J}_1 (0.65) = 0.308 \approx 0.3, \hspace{0.3cm}{\rm J}_2 (0.65) = 0.051 \approx 0\hspace{0.05cm}.$$
From the graph, it can be seen that due to the cosine source signal  $q_2(t)$  and the cosine carrier signal $z(t)$ , the spectral lines at $±3 \ \rm kHz$  are both positive and imaginary.

In the context of this task, we will now investigate the case where the source signal
:$$q(t) = q_1(t) + q_2(t)$$
is applied to the input of the phase modulator.
*It is worth mentioning that  $|q(t)| < q_{\rm max} = 1.45 \ \rm V$ .
*This maximum value is slightly smaller than the sum  $A_1 + A_2$ of the individual amplitudes when a sine and a cosine function with the given amplitudes are added up.

In the following questionnaire,
*$S_{\rm TP}(f)$  denotes the spectral function of the equivalent low-pass signal,
*$S_+(f)$  denotes the spectral functions of the analytic signal,

in both cases assuming that  $q(t) = q_1(t) + q_2(t)$  holds and that the carrier frequency is $f_{\rm T} = 100 \ \rm kHz$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
*The values of the Bessel functions can be found in formula collections in table form.
*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kind| Bessel functions of the first kind]]  to solve this task.

===Questions===

<quiz display=simple>
{Let  $q(t) = q_1(t)+q_2(t)$.  Which geometric figure describes the given locus curve $s_{\rm TP}(t)$?
|type="()"}
- The locus curve is an ellipse.
- The locus curve is a circle.
+ The locus curve is approximately a semi-circle.
- The locus curve is an arc, with an approximate opening angle of  $90^\circ$.

{Calculate the spectral function $S_{\rm TP}(f)$.  Between what frequencies  $f_{\rm min}$  and  $f_{\rm max}$  do the spectral lines lie?
|type="{}"}
$f_{\rm min} \ = \ $ { -5.15--4.85 } $\ \rm kHz$
$f_{\rm max} \ = \ ${ 5 3% } $\ \rm kHz$

{Calculate the weight of the Dirac function at  $f = 0$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0.72 3% }
${\rm Im}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0. }

{Calculate the weight of the Dirac function at  $f = 1\ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.36 3% }
${\rm Im}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.03 3% }

{Calculate the weight of the  $S_+(f)$–Dirac function at  $f = 98 \ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.09 3% }
${\rm Im}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.12 3% }
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The third answer is correct:
*In angle modulation, the complex pointer   $s_{\rm TP}(t)$  always moves on a circular arc with the following opening angle:
:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
*Using the (admittedly very rough) approximation  $166^\circ \approx 180^\circ$  we indeed get a semicircle.

'''(2)'''  In general,   $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.
*Since  $B_1(f)$  is limited to the frequencies   $|f| ≤ 2 \ \rm kHz$  and  $B_2(f)$  is limited to the range   $±3 \ \rm kHz$ , the convolution product is limited to  $|f| ≤ 5 \ \rm kHz$ .
*It follows that:
:$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$
:$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$

'''(3)'''  The convolution product for frequency   $f = 0$  results from multiplying  $B_1(f)$  with  $B_2(f)$  and summing.
*Only for   $f = 0$  are both  $B_1(f)$  and  $B_2(f)$  non-zero.
*Thus, we get:
:$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$

'''(4)'''  Now, before multiplication and summation there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$  erfolgen.  This gives:
:$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
+ B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
= 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$

'''(5)'''  The Dirac line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is
:$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +
B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3: Sum of two Oscillations

2022-04-09T15:00:54Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1084__Mod_A_3_3.png|right|frame|Two different Bessel spectra]]
The equivalent low-pass signal with phase modulation, when normalized to the carrier amplitude $(A_{\rm T} = 1)$ is: 
:$$ s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}K_{\rm PM}\hspace{0.05cm}\cdot \hspace{0.05cm}q(t) }\hspace{0.05cm},$$
The modulator constant is assumed to be  $K_{\rm PM} = \rm 1/V$  throughout the task.

The upper graph shows the corresponding spectral function  $B_1(f)$, when the source signal is
:$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$
The weights of the Bessel-Dirac lines when  $η_1 = 0.9$  are obtained as follows:
:$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm}
{\rm J}_1 (0.9) = 0.406 \approx 0.4,$$
:$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm}
{\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$
Use the approximations given in the graph to simplify the calculations.

The Bessel function  $B_2(f)$  is obtained for the source signal
:$$q_2(t) = 0.65\,{\rm V} \cdot \cos(2 \pi \cdot 3\,{\rm kHz} \cdot t)$$
The numerical values of the Dirac lines are obtained from the following:
:$${\rm J}_0 (0.65) = 0.897 \approx 0.9,\hspace{0.3cm}{\rm J}_1 (0.65) = 0.308 \approx 0.3, \hspace{0.3cm}{\rm J}_2 (0.65) = 0.051 \approx 0\hspace{0.05cm}.$$
From the graph, it can be seen that due to the cosine source signal  $q_2(t)$  and the cosine carrier signal $z(t)$ , the spectral lines at $±3 \ \rm kHz$  are both positive and imaginary.

In the context of this task, we will now investigate the case where the source signal
:$$q(t) = q_1(t) + q_2(t)$$
is applied to the input of the phase modulator.
*It is worth mentioning that  $|q(t)| < q_{\rm max} = 1.45 \ \rm V$ .
*This maximum value is slightly smaller than the sum  $A_1 + A_2$ of the individual amplitudes when a sine and a cosine function with the given amplitudes are added up.

In the following questionnaire,
*$S_{\rm TP}(f)$  denotes the spectral function of the equivalent low-pass signal,
*$S_+(f)$  denotes the spectral functions of the analytic signal,

in both cases assuming that  $q(t) = q_1(t) + q_2(t)$  holds and that the carrier frequency is $f_{\rm T} = 100 \ \rm kHz$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
*The values of the Bessel functions can be found in formula collections in table form.
*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kind| Bessel functions of the first kind]]  to solve this task.

===Questions===

<quiz display=simple>
{Let  $q(t) = q_1(t)+q_2(t)$.  Which geometric figure describes the given locus curve $s_{\rm TP}(t)$?
|type="()"}
- The locus curve is an ellipse.
- The locus curve is a circle.
+ The locus curve is approximately a semi-circle.
- The locus curve is an arc, with an approximate opening angle of  $90^\circ$.

{Calculate the spectral function $S_{\rm TP}(f)$.  Between what frequencies  $f_{\rm min}$  and  $f_{\rm max}$  do the spectral lines lie?
|type="{}"}
$f_{\rm min} \ = \ $ { -5.15--4.85 } $\ \rm kHz$
$f_{\rm max} \ = \ ${ 5 3% } $\ \rm kHz$

{Calculate the weight of the Dirac function at  $f = 0$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0.72 3% }
${\rm Im}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0. }

{Calculate the weight of the Dirac function at  $f = 1\ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.36 3% }
${\rm Im}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.03 3% }

{Calculate the weight of the  $S_+(f)$–Dirac function at  $f = 98 \ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.09 3% }
${\rm Im}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.12 3% }
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The third answer is correct:
*In angle modulation, the complex pointer   $s_{\rm TP}(t)$  always moves on a circular arc with the following opening angle:
:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
*Using the (admittedly very rough) approximation  $166^\circ \approx 180^\circ$  we indeed get a semicircle.

'''(2)'''  In general,   $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.
*Since  $B_1(f)$  is limited to the frequencies   $|f| ≤ 2 \ \rm kHz$  and  $B_2(f)$  is limited to the range   $±3 \ \rm kHz$ , the convolution product is limited to  $|f| ≤ 5 \ \rm kHz$ .
*It follows that:
:$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$
:$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$

'''(3)'''  The convolution product for frequency   $f = 0$  results from multiplying  $B_1(f)$  with  $B_2(f)$  and summing.
*Only for   $f = 0$  are both  $B_1(f)$  and  $B_2(f)$  non-zero.
*Thus, we get:
:$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$

'''(4)'''  Now, before multiplication and summationm there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$  erfolgen.  This gives:
:$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
+ B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
= 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$

'''(5)'''  The Dirac line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is
:$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +
B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1Z: Influence of the Message Phase in Phase Modulation

2022-04-09T14:54:45Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1080__Mod_Z_3_1.png|right|frame|Two PM signal waveforms]]
We will now consider the phase modulation of diverse oscillations
:$$ q(t) = \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.05cm}.$$
The source signal is represented here in normalized form with $($amplitude  $1)$ , so that the phase-modulated signal can be characterised by the modulation index (or phase deviation)  $η$  as follows:
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + \eta \cdot q(t) \big]\hspace{0.05cm}.$$
*The signal  $s_1(t)$  shown in the upper graph is characterized by the parameter values  $ϕ_{\rm N} = -90^\circ$  and  $η_1 = 2$ .
*The frequency  $f_{\rm N}$  of this sinusoidal source signal as well as the carrier frequency  $f_{\rm T}$  can be determined from the signal section of duration  $200 \ \rm µ s$  represented here.

*The signal  $s_2(t)$  possibly differs from  $s_1(t)$  due to a different message phase  $ϕ_{\rm N}$  and modulation index  $η$.  All other system parameters are unchanged from  $s_1(t)$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page [[Modulation_Methods/Phase_Modulation_(PM)#Signal_characteristics_of_phase_modulation|Signal characteristics of phase modulation]].

===Questions===

<quiz display=simple>

{Find the frequency  $f_{\rm N}$  of the message signal.
|type="{}"}
$f_{\rm N} \ = \ $ { 5 3% } $\ \rm kHz$

{What is the carrier frequency $f_{\rm T}$?
|type="{}"}
$f_{\rm T} \ = \ $ { 50 3% } $\ \rm kHz$

{What is the maximum phase deviation  $ϕ_{\rm max}$  between  $z(t)$  and  $s(t)$?
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 0.318 3% } $\ \rm rad$

{What is the maximum time shift of the zero crossings that this phase results in?
|type="{}"}
$Δt_{\rm max} \ = \ $ { 6.37 3% } $\ \rm µ s$

{Determine the modulation index  $η_2$  for the signal  $s_2(t)$.
|type="{}"}
$η_2 \ = \ $ { 2 3% }

{What is the phase  $ϕ_{\rm N2}$  of the underlying source signal  $q(t)$ for  $s_2(t)$ ?
|type="{}"}
$ϕ_{\rm N2} \ = \ $ { -139--131 } $\ \rm Grad$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  It can be seen from the sketch that the represented section of the signal of duration  $200 \ \rm µ s$  corresponds exactly to the period duration of the sinusoidal source signal. From this follows  $f_{\rm N}\hspace{0.15cm}\underline{ = 5 \ \rm kHz}$.
*At times  $t = 0$,  $t = 100 \ \rm µ s$  and  $t = 200 \ \rm µ s$ , the signals   $z(t)$  and  $s(t)$  are synchronous in phase.
*In the first half-wave of   $q(t)$ , the zero crossings of   $s(t)$ come slightly earlier than those of the carrier signal  $z(t)$   ⇒   positive phase.
*In contrast, in the range from  $t = 100 \ \rm µ s$  to  $t = 200 \ \rm µ s$ , the phase  $ϕ(t) < 0$.

'''(2)'''  $f_{\rm T}\hspace{0.15cm}\underline{ = 50 \ \rm kHz}$, holds,
*since exactly   $10$  periods can be counted in the shown section of   $z(t)$ of duration  $200 \ \rm µ s$ .

'''(3)'''  The maximum relative phase deviation is   $ϕ_{\rm max} = η_1/(2π)\hspace{0.15cm}\underline{ ≈ 0.318}$.

'''(4)'''  Since the period of the carrier is   $T_0 = 20 \ \rm µ s$ , we obtain  $Δt_{\rm max} = ϕ_{\rm max} ·T_0\hspace{0.15cm}\underline{ ≈ 6.37 \ \rm µ s}$.

'''(5)'''  The maximum phase deviation (shift in the zero intercepts) is exactly the same for   $s_2(t)$  as for  $s_1(t)$. 
*From this, we can conclude that  $η_2 = η_1\hspace{0.15cm}\underline{ = 2}$ .

'''(6)'''  The signal  $s_2(t)$  is shifted to the right by   $25 \ \rm µ s$  compared to   $s_1(t)$ . Therefore, the same must be true for the source signals:
:$$ q_2(t) = q_1(t - 25\,{\rm \mu s}) = \cos \hspace{-0.1cm} \big[2 \pi f_{\rm N} (t - 25\,{\rm \mu s}) \big ] = \cos (\omega_{\rm N} \cdot t - 0.75 \cdot \pi)\hspace{0.05cm}.$$
*This corresponds to the phase position $ϕ_{\rm N2}\hspace{0.15cm}\underline{ = -135^\circ}$.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1Z: Influence of the Message Phase in Phase Modulation

2022-04-09T14:54:34Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1080__Mod_Z_3_1.png|right|frame|Two PM signal waveforms]]
We will now consider the phase modulation of diverse oscillations
:$$ q(t) = \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.05cm}.$$
The source signal is represented here in normalized form with $($amplitude  $1)$ , so that the phase-modulated signal can be characterised by the modulation index (or phase deviation)  $η$  as follows:
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + \eta \cdot q(t) \big]\hspace{0.05cm}.$$
*The signal  $s_1(t)$  shown in the upper graph is characterized by the parameter values  $ϕ_{\rm N} = -90^\circ$  und  $η_1 = 2$ .
*The frequency  $f_{\rm N}$  of this sinusoidal source signal as well as the carrier frequency  $f_{\rm T}$  can be determined from the signal section of duration  $200 \ \rm µ s$  represented here.

*The signal  $s_2(t)$  possibly differs from  $s_1(t)$  due to a different message phase  $ϕ_{\rm N}$  and modulation index  $η$.  All other system parameters are unchanged from  $s_1(t)$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page [[Modulation_Methods/Phase_Modulation_(PM)#Signal_characteristics_of_phase_modulation|Signal characteristics of phase modulation]].

===Questions===

<quiz display=simple>

{Find the frequency  $f_{\rm N}$  of the message signal.
|type="{}"}
$f_{\rm N} \ = \ $ { 5 3% } $\ \rm kHz$

{What is the carrier frequency $f_{\rm T}$?
|type="{}"}
$f_{\rm T} \ = \ $ { 50 3% } $\ \rm kHz$

{What is the maximum phase deviation  $ϕ_{\rm max}$  between  $z(t)$  and  $s(t)$?
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 0.318 3% } $\ \rm rad$

{What is the maximum time shift of the zero crossings that this phase results in?
|type="{}"}
$Δt_{\rm max} \ = \ $ { 6.37 3% } $\ \rm µ s$

{Determine the modulation index  $η_2$  for the signal  $s_2(t)$.
|type="{}"}
$η_2 \ = \ $ { 2 3% }

{What is the phase  $ϕ_{\rm N2}$  of the underlying source signal  $q(t)$ for  $s_2(t)$ ?
|type="{}"}
$ϕ_{\rm N2} \ = \ $ { -139--131 } $\ \rm Grad$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  It can be seen from the sketch that the represented section of the signal of duration  $200 \ \rm µ s$  corresponds exactly to the period duration of the sinusoidal source signal. From this follows  $f_{\rm N}\hspace{0.15cm}\underline{ = 5 \ \rm kHz}$.
*At times  $t = 0$,  $t = 100 \ \rm µ s$  and  $t = 200 \ \rm µ s$ , the signals   $z(t)$  and  $s(t)$  are synchronous in phase.
*In the first half-wave of   $q(t)$ , the zero crossings of   $s(t)$ come slightly earlier than those of the carrier signal  $z(t)$   ⇒   positive phase.
*In contrast, in the range from  $t = 100 \ \rm µ s$  to  $t = 200 \ \rm µ s$ , the phase  $ϕ(t) < 0$.

'''(2)'''  $f_{\rm T}\hspace{0.15cm}\underline{ = 50 \ \rm kHz}$, holds,
*since exactly   $10$  periods can be counted in the shown section of   $z(t)$ of duration  $200 \ \rm µ s$ .

'''(3)'''  The maximum relative phase deviation is   $ϕ_{\rm max} = η_1/(2π)\hspace{0.15cm}\underline{ ≈ 0.318}$.

'''(4)'''  Since the period of the carrier is   $T_0 = 20 \ \rm µ s$ , we obtain  $Δt_{\rm max} = ϕ_{\rm max} ·T_0\hspace{0.15cm}\underline{ ≈ 6.37 \ \rm µ s}$.

'''(5)'''  The maximum phase deviation (shift in the zero intercepts) is exactly the same for   $s_2(t)$  as for  $s_1(t)$. 
*From this, we can conclude that  $η_2 = η_1\hspace{0.15cm}\underline{ = 2}$ .

'''(6)'''  The signal  $s_2(t)$  is shifted to the right by   $25 \ \rm µ s$  compared to   $s_1(t)$ . Therefore, the same must be true for the source signals:
:$$ q_2(t) = q_1(t - 25\,{\rm \mu s}) = \cos \hspace{-0.1cm} \big[2 \pi f_{\rm N} (t - 25\,{\rm \mu s}) \big ] = \cos (\omega_{\rm N} \cdot t - 0.75 \cdot \pi)\hspace{0.05cm}.$$
*This corresponds to the phase position $ϕ_{\rm N2}\hspace{0.15cm}\underline{ = -135^\circ}$.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1: Phase Modulation Locus Curve

2022-04-09T14:54:07Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1079__Mod_A_3_1.png|right|frame|Two locus curves to choose from]]
The locus curve is generally understood as the plot of the equivalent low-pass signal $s_{\rm TP}(t)$  in the complex plane.
*The graph shows locus curves at the output of two modulators  $\rm M_1$  and  $\rm M_2$.
*The real and imaginary parts are each normalized to $1 \ \rm V$ in this graph.

Let the source signal be the same for both modulators:
$$ q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}
{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.$$
One of the two modulators implements phase modulation, which is characterized by the following equations:
:$$ s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},$$
:$$ s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm},$$
:$$ \phi(t) = K_{\rm PM} \cdot q(t)\hspace{0.05cm}.$$
The maximum value  $ϕ(t)$  is called the   ''modulation index''  $η$.  Often  $η$  is also called   ''phase deviation''  in the literature.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

===Questions===

<quiz display=simple>
{Which modulation method is used by modulator  $\rm M_1$?
|type="()"}
- Double-sideband amplitude modulation.
+ Single sideband amplitude modulation.
- Phase modulation.

{Which modulation method is used by modulator  $\rm M_2$?
|type="()"}
- Double-sideband amplitude modulation.
- Single sideband amplitude modulation.
+ Phase modulation.

{What is the carrier amplitude  $A_{\rm T}$  for the phase modulator?  Note the normalization to  $1 \ \rm V$.
|type="{}"}
$A_{\rm T} \ = \ $ { 1 3% } $\ \rm V$

{What are the values of the modulation index  $η$  and the modulator constant  $K_{\rm PM}$?
|type="{}"}
$η\ = \ $ { 3.1415 3% }
$K_{\rm PM}\ = \ $ { 1.571 3% } $\ \rm 1/V$

{Describe the motion on the locus curve. At what time $t_1$  is the starting point  $s_{\rm TP}(t = 0) = -1 \ \rm V$  first reached again?
|type="{}"}
$t_1\ = \ $ { 100 3% } $ \ \rm µ s$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We are dealing with SSB-AM with a sideband-to-carrier ratio $μ = 1$   ⇒   Answer 2:
*If one moves in the mathematically positive direction on the circle, it is specifically an USB–AM, otherwise it is a LSB–AM.
*The phase function  $ϕ(t)$  as the angle of a point  $s_{\rm TP}(t)$  on the circle (arc) with respect to the coordinate origin can take values between  $±π/2$  and does not show a cosine progression.
*The envelope   $a(t) = |s_{\rm TP}(t)|$  is also not cosine.
*If an envelope demodulator were used for  $\rm M_1$  at the receiver, nonlinear distortions would occur, in contrast to DSB–AM, which has a horizontal straight line for a locus curve.

'''(2)'''  Here, we observe phase modulation   ⇒   Answer 3:
*The envelope   $a(t) = A_{\rm T}$  is constant,
*while the phase  $ϕ(t)$  is cosinusoidal according to the source signal  $q(t)$ .

'''(3)'''  In the case of phase modulation:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.$$
*From the graph, we can read the carrier amplitude   $A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V}$  as the radius of the circle.

'''(4)'''  The source signal  $q(t)$  is at its maximum at time  $t = 0$  and therefore so is the phase function:
:$$ \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.$$
*This gives the modulator constant:
$$K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.$$

'''(5)'''  One moves clockwise along the circular arc.
*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.
*Afterwards, one moves counterclockwise along the arc.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.4: Simple Phase Modulator

2022-04-09T14:51:59Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1086__Mod_A_3_4.png|right|frame|"Approximate phase modulator"]]
The adjacent circuit allows the approximate realization of a phase-modulated signal. 

From the cosinusoidal carrier,  $z(t)$ , the  $90^\circ$ phase shifter forms a sinusoidal signal of the same frequency, such that the modulated signal can be written as:
:$$ s(t) = z(t) + q(t) \cdot \frac{z(t- T_0/4)}{A_{\rm T}}
= A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t) + q(t) \cdot \sin (\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
The second term describes a "DSB–AM without carrier".  Additionally, the carrier, phase-shifted by $90^\circ$ , is added.  Thus, with a cosine source signal $q(t) = A_{\rm N} \cdot \cos (\omega_{\rm N} \cdot t)$ , we get:
:$$s(t) = A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t) + A_{\rm N} \cdot \cos (\omega_{\rm N} \cdot t) \cdot \sin (\omega_{\rm T} \cdot t) $$
:$$\Rightarrow \hspace{0.3cm}s(t) = A_{\rm T} \cdot \big[\cos (\omega_{\rm T} \cdot t) + \eta \cdot \cos (\omega_{\rm N} \cdot t) \cdot \sin (\omega_{\rm T} \cdot t) \big] \hspace{0.05cm}.$$
We refer to the ratio  $η = A_{\rm N}/A_{\rm T}$  as the modulation index;  in the following, the carrier amplitude is set to   $A_{\rm T} = 1$  for simplicity.

*In contrast to  [[Modulation_Methods/Phasenmodulation_(PM)#Signalverl.C3.A4ufe_bei_Phasenmodulation|ideal phase modulation]]  the modulation index  $η$  and the phase deviation  $ϕ_{\rm max}$ may differ in this "approximate phase modulation".
*Additionally, we can see that the envelope  $a(t) ≠ 1$ .  This means that an unwanted amplitude modulation is superimposed on the phase modulation.

From the representation of the equivalent low-pass signal  $s_{\rm TP}(t)$  in the complex plane (locus curve), the following are to be calculated in this task:
*the envelope  $a(t)$  and
*the phase function $ϕ(t)$.

Then, you are to analyse the distortions arising when an ideal PM demodulator, which sets the sink signal  $v(t)$  proportional to the phase  $ϕ(t)$ , is used on the receiving side of this nonideal PM modulator.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

*You can use the following equations to approximate the distortion factor:
:$$\arctan(\gamma) \approx \gamma - {\gamma^3}/{3} \hspace{0.05cm}, \hspace{0.3cm} \cos^3(\gamma) ={3}/{4} \cdot \cos(\gamma) +{1}/{4} \cdot \cos(3 \cdot \gamma) \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Calculate the equivalent low-pass signal. Which statement is true?
|type="()"}
- The locus curve  $s_{\rm TP}(t)$  is a circular arc.
- The locus curve  $s_{\rm TP}(t)$  is a horizontal straight line.
+ The locus curve  $s_{\rm TP}(t)$  is a vertical straight line.

{Calculate the (normalized) envelope $a(t)$  for  $A_{\rm T} = 1$.  What are its minimum and maximum values when  $η = 1$?
|type="{}"}
$a_{\rm min} \ = \ $ { 1 3% }
$a_{\rm max} \ = \ $ { 1.414 3% }

{Calculate the maximum value of the phase $ϕ(t)$  for  $η = 1$  and  $η = 0.5$.
|type="{}"}
$η = 1.0\text{:} \ \ \ ϕ_{\rm max} \ = \ $ { 45 3% } $\ \rm Grad$
$η = 0.5\text{:} \ \ \ ϕ_{\rm max} \ = \ $ { 26.6 3% } $\ \rm Grad$

{What are distortions result after ideal phase demodulation of  $s(t)$?
|type="()"}
- No distortions occur.
- Linear distortions occur.
+ Nonlinear distortions occur.

{Calculate the distortion factor  $K$  considering the trigonometric relationships given on the exercise page.
|type="{}"}
$η = 1.0\text{:} \ \ \ K \ = \ $ { 11.1 3% } $\ \text{%}$
$η = 0.5\text{:} \ \ \ K \ = \ $ { 2.2 3% } $\ \text{%}$

</quiz>

===Solution===
{{ML-Kopf}}
[[File:P_ID1087__Mod_A_3_4_a.png|right|frame|Construction of the "vertical locus" from the pointers]]
'''(1)'''  Answer 3 is correct:
*The equivalent low-pass signal is::$$s_{\rm TP}(t) = A_{\rm T} \cdot \left ( 1 + {\rm j}\cdot \frac {\eta}{2}\cdot \left ({\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}\right) \right)
= A_{\rm T} \cdot \big ( 1 + {\rm j}\cdot {\eta}\cdot \cos (\omega_{\rm N} \cdot t) \big)\hspace{0.05cm}.$$
*The graph illustrates that the locus curve  $s_{\rm TP}(t)$  is now a a vertical straight line in contrast to the ideal PM (circular arc) and DSB–AM  (horizontal straight line). 
*In the following, we set   $A_{\rm T} = 1$ .

'''(2)'''  The envelope is obtained from the time-dependent pointer length as
:$$a(t) = \sqrt{1 + \eta^2 \cdot \cos^2 (\omega_{\rm N} \cdot t)} \hspace{0.3cm}
\Rightarrow \hspace{0.3cm}a_{\rm min} \hspace{0.15cm}\underline { = 1}, \hspace{0.3cm}a_{\rm max} = \sqrt{1 + \eta^2 }\hspace{0.05cm}.$$
*For  $η = 1$  the maximum value becomes  $a_{\rm max} = \sqrt{2}\hspace{0.15cm}\underline { ≈ 1.414}$.

'''(3)'''  The phase function of this simple phase demodulator is given by:
:$$\phi(t) = \arctan \frac{{\rm Im}[s_{\rm TP}(t)]}{{\rm Re}[s_{\rm TP}(t)]} = \arctan (\eta \cdot \cos (\omega_{\rm N} \cdot t)) \hspace{0.05cm}.$$
*The maximum value occurs at time  $t = 0$ , for example, and is   $ϕ_{\rm max} = \arctan(η)$.
:*When  $η = 1$ , one obtains   $ϕ_{\rm max}\hspace{0.15cm}\underline { = 45^\circ}$  $($to compare:  for ideal PM  $57.3^\circ)$,
:*When  $η = 0.5$  one gets   $ϕ_{\rm max}\hspace{0.15cm}\underline { \approx 26.6^\circ}$  $($for ideal PM  $28.7^\circ)$.

'''(4)''' Answer 3 is correct:
*It is  '''not''' true that:      $\arctan\big [η · \cos(γ)\big ] = η · \cos(γ)$.
*This means that the sink signal is not cosine, in contrast to the source signal.
*This points to nonlinear distortions.

'''(5)'''  Using  $γ = η · \cos(ω_N · t)$  and  $\arctan(γ) ≈ γ – γ^3/3$ , we get:
:$$ \phi(t) = \eta \cdot \cos (\omega_{\rm N} \cdot t) - \frac{\eta^3}{3}\cdot \cos^3 (\omega_{\rm N} \cdot t))=
\eta \cdot \cos (\omega_{\rm N} \cdot t) - \frac{\eta^3}{3}\cdot \left [ {3}/{4}\cdot \cos (\omega_{\rm N} \cdot t) + {1}/{4}\cdot \cos (3 \omega_{\rm N} \cdot t)\right ] $$
:$$\Rightarrow \hspace{0.3cm} \phi(t) = \left(\eta - {\eta^3}/{4} \right) \cdot \cos (\omega_{\rm N} \cdot t) - {\eta^3}/{12}\cdot \cos (3\omega_{\rm N} \cdot t) \hspace{0.05cm}.$$
*This means:  using the given series expansion (where 5th and higher order terms are ignored), only the third-order harmonic distortion is non-zero. Thus:
:$$K = K_3 = \frac{\eta^3/12}{\eta-\eta^3/4}= \frac{1}{12/\eta^2 -3} \hspace{0.05cm}.$$
*When  $η = 1$  the numerical value is  $K = 1/9 \hspace{0.15cm}\underline { ≈ 11.1\%}$.
*When  $η = 0.5$  the distortion factor is  $K = 1/45 \hspace{0.15cm}\underline {≈ 2.2\%}$.

A simulation shows that by stopping the series after the third order term, we have made the error of over-estimating the distortion factor:
*The values obtained by simulation are   $K ≈ 6%$  $($for  $η = 1)$  and  $K ≈ 2%$  $($for  $η = 0.5)$.
*Thus, the error increases more than proportionally with increasing  $η$ .

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3Z: Characteristics Determination

2022-04-09T14:51:48Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1085__Mod_Z_3_3.png|right|frame|Spectrum of the analytical signal]]
Let us consider the phase modulation of the harmonic oscillation
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$
which, given a normalized carrier amplitude  $(A_{\rm T} = 1)$ , leads to the following transmitted signal:
:$$ s(t) = \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \big]\hspace{0.05cm}.$$
The spectrum of the corresponding analytical signal $s_{\rm TP}(t)$  is generally:
:$$S_{\rm TP}(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}(\phi_{\rm N}\hspace{0.05cm}+\hspace{0.05cm} 90^\circ) }\cdot \hspace{0.05cm} \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}$$
Here,  $η = K_{\rm PM} · A_{\rm N}$  is called the modulation index.

In the graph, the real and imaginary parts of the spectrum  $S_+(f)$  of the analytical signal  $s_+(t)$ are shown separately. This should be used to determine the characteristics  $f_{\rm T}$,  $f_{\rm N}$,  $ϕ_{\rm N}$  and  $η$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

*For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:
:$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm J}_{2} (\eta)= {2}/{\eta} \cdot {\rm J}_{1} (\eta) - {\rm J}_{0} (\eta) \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{What are the frequencies  $f_{\rm T}$  and  $f_{\rm N}$?
|type="{}"}
$f_{\rm T} \ = \ $ { 40 3% } $\ \rm kHz$
$f_{\rm N} \ = \ $ { 3 3% } $\ \rm kHz$

{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 3 \ \rm kHz)$.
|type="{}"}
$|S_{\rm TP}(f = 3 \ \rm kHz)| \ = \ $ { 0.558 3% }
${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $ { 60 3% } $\ \rm Grad$

{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 6 \ \rm kHz)$.
|type="{}"}
$|S_{\rm TP}(f = 6 \ \rm kHz)| \ = \ $ { 0.232 3% }
${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $ { 120 3% } $\ \rm Grad$

{What is the phase of the source signal  $q(t)$?
|type="{}"}
$ϕ_{\rm N} \ = \ $ { -30.9--29.1 } $\ \rm Grad$

{What is the magnitude of the modulation index  $η$ ?
|type="{}"}
$η \ = \ $ { 1.5 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Regarding   $|S_+(f)|$  there is a symmetry with respect to the carrier frequency  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  The distance between the spectral lines is  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.

'''(2)'''  Considering  $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$ , it holds that:
:$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$
:$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$

'''(3)'''  Analogously to in question  '''(2)''' , at frequency   $f = 6 \ \rm kHz$ we obtain:
:$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$
:$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$

'''(4)'''  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   '''(2)''', the phase is:
:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
*Checkin this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  '''(3)'''  yields the same value:
:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$

'''(5)'''  The given equation can be rewritten as follows:
:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
*Mit  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  und  ${\rm J}_2(η) = 0.232$  erhält man somit:
:$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.3: Sum of two Oscillations

2022-04-09T14:51:38Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1084__Mod_A_3_3.png|right|frame|Two different Bessel spectra]]
The equivalent low-pass signal with phase modulation, when normalized to the carrier amplitude $(A_{\rm T} = 1)$ is: 
:$$ s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}K_{\rm PM}\hspace{0.05cm}\cdot \hspace{0.05cm}q(t) }\hspace{0.05cm},$$
The modulator constant is assumed to be  $K_{\rm PM} = \rm 1/V$  throughout the task.

The upper graph shows the corresponding spectral function  $B_1(f)$, when the source signal is
:$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$
The weights of the Bessel-Dirac lines when  $η_1 = 0.9$  are obtained as follows:
:$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm}
{\rm J}_1 (0.9) = 0.406 \approx 0.4,$$
:$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm}
{\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$
Use the approximations given in the graph to simplify the calculations.

The Bessel function  $B_2(f)$  is obtained for the source signal
:$$q_2(t) = 0.65\,{\rm V} \cdot \cos(2 \pi \cdot 3\,{\rm kHz} \cdot t)$$
The numerical values of the Dirac lines are obtained from the following:
:$${\rm J}_0 (0.65) = 0.897 \approx 0.9,\hspace{0.3cm}{\rm J}_1 (0.65) = 0.308 \approx 0.3, \hspace{0.3cm}{\rm J}_2 (0.65) = 0.051 \approx 0\hspace{0.05cm}.$$
From the graph, it can be seen that due to the cosine source signal  $q_2(t)$  and the cosine carrier signal $z(t)$ , the spectral lines at $±3 \ \rm kHz$  are both positive and imaginary.

In the context of this task, we will now investigate the case where the source signal
:$$q(t) = q_1(t) + q_2(t)$$
is applied to the input of the phase modulator.
*It is worth mentioning that  $|q(t)| < q_{\rm max} = 1.45 \ \rm V$ .
*This maximum value is slightly smaller than the sum  $A_1 + A_2$ of the individual amplitudes when a sine and a cosine function with the given amplitudes are added up.

In the following questionnaire,
*$S_{\rm TP}(f)$  denotes the spectral function of the equivalent low-pass signal,
*$S_+(f)$  denotes the spectral functions of the analytic signal,

in both cases assuming that  $q(t) = q_1(t) + q_2(t)$  holds and that the carrier frequency is $f_{\rm T} = 100 \ \rm kHz$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
*The values of the Bessel functions can be found in formula collections in table form.
*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kind| Bessel functions of the first kind]]  to solve this task.

===Questions===

<quiz display=simple>
{Let  $q(t) = q_1(t)+q_2(t)$.  Which geometric figure describes the given locus curve $s_{\rm TP}(t)$?
|type="()"}
- The locus curve is an ellipse.
- The locus curve is a circle.
+ The locus curve is approximately a semi-circle.
- The locus curve is an arc, with an approximate opening angle of  $90^\circ$.

{Calculate the spectral function $S_{\rm TP}(f)$.  Between what frequencies  $f_{\rm min}$  and  $f_{\rm max}$  do the spectral lines lie?
|type="{}"}
$f_{\rm min} \ = \ $ { -5.15--4.85 } $\ \rm kHz$
$f_{\rm max} \ = \ ${ 5 3% } $\ \rm kHz$

{Calculate the weight of the Dirac function at  $f = 0$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0.72 3% }
${\rm Im}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0. }

{Calculate the weight of the Dirac function at  $f = 1\ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.36 3% }
${\rm Im}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $ { 0.03 3% }

{Calculate the weight of the  $S_+(f)$–Dirac function at  $f = 98 \ \rm kHz$.
|type="{}"}
${\rm Re}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.09 3% }
${\rm Im}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $ { 0.12 3% }
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The third answer is correct:
*In angle modulation, the complex pointer   $s_{\rm TP}(t)$  always moves on a circular arc the following opening angle:
:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
*Using the (admittedly very rough) approximation  $166^\circ \approx 180^\circ$  we indeed get a semicircle.

'''(2)'''  In general,   $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.
*Since  $B_1(f)$  is limited to the frequencies   $|f| ≤ 2 \ \rm kHz$  and  $B_2(f)$  is limited to the range   $±3 \ \rm kHz$ , the convolution product is limited to  $|f| ≤ 5 \ \rm kHz$ .
*It follows that:
:$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$
:$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$

'''(3)'''  The convolution product for frequency   $f = 0$  results from multiplying  $B_1(f)$  with  $B_2(f)$  and summing.
*Only for   $f = 0$  are both  $B_1(f)$  and  $B_2(f)$  non-zero.
*Thus, we get:
:$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$

'''(4)'''  Now, before multiplication and summationm there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$  erfolgen.  This gives:
:$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
+ B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
= 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$

'''(5)'''  The Dirac line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is
:$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +
B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$
:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.2Z: Bessel Spectrum

2022-04-09T14:51:28Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1083__Mod_Z_3_2.png|right|frame|Progression of Bessel functions]]
Consider the complex signal
:$$x(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t) }\hspace{0.05cm}.$$
For example, the equivalent low-pass signal at the output of an angle modulator (PM, FM) can be represented in this form if appropriate normalizations are made.

*When  $T_0 = 2π/ω_0$, the Fourier series representation is:
:$$x(t) = \sum_{n = - \infty}^{+\infty}D_n \cdot{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t }\hspace{0.05cm},$$
:$$ D_n = \frac{1}{T_0}\cdot \int_{- T_0/2}^{+T_0/2}x(t) \cdot{\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
*These complex Fourier coefficients can be expressed using $n$–th order Bessel functions of the first kind:
:$${\rm J}_n (\eta) = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
*These are shown on the graph in the range  $0 ≤ η ≤ 5$ . For negative values of $n$  one obtains:
:$${\rm J}_{-n} (\eta) = (-1)^n \cdot {\rm J}_{n} (\eta)\hspace{0.05cm}.$$
*The series representation of the Bessel functions is:
:$${\rm J}_n (\eta) = \sum\limits_{k=0}^{\infty}\frac{(-1)^k \cdot (\eta/2)^{n \hspace{0.05cm} + \hspace{0.05cm} 2 \hspace{0.02cm}\cdot \hspace{0.05cm}k}}{k! \cdot (n+k)!} \hspace{0.05cm}.$$
*If the function values for  $n = 0$  and  $n = 1$  are known, the Bessel functions for  $n ≥ 2$  can be determined from them by iteration:
:$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.05cm}.$$

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
*The values of the Bessel functions can be found in collections of formulae in table form.
*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kind| Bessel functions of the first kind]]  to solve this task.

===Questions===

<quiz display=simple>

{What are the properties of the signal  $x(t)$?
|type="[]"}
- $x(t)$  is imaginary for all times  $t$ .
+ $x(t)$  is periodic.
- The spectral function $X(f)$  is obtained via the Fourier integral.

{Write the Fourier coefficients  $D_n$  together with the Bessel functions of the first kind   ⇒   ${\rm J}_n(η)$.  What relationships can be seen?
|type="[]"}
- All  $D_n$  are equal to  ${\rm J}_η(0)$.
+  $D_n = {\rm J}_n(η)$ holds.
-  $D_n = -{\rm J}_η(n)$ holds.

{ What are the properties of the Fourier coefficients?
|type="()"}
+ All  $D_n$  are purely real.
- All  $D_n$  are purely imaginary.

{For  $η = 2$ , the coefficients are  $D_0 = 0.224$  and  $D_1 = 0.577$.  From this, calculate the coefficients  $D_2$  and  $D_3$.
|type="{}"}
$D_2 \ = \ $ { 0.353 3% }
$D_3 \ = \ $ { 0.129 3% }

{What are the Fourier coefficients  $D_{-2}$  and  $D_{-3}$ ?
|type="{}"}
$D_{-2} \ = \ $ { 0.353 3% }
$D_{-3} \ = \ $ { -0.133--0.125 }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Only the second answer is correct:
*$x(t)$  is a complex signal that only becomes real in exceptional cases, for example at time   $t = 0$.
*A purely imaginary value (at certain times) can only result when $η ≥ π/2$    ⇒   Answer 1 is incorrect.
*For example, when  $T_0 = 2π/ω_0$ :
:$$ x(t + k \cdot T_0) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} (t \hspace{0.05cm}+ \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm}T_0)) } =
{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} + \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} ) } = x(t)\hspace{0.05cm}.$$
*This signal is periodic.  The Fourier series, not the Fourier integral, must be used to calculate the spectral function.

'''(2)'''  The Fourier coefficients are:
:$$ D_n = \frac{1}{T_0}\cdot \int_{- T_0/2}^{+T_0/2}{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t) }\cdot{\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
*Combining the two terms and after substituting   $α = ω_0 · t$ , we get:
:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm} = {\rm J}_n (\eta) .$$
*Thus, the second answer is correct.

'''(3)'''  Using Euler's theorem, the Fourier coefficients can be represented as follows:
:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha +
\frac{\rm j}{2\pi}\cdot \int_{-\pi}^{+\pi} {\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
*The integrand of the first integral is an even function of  $\alpha$:
:$$I_1 (-\alpha) = {\cos( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\cos( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=
{\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = I_1 (\alpha) \hspace{0.05cm}.$$
*In contrast, the second integrand is an odd function:
:$$I_2 (-\alpha) = {\sin( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\sin( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=
-{\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = -I_2 (\alpha) \hspace{0.05cm}.$$
*Thus, the second integral vanishes and, taking symmetry into account, we obtain:
:$$D_n = \frac{1}{\pi}\cdot \int_{0}^{\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
*Thus, the correct solution is Answer 1.

'''(4)'''  According to the formula for iterative calculation, when  $η = 2$:
:$$ D_2 = D_1 - D_0 = 0.577 - 0.224 \hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$
:$$D_3 = 2 \cdot D_2 - D_1 = 2 \cdot 0.353 - 0.577 \hspace{0.15cm}\underline {= 0.129} \hspace{0.05cm}.$$

'''(5)'''  Due to the given symmetry relation, it further holds that:
:$$ D_{–2} = D_2\hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$
:$$D_{–3} = -D_3 \hspace{0.15cm}\underline {= -0.129} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.2: Spectrum with Angle Modulation

2022-04-09T14:51:16Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1081__Mod_A_3_2.png|right|frame|Table of Bessel functions]]
The following equations are assumed here:
* Source signal:
:$$q(t) = 2\,{\rm V} \cdot \sin(2 \pi \cdot 3\,{\rm kHz} \cdot t)\hspace{0.05cm},$$
* Transmit signal:
:$$s(t) = 1\,{\rm V} \cdot \cos\hspace{-0.1cm}\big[2 \pi \cdot 100\,{\rm kHz} \cdot t + K_{\rm M} \cdot q(t)\big ]\hspace{0.05cm},$$
* Received signal (ideal channel):
:$$r(t) = s(t) = 1\,{\rm V} \cdot \cos\hspace{-0.1cm}\big[2 \pi \cdot 100\,{\rm kHz} \cdot t + \phi(t)\big ]\hspace{0.05cm},$$
* ideal demodulator:
:$$ v(t) = \frac{1}{ K_{\rm M}} \cdot \phi(t)\hspace{0.05cm}.$$
The graphs shows the   $n$–th order Bessel functions of the first kind   ${\rm J}_n (\eta)$  in table form.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the pages  [[Modulation_Methods/Phasenmodulation_(PM)#Spectral_function_of_a_phase-modulated_sine_signal|Spectral function of a phase-modulated sine signal]]  and  [[Modulation_Methods/Phase_Modulation_(PM)#Interpretation_of_the_Bessel_spectrum|Interpretation of the Bessel spectrum]].

===Questions===

<quiz display=simple>
{Which modulation method is used here?
|type="()"}
- Amplitude modulation.
+ Phase modulation.
- Frequency modulation.

{Which modulation method would you choose if the channel bandwidth was only  $B_{\rm K} = 10 \ \rm kHz$ ?
|type="()"}
+ Amplitude modulation.
- Phase modulation.
- Frequency modulation.

{How should one choose the modulator constant $K_{\rm M}$  for a phase deviation of  $η = 1$ ?
|type="{}"}
$K_{\rm M} \ = \ $ { 0.5 3% } $\ \rm 1/V$

{Calculate the spectrum  $S_{\rm TP}(f)$  of the equivalent low-pass signal  $s_{\rm TP}(t)$. 
What are the weights of the spectral lines at  $f = 0$  and  $f = -3 \ \rm kHz$?
|type="{}"}
$S_{\rm TP}(f = 0)\ = \ $ { 0.765 3% } $\ \rm V$
$S_{\rm TP}(f = -3\ \rm kHz) \ = \ $ { -0.453--0.427 } $\ \rm V$

{Calculate the spectra of the analytical signal $s_{\rm +}(t)$  and the physical signal  $s(t)$.  What are the weights of the spectral lines at  $f = 97 \ \rm kHz$?
|type="{}"}
$S_+(f = 97 \ \rm kHz)\ = \ $ { -0.453--0.427 } $\ \rm V$
$S(f = 97 \ \rm kHz)\hspace{0.32cm} = \ $ { -0.226--0.214 } $\ \rm V$

{What is the required channel bandwidth  $B_{\rm K}$  for  $ η = 1$, if one ignores pulse weights smaller (in magnitude) than $0.01$ ?
|type="{}"}
$η = 1\text{:} \ \ \ B_{\rm K}\ = \ $ { 18 3% } $\ \rm kHz$

{What would be the channel bandwidths for  $η = 2$  and  $η = 3$ ?
|type="{}"}
$η = 2\text{:} \ \ \ B_{\rm K}\ = \ $ { 24 3% } $\ \rm kHz$
$η = 3\text{:} \ \ \ B_{\rm K}\ = \ $ { 36 3% } $\ \rm kHz$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The phase  $ϕ(t)$  is proportional to the source signal  $q(t)$   ⇒   this is a phase modulation   ⇒   Answer 2.

'''(2)'''  An angle modulation  (PM, FM)  always results in nonlinear distortion when the channel is bandlimited.
*In contrast, double-sideband amplitude modulation  (DSB-AM)  here enables distortion-free transmission with  $B_{\rm K} = 6 \ \rm kHz$ ; ⇒   Answer 1.

'''(3)'''  The modulation index (or phase deviation) is equal to  $η = K_{\rm M} · A_{\rm N}$ for phase modulation.
*Thus, the modulator constant must be set to  $K_{\rm M} = 1/A_{\rm N}\hspace{0.15cm}\underline { = 0.5 \rm \cdot {1}/{V}}$  to give   $η = 1$ .

'''(4)'''  A so-called Bessel spectrum is present:
:$$ S_{\rm TP}(f) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}.$$
*This is a discrete spectrum with components at   $f = n · f_{\rm N}$, where  $n$  is an integer.
*The weights of the Dirac functions are given by the Bessel functions.  When  $A_{\rm T} = 1\ \rm V$ , one obtains:
[[File:P_ID1082__Mod_A_3_2_d.png|right|frame|PM spectrum in the equivalent low-pass range]]
:$$ S_{\rm TP}(f = 0) = A_{\rm T} \cdot {\rm J}_0 (\eta = 1) \hspace{0.15cm}\underline {= 0.765\,{\rm V}},$$
:$$ S_{\rm TP}(f = f_{\rm N}) = A_{\rm T} \cdot {\rm J}_1 (\eta = 1)\hspace{0.15cm} = 0.440\,{\rm V},$$
:$$ S_{\rm TP}(f = 2 \cdot f_{\rm N}) = A_{\rm T} \cdot {\rm J}_2 (\eta = 1) = 0.115\,{\rm V} \hspace{0.05cm}.$$
*Due to the symmetry   ${\rm J}_{-n} (\eta) = (-1)^n \cdot {\rm J}_{n} (\eta)$ , the spectral line at   $f = -3 \ \rm kHz$ is obtained as:
:$$S_{\rm TP}(f = -f_{\rm N}) = -S_{\rm TP}(f = +f_{\rm N}) =\hspace{-0.01cm}\underline { -0.440\,{\rm V} \hspace{0.05cm}}.$$
''Note'':  For the spectral value at  $f = 0$  we should actually write:
:$$S_{\rm TP}(f = 0) = 0.765\,{\rm V} \cdot \delta (f) \hspace{0.05cm}.$$
*This is therefore infinite due to the Dirac function, and only the weight of the Dirac function is finite.
*The same applies for all discrete spectral lines.

'''(5)'''  $S_+(f)$  is obtained from   $S_{\rm TP}(f)$  by shifting   $f_{\rm T}$ to the right.  Therefore
:$$S_{\rm +}(f = 97\,{\rm kHz}) = S_{\rm TP}(f = -3\,{\rm kHz}) \hspace{0.15cm}\underline {=-0.440\,{\rm V}} \hspace{0.05cm}.$$
*The actual spectrum differs from  $S_+(f)$  by a factor of   $1/2$ at positive frequencies:
:$$S(f = 97\,{\rm kHz}) = {1}/{2} \cdot S_{\rm +}(f = 97\,{\rm kHz}) \hspace{0.15cm}\underline {=-0.220\,{\rm V}} \hspace{0.05cm}.$$
*In general, we can write:
:$$ S(f) = \frac{A_{\rm T}}{2} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta (f \pm (f_{\rm T}+ n \cdot f_{\rm N}))\hspace{0.05cm}.$$

'''(6)'''  Under the suggested conditions, all the Bessel lines  ${\rm J}_{|n|>3}$  can be disregarded.
* This gives  $B_{\rm K} = 2 · 3 · f_{\rm N}\hspace{0.15cm}\underline { = 18 \ \rm kHz}$.

'''(7)'''  The numerical values in the table given on the exercise page show that the following channel bandwidths would now be required:
*für $η = 2$:     $B_{\rm K} \hspace{0.15cm}\underline { = 24 \ \rm kHz}$,
*für $η = 3$:     $B_{\rm K} \hspace{0.15cm}\underline { = 36 \ \rm kHz}$.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1Z: Influence of the Message Phase in Phase Modulation

2022-04-09T14:51:00Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1080__Mod_Z_3_1.png|right|frame|Two PM signal waveforms]]
We will now consider the phase modulation of diverse oscillations
:$$ q(t) = \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.05cm}.$$
The source signal is represented here in normalized form with $($amplitude  $1)$ , so that the phase-modulated signal can be characterised by the modulation index (or phase deviation)  $η$  as follows:
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + \eta \cdot q(t) \big]\hspace{0.05cm}.$$
*The signal  $s_1(t)$  shown in the upper graph is characterized by the paramet values  $ϕ_{\rm N} = -90^\circ$  und  $η_1 = 2$ .
*The frequency  $f_{\rm N}$  of this sinusoidal source signal as well as the carrier frequency  $f_{\rm T}$  can be determined from the signal section of duration  $200 \ \rm µ s$  represented here.

*The signal  $s_2(t)$  possibly differs from  $s_1(t)$  due to a different message phase  $ϕ_{\rm N}$  and modulation index  $η$.  All other system parameters are unchanged from  $s_1(t)$ .

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page [[Modulation_Methods/Phase_Modulation_(PM)#Signal_characteristics_of_phase_modulation|Signal characteristics of phase modulation]].

===Questions===

<quiz display=simple>

{Find the frequency  $f_{\rm N}$  of the message signal.
|type="{}"}
$f_{\rm N} \ = \ $ { 5 3% } $\ \rm kHz$

{What is the carrier frequency $f_{\rm T}$?
|type="{}"}
$f_{\rm T} \ = \ $ { 50 3% } $\ \rm kHz$

{What is the maximum phase deviation  $ϕ_{\rm max}$  between  $z(t)$  and  $s(t)$?
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 0.318 3% } $\ \rm rad$

{What is the maximum time shift of the zero crossings that this phase results in?
|type="{}"}
$Δt_{\rm max} \ = \ $ { 6.37 3% } $\ \rm µ s$

{Determine the modulation index  $η_2$  for the signal  $s_2(t)$.
|type="{}"}
$η_2 \ = \ $ { 2 3% }

{What is the phase  $ϕ_{\rm N2}$  of the underlying source signal  $q(t)$ for  $s_2(t)$ ?
|type="{}"}
$ϕ_{\rm N2} \ = \ $ { -139--131 } $\ \rm Grad$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  It can be seen from the sketch that the represented section of the signal of duration  $200 \ \rm µ s$  corresponds exactly to the period duration of the sinusoidal source signal. From this follows  $f_{\rm N}\hspace{0.15cm}\underline{ = 5 \ \rm kHz}$.
*At times  $t = 0$,  $t = 100 \ \rm µ s$  and  $t = 200 \ \rm µ s$ , the signals   $z(t)$  and  $s(t)$  are synchronous in phase.
*In the first half-wave of   $q(t)$ , the zero crossings of   $s(t)$ come slightly earlier than those of the carrier signal  $z(t)$   ⇒   positive phase.
*In contrast, in the range from  $t = 100 \ \rm µ s$  to  $t = 200 \ \rm µ s$ , the phase  $ϕ(t) < 0$.

'''(2)'''  $f_{\rm T}\hspace{0.15cm}\underline{ = 50 \ \rm kHz}$, holds,
*since exactly   $10$  periods can be counted in the shown section of   $z(t)$ of duration  $200 \ \rm µ s$ .

'''(3)'''  The maximum relative phase deviation is   $ϕ_{\rm max} = η_1/(2π)\hspace{0.15cm}\underline{ ≈ 0.318}$.

'''(4)'''  Since the period of the carrier is   $T_0 = 20 \ \rm µ s$ , we obtain  $Δt_{\rm max} = ϕ_{\rm max} ·T_0\hspace{0.15cm}\underline{ ≈ 6.37 \ \rm µ s}$.

'''(5)'''  The maximum phase deviation (shift in the zero intercepts) is exactly the same for   $s_2(t)$  as for  $s_1(t)$. 
*From this, we can conclude that  $η_2 = η_1\hspace{0.15cm}\underline{ = 2}$ .

'''(6)'''  The signal  $s_2(t)$  is shifted to the right by   $25 \ \rm µ s$  compared to   $s_1(t)$ . Therefore, the same must be true for the source signals:
:$$ q_2(t) = q_1(t - 25\,{\rm \mu s}) = \cos \hspace{-0.1cm} \big[2 \pi f_{\rm N} (t - 25\,{\rm \mu s}) \big ] = \cos (\omega_{\rm N} \cdot t - 0.75 \cdot \pi)\hspace{0.05cm}.$$
*This corresponds to the phase position $ϕ_{\rm N2}\hspace{0.15cm}\underline{ = -135^\circ}$.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1: Phase Modulation Locus Curve

2022-04-09T14:50:38Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1079__Mod_A_3_1.png|right|frame|Two locus curves to choose from]]
The locus curve is generally understood as the plot of the equivalent low-pass signal $s_{\rm TP}(t)$  in the complex plane.
*The graph shows locus curves at the output of two modulators  $\rm M_1$  and  $\rm M_2$.
*The real and imaginary parts are each normalized to $1 \ \rm V$ in this graph.

Let the source signal be the same for both modulators:
$$ q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}
{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.$$
One of the two modulators implements phase modulation, which is characterized by the following equations:
:$$ s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},$$
:$$ s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm},$$
:$$ \phi(t) = K_{\rm PM} \cdot q(t)\hspace{0.05cm}.$$
The maximum value  $ϕ(t)$  is called the   ''modulation index''  $η$.  Often  $η$  is also called   ''phase deviation''  in the literature.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

===Questions===

<quiz display=simple>
{Which modulation method is used by modulator  $\rm M_1$?
|type="()"}
- Double-sideband amplitude modulation.
+ Single sideband amplitude modulation.
- Phase modulation.

{Which modulation method is used by modulator  $\rm M_2$?
|type="()"}
- Double-sideband amplitude modulation.
- Single sideband amplitude modulation.
+ Phase modulation.

{What is the carrier amplitude  $A_{\rm T}$  for the phase modulator?  Note the normalization to  $1 \ \rm V$.
|type="{}"}
$A_{\rm T} \ = \ $ { 1 3% } $\ \rm V$

{What are the values of the modulation index  $η$  and the modulator constant  $K_{\rm PM}$?
|type="{}"}
$η\ = \ $ { 3.1415 3% }
$K_{\rm PM}\ = \ $ { 1.571 3% } $\ \rm 1/V$

{Describe the motion on the locus curve. At what time $t_1$  is the starting point  $s_{\rm TP}(t = 0) = -1 \ \rm V$  first reached again?
|type="{}"}
$t_1\ = \ $ { 100 3% } $ \ \rm µ s$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We are dealing with SSB-AM with a sideband-to-carrier ratio $μ = 1$   ⇒   Answer 2:
*If one moves in the mathematically positive direction on the circle, it is specifically an USB–AM, otherwise it is a LSB–AM.
*The phase function  $ϕ(t)$  as the angle of a point  $s_{\rm TP}(t)$  on the circle (arc) with respect to the coordinate origin can take values between  $±π/2$  and does not show a cosine progression.
*The envelope   $a(t) = |s_{\rm TP}(t)|$  is also not cosine.
*If an envelope demodulator were used for  $\rm M_1$  at the receiver, nonlinear distortions would occur, in contrast to DSB–AM, which has a horizontal straight line for a locus curve.

'''(2)'''  Here, we observe phase modulation   ⇒   Answer 3:
*The envelope   $a(t) = A_{\rm T}$  is constant,
*while the phase  $ϕ(t)$  is cosinusoidal according to the source signal  $q(t)$ .

'''(3)'''  In the case of phase modulation:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.$$
*From the graph, we can read the carrier amplitude   $A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V}$  as the radius of the circle.

'''(4)'''  The source signal  $q(t)$  is at its maximum at time  $t = 0$  and therefore so is the phase function:
:$$ \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.$$
*This gives the modulator constant:
$$K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.$$

'''(5)'''  One moves clockwise along the circular arc.
*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.
*Afterwards, move counterclockwise along the arc.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1: Phase Modulation Locus Curve

2022-04-09T14:50:23Z

Reed:

{{quiz-Header|Buchseite=Modulatio_Methods/Phase_Modulation_(PM)
}}

[[File:P_ID1079__Mod_A_3_1.png|right|frame|Two locus curves to choose from]]
The locus curve is generally understood as the plot of the equivalent low-pass signal $s_{\rm TP}(t)$  in the complex plane.
*The graph shows locus curves at the output of two modulators  $\rm M_1$  and  $\rm M_2$.
*The real and imaginary parts are each normalized to $1 \ \rm V$ in this graph.

Let the source signal be the same for both modulators:
$$ q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}
{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.$$
One of the two modulators implements phase modulation, which is characterized by the following equations:
:$$ s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},$$
:$$ s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm},$$
:$$ \phi(t) = K_{\rm PM} \cdot q(t)\hspace{0.05cm}.$$
The maximum value  $ϕ(t)$  is called the   ''modulation index''  $η$.  Often  $η$  is also called   ''phase deviation''  in the literature.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

===Questions===

<quiz display=simple>
{Which modulation method is used by modulator  $\rm M_1$?
|type="()"}
- Double-sideband amplitude modulation.
+ Single sideband amplitude modulation.
- Phase modulation.

{Which modulation method is used by modulator  $\rm M_2$?
|type="()"}
- Double-sideband amplitude modulation.
- Single sideband amplitude modulation.
+ Phase modulation.

{What is the carrier amplitude  $A_{\rm T}$  for the phase modulator?  Note the normalization to  $1 \ \rm V$.
|type="{}"}
$A_{\rm T} \ = \ $ { 1 3% } $\ \rm V$

{What are the values of the modulation index  $η$  and the modulator constant  $K_{\rm PM}$?
|type="{}"}
$η\ = \ $ { 3.1415 3% }
$K_{\rm PM}\ = \ $ { 1.571 3% } $\ \rm 1/V$

{Describe the motion on the locus curve. At what time $t_1$  is the starting point  $s_{\rm TP}(t = 0) = -1 \ \rm V$  first reached again?
|type="{}"}
$t_1\ = \ $ { 100 3% } $ \ \rm µ s$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We are dealing with SSB-AM with a sideband-to-carrier ratio $μ = 1$   ⇒   Answer 2:
*If one moves in the mathematically positive direction on the circle, it is specifically an USB–AM, otherwise it is a LSB–AM.
*The phase function  $ϕ(t)$  as the angle of a point  $s_{\rm TP}(t)$  on the circle (arc) with respect to the coordinate origin can take values between  $±π/2$  and does not show a cosine progression.
*The envelope   $a(t) = |s_{\rm TP}(t)|$  is also not cosine.
*If an envelope demodulator were used for  $\rm M_1$  at the receiver, nonlinear distortions would occur, in contrast to DSB–AM, which has a horizontal straight line for a locus curve.

'''(2)'''  Here, we observe phase modulation   ⇒   Answer 3:
*The envelope   $a(t) = A_{\rm T}$  is constant,
*while the phase  $ϕ(t)$  is cosinusoidal according to the source signal  $q(t)$ .

'''(3)'''  In the case of phase modulation:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.$$
*From the graph, we can read the carrier amplitude   $A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V}$  as the radius of the circle.

'''(4)'''  The source signal  $q(t)$  is at its maximum at time  $t = 0$  and therefore so is the phase function:
:$$ \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.$$
*This gives the modulator constant:
$$K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.$$

'''(5)'''  One moves clockwise along the circular arc.
*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.
*Afterwards, move counterclockwise along the arc.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Aufgaben:Exercise 3.1: Phase Modulation Locus Curve

2022-04-09T14:49:57Z

Reed:

{{quiz-Header|Buchseite=Modulationsverfahren/Phase Modulation (PM)
}}

[[File:P_ID1079__Mod_A_3_1.png|right|frame|Two locus curves to choose from]]
The locus curve is generally understood as the plot of the equivalent low-pass signal $s_{\rm TP}(t)$  in the complex plane.
*The graph shows locus curves at the output of two modulators  $\rm M_1$  and  $\rm M_2$.
*The real and imaginary parts are each normalized to $1 \ \rm V$ in this graph.

Let the source signal be the same for both modulators:
$$ q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}
{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.$$
One of the two modulators implements phase modulation, which is characterized by the following equations:
:$$ s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},$$
:$$ s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm},$$
:$$ \phi(t) = K_{\rm PM} \cdot q(t)\hspace{0.05cm}.$$
The maximum value  $ϕ(t)$  is called the   ''modulation index''  $η$.  Often  $η$  is also called   ''phase deviation''  in the literature.

''Hints:''
*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].

===Questions===

<quiz display=simple>
{Which modulation method is used by modulator  $\rm M_1$?
|type="()"}
- Double-sideband amplitude modulation.
+ Single sideband amplitude modulation.
- Phase modulation.

{Which modulation method is used by modulator  $\rm M_2$?
|type="()"}
- Double-sideband amplitude modulation.
- Single sideband amplitude modulation.
+ Phase modulation.

{What is the carrier amplitude  $A_{\rm T}$  for the phase modulator?  Note the normalization to  $1 \ \rm V$.
|type="{}"}
$A_{\rm T} \ = \ $ { 1 3% } $\ \rm V$

{What are the values of the modulation index  $η$  and the modulator constant  $K_{\rm PM}$?
|type="{}"}
$η\ = \ $ { 3.1415 3% }
$K_{\rm PM}\ = \ $ { 1.571 3% } $\ \rm 1/V$

{Describe the motion on the locus curve. At what time $t_1$  is the starting point  $s_{\rm TP}(t = 0) = -1 \ \rm V$  first reached again?
|type="{}"}
$t_1\ = \ $ { 100 3% } $ \ \rm µ s$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We are dealing with SSB-AM with a sideband-to-carrier ratio $μ = 1$   ⇒   Answer 2:
*If one moves in the mathematically positive direction on the circle, it is specifically an USB–AM, otherwise it is a LSB–AM.
*The phase function  $ϕ(t)$  as the angle of a point  $s_{\rm TP}(t)$  on the circle (arc) with respect to the coordinate origin can take values between  $±π/2$  and does not show a cosine progression.
*The envelope   $a(t) = |s_{\rm TP}(t)|$  is also not cosine.
*If an envelope demodulator were used for  $\rm M_1$  at the receiver, nonlinear distortions would occur, in contrast to DSB–AM, which has a horizontal straight line for a locus curve.

'''(2)'''  Here, we observe phase modulation   ⇒   Answer 3:
*The envelope   $a(t) = A_{\rm T}$  is constant,
*while the phase  $ϕ(t)$  is cosinusoidal according to the source signal  $q(t)$ .

'''(3)'''  In the case of phase modulation:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.$$
*From the graph, we can read the carrier amplitude   $A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V}$  as the radius of the circle.

'''(4)'''  The source signal  $q(t)$  is at its maximum at time  $t = 0$  and therefore so is the phase function:
:$$ \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.$$
*This gives the modulator constant:
$$K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.$$

'''(5)'''  One moves clockwise along the circular arc.
*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.
*Afterwards, move counterclockwise along the arc.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]

Modulation Methods/Phase Modulation (PM)

2022-04-09T14:45:00Z

Reed:

{{Header
|Untermenü=Angle Modulation and Demodulation
|Vorherige Seite=Further AM Variants
|Nächste Seite=Frequency Modulation (FM)
}}

== # OVERVIEW OF THE THIRD MAIN CHAPTER # ==
 
The third chapter describes  '''angle modulation'''   $\rm (WM)$ (from the German   "Winkelmodulation"). This is a generic term for
*  "phase modulation"  $\rm (PM)$, 
*  "frequency modulation"   $\rm (FM)$. 

In detail,  the chapter covers:
#the  »similarities and differences«  between phase and frequency modulation,
#the  »realization of the associated demodulators«,
#the  »signal characteristics and spectral functions«  of angle-modulated signals and the influence of band limiting,
#the  »signal-to-noise power ratio«  of FM,  which is more favorable than that of AM.

==Similarities between phase and frequency modulation==
 
It has already been pointed out in the chapter  [[Modulation_Methods/General_Model_of_Modulation|General Model of Modulation]]  that there are substantial similarities between phase modulation   $\rm (PM)$  and frequency modulation  $\rm (FM)$.  Therefore,  these two related modulation methods are summarized under the general term "angle modulation".

{{BlaueBox|TEXT=
$\rm Definition\text{:}$  An  '''angle modulation'''  – abbreviated as   $\rm WM$  – is present whenever the modulated signal can be represented as follows:
:$$s(t) = A_{\rm T} \cdot \cos\big[\psi(t)\big] = A_{\rm T} \cdot \cos\hspace{-0.1cm}\big[ω_{\rm T} · t + ϕ(t)\big]
\hspace{0.05cm}.$$
*Here,  as in amplitude modulation,  $A_{\rm T}$  denotes the amplitude of the carrier signal  $z(t)$. 
*However,  all the information about the source signal  $q(t)$  is now captured by the   "angular function"  $ψ(t)$.}}

[[File:EN_Mod_T_3_1_S1a.png|right|frame| Equivalent low-pass signal in angle modulation]]

Based on the plot of the equivalent low-pass signal  $s_{\rm TP}(t)$  (subscript from the German  "Tiefpass"  ⇒   low-pass)  on the complex plane  (we will refer to such a plot as a  "locus curve"),  the following characteristics of angle modulation can be seen:
*The locus curve is an   "arc"  with radius  $A_{\rm T}$.  It follows that the envelope of an angle-modulated signal is always constant:
:$$a(t) = |s_{\rm TP}(t)|= A_{\rm T}= {\rm const.}$$
*The equivalent low-pass signal in angle modulation is always complex and determined by a time-dependent   "phase function"  $ϕ(t)$  (in radians),  which determines the zero intercepts of  $s(t)$:
:$$s_{\rm TP}(t)= A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t)}\hspace{0.05cm}.$$
*For a symmetric source signal,   $ϕ(t)$  can take on all values between  $±ϕ_{\rm max}$,  where  $ϕ_{\rm max}$  indicates the   '''phase deviation'''.  The larger the phase deviation,  the more intense the modulation.
*For a harmonic oscillation,  the phase deviation  $ϕ_{\rm max}$  is equal to the   '''modulation index'''  $η$.  Thus,  the use of  $η$  in what follows also indicates that  $q(t)$  only contains a single frequency.
*The relationship between the source signal  $q(t)$  and the angular function  $ψ(t) = \cos\hspace{-0.1cm}\big[ω_{\rm T} · t + ϕ(t)\big]$ ,  just like the phase function  $ϕ(t)$  that can be derived from it,  differs fundamentally in phase and frequency modulation,  which will be discussed in detail in the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]] .

{{GraueBox|TEXT=
$\text{Example 1:}$  The following graph shows
*on the right,  the transmitted signal  $s(t)$ ⇒   blue waveforms, compared to the carrier signal  $z(t)$   ⇒   red waveforms,
*on the left,  the equivalent low-pass signal  $s_{\rm TP}(t)$  in the complex plane.
[[File:EN_Mod_T_3_1_S1b_v2.png|right|frame|Physical signal and equivalent low-pass signal for angle and amplitude modulation]]

We also refer to the  (left)  plot in the complex plane as the  "locus curve"   ⇒   green waveforms.

The upper graph applies in the case of angle modulation   $\rm (WM)$:
*The equivalent low-pass signal  $s_{\rm TP}(t) = A_{\rm T} · {\rm e}^{ \hspace{0.05cm}{\rm j}\hspace{0.05cm}· \hspace{0.05cm}ϕ(t)}$  describes an arc   ⇒   constant envelope   $a(t) = A_{\rm T}$.
*Thus,  the information about the source signal  $q(t)$  is exclusively found in the zero intercepts of  $s(t)$.
*Should  $ϕ(t) < 0$  hold,  then the zero crossings of  $s(t)$  occur later than those of  $z(t)$.  Otherwise – when  $ϕ(t) > 0$,  the zero crossings of  $s(t)$  come before  $z(t)$.

The lower graph corresponds to   [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|double-sideband amplitude modulation]]  $\rm (DSB-AM)$  as described in Chapter 2,  characterised by
*the time-dependent envelope  $a(t)$  according to the signal  $q(t)$,
*equidistant zero crossings of  $s(t)$  according to the carrier  $z(t)$,  and
*a horizontal straight line as the locus curve  $s_{\rm TP}(t)$. }}

The present third chapter has been structured according to the following considerations:
# Any  $\rm FM$  system can be converted into a corresponding  $\rm PM$  system by simple modifications and vice versa.
# $\rm FM$ is more important for analog systems due to its more favorable noise behaviour.  For this reason,  considerations concerning the realization of the modulator/demodulator will only be dealt with in the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
# Phase modulation  is easier to understand compared to  $\rm FM$.  Therefore,  the basic properties of an angle modulation system are first presented in this chapter using  $\rm PM$  as an example.

==Signal characteristics of phase modulation==
 
Without limiting generality,  the following assumes:
*a cosine carrier signal  $z(t) = A_{\rm T} · \cos(ω_{\rm T} · t)$,  that is,  the carrier phase is always  $ϕ_{\rm T} = 0$,
*a peak-limited source signal between the limits  $\ –q_{\rm max} ≤ q(t) ≤ +q_{\rm max}$.

{{BlaueBox|TEXT=
$\rm Definition\text{:}$  If the phase function $ϕ(t)$  is proportional to the applied source signal  $q(t)$,  we are dealing with  '''phase modulation'''  $\rm (PM)$,  and it holds that:
:$$\phi(t)= K_{\rm PM} \cdot q(t)\hspace{0.05cm}\hspace{0.3cm}\Rightarrow
\hspace{0.3cm}\psi(t)= \omega_{\rm T} \cdot t +
\phi(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s(t) = A_{\rm T}
\cdot \cos \big[\psi(t)\big]\hspace{0.05cm}.$$
Here, $K_{\rm PM}$  denotes the modulator constant with appropriate dimensions.  If  $q(t)$  describes a voltage waveform,  this constant has the unit  $\rm 1/V$. }}

The phase modulation is all the more intensive,
*the larger the modulator constant  $K_{\rm PM}$,  or
*the larger the maximum value  $q_{\rm max}$  of the source signal.

{{BlaueBox|TEXT=
$\rm Definitions\text{:}$ 

'''(1)'''   Quantitatively,  this fact is captured by the  '''phase deviation'''
:$$ \phi_{\rm max} = K_{\rm PM} \cdot q_{\rm max}\hspace{0.05cm}.$$
'''(2)'''   For a harmonic oscillation,  the  "phase deviation"  is also called  '''modulation index'''.  The following holds for the amplitude  $A_{\rm N}$  of the source signal:
:$$\eta = \eta_{\rm PM} = K_{\rm PM} \cdot A_{\rm N}\hspace{0.05cm}.$$}}

The following should be noted about this equation:
*The modulation index  $η$  is comparable to the modulation depth  $m$  in DSB–AM with carrier.
*In the locus curve,  the parameters  $ϕ_{\rm max}$  or  $η$  describe the half angle of the circular arc in radians.
*For other source signals with the same  $η$  $($e.g.:  with different phase  $ϕ_{\rm N})$   the locus curve itself does not change, only the temporal movement along the curve does.
*The modulation index is also used in describing frequency modulation,  though in that case it is to be calculated somewhat differently.
* We therefore distinguish between  $η_{\rm PM}$  and  $η_{\rm FM}$.

{{GraueBox|TEXT=
$\rm Example \ 2\text{:}$   The graph shows a sinusoidal source signal  $q(t)$  with frequency  $f_{\rm N} = 2 \ \rm kHz$  and amplitude  $A_{\rm N}$,  and two phase modulated signals drawn below. 
*These differ by the parameters  $η = 1$  and  $η = 3$, resp.:
[[File: P_ID1070__Mod_T_3_1_S2a_neu.png|right|frame|Signal characteristics for phase modulation with  $η = 1$  resp.  $η = 3$]]
:$$s_\eta(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t +
\eta \cdot \sin (\omega_{\rm N} \cdot t) \big]\hspace{0.05cm}.$$
*Dotted in gray is the cosine carrier signal  $z(t)$,  in each case based on  $f_{\rm T} = 20 \ \rm kHz$ .

*For example,  the modulation index  $η = 1$ , thus the transmitted signal  $s_1(t)$  is given by
:*$A_{\rm N} = 1 \, \rm V $  and  $K_{\rm PM} = \rm 1/V$,
:*but also by parameter values  $A_{\rm N} = 2 \ \rm V$  and  $K_{\rm PM} = \rm 0.5/V$.

From the curves,  one can see:
#The zero crossings of the transmitted signal  $s_1(t)$  and the carrier signal  $z(t)$  coincide exactly when  $q(t) ≈ 0$.
#When  $q(t) = +\hspace{-0.05cm}A_{\rm N}$,  the zero crossings come  $1/(2π) ≈ 0.159$  of a carrier period  $T_0$  earlier  ("leading").
#When  $q(t) = -\hspace{-0.05cm}A_{\rm N}$,  the zero crossings come the same period fraction later ("lagging").
#Increasing the index to  $η = 3$  $($by tripling  $A_{\rm N}$  or  $K_{\rm PM})$, we get qualitatively the same result, but with a more intense PM.
#The zero crossings of  $s_3(t)$  are now shifted relative to those of the carrier by a maximum of  $\rm ±3/(2π) ≈ ±0.5$  of a period, i.e., up to $±T_0/2$. }}

==Equivalent low-pass signal in phase modulation==
 
In preparation for deriving the spectrum $S(f)$  of a phase modulated signal  $s(t)$,  we first analyse the equivalent low-pass signal  $s_{\rm TP}(t)$.  In the folowing,  we assume:
*a sine-shaped source signal  $q(t)$  with amplitude  $A_{\rm N}$  and frequency $f_{\rm N}$,
*a cosine-shaped carrier signal  $z(t)$  with amplitude  $A_{\rm T}$  and frequency  $f_{\rm T}$,
*phase modulation with modulation index  $η = K_{\rm PM} · A_{\rm N}$.

Thus,  the phase modulated signal and the corresponding equivalent low-pass signal are:
:$$s(t) = A_{\rm T} \cdot \cos \big[\omega_{\rm T} \cdot t + \eta
\cdot \sin (\omega_{\rm N} \cdot t) \big]\hspace{0.05cm},$$
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j}
\hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot
\hspace{0.05cm}\sin (\omega_{\rm N} \cdot t) }\hspace{0.05cm}.$$

This signal is periodic and can be represented by a   [[Signal_Representation/Fourier_Series#Complex_Fourier_series|complex Fourier series]].  Thus,  in general one obtains:
:$$s_{\rm TP}(t) = \sum_{n = - \infty}^{+\infty}D_{n} \cdot {\rm
e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot
\hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N}
\hspace{0.05cm}\cdot \hspace{0.05cm} t} \hspace{0.05cm}.$$

In the special case considered here   (sinusoidal source signal,  cosinusoidal carrier)  the typically complex Fourier coefficients  $D_n$  are all real and given by   $n$–th order  '''Bessel functions'''  of the first kind  ${\rm J}_n(η)$  as follows:
:$$D_{n} = A_{\rm T}\cdot {\rm J}_n (\eta) \hspace{0.05cm}. \hspace{1cm} $$

{{BlaueBox|TEXT=
$\text{Important intermediate result:}$   
Now it should be mathematically proven that in the case of phase modulation,  the equivalent low-pass signal can indeed be converted into the following function series:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm N} \cdot t) } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} s_{\rm TP}(t) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot
\hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$}}

{{BlaueBox|TEXT=
$\text{Proof:}$  For simplicity,  we set  $A_{\rm T} = 1$.  Thus,  the given equivalent low-pass signal is:    
:$$s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm N} \cdot t) }\hspace{0.05cm}.$$

'''(1)'''  When  $x = {\rm j} · η · \sin(γ)$  and  $γ = ω_{\rm N} · t$,  the power series expansion of this equation is:
:$$s_{\rm TP}(t) = {\rm e}^{x } = 1 + x + \frac{1}{2!} \cdot x^2 + \frac{1}{3!} \cdot x^3 + \text{...} = 1 + {\rm j} \cdot \eta \cdot \sin (\gamma)+ \frac{1}{2!} \cdot {\rm j}^2 \cdot \eta^2 \cdot \sin^2 (\gamma)+ \frac{1}{3!} \cdot {\rm j}^3 \cdot \eta^3 \cdot \sin^3 (\gamma) + \text{...}$$

'''(2)'''  The individual trigonometric expressions can be rewritten as follows:
:$$ \frac{1}{2!} \cdot {\rm j}^2 \cdot \eta^2 \cdot \sin^2 (\gamma) = \frac{- \eta^2}{2 \cdot 2!} \cdot \big[ 1 - \cos (2\gamma)\big],\hspace{1.0cm} \frac{1}{3!} \cdot {\rm j}^3 \cdot \eta^3 \cdot \sin^3 (\gamma) = \frac{- {\rm j} \cdot \eta^3}{4 \cdot 3!} \cdot \big[ 3 \cdot \sin (\gamma)- \sin (3\gamma)\big],$$
:$$ \frac{1}{4!} \cdot {\rm j}^4 \cdot \eta^4 \cdot \sin^4 (\gamma) = \frac{\eta^4}{8 \cdot 4!} \cdot \left[ 3+ 4 \cdot \cos (2\gamma)+ \cos (4\gamma)\right], \text{...} $$

'''(3)'''  By rearranging using  ${\rm J}_n(η)$,  we obtain the first kind of  $n$–th order Bessel functions:
:$$s_{\rm TP}(t) = 1 \cdot {\rm J}_0 (\eta) + 2 \cdot {\rm j}\cdot {\rm J}_1 (\eta)\cdot \sin (\gamma) \hspace{0.2cm} + 2 \cdot {\rm J}_2 (\eta)\cdot \cos (2\gamma) + 2 \cdot {\rm j}\cdot {\rm J}_3 (\eta)\cdot \sin (3\gamma)+ 2 \cdot {\rm J}_4 (\eta)\cdot \cos (4\gamma) + \text{...} $$

'''(4)'''  Using Euler's theorem,  this can be written as:
:$$s_{\rm TP}(t) = {\rm J}_0 (\eta) + \big[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{ - \rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \big]\cdot {\rm J}_1 (\eta) \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{ - \rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 (\eta)+ \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 3\gamma} - {\rm e}^{\hspace{0.05cm}{ - \rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 3\gamma} \right]\cdot {\rm J}_3 (\eta)+ \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 4\gamma} + {\rm e}^{\hspace{0.05cm}{ - \rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 4\gamma} \right]\cdot {\rm J}_4 (\eta)+\text{...}$$

'''(5)'''  The Bessel functions exhibit the following symmetrical properties:
:$${\rm J}_{-n} (\eta) = ( - 1)^n \cdot {\rm J}_{n} (\eta)\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} {\rm J}_{ - 1} (\eta) = - {\rm J}_{1}
(\eta),\hspace{0.3cm}{\rm J}_{ - 2} (\eta) = {\rm J}_{2}
(\eta),\hspace{0.3cm}{\rm J}_{ - 3} (\eta) = - {\rm J}_{3}
(\eta),\hspace{0.3cm}{\rm J}_{ - 4} (\eta) = {\rm J}_{4} (\eta).$$

'''(6)'''  Considering this fact and the factor  $A_{\rm T}$ omitted so far,  we get the desired result:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$

<div align="right">$\text{q.e.d.}$</div>}}

[[File: P_ID2329__Mod_T_3_1_A1_70neu.png|right|frame|Calculating the Bessel functions]]
These mathematical functions,  introduced as early as 1844 by  [https://en.wikipedia.org/wiki/Friedrich_Bessel Friedrich Wilhelm Bessel]  are defined as

:$${\rm J}_n (\eta) = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha\hspace{0.05cm},$$
and can be approximated by a series according to the next equation:
:$${\rm J}_n (\eta) = \sum\limits_{k=0}^{\infty}\frac{(-1)^k \cdot (\eta/2)^{n \hspace{0.05cm} + \hspace{0.05cm} 2
\hspace{0.02cm}\cdot \hspace{0.05cm}k}}{k! \cdot (n+k)!} \hspace{0.05cm}.$$

The adjacent graph shows the first three summands $(k = 0,\ 1,\ 2)$  of each of the series  ${\rm J}_0(η)$, ... ,  ${\rm J}_3(η).$ 

*For example,  the term outlined in red – valid for  $n = 3$  and  $k = 2$  – is written as:
:$$\frac{(-1)^2 \cdot (\eta/2)^{3 \hspace{0.05cm} + \hspace{0.05cm} 2 \hspace{0.02cm}\cdot \hspace{0.05cm}2}}{2\hspace{0.05cm}! \cdot (3+2)\hspace{0.05cm}!} = \frac{1}{240}\cdot (\frac{\eta}{2})^7 \hspace{0.05cm}.$$
*The Bessel functions  ${\rm J}_n(η)$  can also be found in collections of formulae or with our applet  [[Applets:Bessel_functions_of_the_first_kind|Bessel functions of the first kind]].
*If the function values for $n = 0$  and  $n = 1$  are known,  the Bessel functions for  $n ≥ 2$ can be iteratively determined from them:
:$${\rm J}_n (\eta) ={2 \cdot (n-1)}/{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.05cm}.$$

==Interpretation of the Bessel spectrum==
 
The graph shows the Bessel functions  ${\rm J}_0(η)$, ... ,  ${\rm J}_7(η)$  depending on the modulation index  $η$  in the range   $0 ≤ η ≤ 10$.

[[File:Mod_T_3_1_S3a_version2.png|right|frame|  $n$–th order Bessel functions of the first kind]]

One can also find these in formula collections such as  [BS01]<ref>Bronstein, I.N.; Semendjajew, K.A.:  Taschenbuch der Mathematik.  5. Auflage. Frankfurt: Harry Deutsch, 2001.</ref>  in tabular form.
*That the functions are of the first kind is expressed by the  "$\rm J$",  and
*the order is given by the index $n$.

Using this graph,  the equivalent low-pass signal is given by
:$$s_{\rm TP}(t) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$

The following properties can be derived:

*The equivalent low-pass signal is composed of
:*a pointer at rest  $(n = 0)$ 
:*and infinitely many clockwise  $(n < 0)$ 
:*or counterclockwise  $(n > 0)$ rotating pointers.

*The pointer lengths depend on the modulation index $η$  via the Bessel functions ${\rm J}_n(η)$. 

*The smaller the modulation index $η$,  the more pointers can be ignored for the construction of $s_{\rm TP}(t)$.  For example,  with a modulation index of  $η = 1$,  the following approximation holds:
:$$s_{\rm TP}(t) = {\rm J}_0 (1) + {\rm J}_1 (1)\cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm} t}+ {\rm J}_2 (1)\cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm} t}+ {\rm J}_3 (1)\cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}3 \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm} t}- {\rm J}_1 (1)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm} t}+ {\rm J}_2 (1)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}2 \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm} t}- {\rm J}_3 (1)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}3 \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm} t}\hspace{0.05cm}.$$

*Here,  the symmetrical relationship  ${\rm J}_{–n}(η) = (–1)^n · {\rm J}_n(η)$  is taken into account.  Thus:
:$${\rm J}_{-1}(\eta) = - {\rm J}_{1}(\eta), \hspace{0.3cm}{\rm J}_{-2}(\eta) = {\rm J}_{2}(\eta), \hspace{0.3cm}{\rm J}_{-3}(\eta) = - {\rm J}_{3}(\eta).$$
*Further,  it can be seen from the above equation that with $η = 3$,  the equivalent low-pass signal  $s_{\rm TP}(t)$  is composed of significantly more pointers,  namely those with indices  ${\rm J}_{–6}(\eta)$, ... ,  ${\rm J}_{+6}(\eta)$.

{{GraueBox|TEXT=
[[File:EN_Mod_T_3_1_S3c.png |right|frame| Example of an equivalent low-pass–signal in phase modulation '''Korrektur''']]
$\rm Example\ 3\text{:}$  The Bessel functions yield the following values for the modulation index $η = 1$:
:$${\rm J}_0 = 0.765,\hspace{0.3cm}{\rm J}_1 = - {\rm J}_{ - 1} = 0.440, \hspace{0.3cm}{\rm J}_2 = {\rm J}_{ - 2} = 0.115,\hspace{0.3cm}{\rm J}_3 = - {\rm J}_{ - 3} = 0.020\hspace{0.05cm}.$$

The graph shows the composition of the locus curves from the seven pointers. 
#For simplicity,  set  $A_{\rm T} = 1$.
#The frequency of the sinusoidal source signal is $f_{\rm N} = 2 \ \rm kHz$,  which gives the period 
::$$T_{\rm N} = 1/f_{\rm N} = 500 \ \rm µ s.$$

The left image shows the snapshot at time  $t = 0$.
*Because  ${\rm J}_1 = – {\rm J}_{ – 1}$  and  ${\rm J}_3 = – {\rm J}_{ – 3}$,  the following holds:
:$$s_{\rm TP}(t = 0) = {\rm J}_0 + {\rm J}_{2} + {\rm J}_{ - 2} = 0.765 + 2 \cdot 0.115 = 0.995 \hspace{0.05cm}.$$
*The phase  ${\mathbf ϕ}(t = 0) = 0$   and the magnitude   $a(t = 0) = 1$  follow from the real result.
*The slightly different value  $0.995$  shows that though ${\rm J}_4 = {\rm J}_{ – 4}$  is small  $(≈ 0.002)$, it is not equal to zero.

The right image shows the ratios at time  $t = T_{\rm N}/4 = 125\ \rm µ s$:
*The pointers with lengths  ${\rm J}_{– 1}$  and  ${\rm J}_1$  have rotated clockwise resp. counterclockwise by   $90^\circ$,  and now both point in the direction of the imaginary axis.
*The pointers  ${\rm J}_2$  and  ${\rm J}_{– 2}$  rotate twice as fast as  ${\rm J}_1$  and  ${\rm J}_{– 1}$  and now both point in the direction of the negative real axis.
* ${\rm J}_3$  and  ${\rm J}_{– 3}$  rotate at three times the speed of  ${\rm J}_1$ und ${\rm J}_{– 1}$  and now both point downward.
*This gives:
:$$s_{\rm TP}(t = 125\,{\rm µ s}) = {\rm J}_0 - 2 \cdot {\rm J}_{2} + {\rm j} \cdot (2 \cdot {\rm J}_{1} - 2 \cdot {\rm J}_{3})= 0.535 + {\rm j} \cdot 0.840 $$
:$$ \Rightarrow \hspace{0.3cm} a(t = 125\,{\rm µ s}) = \sqrt{0.535^2 + 0.840^2}= 0.996\hspace{0.05cm},$$
:$$ \Rightarrow \hspace{0.3cm}\phi(t = 125\,{\rm µ s}) = \arctan \frac{0.840}{0.535} = 57.5^\circ \approx 1\,{\rm rad}\hspace{0.05cm}.$$
*At all others times,  the vector sum of the seven pointers each also yields a point on the arc with angle  $ϕ(t)$,  where  $\vert ϕ(t) \vert ≤ η = 1\ \rm rad $. }}

==Spectral function of a phase-modulated sine signal==
 
{{BlaueBox|TEXT=
$\text{Without proof:}$ 
*Based on the equivalent low-pass signal just calculated,  we obtain for the  '''analytical signal''':
:$$s_{\rm +}(t) = s_{\rm TP}(t) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot
\hspace{0.05cm}\omega_{\rm T} \hspace{0.05cm}\cdot \hspace{0.05cm} t}= A_{\rm T} \cdot
\sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}(\omega_{\rm T}\hspace{0.05cm}+\hspace{0.05cm} n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N}) \hspace{0.05cm}\cdot \hspace{0.05cm} t}$$

*By Fourier transform, we get the  '''spectrum of the analytical signal''':
:$$S_{\rm +}(f) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta \big[f - (f_{\rm T}+ n \cdot f_{\rm N})\big]\hspace{0.05cm}.$$

*The   '''spectrum of the physical signal'''  is obtained by expanding to negative frequencies taking into account a factor of  $1/2$:
:$$S(f) = \frac{A_{\rm T} }{2} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta \big[f \pm (f_{\rm T}+ n \cdot f_{\rm N})\big]\hspace{0.05cm}.$$}}

[[File:Mod_T_3_1_S4_version2.png|right|frame|Spectrum of the analytical signal for  $\rm PM$  $($also valid for  $\rm FM)$]]
Based on the graph, the following statements can be made:
*The spectrum $S_+(f)$  of a phase-modulated sinusoidal signal consists of infinitely many discrete lines spaced at the frequency $f_{\rm N}$  of the sine signal.  In principle,  it is infinitely extended.
*The heights  (weights)  of the spectral lines at $f_{\rm T} + n · f_{\rm N}$  ($n$  is an integer)  are determined by the modulation index $η$  via the Bessel functions ${\rm J}_n(η)$.
*The  ${\rm J}_n(η)$  values show that in practice the spectrum is barely changed by bandlimiting.  However,  the resulting error grows as  $η$  increases.
*The spectral lines are real for a sinusoidal source signal and cosinusoidal carrier and symmetric about  $f_{\rm T}$ for even values of  $n$ .  When  $n$  is odd, a sign change must be taken into account.
*The PM of an oscillation with a different phase of source and/or carrier signal yields the same magnitude spectrum and differs only with respect to the phase function.

If the source signal is composed of several oscillations,  the spectrum calculation becomes difficult,  namely:       '''Convolution of the single spectra'''   $($see next section and  [[Aufgaben:Exercise_3.3:_Sum_of_two_Oscillations| Exercise 3.3)]].

==Phase modulation of the sum of two sinusoidal oscillations==
 
If the source signal is composed of the sum of two sinusoidal oscillations,  the signals at the output of the phase modulator are:
:$$s(t) = A_{\rm T} \cdot \cos \big[\omega_{\rm T} \cdot t + \eta_1 \cdot \sin (\omega_{\rm 1} \cdot t) + \eta_2 \cdot \sin(\omega_{\rm 2} \cdot t)\big]\hspace{0.05cm},$$
:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot [\hspace{0.05cm}\eta_1 \hspace{0.05cm}\cdot \hspace{0.08cm}\sin (\omega_{\rm 1} \hspace{0.05cm}\cdot
\hspace{0.05cm} t) \hspace{0.05cm}+ \hspace{0.05cm}\eta_2 \hspace{0.05cm}\cdot \hspace{0.08cm}\sin (\omega_{\rm 2} \hspace{0.05cm}\cdot \hspace{0.05cm}t)]}\hspace{0.05cm}.$$

*For ease of representation,  we now set  $A_{\rm T} = 1$  and get:
:$$s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta_1 \hspace{0.05cm}\cdot \hspace{0.08cm}\sin (\omega_{\rm 1} \hspace{0.05cm}\cdot \hspace{0.05cm} t) } \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta_2 \hspace{0.05cm}\cdot \hspace{0.08cm}\sin (\omega_{\rm 2} \hspace{0.05cm}\cdot \hspace{0.05cm} t) }\hspace{0.05cm}.$$

*The spectral functions of the two terms are:
:$${\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta_1 \hspace{0.05cm}\cdot
\hspace{0.08cm}\sin (2 \pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm 1} \hspace{0.01cm}\cdot \hspace{0.05cm} t) } \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} B_1(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta_1) \cdot \delta (f - n \cdot f_{\rm 1})\hspace{0.05cm},$$
:$${\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta_2 \hspace{0.05cm}\cdot
\hspace{0.08cm}\sin (2 \pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm 2} \hspace{0.01cm}\cdot
\hspace{0.05cm} t) } \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} B_2(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta_2) \cdot \delta (f - n \cdot f_{\rm 2})\hspace{0.05cm}.$$

*The Bessel functions $B_1(f)$  and  $B_2(f)$  describe line spectra spaced in frequency at  $f_1$  and  $f_2$, whose weights are determined by  $η_1$  and  $η_2$.
*Due to multiplication in the time domain,  the spectral function is given by the convolution:
:$$S_{\rm TP}(f) = B_1(f) \star B_2(f)= S_{\rm +}(f + f_{\rm T}) \hspace{0.05cm}.$$

{{GraueBox|TEXT=
$\rm Example\ 4\text{:}$  The left graph shows the Bessel function  $B_1(f)$  for  $η_1 = 0.64$  and  $f_1 = 1 \ \rm kHz$.  The much smaller lines at  $f = ±2 \ \rm kHz$  with weights  $0.05$  are left out for clarity.

[[File:P_ID1076__Mod_T_3_1_S5_Ganz_neu.png|right|frame|Equivalent low-pass spectrum as convolution of two Bessel spectra]]
*The function  $B_2(f)$  is valid for the same modulation index $η_2 = η_1$,  but at the signal frequency  $f_2 = 4 \ \rm kHz$.

*The low-pass spectrum  $S_{\rm TP}(f) = B_1(f) \star B_2(f)$  consists of nine Dirac lines and is sketched in the diagram on the right.

*By shifting the frequency   $f_{\rm T}$  to the right,  we obtain the spectrum  $S_+(f)$  of the analytical signal  $s_+(t)$.  Thus:
:$$S_+(f = f_{\rm T}) = S_{\rm TP}(f = 0) = 0.81.$$ }}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_3.1:_Phase_Modulation_Locus_Curve|Exercise 3.1: Phase Modulation Locus Curve]]

[[Aufgaben:Exercise_3.1Z:_Influence_of_the_Message_Phase_in_Phase_Modulation|Exercise 3.1Z: Influence of the Message Phase in Phase Modulation]]

[[Aufgaben:Exercise_3.2:_Spectrum_with_Angle_Modulation|Exercise 3.2: Spectrum with Angle Modulation]]

[[Aufgaben:Exercise_3.2Z:_Bessel_Spectrum|Exercise 3.2Z: Bessel Spectrum]]

[[Aufgaben:Exercise_3.3:_Sum_of_two_Oscillations|Exercise 3.3: Sum of two Oscillations]]

[[Aufgaben:Exercise_3.3Z:_Characteristics_Determination|Exercise 3.3Z: Characteristics Determination]]

[[Aufgaben:Exercise_3.4:_Simple_Phase_Modulator|Exercise 3.4: Simple Phase Modulator]]

==References==
 
<references/>

{{Display}}

Aufgaben:Exercise 2.13: Quadrature Amplitude Modulation

2022-04-09T14:27:13Z

Reed:

{{quiz-Header|Buchseite=Modulation Methods/Further AM Variants
}}

[[File:EN_Mod_A_2_11.png|right|frame|QAM model under consideration]]
The  "quadrature amplitude modulation"  $\rm (QAM)$  explained by the diagram allows the transmission of two source signals  $q_1(t)$  and  $q_2(t)$  over the same channel
*under certain boundary conditions,
*which are to be determined in this exercise.

In this exercise,  with  $A_1 = A_2 = 2\ \rm V$:
:$$q_1(t) = A_1 \cdot \cos(2 \pi \cdot f_{\rm 1} \cdot t),$$
:$$q_2(t) = A_2 \cdot \sin(2 \pi \cdot f_{\rm 2} \cdot t)\hspace{0.05cm}.$$
For  $ω_{\rm T} = 2π · 25\ \rm kHz$,  the four carrier signals shown in the diagram are:
:$$z_1(t) = \cos(\omega_{\rm T} \cdot t),$$
:$$ z_2(t) = \sin(\omega_{\rm T} \cdot t),$$
:$$ z_{1,\hspace{0.05cm}{\rm E}}(t) = 2 \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}),$$
:$$ z_{2,\hspace{0.05cm}{\rm E}}(t) = 2 \cdot \sin(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T})\hspace{0.05cm}.$$

Both lowpass filters   $\rm LP_1$  and  $\rm LP_2$  with input signals  $b_1(t)$  resp.  $b_2(t)$ , remove all frequency components  $|f| > f_{\rm T}$.

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Further_AM_Variants|Further AM Variants]].
*Particular reference is made to the page  [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29|Quadrature Amplitude Modulation (QAM)]].
*It is worth noting that the carrier signals  $z_2(t)$  and  $z_{2,\hspace{0.05cm}{\rm E}}(t)$  are applied with positive signs here.
*Often – as in the theory section – these carrier signals are given as  "minus-sine".
*The following trigonometric transformations are given:
:$$ \cos(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$
:$$ \sin(\alpha) \cdot \sin(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)- \cos(\alpha + \beta) \big],$$
:$$ \sin(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \sin(\alpha - \beta)+ \sin(\alpha + \beta) \big] \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Calculate the transmitted signal  $s(t)$  in the case that  $f_1 ≠ f_2$.  Which of the following statements apply?
|type="()"}
- $s(t)$  is composed of two cosine and two sine oscillations.
+ $s(t)$  is composed of four cosine oscillations.
- $s(t)$  is composed of four sine oscillations.

{What is  $s(t)$  with  $f_1 = f_2 = 5 \ \rm kHz$?  What signal value arises for  $t = 50 \ \rm µ s$ ?
|type="{}"}
$s(t = 50 \ \rm µ s) \ = \ $ { 2 3% } $\ \rm V$

{Calculate the sink signals  $v_1(t)$  and  $v_2(t)$ for  $f_1 = f_2$  and  $Δϕ_{\rm T} = 0$  (no phase offset).  Which statements are true?
|type="()"}
+ $v_1(t) = q_1(t)$  and  $v_2(t) = q_2(t)$  both hold.
- Linear distortions occur.
- Nonlinear distortions occur.

{Calculate the sink signals  $v_1(t)$  and  $v_2(t)$  for  $f_1 = f_2$  and a phase offset  $Δϕ_{\rm T} = 30^\circ$.  Which statements are true?
|type="()"}
- $v_1(t) = q_1(t)$  and  $v_2(t) = q_2(t)$  both hold.
+ Linear distortions occur.
- Nonlinear distortions occur.

{Which of the following statements apply when  $f_1 ≠ f_2$  and  $Δϕ_{\rm T} ≠ 0$  (with an arbitrary phase offset)?
|type="()"}
- $v_1(t) = q_1(t)$  and  $v_2(t) = q_2(t)$  both hold.
- Linear distortions occur.
+ Nonlinear distortions occur.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  With the given trigonometric transformations we get:
:$$s(t) = A_1 \cdot \cos(\omega_{\rm 1} \cdot t)\cdot \cos(\omega_{\rm T} \cdot t) + A_2 \cdot \sin(\omega_{\rm 2} \cdot t)\cdot \sin(\omega_{\rm T} \cdot t) $$
:$$\Rightarrow \hspace{0.3cm}s(t) = \frac{A_1}{2}\cdot \cos((\omega_{\rm T} - \omega_{\rm 1})\cdot t) + \frac{A_1}{2}\cdot \cos((\omega_{\rm T} + \omega_{\rm 1})\cdot t) +
\frac{A_2}{2}\cdot \cos((\omega_{\rm T} - \omega_{\rm 2})\cdot t) - \frac{A_2}{2}\cdot \cos((\omega_{\rm T} + \omega_{\rm 2})\cdot t)\hspace{0.05cm}.$$
*The second answer is correct.

'''(2)'''  With $A_1 = A_2 = 2 \ \rm V$ and $f_1 = f_2 = 5\ \rm kHz$, the first and third cosine oscillations constructively overlap and the other two cancel completely.
*Thus,  the following simple result is obtained:
:$$ s(t) = 2\,{\rm V} \cdot \cos(2 \pi \cdot 20\,{\rm kHz} \cdot t) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} s(t = 50\,{\rm µ s}) \hspace{0.15cm}\underline {= 2\,{\rm V}} \hspace{0.05cm}.$$

'''(3)'''  The  first answer  is correct:
*For phase-synchronous demodulation  $(Δϕ_T = 0)$,  the signals before the low-pass filters according to subtask  '''(2)''' are obtained as:
:$$b_1(t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \cos(\omega_{\rm 45} \cdot t),$$
:$$ b_2(t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \sin(\omega_{\rm 45} \cdot t)\hspace{0.05cm}.$$
*Thus,  after eliminating the respective   $45\ \rm kHz$ components,  we get   $v_1(t) = q_1(t)$  and  $v_2(t) = q_2(t)$.

'''(4)'''  Analogously to subtask  '''(3)'''  it now holds that:
:$$ b_1(t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})=
2\,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )},$$
:$$b_2(t)= 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})=
2\,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )}\hspace{0.05cm}.$$
*The sink signals  $v_1(t)$  and  $v_2(t)$  in this constellation exhibit delays and thus phase distortions compared with   $q_1(t)$  and  $q_2(t)$.
*These belong to the class of linear distortions   ⇒   Answer 2.

'''(5)'''  In general,  it holds for the received signal:
:$$r(t) = s(t) = q_1(t) \cdot \cos(\omega_{\rm T} \cdot t) + q_2(t) \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
Multiplication by the receiver-side carrier signals  $z_{1,\hspace{0.05cm}{\rm E}}(t)$  and  $z_{2,\hspace{0.05cm}{\rm E}}(t)$  and band-limiting leads to the signals
:$$v_1(t) = \cos(\Delta \phi_{\rm T}) \cdot q_1(t) - \sin(\Delta \phi_{\rm T}) \cdot q_2(t),$$
:$$ v_2(t) = \sin(\Delta \phi_{\rm T}) \cdot q_1(t) + \cos(\Delta \phi_{\rm T}) \cdot q_2(t) \hspace{0.05cm}.$$
From this it can be seen:
*With a phase offset of   $Δϕ_{\rm T} = 30^\circ$,  the sink signal  $v_1(t)$  includes not only the signal   $q_1(t)$ attenuated by about   $\cos(30^\circ) = 0.866$,  but also the frequency   $f_2$ is contained in   $q_2(t)$.
*This is weighted by the factor   $\sin(30^\circ) = 0.5$.
*Thus,  nonlinear distortions are present   ⇒   Answer 3.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.5 Other AM Variants^]]

Aufgaben:Exercise 2.12: Non-coherent Demodulation

2022-04-09T14:26:49Z

Reed:

{{quiz-Header|Buchseite=Modulation Methods/Further AM Variants
}}

[[File:EN_Mod_A_2_12.png|right|frame|ASK Demodulation (non-coherent) ]]
Consider an amplitude modulated signal:
:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
Reaching the receiver based on the channel propagation time,  the signal is
:$$ r(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \hspace{0.05cm}.$$
The arrangement shown here allows perfect demodulation – that is:  $v(t) = q(t)$ – without knowledge of the phase  $Δϕ_T$,  but only if the source signal  $q(t)$  satisfies certain conditions.

The two receiver-side carrier signals are:
:$$ z_{\rm 1, \hspace{0.08cm}E}(t) = 2 \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm},$$
:$$ z_{\rm 2, \hspace{0.08cm}E}(t) = -2 \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

$\rm LP_1$  and  $\rm LP_2$  denote two ideal  (rectangular)  low-pass filters,  each with cutoff frequency equal to the carrier frequency  $f_{\rm T}$.

We consider as (digital) source signals:
# the unipolar square wave signal  $q_1(t)$  with dimensionless amplitude values  $0$  and  $3$,
# the bipolar square wave signal  $q_2(t)$  with the dimensionless amplitude values  $±3$.

With respect to  $s(t)$,  these two signals result in
#an  [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying|ASK signal]], 
#a  [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|BPSK signal]].

The nonlinear function  $v = g(b)$  is to be determined in this exercise.

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Further_AM_Variants|Further AM Variants]].
*Particular reference is made to the page  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation|Incoherent (non-coherent) Demodulation]].

*The following trigonometric transformations are given:
:$$ \cos(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$
:$$ \sin(\alpha) \cdot \sin(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)- \cos(\alpha + \beta) \big],$$
:$$ \sin(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \sin(\alpha - \beta)+ \sin(\alpha + \beta) \big] \hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{What are the signals  $b_1(t)$  and  $b_2(t)$  in both branches – after multiplier and low-pass respectively?  Which statements apply?
|type="[]"}
+ $b_1(t) = q(t) · \cos(Δϕ_{\rm T})$.
- $b_2(t) = q(t) · \cos(Δϕ_{\rm T})$.
- $b_1(t) = q(t) · \sin(Δϕ_{\rm T})$.
+ $b_2(t) = q(t) · \sin(Δϕ_{\rm T})$.
- $b_1(t) = b_2(t) = q(t)$.

{What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on,  when the unipolar source signal  $q_1(t)$  is applied to the input?
|type="{}"}
$b_{\rm min} \ = \ $ { 0. }
$b_{\rm max} \ = \ $ { 9 3% }

{How should the characteristic curve  $v = g(b)$  be chosen,  so that  $v(t) = q(t)$  holds?
|type="()"}
- $v=g(b) = b^2$.
+ $v=g(b) = \sqrt{b}$.
- $v=g(b) = \arctan(b).$

{What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on,  when the bipolar source signal  $q_2(t)$  is applied to the input?
|type="{}"}
$b_{\rm min} \ = \ $ { 9 3% }
$b_{\rm max} \ = \ $ { 9 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Applying the trigonometric transformations given on the exercise page and taking into account the two low-pass filters  (the components around twice the carrier frequency are removed),  we obtain:
:$$b_1(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$
:$$ b_2(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$
*Thus,  the first and fourth answers  are correct.

'''(2)'''  The sum of the squares of the two partial signals gives:
:$$ b(t) = b_1^2(t) + b_2^2(t)= q^2(t) \cdot \left( \cos^2(\Delta \phi_{\rm T})+ \sin^2(\Delta \phi_{\rm T})\right) = q^2(t)\hspace{0.05cm}.$$
The possible amplitude values are thus:
:$$b_{\rm min}\hspace{0.15cm}\underline{ = 0},$$
:$$ b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$

'''(3)'''  The second answer is correct:
:$$v=g(b) = \sqrt{b} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t) = \sqrt{ q^2(t) } = q(t)\hspace{0.05cm}.$$

'''(4)'''  The result  $b(t) = q^2(t)$ – see subtask '''(2)'''  – leads here to the result:
:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$
:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$

This shows that the demodulator considered here only functions,
*if at all times   $q(t) ≥ 0$   or   $q(t) ≤ 0$   holds,
*and this is known at the receiver.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.5 Other AM Variants^ ]]

Modulation Methods/Further AM Variants

2022-04-09T14:21:53Z

Reed:

{{Header
|Untermenü=Amplitude Modulation and Demodulation
|Vorherige Seite=Single-Sideband Modulation
|Nächste Seite=Phase Modulation (PM)
}}
==Vestigial sideband amplitude modulation==
 
When transmitting signals using single-sideband modulation  $\rm (SSB–AM)$  the following problems occur:
*To suppress the unwanted sideband,  a filter with a very high edge slope must be used.
*Such steep-edged filters exhibit strong group delay distortions,  especially at the limit of the passband.

[[File:EN_Mod_T_2_5_S1.png|right|frame|Spectrum  (of the analytical signal)  for vestigial sideband amplitude modulation]]
The problem can be greatly mitigated if instead of single-sideband AM one uses   "vestigial sideband amplitude modulation"  $\rm (VSB–AM)$,  as shown in the adjacent graph.

The present description is based on the textbook   [Mäu88]<ref>Mäusl, R.:  Analoge Modulationsverfahren.  Heidelberg: Dr. Hüthig, 1988.</ref>.  According to it,  the VSB-AM can be characterized as follows:
*A certain frequency range of the actually suppressed sideband – in the considered example of the LSB – is additionally used with a relatively flat decreasing transfer function.
*On the receiver side,  a selection curve linearly increasing in frequency with a so-called  "Nyquist edge"  is used in the transition range from the suppressed sideband to the transmitted sideband.
*The demodulation performs a convolution of the sidebands around the carrier,  so that as a result the message content of a band with the same amplitude for all frequencies is obtained.

{{GraueBox|TEXT=
$\text{Example 1:}$  The vestigial sideband method is used for  (analog)  color television,  whose frequency spectrum according to the CCIR standard is shown in the graphic.  The frequencies given refer to the  [https://en.wikipedia.org/wiki/PAL PAL–B/G Television format]  used in Germany.

[[File:EN_Mod_T_2_5_S1b.png|right|frame|Illustrating the Nyquist edge for PAL–B/G Television]]
In this schematic representation,  one recognises:
*The radiated spectrum  (only positive frequencies are drawn)  ranges from  $f_{\rm T} - 1.25 \ \rm MHz$  to  $f_{\rm T} + 5.75 \ \rm MHz$.  The lower vestigial sideband including Nyquist edge is  $\approx 1.25 \ \rm MHz$  wide.
*The green dashed line shows the receiver passband.  The image carrier  '''(B)'''  – abbreviation  "B"  from German "Bild"   ⇒   "image") –   at carrier frequency  $f_{\rm T}$  is centered on the Nyquist edge.
*The luminance signal   '''(L)'''  goes up to about  $5 \ \rm MHz$.  It contains the information for the image brightness and for the color  "green".
*The chrominance signal   '''(C)'''  is embedded in the upper part. Two orthogonal carriers are [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29|QAM-modulated]]  at  $4.43 \ \rm MHz$  for  "red"  and  "blue";  the carrier is suppressed.
*The audio carrier   '''(T)'''  – abbreviation  "T"  from German "Ton"   ⇒   "sound/audio") –   is at  $f_{\rm T} + 5.5 \ \rm MHz$  and is  $12 \ \rm dB$  lower than the image carrier.
*If there is a two-channel sound  (stereo)  transmission   ⇒   a second   '''(T)'''  carrier at  $5.75 \ \rm MHz$.}}

==Quadrature Amplitude Modulation (QAM)==
 
By exploiting the orthogonality of cosine and sine functions,  a channel can be used twice for simultaneous transmission of two source signals  $q_1(t)$  and  $q_2(t)$  without mutual interference.  This method is called  "quadrature amplitude modulation"  $\rm (QAM)$.

[[File:P_ID1053__Mod_T_2_5_S2_neu.png |right|frame|Model of quadrature amplitude modulation]]

The QAM system has the following characteristics:
*The transmitted signal is composed of two mutually orthogonal components:
:$$s(t) = q_1(t) \cdot \cos (\omega_{\rm T}\cdot t) - q_2(t) \cdot \sin (\omega_{\rm T}\cdot t)\hspace{0.05cm}.$$
*For frequency and phase synchronous demodulation,  the signal in the upper branch before the low-pass  $H_{\rm E1}(f)$  is:
:$$b_1(t) = q_1(t) \cdot 2 \cdot \cos^2 (\omega_{\rm T}\cdot t) - q_2(t) \cdot 2 \cdot
\cos (\omega_{\rm T}\cdot t)\cdot \sin (\omega_{\rm T}\cdot t) $$
:$$\Rightarrow \hspace{0.3cm} b_1(t) = q_1(t)\cdot \big[ 1 + \cos (2 \omega_{\rm T}\cdot t) \big] - q_2(t)\cdot \sin (2 \omega_{\rm T}\cdot t) \hspace{0.05cm}.$$
*Thus,  by limiting to frequencies  $|f| < f_{\rm T}$,  we obtain in the upper and lower branches,  resp.:
:$$v_1(t) = q_1(t),$$
:$$v_2(t) = q_2(t)\hspace{0.05cm}.$$
*When there is a phase offset  $Δϕ_{\rm T}$  between the transmitted and received carrier signals,  in addition to attenuation of the intended participant, crosstalk from the second participant occurs,  resulting in nonlinear distortion:
:$$v_1(t) = \alpha_{11} \cdot q_1(t)+ \alpha_{12} \cdot q_2(t) \hspace{0.05cm}, $$
:$$v_2(t) = \alpha_{21} \cdot q_1(t)+ \alpha_{22} \cdot q_2(t)$$
:$$\Rightarrow\hspace{0.3cm}\alpha_{11} = \alpha_{22} = \cos(\Delta \phi_{\rm T}) \hspace{0.05cm}, \hspace{0.3cm} \alpha_{12} = -\alpha_{21} = \sin(\Delta \phi_{\rm T}) \hspace{0.05cm}.$$

==Incoherent (non-coherent) Demodulation==
 
{{BlaueBox|TEXT=
$\text{Definition:}$  Demodulators can be classified in the following way:
*A demodulator is said to be  '''coherent''',  if in addition to the required frequency synchronization,  it requires accurate information about the phase of the transmit-side carrier signal  $z(t)$  to reconstruct the source signal.

*If this phase information is not required,  the demodulator is said to be an  '''incoherent demodulator'''  or  "non-coherent demodulator".}}

An example of an incoherent demodulator is the  [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]].

{{GraueBox|TEXT=
$\text{Example 2:}$  A second example is shown in the following block diagram.  In contrast to quadrature amplitude modulation,  here the orthogonality between cosine and sine functions is not used for the simultaneous transmission of a second source signal,  but rather to simplify the receiver-side device.

[[File:P_ID1054__Mod_T_2_5_S3_neu.png |right|frame|Incoherent demodulation with  $\rm DSB-AM$]]

It should be further noted regarding this arrangement:
*The receiver-side carrier signals can have an arbitrary and also time-dependent phase offset  $Δϕ_{\rm T}$  with respect to the carrier signals at the transmitter, as long as the phase difference between the two branches remains exactly  $90^\circ$ .
*For the signals in the upper and lower branches – after the multiplier and low-pass filtering, respectively – it holds:
:$$b_1(t) = \cos(\Delta \phi_{\rm T}) \cdot q(t), $$
:$$b_2(t) = -\sin(\Delta \phi_{\rm T}) \cdot q(t).$$
*This ensures that the sink signal  $v(t)$  matches the source signal  $q(t)$  in magnitude,  regardless of the phase offset  $Δϕ_{\rm T}$:
:$$v(t) = \sqrt{ b_1^2(t) + b_2^2(t)} = \sqrt{ q^2(t) } = \vert q(t) \vert \hspace{0.05cm}.$$
*A prerequisite for operability  – that is,  for the result  $v(t) = q(t)$  –  is that at all times  $q(t) ≥ 0$.  In an analog transmission system,  this fact could be enforced using the  "DSB-AM with carrier"  modulation method,  for example.
*This form of non-coherent demodulation  (or modifications of it)  is mainly applied in some  '''digital modulation methods''',  which are discussed in detail in the fourth chapter of  [[Modulation_Methods|this book]].}}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_2.12:_Non-coherent_Demodulation|Exercise 2.12: Non-coherent Demodulation]]

[[Aufgaben:Exercise_2.13:_Quadrature_Amplitude_Modulation|Exercise 2.13: Quadrature Amplitude Modulation (QAM)]]

==References==
<references/>

{{Display}}

Modulation Methods/Further AM Variants

2022-04-09T14:21:31Z

Reed:

{{Header
|Untermenü=Amplitude Modulation and Demodulation
|Vorherige Seite=Single-Sideband Modulation
|Nächste Seite=Phase Modulation (PM)
}}
==Vestigial sideband amplitude modulation==
 
When transmitting signals using single-sideband modulation  $\rm (SSB–AM)$  the following problems occur:
*To suppress the unwanted sideband,  a filter with a very high edge slope must be used.
*Such steep-edged filters exhibit strong group delay distortions,  especially at the limit of the passband.

[[File:EN_Mod_T_2_5_S1.png|right|frame|Spectrum  (of the analytical signal)  for vestigial sideband amplitude modulation]]
The problem can be greatly mitigated if instead single-sideband AM one uses   "vestigial sideband amplitude modulation"  $\rm (VSB–AM)$,  as shown in the adjacent graph.

The present description is based on the textbook   [Mäu88]<ref>Mäusl, R.:  Analoge Modulationsverfahren.  Heidelberg: Dr. Hüthig, 1988.</ref>.  According to it,  the VSB-AM can be characterized as follows:
*A certain frequency range of the actually suppressed sideband – in the considered example of the LSB – is additionally used with a relatively flat decreasing transfer function.
*On the receiver side,  a selection curve linearly increasing in frequency with a so-called  "Nyquist edge"  is used in the transition range from the suppressed sideband to the transmitted sideband.
*The demodulation performs a convolution of the sidebands around the carrier,  so that as a result the message content of a band with the same amplitude for all frequencies is obtained.

{{GraueBox|TEXT=
$\text{Example 1:}$  The vestigial sideband method is used for  (analog)  color television,  whose frequency spectrum according to the CCIR standard is shown in the graphic.  The frequencies given refer to the  [https://en.wikipedia.org/wiki/PAL PAL–B/G Television format]  used in Germany.

[[File:EN_Mod_T_2_5_S1b.png|right|frame|Illustrating the Nyquist edge for PAL–B/G Television]]
In this schematic representation,  one recognises:
*The radiated spectrum  (only positive frequencies are drawn)  ranges from  $f_{\rm T} - 1.25 \ \rm MHz$  to  $f_{\rm T} + 5.75 \ \rm MHz$.  The lower vestigial sideband including Nyquist edge is  $\approx 1.25 \ \rm MHz$  wide.
*The green dashed line shows the receiver passband.  The image carrier  '''(B)'''  – abbreviation  "B"  from German "Bild"   ⇒   "image") –   at carrier frequency  $f_{\rm T}$  is centered on the Nyquist edge.
*The luminance signal   '''(L)'''  goes up to about  $5 \ \rm MHz$.  It contains the information for the image brightness and for the color  "green".
*The chrominance signal   '''(C)'''  is embedded in the upper part. Two orthogonal carriers are [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29|QAM-modulated]]  at  $4.43 \ \rm MHz$  for  "red"  and  "blue";  the carrier is suppressed.
*The audio carrier   '''(T)'''  – abbreviation  "T"  from German "Ton"   ⇒   "sound/audio") –   is at  $f_{\rm T} + 5.5 \ \rm MHz$  and is  $12 \ \rm dB$  lower than the image carrier.
*If there is a two-channel sound  (stereo)  transmission   ⇒   a second   '''(T)'''  carrier at  $5.75 \ \rm MHz$.}}

==Quadrature Amplitude Modulation (QAM)==
 
By exploiting the orthogonality of cosine and sine functions,  a channel can be used twice for simultaneous transmission of two source signals  $q_1(t)$  and  $q_2(t)$  without mutual interference.  This method is called  "quadrature amplitude modulation"  $\rm (QAM)$.

[[File:P_ID1053__Mod_T_2_5_S2_neu.png |right|frame|Model of quadrature amplitude modulation]]

The QAM system has the following characteristics:
*The transmitted signal is composed of two mutually orthogonal components:
:$$s(t) = q_1(t) \cdot \cos (\omega_{\rm T}\cdot t) - q_2(t) \cdot \sin (\omega_{\rm T}\cdot t)\hspace{0.05cm}.$$
*For frequency and phase synchronous demodulation,  the signal in the upper branch before the low-pass  $H_{\rm E1}(f)$  is:
:$$b_1(t) = q_1(t) \cdot 2 \cdot \cos^2 (\omega_{\rm T}\cdot t) - q_2(t) \cdot 2 \cdot
\cos (\omega_{\rm T}\cdot t)\cdot \sin (\omega_{\rm T}\cdot t) $$
:$$\Rightarrow \hspace{0.3cm} b_1(t) = q_1(t)\cdot \big[ 1 + \cos (2 \omega_{\rm T}\cdot t) \big] - q_2(t)\cdot \sin (2 \omega_{\rm T}\cdot t) \hspace{0.05cm}.$$
*Thus,  by limiting to frequencies  $|f| < f_{\rm T}$,  we obtain in the upper and lower branches,  resp.:
:$$v_1(t) = q_1(t),$$
:$$v_2(t) = q_2(t)\hspace{0.05cm}.$$
*When there is a phase offset  $Δϕ_{\rm T}$  between the transmitted and received carrier signals,  in addition to attenuation of the intended participant, crosstalk from the second participant occurs,  resulting in nonlinear distortion:
:$$v_1(t) = \alpha_{11} \cdot q_1(t)+ \alpha_{12} \cdot q_2(t) \hspace{0.05cm}, $$
:$$v_2(t) = \alpha_{21} \cdot q_1(t)+ \alpha_{22} \cdot q_2(t)$$
:$$\Rightarrow\hspace{0.3cm}\alpha_{11} = \alpha_{22} = \cos(\Delta \phi_{\rm T}) \hspace{0.05cm}, \hspace{0.3cm} \alpha_{12} = -\alpha_{21} = \sin(\Delta \phi_{\rm T}) \hspace{0.05cm}.$$

==Incoherent (non-coherent) Demodulation==
 
{{BlaueBox|TEXT=
$\text{Definition:}$  Demodulators can be classified in the following way:
*A demodulator is said to be  '''coherent''',  if in addition to the required frequency synchronization,  it requires accurate information about the phase of the transmit-side carrier signal  $z(t)$  to reconstruct the source signal.

*If this phase information is not required,  the demodulator is said to be an  '''incoherent demodulator'''  or  "non-coherent demodulator".}}

An example of an incoherent demodulator is the  [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]].

{{GraueBox|TEXT=
$\text{Example 2:}$  A second example is shown in the following block diagram.  In contrast to quadrature amplitude modulation,  here the orthogonality between cosine and sine functions is not used for the simultaneous transmission of a second source signal,  but rather to simplify the receiver-side device.

[[File:P_ID1054__Mod_T_2_5_S3_neu.png |right|frame|Incoherent demodulation with  $\rm DSB-AM$]]

It should be further noted regarding this arrangement:
*The receiver-side carrier signals can have an arbitrary and also time-dependent phase offset  $Δϕ_{\rm T}$  with respect to the carrier signals at the transmitter, as long as the phase difference between the two branches remains exactly  $90^\circ$ .
*For the signals in the upper and lower branches – after the multiplier and low-pass filtering, respectively – it holds:
:$$b_1(t) = \cos(\Delta \phi_{\rm T}) \cdot q(t), $$
:$$b_2(t) = -\sin(\Delta \phi_{\rm T}) \cdot q(t).$$
*This ensures that the sink signal  $v(t)$  matches the source signal  $q(t)$  in magnitude,  regardless of the phase offset  $Δϕ_{\rm T}$:
:$$v(t) = \sqrt{ b_1^2(t) + b_2^2(t)} = \sqrt{ q^2(t) } = \vert q(t) \vert \hspace{0.05cm}.$$
*A prerequisite for operability  – that is,  for the result  $v(t) = q(t)$  –  is that at all times  $q(t) ≥ 0$.  In an analog transmission system,  this fact could be enforced using the  "DSB-AM with carrier"  modulation method,  for example.
*This form of non-coherent demodulation  (or modifications of it)  is mainly applied in some  '''digital modulation methods''',  which are discussed in detail in the fourth chapter of  [[Modulation_Methods|this book]].}}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_2.12:_Non-coherent_Demodulation|Exercise 2.12: Non-coherent Demodulation]]

[[Aufgaben:Exercise_2.13:_Quadrature_Amplitude_Modulation|Exercise 2.13: Quadrature Amplitude Modulation (QAM)]]

==References==
<references/>

{{Display}}

Modulation Methods/Further AM Variants

2022-04-09T14:21:18Z

Reed:

{{Header
|Untermenü=Amplitude Modulation and Demodulation
|Vorherige Seite=Single-Sideband Modulation
|Nächste Seite=Phase Modulation (PM)
}}
==Vestigial sideband amplitude modulation==
 
When transmitting signals using single-sideband modulation  $\rm (SSB–AM)$  the following problems occur:
*To suppress the unwanted sideband,  a filter with a very high edge slope must be used.
*Such a steep-edged filters exhibit strong group delay distortions,  especially at the limit of the passband.

[[File:EN_Mod_T_2_5_S1.png|right|frame|Spectrum  (of the analytical signal)  for vestigial sideband amplitude modulation]]
The problem can be greatly mitigated if instead single-sideband AM one uses   "vestigial sideband amplitude modulation"  $\rm (VSB–AM)$,  as shown in the adjacent graph.

The present description is based on the textbook   [Mäu88]<ref>Mäusl, R.:  Analoge Modulationsverfahren.  Heidelberg: Dr. Hüthig, 1988.</ref>.  According to it,  the VSB-AM can be characterized as follows:
*A certain frequency range of the actually suppressed sideband – in the considered example of the LSB – is additionally used with a relatively flat decreasing transfer function.
*On the receiver side,  a selection curve linearly increasing in frequency with a so-called  "Nyquist edge"  is used in the transition range from the suppressed sideband to the transmitted sideband.
*The demodulation performs a convolution of the sidebands around the carrier,  so that as a result the message content of a band with the same amplitude for all frequencies is obtained.

{{GraueBox|TEXT=
$\text{Example 1:}$  The vestigial sideband method is used for  (analog)  color television,  whose frequency spectrum according to the CCIR standard is shown in the graphic.  The frequencies given refer to the  [https://en.wikipedia.org/wiki/PAL PAL–B/G Television format]  used in Germany.

[[File:EN_Mod_T_2_5_S1b.png|right|frame|Illustrating the Nyquist edge for PAL–B/G Television]]
In this schematic representation,  one recognises:
*The radiated spectrum  (only positive frequencies are drawn)  ranges from  $f_{\rm T} - 1.25 \ \rm MHz$  to  $f_{\rm T} + 5.75 \ \rm MHz$.  The lower vestigial sideband including Nyquist edge is  $\approx 1.25 \ \rm MHz$  wide.
*The green dashed line shows the receiver passband.  The image carrier  '''(B)'''  – abbreviation  "B"  from German "Bild"   ⇒   "image") –   at carrier frequency  $f_{\rm T}$  is centered on the Nyquist edge.
*The luminance signal   '''(L)'''  goes up to about  $5 \ \rm MHz$.  It contains the information for the image brightness and for the color  "green".
*The chrominance signal   '''(C)'''  is embedded in the upper part. Two orthogonal carriers are [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29|QAM-modulated]]  at  $4.43 \ \rm MHz$  for  "red"  and  "blue";  the carrier is suppressed.
*The audio carrier   '''(T)'''  – abbreviation  "T"  from German "Ton"   ⇒   "sound/audio") –   is at  $f_{\rm T} + 5.5 \ \rm MHz$  and is  $12 \ \rm dB$  lower than the image carrier.
*If there is a two-channel sound  (stereo)  transmission   ⇒   a second   '''(T)'''  carrier at  $5.75 \ \rm MHz$.}}

==Quadrature Amplitude Modulation (QAM)==
 
By exploiting the orthogonality of cosine and sine functions,  a channel can be used twice for simultaneous transmission of two source signals  $q_1(t)$  and  $q_2(t)$  without mutual interference.  This method is called  "quadrature amplitude modulation"  $\rm (QAM)$.

[[File:P_ID1053__Mod_T_2_5_S2_neu.png |right|frame|Model of quadrature amplitude modulation]]

The QAM system has the following characteristics:
*The transmitted signal is composed of two mutually orthogonal components:
:$$s(t) = q_1(t) \cdot \cos (\omega_{\rm T}\cdot t) - q_2(t) \cdot \sin (\omega_{\rm T}\cdot t)\hspace{0.05cm}.$$
*For frequency and phase synchronous demodulation,  the signal in the upper branch before the low-pass  $H_{\rm E1}(f)$  is:
:$$b_1(t) = q_1(t) \cdot 2 \cdot \cos^2 (\omega_{\rm T}\cdot t) - q_2(t) \cdot 2 \cdot
\cos (\omega_{\rm T}\cdot t)\cdot \sin (\omega_{\rm T}\cdot t) $$
:$$\Rightarrow \hspace{0.3cm} b_1(t) = q_1(t)\cdot \big[ 1 + \cos (2 \omega_{\rm T}\cdot t) \big] - q_2(t)\cdot \sin (2 \omega_{\rm T}\cdot t) \hspace{0.05cm}.$$
*Thus,  by limiting to frequencies  $|f| < f_{\rm T}$,  we obtain in the upper and lower branches,  resp.:
:$$v_1(t) = q_1(t),$$
:$$v_2(t) = q_2(t)\hspace{0.05cm}.$$
*When there is a phase offset  $Δϕ_{\rm T}$  between the transmitted and received carrier signals,  in addition to attenuation of the intended participant, crosstalk from the second participant occurs,  resulting in nonlinear distortion:
:$$v_1(t) = \alpha_{11} \cdot q_1(t)+ \alpha_{12} \cdot q_2(t) \hspace{0.05cm}, $$
:$$v_2(t) = \alpha_{21} \cdot q_1(t)+ \alpha_{22} \cdot q_2(t)$$
:$$\Rightarrow\hspace{0.3cm}\alpha_{11} = \alpha_{22} = \cos(\Delta \phi_{\rm T}) \hspace{0.05cm}, \hspace{0.3cm} \alpha_{12} = -\alpha_{21} = \sin(\Delta \phi_{\rm T}) \hspace{0.05cm}.$$

==Incoherent (non-coherent) Demodulation==
 
{{BlaueBox|TEXT=
$\text{Definition:}$  Demodulators can be classified in the following way:
*A demodulator is said to be  '''coherent''',  if in addition to the required frequency synchronization,  it requires accurate information about the phase of the transmit-side carrier signal  $z(t)$  to reconstruct the source signal.

*If this phase information is not required,  the demodulator is said to be an  '''incoherent demodulator'''  or  "non-coherent demodulator".}}

An example of an incoherent demodulator is the  [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]].

{{GraueBox|TEXT=
$\text{Example 2:}$  A second example is shown in the following block diagram.  In contrast to quadrature amplitude modulation,  here the orthogonality between cosine and sine functions is not used for the simultaneous transmission of a second source signal,  but rather to simplify the receiver-side device.

[[File:P_ID1054__Mod_T_2_5_S3_neu.png |right|frame|Incoherent demodulation with  $\rm DSB-AM$]]

It should be further noted regarding this arrangement:
*The receiver-side carrier signals can have an arbitrary and also time-dependent phase offset  $Δϕ_{\rm T}$  with respect to the carrier signals at the transmitter, as long as the phase difference between the two branches remains exactly  $90^\circ$ .
*For the signals in the upper and lower branches – after the multiplier and low-pass filtering, respectively – it holds:
:$$b_1(t) = \cos(\Delta \phi_{\rm T}) \cdot q(t), $$
:$$b_2(t) = -\sin(\Delta \phi_{\rm T}) \cdot q(t).$$
*This ensures that the sink signal  $v(t)$  matches the source signal  $q(t)$  in magnitude,  regardless of the phase offset  $Δϕ_{\rm T}$:
:$$v(t) = \sqrt{ b_1^2(t) + b_2^2(t)} = \sqrt{ q^2(t) } = \vert q(t) \vert \hspace{0.05cm}.$$
*A prerequisite for operability  – that is,  for the result  $v(t) = q(t)$  –  is that at all times  $q(t) ≥ 0$.  In an analog transmission system,  this fact could be enforced using the  "DSB-AM with carrier"  modulation method,  for example.
*This form of non-coherent demodulation  (or modifications of it)  is mainly applied in some  '''digital modulation methods''',  which are discussed in detail in the fourth chapter of  [[Modulation_Methods|this book]].}}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_2.12:_Non-coherent_Demodulation|Exercise 2.12: Non-coherent Demodulation]]

[[Aufgaben:Exercise_2.13:_Quadrature_Amplitude_Modulation|Exercise 2.13: Quadrature Amplitude Modulation (QAM)]]

==References==
<references/>

{{Display}}

Aufgaben:Exercise 2.11Z: Once again SSB-AM and Envelope Demodulator

2022-04-09T14:19:33Z

Reed: /* Solution */

{{quiz-Header|Buchseite=Modulation_Methods/Single-Sideband_Modulation
}}

[[File:P_ID1050__Mod_Z_2_10.png|right|frame|Equivalent low-pass signal in single-sideband AM]]
The adjacent graph shows the locus curve – i.e., the representation of the equivalent low-pass signal  (German:  "äquivalentes Tiefpass-Signal"   ⇒   subscript:  "TP")  in the complex plane – for a  single-sideband amplitude modulation  $\text{(SSB-AM)}$  system.

It is further given that the carrier frequency is  $f_{\rm T} = 100 \ \rm kHz$  and the channel is ideal:
:$$ r(t) = s(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} r_{\rm TP}(t) = s_{\rm TP}(t) \hspace{0.05cm}.$$

An ideal envelope demodulator is used at the receiver.

The following values are used in these exercises:
*the sideband-to-carrier ratio
:$$\mu = \frac{A_{\rm N}/2}{A_{\rm T}}\hspace{0.05cm},$$
*the envelope
:$$a(t) = |s_{\rm TP}(t)| \hspace{0.05cm},$$
*the maximum deviation  $τ_{\rm max}$  of the zero crossings between the transmitted signal  $s(t)$  and the carrier signal  $z(t)$.

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Single-Sideband_Modulation|Single-sideband Modulation]].
*Particular reference is made to the page  [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|Sideband-to-carrier ratio]].
*In this exercise,  apply the same assumptions as in  [[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_ESB_Signal|Exercise 2.11]].

===Questions===

<quiz display=simple>
{Find the equivalent low-pass signal  $s_{\rm TP}(t)$  in analytical form and choose which statements apply.
|type="[]"}
- We are dealing with an  "upper-sideband amplitude modulation"  $\text{(USB-AM)}$.
+ We are dealing with a  "lower-sideband amplitude modulation"  $\text{(LSB-AM)}$.
- The source signal  $q(t)$  is cosine-shaped.
+ The source signal  $q(t)$  is sine-shaped.

{Find the amplitude  $A_{\rm N}$  and the frequency  $f_{\rm N}$  of the source signal.  Take into account that we are dealing with a SSB-AM.
|type="{}"}
$A_{\rm N} \ = \ $ { 2 3% } $\ \rm V$
$f_{\rm N} \ = \ $ { 5 3% } $\ \rm kHz$

{Which value results for the sideband-to-carrier ratio   $μ$?  Use this value to describe  $s_{\rm TP}(t)$.
|type="{}"}
$μ \ = \ $ { 1 3% }

{Calculate the time course of the envelope  $a(t)$.  Which values arise for  $t = 50 \ \rm µ s$,  $t = 100 \ \rm µ s$  and  $t = 150 \ \rm µ s$?
|type="{}"}
$a(t = 50 \ \rm µ s) \hspace{0.32cm} = \ $ { 2 3% } $\ \rm V$
$a(t = 100 \ \rm µ s) \ = \ $ { 1.414 3% } $\ \rm V$
$a(t = 150 \ \rm µ s) \ = \ $ { 0. } $\ \rm V$

{By what time difference   $τ_{\rm max}$  (in magnitude) are the zero crossings of  $s(t)$  maximally shifted with respect to  $z(t)$ ?
|type="{}"}
$τ_{\rm max} \ = \ $ { 2.5 3% } $\ \rm µ s$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Answers 2 and 4  are correct:
*The equivalent low-pass signal is:
:$$ s_{\rm TP}(t) = 1\,{\rm V} + {\rm j}\cdot 1\,{\rm V}\cdot {\rm e}^{-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \hspace{0.05cm}.$$
*The locus curve is a circle with its center at  $A_{\rm T} = 1 \ \rm V$.
*Since the rotation is clockwise,  we are dealing with a  "LSB-AM".
*At the start time  $t = 0$,  the green pointer is in the direction of the imaginary axis.
*It follows that the source signal is characterized by   $q(t) = A_{\rm N} \cdot \sin(\omega_{\rm N} \cdot t).$

'''(2)'''  For the LSB,  only the lower sideband is transmitted,  with pointer length   $A_{\rm N}/2 = 1 \ \rm V$ .
*This results in  $A_{\rm N}\hspace{0.15cm}\underline { = 2 \ \rm V}$.
*For one revolution in the locus,  the pointer needs the time   $200 \ \rm µ s$.
*The reciprocal of this is the frequency   $f_{\rm N}\hspace{0.15cm}\underline { = 5 \ \rm kHz}$.

'''(3)'''  According to the definition on the exercise page and the results in subtasks  '''(1)'''  and  '''(2)''',  the following holds:
:$$ \mu = \frac{A_{\rm N}/2}{A_{\rm T}}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$
*Thus,  the equivalent low-pass signal can also be written as:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + {\rm j} \cdot \mu \cdot {\rm e}^{-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right),\hspace{0.3cm}{\rm here}\hspace{0.15cm}\mu = 1 \hspace{0.05cm}.$$

'''(4)'''  Splitting the complex exponential function into real and imaginary parts using Euler's theorem,  we get:
:$$s_{\rm TP}(t) = A_{\rm T} \cdot \big( 1 + \sin(\omega_{\rm N}\cdot t) + {\rm j} \cos(\omega_{\rm N}\cdot t)\big) \hspace{0.05cm}.$$

*By applying the  "Pythagorean Theorem",  this can also be written as:
:$$a(t) = |s_{\rm TP}(t)| = A_{\rm T} \cdot \sqrt{ (1 + \sin(\omega_{\rm N}\cdot t))^2 + \cos^2(\omega_{\rm N}\cdot t)} =
A_{\rm T} \cdot \sqrt{ 2 + 2 \cdot \sin(2\omega_{\rm N}\cdot t)} \hspace{0.05cm}.$$
*The retrieved values when  $A_{\rm T} = 1\ \rm V$  are:
:$$ a(t = 50\,{\rm µ s}) \hspace{0.15cm}\underline {= 2\,{\rm V}},\hspace{0.3cm}a(t = 100\,{\rm µ s}) \hspace{0.15cm}\underline {= 1.414\,{\rm V}},\hspace{0.3cm}a(t = 150\,{\rm µ s}) \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$$
*These results can be directly read off the graph on the exercise page.

'''(5)'''  A hint for the location of the zero crossings of   $s(t)$  with respect to the grid given by the carrier signal   $z(t)$  is provided by the phase function  $ϕ(t)$.
*For the given locus,  these can take on values between  $±π/2\ (±90^\circ)$ .
*For example,  these maximum values arise in the region around   $t ≈ 150 \ \rm µ s$,  since a phase jump occurs there.
*The relationship between  $τ_{\rm max}$  and  $\Delta ϕ_{\rm max}$  is:
:$$ \tau_{\rm max} = \frac {\Delta \phi_{\rm max}}{2 \pi }\cdot \frac{1 }{f_{\rm T}} = \frac {1}{4}\cdot 10\,{\rm µ s} \hspace{0.15cm}\underline {= 2.5\,{\rmµ s}} \hspace{0.05cm}.$$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.4 Single Sideband Amplitude Modulation^]]

Aufgaben:Exercise 2.11: Envelope Demodulation of an SSB Signal

2022-04-09T14:17:34Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Single-Sideband_Modulation
}}

[[File:P_ID1047__Mod_A_2_10.png|right|frame|(Normalized) envelope in Single-sideband modulation]]
Let us consider the transmission of the cosine signal
:$$ q(t) = A_{\rm N} \cdot \cos(\omega_{\rm N} \cdot t)$$
according to the modulation method  $\rm USB–AM$  ("upper-sideband amplitude modulation")  with carrier.  At the receiver,  the high frequency range (HF) is reset to the low frquency range (LF) with an   [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]].

The channel is assumed to be ideal such that the received signal  $r(t)$  is identical to the transmitted signal  $s(t)$ .  With the sideband-to-carrier ratio
:$$ \mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}}$$
the equivalent low-pass signal  (German:  "äquivalentes Tiefpass-Signal"   ⇒   subscript:  "TP")  can be written as:
:$$r_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + \mu \cdot {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right) \hspace{0.05cm}$$

The envelope – i.e.,  the magnitude of this complex signal – can be determined by geometric considerations.  Independent of the parameter  $μ$,  one obtains:
:$$a(t ) = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu \cdot \cos(\omega_{\rm N} \cdot t)}\hspace{0.05cm}.$$
The time-independent envelope  $a(t)$  for  $μ = 1$  and  $μ = 0.5$  is shown in the graph.  In each case,  the amplitude-matched cosine oscillations,  which would be a prerequisite for distortion-free demodulation,  are plotted as dashed comparison curves.

*The periodic signal  $a(t)$  can be approximated by a  [[Signal_Representation/Fourier_Series|Fourier series]] :
:$$a(t ) = A_{\rm 0} + A_{\rm 1} \cdot \cos(\omega_{\rm N} \cdot t) + A_{\rm 2} \cdot \cos(2\omega_{\rm N} \cdot t)+ A_{\rm 3} \cdot \cos(3\omega_{\rm N} \cdot t)\hspace{0.05cm}+\text{...}$$
*The Fourier coefficients were determined using a simulation program.   With  $μ = 1$  the following values were obtained:
:$$A_{\rm 0} = 1.273\,{\rm V},\hspace{0.3cm} A_{\rm 1} = 0.849\,{\rm V},\hspace{0.3cm}A_{\rm 2} = -0.170\,{\rm V},\hspace{0.3cm} A_{\rm 3} = 0.073\,{\rm V},\hspace{0.3cm}A_{\rm 4} = 0.040\,{\rm V} \hspace{0.05cm}.$$
*Accordingly,  for  $μ = 0.5$,  the simulation yielded:
:$$A_{\rm 0} = 1.064\,{\rm V},\hspace{0.3cm} A_{\rm 1} = 0.484\,{\rm V},\hspace{0.3cm}A_{\rm 2} = 0.058\,{\rm V} \hspace{0.05cm}.$$
:The values not given here can be ignored when calculating of the distortion factor.
*The sink signal  $v(t)$  is obtained from  $a(t)$  as follows:
:$$v(t) = 2 \cdot \big [a(t ) - A_{\rm 0} \big ] \hspace{0.05cm}.$$
:The factor of  $2$  corrects for the amplitude loss due to  "single-sideband amplitude modulation",  while the subtraction of the DC signal coefficient  $A_0$  takes into account the influence of the high-pass within the envelope demodulator.
*In questions  '''(1)'''  to  '''(3)''',  it is assumed that  $A_{\rm N} = 2 \ \rm V$, $A_{\rm T} = 1 \ \rm V$   ⇒   $μ = 1$,  whereas from question  '''(4)''',   $A_{\rm N} = A_{\rm T} = 1 \ \rm V$  should apply for the parameter  $μ = 0.5$.

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Single-Sideband_Modulation|Single-Sideband Modulation]].
*Particular reference is made to the page   [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|Sideband-to-carrier ratio]].
*Also compare your results to the rule of thumb which states that "for the envelope demodulation of an SSB-AM signal with sideband-to-carrier ratio  $μ$,  the distortion factor is  $K ≈ μ/4$".

===Questions===

<quiz display=simple>
{Give the maximum and minimum values of the sink signal  $v(t)$  when  $μ = 1$.
|type="{}"}
$v_{\rm max} \ = \ $ { 1.454 3% } $\ \rm V$
$v_{\rm min} \ = \ $ { -2.62--2.48 } $\ \rm V$

{Calculate the distortion factor when  $μ = 1$.
|type="{}"}
$K \ = \ $ { 22.3 3% } $\ \text{%}$

{How can you recognize the nonlinear distortions in the signal  $v(t)$?
|type="[]"}
+ The lower cosine half-wave is more peaked than the upper one.
- The DC component  ${\rm Ε}\big[v(t)\big ] = 0$.

{Give the maximum and minimum values of the sink signal  $v(t)$  when  $μ = 0.5$.
|type="{}"}
$v_{\rm max} \ = \ $ { 0.872 3% } $\ \rm V$
$v_{\rm min} \ = \ $ { -2.19--2.07 } $\ \rm V$

{Calculate the distortion factor when  $μ = 0.5$.
|type="{}"}
$K \ = \ $ { 12 3% } $\ \text{%}$

{What is the upper bound  $K_{\rm max}$  of the distortion factor in  "double-sideband amplitude modulation"  $\text{(DSB-AM)}$  with  $m = 0.5$  and envelope demodulation,  if one sideband is completely damped by the channel.
|type="{}"}
$K_{\rm max} \ = \ ${ 6.25 3% } $\ \text{%}$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The maximum value   $a_{\rm max} = 2\ \rm V$  and the minimum value   $a_{\rm min} = 0$  can be read off the graph or calculated using the equation given:
:$$ a_{\rm max} = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu}= A_{\rm T} \cdot (1+ \mu) = 2\,{\rm V} \hspace{0.05cm},$$
:$$a_{\rm min} = A_{\rm T} \cdot \sqrt{1+ \mu^2 - 2 \mu}= A_{\rm T} \cdot (1- \mu) = 0 \hspace{0.05cm}.$$
*For the two extreme values of the sink signal it follows:
:$$ v_{\rm max} = 2 \cdot [a_{\rm max} - A_{\rm 0}] = 2 \cdot [2\,{\rm V} - 1.273\,{\rm V}] \hspace{0.15cm}\underline {=1.454\,{\rm V}}\hspace{0.05cm},$$
:$$ v_{\rm min} = -2 \cdot A_{\rm 0} \hspace{0.15cm}\underline {= -2.546\,{\rm V}}\hspace{0.05cm}.$$

'''(2)'''  Ignoring the Fourier coefficients   $A_5$,  $A_6$,  etc., we obtain:
:$$K = \frac{\sqrt{A_2^2 + A_3^2+ A_4^2 }}{A_1}= \frac{\sqrt{0.170^2 + 0.073^2 + 0.040^2 }{\,\rm V}}{0.849\,{\rm V}}\hspace{0.15cm}\underline { \approx 22.3 \%}.$$
*Here,  the approximation  $K ≈ μ/4$  yields the value $25\%$.

'''(3)'''  Only the  first answer  is correct.
*Due to the high-pass within the envelope demodulator,  the DC signal component would also be equal to zero if no distortions were present.

'''(4)'''  Like in subtask  '''(1)'''  here it holds that:
:$$v_{\rm max} = 2 \cdot [a_{\rm max} - A_{\rm 0}] = 2 \cdot [1.5\,{\rm V} - 1.064\,{\rm V}] \hspace{0.15cm}\underline {= 0.872\,{\rm V}}\hspace{0.05cm},$$
:$$ v_{\rm min} = -2 \cdot A_{\rm 0} \hspace{0.15cm}\underline {= -2.128\,{\rm V}}\hspace{0.05cm}.$$

'''(5)'''  A smaller sideband-to-carrier ratio also results in a smaller distortion factor:
:$$K = \frac{0.058{\,\rm V}}{0.484\,{\rm V}}\hspace{0.15cm}\underline { \approx 12 \%}.$$
*The simple approximation   $K ≈ μ/4$  here yields  $12.5\%$.
*It can be concluded that the above rule of thumb is more accurate for smaller values of   $μ$.

'''(6)'''  Thus,  the distortion factor is largest when one of the sidebands is entirely cut out.
*However,  since the envelope demodulator has no information to distinguish between
#a SSB–AM, or
#a DSB-AM which has been extremely affected by the channel,

:  $K_{\rm max} ≈ μ/4$  simultaneously gives an upper bound for the DSB-AM.

*A comparison of the parameters  $m = A_{\rm N}/A_{\rm T}$  and  $μ = A_{\rm N}/(2A_{\rm T})$  leads to the result:
:$$K_{\rm max} = \frac{\mu}{4} = \frac{m}{8} \hspace{0.15cm}\underline {=6.25 \%}.$$
{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.4 Single Sideband Amplitude Modulation^]]

Aufgaben:Exercise 2.11: Envelope Demodulation of an SSB Signal

2022-04-09T14:16:38Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Single-Sideband_Modulation
}}

[[File:P_ID1047__Mod_A_2_10.png|right|frame|(Normalized) envelope in Single-sideband modulation]]
Let us consider the transmission of the cosine signal
:$$ q(t) = A_{\rm N} \cdot \cos(\omega_{\rm N} \cdot t)$$
according to the modulation method  $\rm USB–AM$  ("upper-sideband amplitude modulation")  with carrier.  At the receiver,  the high frequency range (HF) is reset to the low frquency range (LF) with an   [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]].

The channel is assumed to be ideal such that the received signal  $r(t)$  is identical to the transmitted signal  $s(t)$ .  With the sideband-to-carrier ratio
:$$ \mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}}$$
the equivalent low-pass signal  (German:  "äquivalentes Tiefpass-Signal"   ⇒   subscript:  "TP")  can be written as:
:$$r_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + \mu \cdot {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right) \hspace{0.05cm}$$

The envelope – i.e.,  the magnitude of this complex signal – can be determined by geometric considerations.  Independent of the parameter  $μ$,  one obtains:
:$$a(t ) = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu \cdot \cos(\omega_{\rm N} \cdot t)}\hspace{0.05cm}.$$
The time-independent envelope  $a(t)$  for  $μ = 1$  and  $μ = 0.5$  is shown in the graph.  In each case,  the amplitude-matched cosine oscillations,  which would be a prerequisite for distortion-free demodulation,  are plotted as dashed comparison curves.

*The periodic signal  $a(t)$  can be approximated by a  [[Signal_Representation/Fourier_Series|Fourier series]] :
:$$a(t ) = A_{\rm 0} + A_{\rm 1} \cdot \cos(\omega_{\rm N} \cdot t) + A_{\rm 2} \cdot \cos(2\omega_{\rm N} \cdot t)+ A_{\rm 3} \cdot \cos(3\omega_{\rm N} \cdot t)\hspace{0.05cm}+\text{...}$$
*The Fourier coefficients were determined using a simulation program.   With  $μ = 1$  the following values were obtained:
:$$A_{\rm 0} = 1.273\,{\rm V},\hspace{0.3cm} A_{\rm 1} = 0.849\,{\rm V},\hspace{0.3cm}A_{\rm 2} = -0.170\,{\rm V},\hspace{0.3cm} A_{\rm 3} = 0.073\,{\rm V},\hspace{0.3cm}A_{\rm 4} = 0.040\,{\rm V} \hspace{0.05cm}.$$
*Accordingly,  for  $μ = 0.5$,  the simulation yielded:
:$$A_{\rm 0} = 1.064\,{\rm V},\hspace{0.3cm} A_{\rm 1} = 0.484\,{\rm V},\hspace{0.3cm}A_{\rm 2} = 0.058\,{\rm V} \hspace{0.05cm}.$$
:The values not given here can be ignored when calculating of the distortion factor.
*The sink signal  $v(t)$  is obtained from  $a(t)$  as follows:
:$$v(t) = 2 \cdot \big [a(t ) - A_{\rm 0} \big ] \hspace{0.05cm}.$$
:The factor of  $2$  corrects for the amplitude loss due to  "single-sideband amplitude modulation",  while the subtraction of the DC signal coefficient  $A_0$  takes into account the influence of the high-pass within the envelope demodulator.
*In questions  '''(1)'''  to  '''(3)''',  it is assumed that  $A_{\rm N} = 2 \ \rm V$, $A_{\rm T} = 1 \ \rm V$   ⇒   $μ = 1$,  whereas from question  '''(4)''',   $A_{\rm N} = A_{\rm T} = 1 \ \rm V$  should apply for the parameter  $μ = 0.5$.

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Single-Sideband_Modulation|Single-Sideband Modulation]].
*Particular reference is made to the page   [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|Sideband-to-carrier ratio]].
*Also compare your results to the rule of thumb which states that "for the envelope demodulation of an SSB-AM signal with sideband-to-carrier ratio  $μ$,  the distortion factor is  $K ≈ μ/4$".

===Questions===

<quiz display=simple>
{Give the maximum and minimum values of the sink signal  $v(t)$  when  $μ = 1$.
|type="{}"}
$v_{\rm max} \ = \ $ { 1.454 3% } $\ \rm V$
$v_{\rm min} \ = \ $ { -2.62--2.48 } $\ \rm V$

{Calculate the distortion factor when  $μ = 1$.
|type="{}"}
$K \ = \ $ { 22.3 3% } $\ \text{%}$

{How can you recognize the nonlinear distortions in the signal  $v(t)$?
|type="[]"}
+ The lower cosine half-wave is more peaked than the upper one.
- The DC component  ${\rm Ε}\big[v(t)\big ] = 0$.

{Give the maximum and minimum values of the sink signal  $v(t)$  when  $μ = 0.5$.
|type="{}"}
$v_{\rm max} \ = \ $ { 0.872 3% } $\ \rm V$
$v_{\rm min} \ = \ $ { -2.19--2.07 } $\ \rm V$

{Calculate the distortion factor when  $μ = 0.5$.
|type="{}"}
$K \ = \ $ { 12 3% } $\ \text{%}$

{What is the upper bound  $K_{\rm max}$  of the distortion factor in  "double-sideband amplitude modulation"  $\text{(DSB-AM)}$  with  $m = 0.5$  and envelope demodulation,  if one sideband is completely damped by the channel.
|type="{}"}
$K_{\rm max} \ = \ ${ 6.25 3% } $\ \text{%}$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The maximum value   $a_{\rm max} = 2\ \rm V$  and the minimum value   $a_{\rm min} = 0$  can be read off the graph or calculated using the equation given:
:$$ a_{\rm max} = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu}= A_{\rm T} \cdot (1+ \mu) = 2\,{\rm V} \hspace{0.05cm},$$
:$$a_{\rm min} = A_{\rm T} \cdot \sqrt{1+ \mu^2 - 2 \mu}= A_{\rm T} \cdot (1- \mu) = 0 \hspace{0.05cm}.$$
*For the two extreme values of the sink signal it follows:
:$$ v_{\rm max} = 2 \cdot [a_{\rm max} - A_{\rm 0}] = 2 \cdot [2\,{\rm V} - 1.273\,{\rm V}] \hspace{0.15cm}\underline {=1.454\,{\rm V}}\hspace{0.05cm},$$
:$$ v_{\rm min} = -2 \cdot A_{\rm 0} \hspace{0.15cm}\underline {= -2.546\,{\rm V}}\hspace{0.05cm}.$$

'''(2)'''  Ignoring the Fourier coefficients   $A_5$,  $A_6$,  etc., we obtain:
:$$K = \frac{\sqrt{A_2^2 + A_3^2+ A_4^2 }}{A_1}= \frac{\sqrt{0.170^2 + 0.073^2 + 0.040^2 }{\,\rm V}}{0.849\,{\rm V}}\hspace{0.15cm}\underline { \approx 22.3 \%}.$$
*Here,  the approximation  $K ≈ μ/4$  yields the value $25\%$.

'''(3)'''  Only the  first answer  is correct.
*Due to the high-pass within the envelope demodulator,  the DC signal component would also be equal to zero if no distortions were present.

'''(4)'''  Like in subtask  '''(1)'''  here it holds that:
:$$v_{\rm max} = 2 \cdot [a_{\rm max} - A_{\rm 0}] = 2 \cdot [1.5\,{\rm V} - 1.064\,{\rm V}] \hspace{0.15cm}\underline {= 0.872\,{\rm V}}\hspace{0.05cm},$$
:$$ v_{\rm min} = -2 \cdot A_{\rm 0} \hspace{0.15cm}\underline {= -2.128\,{\rm V}}\hspace{0.05cm}.$$

'''(5)'''  A smaller sideband-to-carrier ratio also results in a smaller distortion factor:
:$$K = \frac{0.058{\,\rm V}}{0.484\,{\rm V}}\hspace{0.15cm}\underline { \approx 12 \%}.$$
*The simple approximation   $K ≈ μ/4$  yields here  $12.5\%$.
*It can be concluded that the above rule of thumb is more accurate for smaller values of   $μ$.

'''(6)'''  Thus,  the distortion factor is largest when one of the sidebands is entirely cut out.
*However,  since the envelope demodulator has no information to distinguish between
#a SSB–AM, or
#a DSB-AM which has been extremely affected by the channel,

:  $K_{\rm max} ≈ μ/4$  simultaneously gives an upper bound for the DSB-AM.

*A comparison of the parameters  $m = A_{\rm N}/A_{\rm T}$  and  $μ = A_{\rm N}/(2A_{\rm T})$  leads to the result:
:$$K_{\rm max} = \frac{\mu}{4} = \frac{m}{8} \hspace{0.15cm}\underline {=6.25 \%}.$$
{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.4 Single Sideband Amplitude Modulation^]]

Aufgaben:Exercise 2.10Z: Noise with DSB-AM and SSB-AM

2022-04-09T14:13:30Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Single-Sideband_Modulation
}}

[[File:EN_Mod_Z_2_9_c.png|right|frame|Shared block diagram for DSB-AM and SSB-AM]]
Now the influence of noise on the sink-to-noise power ratio  $10 · \lg ρ_v$  for both  $\rm DSB–AM$  and  $\rm SSB–AM$ transmission  will be compared.   The illustration shows the underlying block diagram.

The differences between the two system variants are highlighted in red on the image, namely the modulator  (DSB or SSB)  as well as the dimensionless constant
:$$ K = \left\{ \begin{array}{c} 2/\alpha_{\rm K} \\ 4/\alpha_{\rm K} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\rm DSB} \hspace{0.05cm}, \\ {\rm SSB} \hspace{0.05cm} \\ \end{array}$$
of the receiver-side carrier signal  $z_{\rm E}(t) = K · \cos(ω_{\rm T} · t)$,  which is assumed to be frequency and phase synchronous with the carrier signal  $z(t)$  at the transmitter.

The system characteristics
*frequency-independent channel transmission factor  $α_{\rm K}$,
*transmission power  $P_{\rm S}$,
*one–sided noise power density  $N_{\rm 0}$,
*bandwidth  $B_{\rm NF}$  of the source signal,

captured by the shared performance parameter are labelled in green:
:$$\xi = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}$$

Further note:
* The cosine signal  $q(t)$  with frequency  $B_{\rm NF}$  stands for a source signal with bandwidth  $B_{\rm NF}$ composed of multiple frequencies.
*The relationship between the transmission power  $P_{\rm S}$  and the power  $P_{q}$  of the source signal depends,  among other things,  on the modulation method.
* "DSB–AM with carrier"  is parameterized by the modulation depth  $m = A_{\rm N}/A_{\rm T}$,  while  "SSB-AM"  is determined by the sideband-to-carrier ratio  $μ = A_{\rm N}/(2 · A_{\rm T})$.
* The frequency-independent channel transmission factor  $α_{\rm K}$  is balanced by the constant  $K$,  so that in the noise-free case  $(N_0 = 0)$,  the sink signal  $v(t)$  matches the source signal  $q(t)$.
* The sink SNR can thus be given as follows  $(T_0$ indicates the period of the source signal$)$:
:$$ \rho_{v } = \frac{P_{q}}{P_{\varepsilon }}\hspace{0.5cm}{\rm with}\hspace{0.5cm} P_{q} = \frac{1}{T_{\rm 0}}\cdot\int_{0}^{ T_{\rm 0}} {q^2(t)}\hspace{0.1cm}{\rm d}t, \hspace{0.5cm}P_{\varepsilon} = \int_{-B_{\rm NF}}^{ +B_{\rm NF}} \hspace{-0.1cm}{\it \Phi_{\varepsilon}}(f)\hspace{0.1cm}{\rm d}f\hspace{0.05cm}.$$

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Single-Sideband_Modulation|"Single-Sideband Modulation"]].
*Particular reference is made to the section  [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|"Sideband-to-carrier ratio"]].
*The results for DSB–AM can be found in the section  [[Modulation_Methods/Synchronous_Demodulation#Sink_SNR_and_the_performance_parameter|"Sink SNR and the performance parameter"]].

===Questions===

<quiz display=simple>
{Which kind of demodulation is considered here?
|type="()"}
+ Synchronous demodulation.
- Envelope demodulation.

{Which relationship holds between the quantities  $ρ_v$  and  $ξ$  for  double-sideband AM without carrier  $(m → ∞)$?
|type="[]"}
-  $ρ_v = 2 · ξ$.
+  $ρ_v = ξ$.
-  $ρ_v = ξ/2$.

{Which relationship holds between  $ρ_v$  and  $ξ$  for  single-sideband AM without carrier  as $(μ → ∞)$?
|type="[]"}
-  $ρ_v = 2 · ξ$.
+  $ρ_v = ξ$.
-  $ρ_v = ξ/2$.

{Let  $ξ = 10^4$.  Calculate the sink-to-noise ratio of   double-sideband AM  for modulation depths  $m = 0.5$  and  $m = 1$.
|type="{}"}
$m = 0.5\text{:} \ \ 10 · \lg \ ρ_v \ = \ $ { 30.46 3% } $\ \rm dB$
$m = 1.0\text{:} \ \ 10 · \lg \ ρ_v \ = \ $ { 35.23 3% } $\ \rm dB$

{Further let  $ξ = 10^4$.  Calculate the sink-to-noise ratio of   single-sideband AM  for the parameters  $μ = 0.354$  and  $μ = 0.707$.
|type="{}"}
$μ = 0.354\text{:} \ \ \ 10 · \lg \ ρ_v \ = \ $ { 30.46 3% } $\ \rm dB$
$μ = 0.707\text{:} \ \ \ 10 · \lg \ ρ_v \ = \ $ { 35.23 3% } $\ \rm dB$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We are dealing with a   synchronous demodulator.  Answer 1  is correct.

'''(2)'''  Answer 2  is correct:
*For DSB–AM without a carrier,   $P_{\rm S} = P_q/2$.  This is simultaneously the power of the useful component of the sink signal  $v(t)$.
*The power-spectral density   ${\it Φ}_ε(f)$  of the   $v(t)$  noise component results from the convolution:
[[File:P_ID1048__Mod_Z_2_9_b.png|right|frame|Noise power density in double-sideband AM]]
[[File:P_ID1049__Mod_Z_2_9_c.png|right|frame|Noise power density in  upper-sideband AM]]

:$${\it \Phi}_\varepsilon(f) = {\it \Phi}_{z{\rm E} }(f) \star {\it \Phi}_n (f) = \frac{1}{\alpha_{\rm K}^2} \cdot \big[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \big]\star {\it \Phi}_n (f) \hspace{0.05cm}.$$
*The expression  $\big[$ ... $\big]$  describes the power-spectral density of a cosine signal with the signal amplitude   $K = 2$.
*The correction of channel attenuation is considered with  $1/α_K^2$ .
*Thus,  taking   ${\it \Phi}_n(f) = N_0/2$  into account,  we get:
:$${\it \Phi}_\varepsilon(f) = \frac{N_0}{\alpha_{\rm K}^2}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} P_\varepsilon = \int_{-B_{\rm NF}}^{+B_{\rm NF}} {{\it \Phi}_\varepsilon(f) }\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0 \cdot B_{\rm NF}}{\alpha_{\rm K}^2}\hspace{0.05cm}.$$
*From this,  it follows for the the signal-to-noise power ratio  $\rm (SNR)$:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}\hspace{0.15cm}\underline { = \xi} \hspace{0.05cm}.$$

'''(3)'''  Answer 2  is correct:
*In contrast to DSB,  $P_S = P_q/4$  holds for SSB,  as well as
:$${\it \Phi}_\varepsilon(f) = {\it \Phi}_{z{\rm E} }(f) \star {\it \Phi}_n (f) = \frac{4}{\alpha_{\rm K}^2} \cdot \big[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \big]\star {\it \Phi}_n (f) \hspace{0.05cm}.$$
*Taking   $B_{\rm HF} = B_{\rm NF}$  into account  (see adjacent diagram for USB modulation),  we now get:
:$${\it \Phi}_\varepsilon(f) = \frac{2 \cdot N_0}{\alpha_{\rm K}^2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} P_\varepsilon = \frac{4 \cdot N_0 \cdot B_{\rm NF}}{\alpha_{\rm K}^2}\hspace{0.05cm}.$$

*This means:  When without carrier,  single-sideband modulation demonstrates the same noise behaviour as DSB-AM.

'''(4)'''  Assuming a cosine carrier with amplitude  $A_{\rm T}$  and a similarly cosine source signal  $q(t)$,  we get for DSB with carrier:
:$$ s(t) = \big (q(t) + A_{\rm T}\big ) \cdot \cos( \omega_{\rm T} \cdot t)
= A_{\rm T} \cdot \cos( \omega_{\rm T} \cdot t) + \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}+ \omega_{\rm N}) \cdot t \big)+ \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}- \omega_{\rm N}) \cdot t\big)\hspace{0.05cm}.$$
*The transmission power is thus given by
:$$ P_{\rm S}= \frac{A_{\rm T}^2}{2} + 2 \cdot \frac{(A_{\rm N}/2)^2}{2} = \frac{A_{\rm T}^2}{2} + \frac{A_{\rm N}^2}{4} \hspace{0.05cm}.$$
*Taking  $P_q = A_{\rm N}^2/2$  and  $m = A_{\rm N}/A_{\rm T}$  into account,  this can also be written as:
:$$P_{\rm S}= \frac{A_{\rm N}^2}{4} \cdot \left[ 1 + \frac{2 \cdot A_{\rm T}^2}{A_{\rm N}^2}\right] = \frac{P_q}{2} \cdot \left[ 1 + {2 }/{m^2}\right]\hspace{0.05cm}.$$
*With the noise power  $P_ε$  according to subtask  '''(2)'''  we thus obtain:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}\cdot (1 + 2/m^2)}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}} \cdot \frac{1}{1 +{2 }/{m^2}} \hspace{0.05cm}.$$
*And in logarithmic representation:
:$$ 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi - 10 \cdot {\rm lg} \hspace{0.15cm}\left[{1 +{2 }/{m^2}}\right] \hspace{0.05cm}.$$
:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \ (m = 0.5) = 40 \,{\rm dB} - 10 \cdot {\rm lg} (9) \hspace{0.15cm}\underline {= 30.46\, {\rm dB}}$$
:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \ (m = 1.0) = 40 \,{\rm dB} - 10 \cdot {\rm lg} (3) \hspace{0.15cm}\underline {= 35.23\, {\rm dB} \hspace{0.05cm}}.$$

'''(5)'''  In  "SSB–AM"  there is only one sideband.
*Therefore,  considering the sideband-to-carrier ratio   $μ = A_{\rm N}/(2A_{\rm T})$  gives:
:$$ P_{\rm S}= \frac{A_{\rm T}^2}{2} + \frac{(A_{\rm N}/2)^2}{2} = {A_{\rm N}^2}/{8} \cdot \big[ 1 + {4 \cdot A_{\rm T}^2}/{A_{\rm N}^2}\big] = {P_q}/{4} \cdot \big[ 1 + {1 }/{\mu^2}\big] \hspace{0.05cm}.$$
*Thus,  with the noise power from subtask  '''(3)''' we obtain:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{4 \cdot P_{\rm S}\cdot (1 + 1/\mu^2)}{4 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}} \cdot \frac{1}{1 +{1 }/{\mu^2}}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi - 10 \cdot {\rm lg} \hspace{0.15cm}\big[{1 +{1 }/{\mu^2}}\big] \hspace{0.05cm}.$$
*So we get the same result with SSB-AM as in DSB-AM with a modulation depth of  $m = \sqrt{2} · μ$.  From this,  it further follows:
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {0.5}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=0.5) \hspace{0.15cm}\underline {=30.46\,{\rm dB}},$$
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {1.0}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=1.0) \hspace{0.15cm}\underline {=35.23\,{\rm dB}}\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.4 Single Sideband Amplitude Modulation^]]

Aufgaben:Exercise 2.10Z: Noise with DSB-AM and SSB-AM

2022-04-09T14:11:24Z

Reed:

{{quiz-Header|Buchseite=Modulation_Methods/Single-Sideband_Modulation
}}

[[File:EN_Mod_Z_2_9_c.png|right|frame|Shared block diagram for DSB-AM and SSB-AM]]
Now the influence of noise on the sink-to-noise power ratio  $10 · \lg ρ_v$  for both  $\rm DSB–AM$  and  $\rm SSB–AM$ transmission  will be compared.   The illustration shows the underlying block diagram.

The differences between the two system variants are highlighted in red on the image, namely the modulator  (DSB or SSB)  as well as the dimensionless constant
:$$ K = \left\{ \begin{array}{c} 2/\alpha_{\rm K} \\ 4/\alpha_{\rm K} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\rm DSB} \hspace{0.05cm}, \\ {\rm SSB} \hspace{0.05cm} \\ \end{array}$$
of the receiver-side carrier signal  $z_{\rm E}(t) = K · \cos(ω_{\rm T} · t)$,  which is assumed to be frequency and phase synchronous with the carrier signal  $z(t)$  at the transmitter.

The system characteristics
*frequency-independent channel transmission factor  $α_{\rm K}$,
*transmission power  $P_{\rm S}$,
*one–sided noise power density  $N_{\rm 0}$,
*bandwidth  $B_{\rm NF}$  of the source signal,

captured by the shared performance parameter are labelled in green:
:$$\xi = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}$$

Further note:
* The cosine signal  $q(t)$  with frequency  $B_{\rm NF}$  stands for a source signal with bandwidth  $B_{\rm NF}$ composed of multiple frequencies.
*The relationship between the transmission power  $P_{\rm S}$  and the power  $P_{q}$  of the source signal depends,  among other things,  on the modulation method.
* "DSB–AM with carrier"  is parameterized by the modulation depth  $m = A_{\rm N}/A_{\rm T}$,  while  "SSB-AM"  is determined by the sideband-to-carrier ratio  $μ = A_{\rm N}/(2 · A_{\rm T})$.
* The frequency-independent channel transmission factor  $α_{\rm K}$  is balanced by the constant  $K$,  so that in the noise-free case  $(N_0 = 0)$,  the sink signal  $v(t)$  matches the source signal  $q(t)$.
* The sink SNR can thus be given as follows  $(T_0$ indicates the period of the source signal$)$:
:$$ \rho_{v } = \frac{P_{q}}{P_{\varepsilon }}\hspace{0.5cm}{\rm with}\hspace{0.5cm} P_{q} = \frac{1}{T_{\rm 0}}\cdot\int_{0}^{ T_{\rm 0}} {q^2(t)}\hspace{0.1cm}{\rm d}t, \hspace{0.5cm}P_{\varepsilon} = \int_{-B_{\rm NF}}^{ +B_{\rm NF}} \hspace{-0.1cm}{\it \Phi_{\varepsilon}}(f)\hspace{0.1cm}{\rm d}f\hspace{0.05cm}.$$

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Single-Sideband_Modulation|"Single-Sideband Modulation"]].
*Particular reference is made to the section  [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|"Sideband-to-carrier ratio"]].
*The results for DSB–AM can be found in the section  [[Modulation_Methods/Synchronous_Demodulation#Sink_SNR_and_the_performance_parameter|"Sink SNR and the performance parameter"]].

===Questions===

<quiz display=simple>
{Which kind of demodulation is considered here?
|type="()"}
+ Synchronous demodulation.
- Envelope demodulation.

{Which relationship holds between the quantities  $ρ_v$  and  $ξ$  for  double-sideband AM without carrier  $(m → ∞)$?
|type="[]"}
-  $ρ_v = 2 · ξ$.
+  $ρ_v = ξ$.
-  $ρ_v = ξ/2$.

{Which relationship holds between  $ρ_v$  and  $ξ$  for  single-sideband AM without carrier  as $(μ → ∞)$?
|type="[]"}
-  $ρ_v = 2 · ξ$.
+  $ρ_v = ξ$.
-  $ρ_v = ξ/2$.

{Let  $ξ = 10^4$.  Calculate the sink-to-noise ratio of   double-sideband AM  for modulation depths  $m = 0.5$  and  $m = 1$.
|type="{}"}
$m = 0.5\text{:} \ \ 10 · \lg \ ρ_v \ = \ $ { 30.46 3% } $\ \rm dB$
$m = 1.0\text{:} \ \ 10 · \lg \ ρ_v \ = \ $ { 35.23 3% } $\ \rm dB$

{Further let  $ξ = 10^4$.  Calculate the sink-to-noise ratio of   single-sideband AM  for the parameters  $μ = 0.354$  and  $μ = 0.707$.
|type="{}"}
$μ = 0.354\text{:} \ \ \ 10 · \lg \ ρ_v \ = \ $ { 30.46 3% } $\ \rm dB$
$μ = 0.707\text{:} \ \ \ 10 · \lg \ ρ_v \ = \ $ { 35.23 3% } $\ \rm dB$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We are dealing with a   synchronous demodulator.  Answer 1  is correct.

'''(2)'''  Answer 2  is correct:
*For DSB–AM without a carrier,   $P_{\rm S} = P_q/2$.  This is simultaneously the power of the useful component of the sink signal  $v(t)$.
*The power-spectral density   ${\it Φ}_ε(f)$  of the   $v(t)$  noise component results from the convolution:
[[File:P_ID1048__Mod_Z_2_9_b.png|right|frame|Noise power density in double-sideband AM]]
[[File:P_ID1049__Mod_Z_2_9_c.png|right|frame|Noise power density in  upper-sideband AM]]

:$${\it \Phi}_\varepsilon(f) = {\it \Phi}_{z{\rm E} }(f) \star {\it \Phi}_n (f) = \frac{1}{\alpha_{\rm K}^2} \cdot \big[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \big]\star {\it \Phi}_n (f) \hspace{0.05cm}.$$
*The expression  $\big[$ ... $\big]$  describes the power-spectral density of a cosine signal with the signal amplitude   $K = 2$.
*The correction of channel attenuation is considered with  $1/α_K^2$ .
*Thus,  taking   ${\it \Phi}_n(f) = N_0/2$  into account,  we get:
:$${\it \Phi}_\varepsilon(f) = \frac{N_0}{\alpha_{\rm K}^2}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} P_\varepsilon = \int_{-B_{\rm NF}}^{+B_{\rm NF}} {{\it \Phi}_\varepsilon(f) }\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0 \cdot B_{\rm NF}}{\alpha_{\rm K}^2}\hspace{0.05cm}.$$
*From this,  it follows for the the signal-to-noise power ratio  $\rm (SNR)$:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}\hspace{0.15cm}\underline { = \xi} \hspace{0.05cm}.$$

'''(3)'''  Answer 2  is correct:
*In contrast to DSB,  $P_S = P_q/4$  holds for SSB,  as well as
:$${\it \Phi}_\varepsilon(f) = {\it \Phi}_{z{\rm E} }(f) \star {\it \Phi}_n (f) = \frac{4}{\alpha_{\rm K}^2} \cdot \big[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \big]\star {\it \Phi}_n (f) \hspace{0.05cm}.$$
*Taking   $B_{\rm HF} = B_{\rm NF}$  into account  (see adjacent diagram for USB modulation),  we now get:
:$${\it \Phi}_\varepsilon(f) = \frac{2 \cdot N_0}{\alpha_{\rm K}^2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} P_\varepsilon = \frac{4 \cdot N_0 \cdot B_{\rm NF}}{\alpha_{\rm K}^2}\hspace{0.05cm}.$$

*This means:  When wittout carrier,  single-sideband modulation demonstrates the same noise behaviour as DSB-AM.

'''(4)'''  Assuming a cosine carrier with amplitude  $A_{\rm T}$  and a similarly cosine source signal  $q(t)$,  we get for DSB with carrier:
:$$ s(t) = \big (q(t) + A_{\rm T}\big ) \cdot \cos( \omega_{\rm T} \cdot t)
= A_{\rm T} \cdot \cos( \omega_{\rm T} \cdot t) + \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}+ \omega_{\rm N}) \cdot t \big)+ \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}- \omega_{\rm N}) \cdot t\big)\hspace{0.05cm}.$$
*The transmission power is thus given by
:$$ P_{\rm S}= \frac{A_{\rm T}^2}{2} + 2 \cdot \frac{(A_{\rm N}/2)^2}{2} = \frac{A_{\rm T}^2}{2} + \frac{A_{\rm N}^2}{4} \hspace{0.05cm}.$$
*Taking  $P_q = A_{\rm N}^2/2$  and  $m = A_{\rm N}/A_{\rm T}$  into account,  this can also be written as:
:$$P_{\rm S}= \frac{A_{\rm N}^2}{4} \cdot \left[ 1 + \frac{2 \cdot A_{\rm T}^2}{A_{\rm N}^2}\right] = \frac{P_q}{2} \cdot \left[ 1 + {2 }/{m^2}\right]\hspace{0.05cm}.$$
*With the noise power  $P_ε$  according to subtask  '''(2)'''  we thus obtain:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}\cdot (1 + 2/m^2)}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}} \cdot \frac{1}{1 +{2 }/{m^2}} \hspace{0.05cm}.$$
*And in logarithmic representation:
:$$ 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi - 10 \cdot {\rm lg} \hspace{0.15cm}\left[{1 +{2 }/{m^2}}\right] \hspace{0.05cm}.$$
:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \ (m = 0.5) = 40 \,{\rm dB} - 10 \cdot {\rm lg} (9) \hspace{0.15cm}\underline {= 30.46\, {\rm dB}}$$
:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \ (m = 1.0) = 40 \,{\rm dB} - 10 \cdot {\rm lg} (3) \hspace{0.15cm}\underline {= 35.23\, {\rm dB} \hspace{0.05cm}}.$$

'''(5)'''  In  "SSB–AM"  there is only one sideband.
*Therefore,  considering the sideband-to-carrier ratio   $μ = A_{\rm N}/(2A_{\rm T})$  gives:
:$$ P_{\rm S}= \frac{A_{\rm T}^2}{2} + \frac{(A_{\rm N}/2)^2}{2} = {A_{\rm N}^2}/{8} \cdot \big[ 1 + {4 \cdot A_{\rm T}^2}/{A_{\rm N}^2}\big] = {P_q}/{4} \cdot \big[ 1 + {1 }/{\mu^2}\big] \hspace{0.05cm}.$$
*Thus,  with the noise power from subtask  '''(3)''' we obtain:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{4 \cdot P_{\rm S}\cdot (1 + 1/\mu^2)}{4 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}} \cdot \frac{1}{1 +{1 }/{\mu^2}}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi - 10 \cdot {\rm lg} \hspace{0.15cm}\big[{1 +{1 }/{\mu^2}}\big] \hspace{0.05cm}.$$
*So we get the same result with SSB-AM as in DSB-AM with a modulation depth of  $m = \sqrt{2} · μ$.  From this,  it further follows:
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {0.5}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=0.5) \hspace{0.15cm}\underline {=30.46\,{\rm dB}},$$
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {1.0}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=1.0) \hspace{0.15cm}\underline {=35.23\,{\rm dB}}\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^2.4 Single Sideband Amplitude Modulation^]]

Modulation Methods/Single-Sideband Modulation

2022-04-09T14:07:24Z

Reed:

{{Header
|Untermenü=Amplitude Modulation and Demodulation
|Vorherige Seite=Envelope Demodulation
|Nächste Seite=Further AM Variants
}}
==Description in the frequency domain==
 
Double sideband amplitude modulation  $\rm (DSB–AM)$  – both with and without a carrier –  has the following characteristics:
*The modulated signal  $s(t)$  requires twice the bandwidth of the source signal  $q(t)$.
*The complete information about  $q(t)$  is in both the upper sideband   $\rm (USB)$  as well as in the lower sideband  $\rm (LSB)$.

The so-called  '''single-sideband amplitude modulation'''  $\rm (SSB–AM)$  takes advantage of this property by transmitting only one of these sidebands,  either the upper sideband (USB) or the lower sideband (LSB).  This reduces the required bandwidth by half compared to DSB-AM.

The graph illustrates single-sideband amplitude modulation in the frequency domain and simultaneously presents a realization of a SSB modulator. 

In this representation,  one can see:

[[File:EN_Mod_T_2_4_S1.png|right|frame| Spectra for double sideband  $\rm (DSB)$  and single-sideband  $\rm (USB, LSB)$  modulation]]

*The SSB spectrum results from the DSB spectrum by filtering with a band-pass which displays asymmetrical behaviour with respect to the carrier frequency  $f_{\rm T}$.
*For upper sideband modulation  $\rm (USB)$,  the lower cutoff frequency is chosen at  $f_{\rm U} = f_{\rm T} - f_ε$  and  $f_{\rm O} ≥ f_{\rm T} + B_{\rm NF}$  $($subscripts from the German:  "U"  ⇒  "untere"  ⇒  "lower",  and  "O"  ⇒  "obere" ⇒  "upper"$)$.  Here,  $f_ε$  denotes an arbitrarily small positive frequency.
*Thus,  the USB spectrum only contains the upper sideband and the carrier  (though not necessarily the latter).  To generate an USB modulation,  the upper and lower cutoff frequencies of the band-pass must be set as follows:
:$$f_{\rm U} ≤ f_{\rm T} \ - \ B_{\rm NF}, \hspace{0.5cm}f_{\rm O} = f_{\rm T} + f_ε.$$
 
{{GraueBox|TEXT=
$\text{Example 1:}$  For  $f_{\rm T} = 100 \ \rm kHz$,  $B_{\rm NF} = 3 \ \rm kHz$ , DSB gives us an  "USB-AM with carrier"  if the filter cuts out all frequencies below  $99.999\text{...} \ \rm kHz$ .
*If the lower cut-off frequency  $f_{\rm U}$  is larger than  $f_{\rm T}$  by an  (arbitrarily)  small  "epsilon",  we get an  "USB-AM without carrier".
*A  "LSB with/without carrier"  can be realised accordingly with  $f_{\rm O} =100 \ \rm kHz$  and  $f_{\rm O} = 99.999\text{...} \ \rm kHz$,  resp. }}

{{GraueBox|TEXT=
$\text{Example 2:}$  To save bandwidth,  single-sideband technology was already used in the 1960s for the analog transmission of telephone calls.
*In accordance with a hierarchical structure,  three telephone channels  – each band-limited to the range from  $300\ \rm Hz$  to  $3.4 \ \rm kHz$  –  were initially combined to form a preliminary grouping with a bandwidth of  $12 \ \rm kHz$ .
*In order to be able to accommodate three channels in  $12 \ \rm kHz$  with a safety margin,  only one sideband  (LSB or USB)  of each telephone channel was considered.
*Using further combinations,  the long distance traffic system  '''V10800'''  with up to  $10800$  voice channels and a total bandwidth of  $60 \ \rm MHz$  was thus realized.}}

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
*The great advantage of a  $\rm SSB-AM$  is that it has '''half the bandwidth''' compared to  $\rm DSB-AM$.
*The trade-off with disadvantages required for this advantage is explained in the next sections.}}

==Synchronous demodulation of a SSB-AM signal==
 
Let us now consider a SSB-AM modulated signal and a  [[Modulation_Methods/Synchronous_Demodulation|synchronous demodulator]]  at the receiver.
*Perfect frequency and phase synchronization will be assumed.
*Without affecting generality,  in the rest of this section we will always let  ${ ϕ}_{\rm T} = 0$  ⇒   cosine carrier.

[[File:EN_Mod_T_2_4_S2_neu.png |right|frame| Synchronous demodulation of a SSB-AM signal]]
 
A comparison with the  [[Modulation_Methods/Synchronous_Demodulation|characteristics of the synchronous demodulator for DSB-AM]]  shows the follows similarities and differences:

*The spectrum  $V(f)$  of the sink signal results in both cases from the convolution of the spectra  $R(f)$  and  $Z_{\rm E}(f)$,  the latter being composed of two Dirac delta functions at $±f_{\rm T}$.

*For SSB–AM the convolution products overlap with the left and the right Dirac delta function at every frequency.  These are denoted by  "+"  and  "–"  respectively in the [[Modulation_Methods/Synchronous_Demodulation#Description_in_the_frequency_domain|corresponding DSB graph]].

*In contrast,  for USB,  only the convolution with the Dirac line at  $ -f_{\rm T}$  yields the  $V(f)$  component for positive frequencies,  and for LSB modulation it is the convolution with the Dirac function  $δ(f - f_{\rm T})$.

*In the case of DSB–AM,  $v(t) = q(t)$  is reached with the receiver-side carrier signal  $z_{\rm E}(t) = 2 · \cos(ω_{\rm T} · t)$.  In contrast,  the carrier amplitude must be increased to  $A_{\rm T} = 4$   for SSB-AM.
 
{{BlaueBox|TEXT=
$\text{Conclusion:}$  If there is a frequency offset between the carrier signals  $z(t)$  and  $z_{\rm E}(t)$,  strong nonlinear distortion always occurs, i.e.,  regardless of whether DSB-AM or SSB-AM is present.  For the   '''implementation of a synchronous demodulator'''  a   '''perfect frequency synchronization'''  is essential. }}

==Influence of a phase offset for SSB-AM==
 
Let us now consider the influence of a phase offset  $Δ{\mathbf ϕ}_{\rm T}$  between the transmitter and receiver side carrier signals,  using an example source signal
:$$q(t) = A_1 \cdot \cos(\omega_1 \cdot t ) + A_2 \cdot \cos(\omega_2 \cdot t)\hspace{0.05cm}.$$
*For DSB–AM,  such a phase offset only leads to frequency-independent attenuation,  but not to distortions:
:$$v(t) = \cos (\Delta \phi_{\rm T}) \cdot q(t) = \cos (\Delta \phi_{\rm T}) \cdot A_1 \cdot \cos(\omega_1 \cdot t ) + \cos (\Delta \phi_{\rm T}) \cdot A_2 \cdot \cos(\omega_2 \cdot t ) \hspace{0.05cm}.$$
*In contrast,  for USB-AM we get:
:$$v(t) = A_1 \cdot \cos(\omega_1 \cdot t - \Delta \phi_{\rm T}) + A_2
\cdot \cos(\omega_2 \cdot t - \Delta \phi_{\rm T})= A_1 \cdot \cos(\omega_1 \cdot (t - \tau_1)) + A_2 \cdot
\cos(\omega_2 \cdot (t - \tau_2))\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{From this equation we can see:}$
*The two delay times  $τ_1 = Δ{\mathbf ϕ}_{\rm T}/ω_1$  and  $τ_2 = Δ{\mathbf ϕ}_{\rm T}/ω_2$  are different.
*This means that a phase offset in single sideband amplitude modulation  (USB–AM or LSB–AM)  leads to  '''phase distortions'''  (i.e., to linear distortions).
*A positive value of  $Δ{\mathbf ϕ}_{\rm T}$  will cause
**positive values of  $τ_1$  and  $τ_2$  (i.e., lagging signals with respect to the cosine)  for USB,  and
**negative  $τ_1$  and  $τ_2$ values  (leading signals)  for LSB.}}

The effects of phase distortions on a signal composed two cosine oscillations is illustrated with the HTML 5/JavaScript applet  [[Applets:Linear_Distortions_of_Periodic_Signals|"Linear Distortions of Periodic Signals"]].

==Sideband-to-carrier ratio==
 
An important parameter for DSB–AM is the  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier|modulation depth]]  $m = q_{\rm max}/A_{\rm T}$.  In the special case of a harmonic oscillation,  $m = A_{\rm N}/A_{\rm T}$  holds and one obtains the spectrum  $S_+(f)$  of the analytical signal corresponding to the upper graph.  Please note the normalization with respect to  $A_{\rm T}$.

[[File:EN_Mod_T_2_4_S4a.png|right|frame|Spectra for DSB-AM,  USB-AM]]

For SSB–AM,  the application of the parameter  $m$  is possible in principle,  but it is not practical.  

For example,  the following holds for the time-domain representation of USB-AM with the spectrum  $S_+(f)$  corresponding to the lower graph:
:$$s_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.05cm} \omega_{\rm T} \hspace{0.03cm}\cdot \hspace{0.05cm}t } + {A_{\rm N}}/{2} \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm} (\omega_{\rm T} + \omega_{\rm N}) \hspace{0.05cm}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$
In a similar way,  this can be written as
:$$s_+(t) = A_{\rm T} \cdot \left({\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.05cm} \omega_{\rm T} \hspace{0.03cm}\cdot \hspace{0.05cm}t } + \mu \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm} (\omega_{\rm T} + \omega_{\rm N}) \hspace{0.03cm}\cdot \hspace{0.05cm}t }\right)$$
now using the   '''sideband–to–carrier ratio''' :
:$$\mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}} \hspace{0.05cm}.$$
If the source signal is not a harmonic oscillation,  it is difficult to specify this quantity.
*Here one can use the following approximation:
:$$\mu \approx \frac{q_{\rm max}}{2 \cdot A_{\rm T}} \hspace{0.3cm}{\rm with}\hspace{0.3cm} q_{\rm max} = \max_{t} \hspace{0.05cm} |q(t)| \hspace{0.05cm}.$$
*Thus,  $\mu \approx m/2.$  However,  a comparison of DSB-AM to SSB-AM should always be made for the same numerical value of  $m$ and  $,\mu$  resp.

{{GraueBox|TEXT=
[[File:EN_Mod_T_2_4_S4b.png |right|frame| Signal waveforms for DSB-AM and USB-AM]]
$\text{Example 3:}$ 
*The upper graph shows the DSB-AM signal for a modulation depth of  $m = q_{\rm max}/A_{\rm T} = 1$   ⇒   limit case for the application of envelope demodulation with DSB-AM   ⇒   The signal  $q(t)$  is then just contained in the envelope $a(t)$.

*For the USB-AM signal in the middle graph,  $A_{\rm T} = q_{\rm max}$  was also chosen  which corresponds to the numerical values  $m = 1$  and  $\mu = 0.5$  as defined above   ⇒   Since the LSB magnitude is missing, the envelope  $a(t)$  is significantly different to  $q(t) + A_{\rm T}.$

*In contrast,  $q_{\rm max} = 2 · A_{\rm T}$  was chosen for the lower signal waveform  ⇒   Here, the sideband-to-carrier ratio has the numerical value  $\mu =1$.

This graph makes the following clear:
#There are more similarities between the upper and lower signal than between the first two,  because
#The comparison of a DSB-AM and a SSB-AM should be made for the same numerical value of  $m$  and  $\mu$  if possible.
#For each value  $\mu$  of the sideband-to-carrier ratio,  the envelope demodulation with SSB-AM leads to severe distortions.
#These are of a nonlinear nature and therefore irreversible. }}

==Summarized evaluation of SSB–AM==
 
The key advantage of  $\text{SSB–AM}$  compared to  $\text{DSB–AM}$  is the bandwidth requirement,  smaller by a factor of  $2$.  However,  for the half bandwidth in SSB-AM,  some disadvantages have to be accepted,  which will be investigated in the exercises for this section:
*The information about the source signal  $q(t)$  is,  in contrast to DSB-AM,  no longer exclusively in the amplitude,  but equally in the phase  (see [[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_SSB_Signal| Exercise 2.11]]).
*Synchronous demodulation of a SSB–AM signal leads to phase distortions if there is a phase offset between the carrier signals at the transmitter and the receiver.
*The application of  $\text{envelope demodulation}$  with  "USB–AM"  or  "LSB–AM"  thus always leads to strong nonlinear distortions  (see [[Aufgaben:Exercise_2.11Z:_Once_again_SSB-AM_and_Envelope_Demodulator|Exercise 2.11Z]]).

As with DSB–AM and synchronous demodulation,  the following statements also apply here:
*Attenuation distortions of the channel also lead only to (linear)  attenuation distortions with respect to the sink signal.  No nonlinear distortions arise  (see  [[Aufgaben:Exercise_2.10:_SSB-AM_with_Channel_Distortions|Exercise A2.10]]).
*SSB–AM without a carrier shows the exact same noise behaviour as DSB-AM without a carrier.  The advantage of the smaller RF bandwidth is cancelled out by the necessary level matching.
*A SSB–AM with a sideband-to-carrier ratio  $\mu$  shows similar noise behaviour as in DSB-AM with modulation depth  $m = \sqrt{2} · \mu$  (see  [[Aufgaben:Exercise_2.10Z:_Noise_with_DSB-AM_and_SSB-AM|Exercise 2.10Z]]).
*In any case,  it should be noted that SSB-AM with carrier is not very useful due to the nonlinear distortions brought about with envelope demodulation.

==Exercises for the chapter==
 
[[Aufgaben:Exercise_2.10:_SSB-AM_with_Channel_Distortions|Exercise 2.10: SSB-AM with Channel Distortions]]

[[Aufgaben:Exercise_2.10Z:_Noise_with_DSB-AM_and_SSB-AM|Exercise 2.10Z: Noise with DSB-AM and SSB-AM]]

[[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_SSB_Signal|Exercise 2.11: Envelope Demodulation of an SSB signal]]

[[Aufgaben:Exercise_2.11Z:_Once_again_SSB-AM_and_Envelope_Demodulator|Exercise 2.11Z: Once again SSB-AM and Envelope Demodulator]]

{{Display}}

Modulation Methods/Single-Sideband Modulation

2022-04-09T14:05:37Z

Reed:

{{Header
|Untermenü=Amplitude Modulation and Demodulation
|Vorherige Seite=Envelope Demodulation
|Nächste Seite=Further AM Variants
}}
==Description in the frequency domain==
 
Double sideband amplitude modulation  $\rm (DSB–AM)$  – both with and without a carrier –  has the following characteristics:
*The modulated signal  $s(t)$  requires twice the bandwidth of the source signal  $q(t)$.
*The complete information about  $q(t)$  is in both the upper sideband   $\rm (USB)$  as well as in the lower sideband  $\rm (LSB)$.

The so-called  '''single-sideband amplitude modulation'''  $\rm (SSB–AM)$  takes advantage of this property by transmitting only one of these sidebands,  either the upper sideband (USB) or the lower sideband (LSB).  This reduces the required bandwidth by half compared to DSB-AM.

The graph illustrates single-sideband amplitude modulation in the frequency domain and simultaneously presents a realization of a SSB modulator. 

In this representation,  one can see:

[[File:EN_Mod_T_2_4_S1.png|right|frame| Spectra for double sideband  $\rm (DSB)$  and single-sideband  $\rm (USB, LSB)$  modulation]]

*The SSB spectrum results from the DSB spectrum by filtering with a band-pass which displays asymmetrical behaviour with respect to the carrier frequency  $f_{\rm T}$.
*For upper sideband modulation  $\rm (USB)$,  the lower cutoff frequency is chosen at  $f_{\rm U} = f_{\rm T} - f_ε$  and  $f_{\rm O} ≥ f_{\rm T} + B_{\rm NF}$  $($subscripts from the German:  "U"  ⇒  "untere"  ⇒  "lower",  and  "O"  ⇒  "obere" ⇒  "upper"$)$.  Here,  $f_ε$  denotes an arbitrarily small positive frequency.
*Thus,  the USB spectrum only contains the upper sideband and the carrier  (though not necessarily the latter).  To generate an USB modulation,  the upper and lower cutoff frequencies of the band-pass must be set as follows:
:$$f_{\rm U} ≤ f_{\rm T} \ - \ B_{\rm NF}, \hspace{0.5cm}f_{\rm O} = f_{\rm T} + f_ε.$$
 
{{GraueBox|TEXT=
$\text{Example 1:}$  For  $f_{\rm T} = 100 \ \rm kHz$,  $B_{\rm NF} = 3 \ \rm kHz$ , DSB gives us an  "USB-AM with carrier"  if the filter cuts out all frequencies below  $99.999\text{...} \ \rm kHz$ .
*If the lower cut-off frequency  $f_{\rm U}$  is larger than  $f_{\rm T}$  by an  (arbitrarily)  small  "epsilon",  we get an  "USB-AM without carrier".
*A  "LSB with/without carrier"  can be realised accordingly with  $f_{\rm O} =100 \ \rm kHz$  and  $f_{\rm O} = 99.999\text{...} \ \rm kHz$,  resp. }}

{{GraueBox|TEXT=
$\text{Example 2:}$  To save bandwidth,  single-sideband technology was already used in the 1960s for the analog transmission of telephone calls.
*In accordance with a hierarchical structure,  three telephone channels  – each band-limited to the range from  $300\ \rm Hz$  to  $3.4 \ \rm kHz$  –  were initially combined to form a preliminary grouping with a bandwidth of  $12 \ \rm kHz$ .
*In order to be able to accommodate three channels in  $12 \ \rm kHz$  with a safety margin,  only one sideband  (LSB or USB)  of each telephone channel was considered.
*Using further combinations,  the long distance traffic system  '''V10800'''  with up to  $10800$  voice channels and a total bandwidth of  $60 \ \rm MHz$  was thus realized.}}

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
*The great advantage of a  $\rm SSB-AM$  is that it has '''half the bandwidth''' compared to  $\rm DSB-AM$.
*The trade-off with disadvantages required for this advantage is explained in the next sections.}}

==Synchronous demodulation of a SSB-AM signal==
 
Let us now consider a SSB-AM modulated signal and a  [[Modulation_Methods/Synchronous_Demodulation|synchronous demodulator]]  at the receiver.
*Perfect frequency and phase synchronization will be assumed.
*Without affecting generality,  in the rest of this section we will always let  ${ ϕ}_{\rm T} = 0$  ⇒   cosine carrier.

[[File:EN_Mod_T_2_4_S2_neu.png |right|frame| Synchronous demodulation of a SSB-AM signal]]
 
A comparison with the  [[Modulation_Methods/Synchronous_Demodulation|characteristics of the synchronous demodulator for DSB-AM]]  shows the follows similarities and differences:

*The spectrum  $V(f)$  of the sink signal results in both cases from the convolution of the spectra  $R(f)$  and  $Z_{\rm E}(f)$,  the latter being composed of two Dirac delta functions at $±f_{\rm T}$.

*For SSB–AM the convolution products overlap with the left and the right Dirac delta function at every frequency.  These are denoted by  "+"  and  "–"  respectively in the [[Modulation_Methods/Synchronous_Demodulation#Description_in_the_frequency_domain|corresponding DSB graph]].

*In contrast,  for USB,  only the convolution with the Dirac line at  $ -f_{\rm T}$  yields the  $V(f)$  component for positive frequencies,  and for LSB modulation it is the convolution with the Dirac function  $δ(f - f_{\rm T})$.

*In the case of DSB–AM,  $v(t) = q(t)$  is reached with the receiver-side carrier signal  $z_{\rm E}(t) = 2 · \cos(ω_{\rm T} · t)$.  In contrast,  the carrier amplitude must be increased to  $A_{\rm T} = 4$   for SSB-AM.
 
{{BlaueBox|TEXT=
$\text{Conclusion:}$  If there is a frequency offset between the carrier signals  $z(t)$  and  $z_{\rm E}(t)$,  strong nonlinear distortion always occurs, i.e.,  regardless of whether DSB-AM or SSB-AM is present.  For the   '''implementation of a synchronous demodulator'''  a   '''perfect frequency synchronization'''  is essential. }}

==Influence of a phase offset for SSB-AM==
 
Let us now consider the influence of a phase offset  $Δ{\mathbf ϕ}_{\rm T}$  between the transmitter and receiver side carrier signals,  using an example source signal
:$$q(t) = A_1 \cdot \cos(\omega_1 \cdot t ) + A_2 \cdot \cos(\omega_2 \cdot t)\hspace{0.05cm}.$$
*For DSB–AM,  such a phase offset only leads to frequency-independent attenuation,  but not to distortions:
:$$v(t) = \cos (\Delta \phi_{\rm T}) \cdot q(t) = \cos (\Delta \phi_{\rm T}) \cdot A_1 \cdot \cos(\omega_1 \cdot t ) + \cos (\Delta \phi_{\rm T}) \cdot A_2 \cdot \cos(\omega_2 \cdot t ) \hspace{0.05cm}.$$
*In contrast,  for USB-AM we get:
:$$v(t) = A_1 \cdot \cos(\omega_1 \cdot t - \Delta \phi_{\rm T}) + A_2
\cdot \cos(\omega_2 \cdot t - \Delta \phi_{\rm T})= A_1 \cdot \cos(\omega_1 \cdot (t - \tau_1)) + A_2 \cdot
\cos(\omega_2 \cdot (t - \tau_2))\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{From this equation we can see:}$
*The two delay times  $τ_1 = Δ{\mathbf ϕ}_{\rm T}/ω_1$  and  $τ_2 = Δ{\mathbf ϕ}_{\rm T}/ω_2$  are different.
*This means that a phase offset in single sideband amplitude modulation  (USB–AM or LSB–AM)  leads to  '''phase distortions'''  (i.e., to linear distortions).
*A positive value of  $Δ{\mathbf ϕ}_{\rm T}$  will cause
**positive values of  $τ_1$  and  $τ_2$  (i.e., lagging signals with respect to the cosine)  for USB,  and
**negative  $τ_1$  and  $τ_2$ values  (leading signals)  for LSB.}}

The effects of phase distortions on a signal composed two cosine oscillations is illustrated with the HTML 5/JavaScript applet  [[Applets:Linear_Distortions_of_Periodic_Signals|"Linear Distortions of Periodic Signals"]].

==Sideband-to-carrier ratio==
 
An important parameter for DSB–AM is the  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier|modulation depth]]  $m = q_{\rm max}/A_{\rm T}$.  In the special case of a harmonic oscillation,  $m = A_{\rm N}/A_{\rm T}$  holds and one obtains the spectrum  $S_+(f)$  of the analytical signal corresponding to the upper graph.  Please note the normalization with respect to  $A_{\rm T}$.

[[File:EN_Mod_T_2_4_S4a.png|right|frame|Spectra for DSB-AM,  USB-AM]]

For SSB–AM,  the application of the parameter  $m$  is possible in principle,  but it is not practical.  

For example,  the following holds for the time-domain representation of USB-AM with the spectrum  $S_+(f)$  corresponding to the lower graph:
:$$s_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.05cm} \omega_{\rm T} \hspace{0.03cm}\cdot \hspace{0.05cm}t } + {A_{\rm N}}/{2} \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm} (\omega_{\rm T} + \omega_{\rm N}) \hspace{0.05cm}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$
In a similar way,  this can be written as
:$$s_+(t) = A_{\rm T} \cdot \left({\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.05cm} \omega_{\rm T} \hspace{0.03cm}\cdot \hspace{0.05cm}t } + \mu \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm} (\omega_{\rm T} + \omega_{\rm N}) \hspace{0.03cm}\cdot \hspace{0.05cm}t }\right)$$
now using the   '''sideband–to–carrier ratio''' :
:$$\mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}} \hspace{0.05cm}.$$
If the source signal is not a harmonic oscillation,  it is difficult to specify this quantity.
*Here one can use the following approximation:
:$$\mu \approx \frac{q_{\rm max}}{2 \cdot A_{\rm T}} \hspace{0.3cm}{\rm with}\hspace{0.3cm} q_{\rm max} = \max_{t} \hspace{0.05cm} |q(t)| \hspace{0.05cm}.$$
*Thus,  $\mu \approx m/2.$  However,  a comparison of DSB-AM to SSB-AM should always be made for the same numerical value of  $m$ and  $,\mu$  resp.

{{GraueBox|TEXT=
[[File:EN_Mod_T_2_4_S4b.png |right|frame| Signal waveforms for DSB-AM and USB-AM]]
$\text{Example 3:}$ 
*The upper graph shows the DSB-AM signal for a modulation depth of  $m = q_{\rm max}/A_{\rm T} = 1$   ⇒   limit case for the application of envelope demodulation with DSB-AM   ⇒   The signal  $q(t)$  is then just contained in the envelope $a(t)$.

*For the USB-AM signal in the middle graph,  $A_{\rm T} = q_{\rm max}$  was also chosen  which corresponds to the numerical values  $m = 1$  and  $\mu = 0.5$  as defined above   ⇒   Since the LSB magnitude is missing, the envelope  $a(t)$  is significantly different to  $q(t) + A_{\rm T}.$

*In contrast,  $q_{\rm max} = 2 · A_{\rm T}$  was chosen for the lower signal waveform  ⇒   Here, the sideband-to-carrier ratio has the numerical value  $\mu =1$.

This graph makes the following clear:
#There are more similarities between the upper and lower signal than between the first two,  because
#The comparison of a DSB-AM and a SSB-AM should be made for the same numerical value of  $m$  and  $\mu$  if possible.
#For each value  $\mu$  of the sideband-to-carrier ratio,  the envelope demodulation with SSB-AM leads to severe distortions.
#These are of a nonlinear nature and therefore irreversible. }}

==Summarized evaluation of SSB–AM==
 
The key advantage of  $\text{SSB–AM}$  compared to  $\text{DSB–AM}$  is the bandwidth requirement,  smaller by a factor of  $2$.  However,  for the half bandwidth in SSB-AM,  some disadvantages have to be accepted,  which will be investigated in the exercises for this section:
*The information about the source signal  $q(t)$  is,  in contrast to DSB-AM,  no longer exclusively in the amplitude,  but equally in the phase  (see [[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_SSB_Signal| Exercise 2.11]]).
*Synchronous demodulation of a SSB–AM signal leds to phase distoritions if there is a phase offset between the carrier signals at the transmitter and the receiver.
*The application of  $\text{envelope demodulation}$  with  "USB–AM"  or  "LSB–AM"  thus always leads to strong nonlinear distortions  (see [[Aufgaben:Exercise_2.11Z:_Once_again_SSB-AM_and_Envelope_Demodulator|Exercise 2.11Z]]).

As with DSB–AM and synchronous demodulation,  the following statements also apply here:
*Attenuation distortions of the channel also lead only to (linear)  attenuation distortions with respect to the sink signal.  No nonlinear distortions arise  (see  [[Aufgaben:Exercise_2.10:_SSB-AM_with_Channel_Distortions|Exercise A2.10]]).
*SSB–AM without a carrier shows the exact same noise behaviour as DSB-AM without a carrier.  The advantage of the smaller RF bandwidth is cancelled out by the necessary level matching.
*A SSB–AM with a sideband-to-carrier ratio  $\mu$  shows similar noise behaviour as in DSB-AM with modulation depth  $m = \sqrt{2} · \mu$  (see  [[Aufgaben:Exercise_2.10Z:_Noise_with_DSB-AM_and_SSB-AM|Exercise 2.10Z]]).
*In any case,  it should be noted that SSB-AM with carrier is not very useful due to the nonlinear distortions brought about with envelope demodulation.

==Exercises for the chapter==
 
[[Aufgaben:Exercise_2.10:_SSB-AM_with_Channel_Distortions|Exercise 2.10: SSB-AM with Channel Distortions]]

[[Aufgaben:Exercise_2.10Z:_Noise_with_DSB-AM_and_SSB-AM|Exercise 2.10Z: Noise with DSB-AM and SSB-AM]]

[[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_SSB_Signal|Exercise 2.11: Envelope Demodulation of an SSB signal]]

[[Aufgaben:Exercise_2.11Z:_Once_again_SSB-AM_and_Envelope_Demodulator|Exercise 2.11Z: Once again SSB-AM and Envelope Demodulator]]

{{Display}}

Modulation Methods/Single-Sideband Modulation

2022-04-09T14:03:54Z

Reed:

{{Header
|Untermenü=Amplitude Modulation and Demodulation
|Vorherige Seite=Envelope Demodulation
|Nächste Seite=Further AM Variants
}}
==Description in the frequency domain==
 
Double sideband amplitude modulation  $\rm (DSB–AM)$  – both with and without a carrier –  has the following characteristics:
*The modulated signal  $s(t)$  requires twice the bandwidth of the source signal  $q(t)$.
*The complete information about  $q(t)$  is in both the upper sideband   $\rm (USB)$  as well as in the lower sideband  $\rm (LSB)$.

The so-called  '''single-sideband amplitude modulation'''  $\rm (SSB–AM)$  takes advantage of this property by transmitting only one of these sidebands,  either the upper sideband (USB) or the lower sideband (LSB).  This reduces the required bandwidth by half compared to DSB-AM.

The graph illustrates single-sideband amplitude modulation in the frequency domain and simultaneously presents a realization of a SSB modulator. 

In this representation,  one can see:

[[File:EN_Mod_T_2_4_S1.png|right|frame| Spectra for double sideband  $\rm (DSB)$  and single-sideband  $\rm (USB, LSB)$  modulation]]

*The SSB spectrum results from the DSB spectrum by filtering with a band-pass which displays asymmetrical behaviour with respect to the carrier frequency  $f_{\rm T}$.
*For upper sideband modulation  $\rm (USB)$,  the lower cutoff frequency is chosen at  $f_{\rm U} = f_{\rm T} - f_ε$  and  $f_{\rm O} ≥ f_{\rm T} + B_{\rm NF}$  $($subscripts from the German:  "U"  ⇒  "untere"  ⇒  "lower",  and  "O"  ⇒  "obere" ⇒  "upper"$)$.  Here,  $f_ε$  denotes an arbitrarily small positive frequency.
*Thus,  the USB spectrum only contains the upper sideband and the carrier  (though not necessarily the latter).  To generate an USB modulation,  the upper and lower cutoff frequencies of the band-pass must be set as follows:
:$$f_{\rm U} ≤ f_{\rm T} \ - \ B_{\rm NF}, \hspace{0.5cm}f_{\rm O} = f_{\rm T} + f_ε.$$
 
{{GraueBox|TEXT=
$\text{Example 1:}$  For  $f_{\rm T} = 100 \ \rm kHz$,  $B_{\rm NF} = 3 \ \rm kHz$ , DSB gives us an  "USB-AM with carrier"  if the filter cuts out all frequencies below  $99.999\text{...} \ \rm kHz$ .
*If the lower cut-off frequency  $f_{\rm U}$  is larger than  $f_{\rm T}$  by an  (arbitrarily)  small  "epsilon",  we get an  "USB-AM without carrier".
*A  "LSB with/without carrier"  can be realised accordingly with  $f_{\rm O} =100 \ \rm kHz$  and  $f_{\rm O} = 99.999\text{...} \ \rm kHz$,  resp. }}

{{GraueBox|TEXT=
$\text{Example 2:}$  To save bandwidth,  single-sideband technology was already used in the 1960s for the analog transmission of telephone calls.
*In accordance with a hierarchical structure,  three telephone channels  – each band-limited to the range from  $300\ \rm Hz$  to  $3.4 \ \rm kHz$  –  were initially combined to form a preliminary grouping with a bandwidth of  $12 \ \rm kHz$ .
*In order to be able to accommodate three channels in  $12 \ \rm kHz$  with a safety margin,  only one sideband  (LSB or USB)  of each telephone channel was considered.
*Using further combinations,  the long distance traffic system  '''V10800'''  with up to  $10800$  voice channels and a total bandwidth of  $60 \ \rm MHz$  was thus realized.}}

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
*The great advantage of a  $\rm SSB-AM$  is that it has '''half the bandwidth''' compared to  $\rm DSB-AM$.
*The trade-off with disadvantages required for this advantage is explained in the next sections.}}

==Synchronous demodulation of a SSB-AM signal==
 
Let us now consider a SSB-AM modulated signal and a  [[Modulation_Methods/Synchronous_Demodulation|synchronous demodulator]]  at the receiver.
*Perfect frequency and phase synchronization will be assumed.
*Without affecting generality,  in the rest of this section we will always let  ${ ϕ}_{\rm T} = 0$  ⇒   cosine carrier.

[[File:EN_Mod_T_2_4_S2_neu.png |right|frame| Synchronous demodulation of a SSB-AM signal]]
 
A comparison with the  [[Modulation_Methods/Synchronous_Demodulation|characteristics of the synchronous demodulator for DSB-AM]]  shows the follows similarities and differences:

*The spectrum  $V(f)$  of the sink signal results in both cases from the convolution of the spectra  $R(f)$  and  $Z_{\rm E}(f)$,  the latter being composed of two Dirac delta functions at $±f_{\rm T}$.

*For SSB–AM the convolution products overlap with the left and the right Dirac delta function at every frequency.  These are denoted by  "+"  and  "–"  respectively in the [[Modulation_Methods/Synchronous_Demodulation#Description_in_the_frequency_domain|corresponding DSB graph]].

*In contrast,  for USB,  only the convolution with the Dirac line at  $ -f_{\rm T}$  yields the  $V(f)$  component for positive frequencies,  and for LSB modulation it is the convolution with the Dirac function  $δ(f - f_{\rm T})$.

*In the case of DSB–AM,  $v(t) = q(t)$  is reached with the receiver-side carrier signal  $z_{\rm E}(t) = 2 · \cos(ω_{\rm T} · t)$.  In contrast,  the carrier amplitude must be increased to  $A_{\rm T} = 4$   for SSB-AM.
 
{{BlaueBox|TEXT=
$\text{Conclusion:}$  If there is a frequency offset between the carrier signals  $z(t)$  and  $z_{\rm E}(t)$,  strong nonlinear distortion always occurs, i.e.,  regardless of whether DSB-AM or SSB-AM is present.  For the   '''implementation of a synchronous demodulator'''  a   '''perfect frequency synchronization'''  is essential. }}

==Influence of a phase offset for SSB-AM==
 
Let us now consider the influence of a phase offset  $Δ{\mathbf ϕ}_{\rm T}$  between the transmitter and receiver side carrier signals,  using an example source signal
:$$q(t) = A_1 \cdot \cos(\omega_1 \cdot t ) + A_2 \cdot \cos(\omega_2 \cdot t)\hspace{0.05cm}.$$
*For DSB–AM,  such a phase offset only leads to frequency-independent attenuation,  but not to distortions:
:$$v(t) = \cos (\Delta \phi_{\rm T}) \cdot q(t) = \cos (\Delta \phi_{\rm T}) \cdot A_1 \cdot \cos(\omega_1 \cdot t ) + \cos (\Delta \phi_{\rm T}) \cdot A_2 \cdot \cos(\omega_2 \cdot t ) \hspace{0.05cm}.$$
*In contrast,  for USB-AM we get:
:$$v(t) = A_1 \cdot \cos(\omega_1 \cdot t - \Delta \phi_{\rm T}) + A_2
\cdot \cos(\omega_2 \cdot t - \Delta \phi_{\rm T})= A_1 \cdot \cos(\omega_1 \cdot (t - \tau_1)) + A_2 \cdot
\cos(\omega_2 \cdot (t - \tau_2))\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{From this equation we can see:}$
*The two delay times  $τ_1 = Δ{\mathbf ϕ}_{\rm T}/ω_1$  and  $τ_2 = Δ{\mathbf ϕ}_{\rm T}/ω_2$  are different.
*This means that a phase offset in single sideband amplitude modulation  (USB–AM or LSB–AM)  leads to  '''phase distortions'''  (i.e., to linear distortions).
*A positive value of  $Δ{\mathbf ϕ}_{\rm T}$  will cause
**positive values of  $τ_1$  and  $τ_2$  (i.e., lagging signals with respect to the cosine)  for USB,  and
**negative  $τ_1$  and  $τ_2$ values  (leading signals)  for LSB.}}

The effects of phase distortions on a signal composed two cosine oscillations is illustrated with the HTML 5/JavaScript applet  [[Applets:Linear_Distortions_of_Periodic_Signals|"Linear Distortions of Periodic Signals"]].

==Sideband-to-carrier ratio==
 
An important parameter for DSB–AM is the  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier|modulation depth]]  $m = q_{\rm max}/A_{\rm T}$.  In the special case of a harmonic oscillation,  $m = A_{\rm N}/A_{\rm T}$  holds and one obtains the spectrum  $S_+(f)$  of the analytical signal corresponding to the upper graph.  Please note the normalization with respect to  $A_{\rm T}$.

[[File:EN_Mod_T_2_4_S4a.png|right|frame|Spectra for DSB-AM,  USB-AM]]

For SSB–AM,  the application of the parameter  $m$  is possible in principle,  but it is not practical.  

For example,  the following holds for the time-domain representation of USB-AM with the spectrum  $S_+(f)$  corresponding to the lower graph:
:$$s_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.05cm} \omega_{\rm T} \hspace{0.03cm}\cdot \hspace{0.05cm}t } + {A_{\rm N}}/{2} \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm} (\omega_{\rm T} + \omega_{\rm N}) \hspace{0.05cm}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$
In a similar way,  this can be written as
:$$s_+(t) = A_{\rm T} \cdot \left({\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.05cm} \omega_{\rm T} \hspace{0.03cm}\cdot \hspace{0.05cm}t } + \mu \cdot {\rm e}^{{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm} (\omega_{\rm T} + \omega_{\rm N}) \hspace{0.03cm}\cdot \hspace{0.05cm}t }\right)$$
now using the   '''sideband–to–carrier ratio''' :
:$$\mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}} \hspace{0.05cm}.$$
If the source signal is not a harmonic oscillation,  it is difficult to specify this quantity.
*Here one can use the following approximation:
:$$\mu \approx \frac{q_{\rm max}}{2 \cdot A_{\rm T}} \hspace{0.3cm}{\rm with}\hspace{0.3cm} q_{\rm max} = \max_{t} \hspace{0.05cm} |q(t)| \hspace{0.05cm}.$$
*Thus,  $\mu \approx m/2.$  However,  a comparison of DSB-AM to SSB-AM should always be made for the same numerical value of  $m$ and  $,\mu$  resp..

{{GraueBox|TEXT=
[[File:EN_Mod_T_2_4_S4b.png |right|frame| Signal waveforms for DSB-AM and USB-AM]]
$\text{Example 3:}$ 
*The upper graph shows the DSB-AM signal for a modulation depth of  $m = q_{\rm max}/A_{\rm T} = 1$   ⇒   limit case for the application of envelope demodulation with DSB-AM   ⇒   The signal  $q(t)$  is then just contained in the envelope $a(t)$.

*For the USB-AM signal in the middle graph,  $A_{\rm T} = q_{\rm max}$  was also chosen  which corresponds to the numerical values  $m = 1$  and  $\mu = 0.5$  as defined above   ⇒   Since the LSB magnitude is missing, the envelope  $a(t)$  is significantly different to  $q(t) + A_{\rm T}.$

*In contrast,  $q_{\rm max} = 2 · A_{\rm T}$  was chosen for the lower signal waveform  ⇒   Here, the sideband-to-carrier ratio has the numerical value  $\mu =1$.

This graph makes the following clear:
#There are more similarities between the upper and lower signal than between the first two,  because
#The comparison of a DSB-AM and a SSB-AM should be made for the same numerical value of  $m$  and  $\mu$  if possible.
#For each value  $\mu$  of the sideband-to-carrier ratio,  the envelope demodulation with SSB-AM leads to severe distortions.
#These are of a nonlinear nature and therefore irreversible. }}

==Summarized evaluation of SSB–AM==
 
The key advantage of  $\text{SSB–AM}$  compared to  $\text{DSB–AM}$  is the bandwidth requirement,  smaller by a factor of  $2$.  However,  for the half bandwidth in SSB-AM,  some disadvantages have to be accepted,  which will be investigated in the exercises for this section:
*The information about the source signal  $q(t)$  is,  in contrast to DSB-AM,  no longer exclusively in the amplitude,  but equally in the phase  (see [[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_SSB_Signal| Exercise 2.11]]).
*Synchronous demodulation of a SSB–AM signal leds to phase distoritions if there is a phase offset between the carrier signals at the transmitter and the receiver.
*The application of  $\text{envelope demodulation}$  with  "USB–AM"  or  "LSB–AM"  thus always leads to strong nonlinear distortions  (see [[Aufgaben:Exercise_2.11Z:_Once_again_SSB-AM_and_Envelope_Demodulator|Exercise 2.11Z]]).

As with DSB–AM and synchronous demodulation,  the following statements also apply here:
*Attenuation distortions of the channel also lead only to (linear)  attenuation distortions with respect to the sink signal.  No nonlinear distortions arise  (see  [[Aufgaben:Exercise_2.10:_SSB-AM_with_Channel_Distortions|Exercise A2.10]]).
*SSB–AM without a carrier shows the exact same noise behaviour as DSB-AM without a carrier.  The advantage of the smaller RF bandwidth is cancelled out by the necessary level matching.
*A SSB–AM with a sideband-to-carrier ratio  $\mu$  shows similar noise behaviour as in DSB-AM with modulation depth  $m = \sqrt{2} · \mu$  (see  [[Aufgaben:Exercise_2.10Z:_Noise_with_DSB-AM_and_SSB-AM|Exercise 2.10Z]]).
*In any case,  it should be noted that SSB-AM with carrier is not very useful due to the nonlinear distortions brought about with envelope demodulation.

==Exercises for the chapter==
 
[[Aufgaben:Exercise_2.10:_SSB-AM_with_Channel_Distortions|Exercise 2.10: SSB-AM with Channel Distortions]]

[[Aufgaben:Exercise_2.10Z:_Noise_with_DSB-AM_and_SSB-AM|Exercise 2.10Z: Noise with DSB-AM and SSB-AM]]

[[Aufgaben:Exercise_2.11:_Envelope_Demodulation_of_an_SSB_Signal|Exercise 2.11: Envelope Demodulation of an SSB signal]]

[[Aufgaben:Exercise_2.11Z:_Once_again_SSB-AM_and_Envelope_Demodulator|Exercise 2.11Z: Once again SSB-AM and Envelope Demodulator]]

{{Display}}

Modulation Methods/Further OFDM Applications

2022-04-09T14:02:48Z

Reed:

{{LastPage}}
{{Header
|Untermenü=Multiple Access Methods
|Vorherige Seite=Further OFDM Applications
|Nächste Seite=
}}
==OFDM for DVB–T==
 
$\rm DVB–T$   ("Digital Video Broadcasting Terrestrial")   is one of several variants derived from the 1997 DVB standard for the distribution of television signals in digital form.  Other related standards are:
*$\rm DVB–S$   ⇒   for transmission via  $\rm S\hspace{0.03cm}$atellite,
*$\rm DVB–C$   ⇒   for transmission via the cable network  $\rm (C\hspace{0.03cm}$able$)$,
*$\rm DVB–H$   ⇒   for mobile devices  $\rm (H\hspace{0.03cm}$andhelds$)$.

In Germany,  more than  $90\%$  of the population was already able to receive at least a selection of public broadcasters via DVB-T as of the end of 2008.  Since 2017,  analog TV reception via antenna has finally been no longer possible.  The frequency bands freed up by the shutdown of analog transmission were directly reused by DVB-T,  so that parallel operation and thus a  "smooth transition"  was not possible.

*DVB-T uses the modulation method  "Coded Orthogonal Frequency Division Multiplex"  – abbreviated to  $\rm COFDM$.  The addition of  "Forward Error Correction"  $($ $\rm FEC)$  in combination with the guard interval technique already introduced is necessary to counteract interference caused by echoes  ("multipath reception")  and receiver movement  ("Doppler effect").

*Basically,  DVB-T is a  "common wave network".  In contrast to conventional mobile communications networks  – such as the GSM network –  the same frequencies are reused even by directly adjacent transmitting stations.  This leads to interference between similarly strong signals at the cell boundaries,  which can,  however,  exhibit considerable differences in propagation time.

*The effects of such a frequency-selective channel can be counteracted by extending the guard interval.  However,  on the one hand,  this reduces the bandwidth efficiency,  and it also increases the susceptibility to time-varying effects.

==System parameters of DVB-T==
 
The old analog television channels each had a bandwidth of
[[File:EN_Mod_T_5_8_S2a_neu.png|right|frame| Parameters for DVB-T]]

*$\text{7 MHz}$  $\rm (VHF$, "Very High Frequency"   ⇒   "Ultra-short Wave$")$,
*$\text{8 MHz}$  $\rm (UHF$, "Ultra High Frequency"  ⇒  "Decimetric Wave"$)$.

In DVB-T,  each of these channels is usually occupied by four programs.  For this purpose,  the available spectrum is divided among the OFDM subcarriers.

The table shows possible parameter constellations of a DVB-T system.  "Mode"  is the number of subcarriers used for FFT/IFFT,  i.e.  $\text{2048 (2K)}$  or  $\text{8192 (8K)}$.  However,  not all carriers are actually used.

A comparison of the two modes shows that
*the  $\rm 2K$  mode is well suited for time-variant operating conditions because of the shorter core symbol duration  $T$, 
*but due to the rather short guard interval duration  $T_{\rm G}$,  there are narrow limits to the tolerable channel impulse response.

In the  $\rm 8K$  mode,  this is exactly the opposite.  Because of the requirement of a common wave network,  8K operation is therefore preferable to 2K operation,  although the disadvantage of a more complex implementation of the FFT/IFFT must be accepted.

Three QAM variants are available for DVB-T subcarrier modulation:
*$\rm 4–QAM$  (this method can also be understood as  $\rm 4–PSK$ ),
*$\rm 16–QAM$  (optionally with asymmetrical signal space constellation),
*$\rm 64–QAM$  (also optionally with asymmetrical signal space constellation).

An asymmetrical signal space constellation allows hierarchical source coding.  In poor transmission conditions,  only the quadrant can be detected instead of the exact signal space point.  Reception is thus still possible,  albeit with  (greatly)  reduced image quality.

[[File:EN_Mod_T_5_8_S2b_neu.png|right|frame| Net bit rates for DVB-T. Source:   https://en.wikipedia.org/wiki/DVB-T#Technik]]
The table from WIKIPEDIA shows some resulting data rates for DVB-T depending on the channel coding rate  $R_{\rm C}$  and the duration  $T_{\rm G}$  of the guard interval.  The relative code redundancy is  $1 - R_{\rm C}$.  This means that the larger the code rate  $R_{\rm C}$,  the less redundancy is added.

The numerical values can be interpreted as follows:
*The effective net bit rates vary greatly between individual transmission areas because different system parameters must be used depending on the region.  On average, approximately  $12$ ...  $\text{20 Mbit/s}$  are achieved.  This is now used to transmit an  $\rm MPEG–2$  data stream, which usually contains four programs.

*The video data streams for DVB-T in Germany are also MPEG-2 encoded and achieve a data rate of about  $\text{3.5 Mbit/s}$.  In comparison,  a digitized PAL signal would already correspond to a data rate of  $3$ ... $\text{5 Mbit/s},$  and DVD would already reach a data rate of approx.  $\text{9.8 Mbit/s}$.

*Since 2009,   $\rm DVB–T2$  has been a standard that,  among other things,  uses  $\rm MPEG–4$  instead of  "MPEG–2".  Compared to  "DVB-T",  the standard  "DVB-T2"  achieves a significantly better image quality.
 

==A brief description of DSL - Digital Subscriber Line==
 
As another example of the use of OFDM,  a brief overview of  [https://en.wikipedia.org/wiki/Digital_subscriber_line $\rm DSL$]  ("Digital Subscriber Line")  will now be given:
*DSL technology allows a considerable increase in data transmission rate compared to conventional telephone connections such as  [https://en.wikipedia.org/wiki/Plain_old_telephone_service $\rm POTS$]  ("Plain Old Telephone Service")  or  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|$\rm ISDN$]]  ("Integrated Services Digital Network"),  without having to replace the copper pairs of the subscriber line for this new technology.
*POTS or ISDN can also be operated in parallel with DSL without any restrictions,  since the frequency bands used do not overlap if the appropriate adjustments are made  (see diagram).

[[File:P_ID2967__Mod_T_5_8_S3_Ganz_neu.png |right|frame| "Asymmetric Digital Subscriber Line"  in an ISDN network]]
 
The following main xDSL variants exist ("x" is to be understood as a placeholder here):
*$\rm ADSL$  ("Asymmetric Digital Subscriber Line"),
*$\rm ADSL+$,  an extension to ADSL,
*$\rm VDSL/VDSL2$  ("Very High Data Rate Digital Subscriber Line").

VDSL2 is designed for maximum symmetrical data rates  $($ $\text{210 Mbit/s}$  each upstream and downstream$)$,  but can only be implemented over short lines.

OFDM is always used as the modulation method for all of these xDSL variants,  which is often also called  $\rm DMT$  ("Discrete Multitone Transmission")  in this context.

We now consider DMT in ADSL with the following characteristics:
*Base frequency  $f_0 = 4.3125\ \rm kHz$,
*maximum number of useful carriers:  $N = 255$,
*DC signal free,  since  $S(f = 0) = 0$,
*the Nyquist tone   ⇒   $S(256 · f_0)$  is also set to zero,
*maximum frequency:   $256 · 4.3125 \ \rm kHz = 1104 \ \rm kHz$.

This results in the spectrum occupancy shown in the figure.  A detailed system description can be found in the chapter  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_DSL |"$\rm DSL$"]]  of the LNTww book  "Examples of Communication Systems".

The DSL spectrum shown can be described very briefly as follows:
*The  $224$  carriers available for DMT  - these are often referred to as  "bins" in the context of DMT –  are used
:*either for  "upstream"  $($transmission into the network,  $32$ bins,  $\text{138 kHz)}$ 
:*or for  "downstream"  $($transmission out of the network,  $192$ bins,  $\text{828 kHz)}$. 
*All ADSL lines from  "Deutsche Telekom"  are switched according to theeither "ADSL-over-ISDN"either standard,  so that parallel operation with ISDN is possible.  For this purpose,  the frequency range up to  $\text{138 kHz}$  is not used for ADSL and the corresponding bins are set to zero.
*"ADSL-over-ISDN"  and not  "ADSL-over-POTS"  is also used for analog connections,  although this wastes precisely that frequency range which has a particularly favorable  (low)  cable attenuation.

The main advantage of DSL or DMT is again its adaptability to the channel.  The copper twisted pairs used in the access network do not have an ideal attenuation and phase response,  among other things,  because of the skin effect.  DMT again offers the possibility,  depending on the quality of the frequency range of the bins,  to adapt the respective modulation method or even to dispense with the use of a carrier altogether.

DSL also uses the guard interval technique in the form of the cyclic prefix.    "Interleaving"  can also be used to avoid errors,  but this delays the transmission.  The term  "fast path"  is used to describe the omission of this technique.

{{BlaueBox|TEXT=
$\text{Conclusion:}$  In the field of wireline transmission technology,  too,      '''it was only the transition from a single-carrier system to a multi-carrier system that enabled the high transmission rates common today'''.}}

==Differences between DMT and the described OFDM==
 
The OFDM system descriptions in the chapters  [[Modulation_Methods/Allgemeine_Beschreibung_von_OFDM |"General Description of OFDM"]]  as well as  [[Modulation_Methods/Realisierung_von_OFDM-Systemen |"Implementation of OFDM Systems"]]  apply only to band-pass systems,  which were always considered here in the equivalent low-pass range:
*The low-pass transform removes all components at negative frequencies,  resulting in an  "unbalanced spectral function"  and thus in a  "complex-valued time signal".
*This low-pass transformation is not necessary for DSL because,  in contrast to mobile communications systems,  it is operated in the baseband.

This leads to a different formula representation.  For a clear distinction between the two approaches,
[[File:P_ID1658__Mod_T_5_8_S4_neu.png |right|frame|DMT spectrum]]
*we will use  "OFDM"  in the following only for a bandpass system,
*while the term  "DMT"  always refers to a baseband system.

The diagram shows the spectrum of a DMT signal with  $255$  used bins.
*The useful carrier spectrum from  $1$  to  $255$  is supplemented by the corresponding negative frequency components to obtain a period of the "finite spectrum" required for IDFT, see   [Han08]<ref>Hanik, N.:  Leitungsgebundene Übertragungstechnik.  Lecture manuscript. Lehrstuhl für Nachrichtentechnik, TU München, 2017.</ref>:
:$$S(-\mu \cdot f_0 ) = S\big[(N - \mu ) \cdot f_0 \big] = S(\mu \cdot f_0)^*.$$

:In general  $0 < \mu < N/2$  and for the diagram  $N = 512$.

*According to the  $\rm IDFT$,  the thus purely  "real time signal"  at the sampling times  $ν · T_{\rm A}$  with  $0 ≤ ν < N$  and  $T_{\rm A} = T/N$  results to:
:$$s_\nu = \sum\limits_{\mu = 0}^{N - 1} {S(\mu \cdot f_0 ) \cdot {\rm{e}}^{{\kern 1pt} {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } \hspace{0.03cm}\cdot
\hspace{0.03cm}\nu \hspace{0.03cm}\cdot \hspace{0.03cm}\mu /{N}} }.$$
 
{{BlaueBox|TEXT=
$\text{Conclusion:}$  The required number of frequency support points and samples of the  $\rm IDFT/DFT$  corresponds to twice the usable carriers in the DMT method,
:*if the DC component as well as the  "Nyquist tone"  are neglected,
:*and is thus also twice as large as in an OFDM band-pass system.}}

{{GraueBox|TEXT=
$\text{Example 1:}$ 
Let  $N = 512$  and  $S(64 · f_0) = (a + {\rm j} · b)/2$  and  $S(448 · f_0) = S^{\star}(64 · f_0) = (a - {\rm j} · b)/2$.  Let all other spectral coefficients be zero.

'''(1)'''   We now calculate the time domain coefficients using IDFT according to the given formula:
:$$\begin{align*}s_\nu & = \sum\limits_{\mu = 0}^{N - 1} {S (\mu \cdot f_0 ) \cdot {\rm{e} }^{ {\rm{j \hspace{0.03cm}\cdot \hspace{0.03cm}2 \pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} \nu \hspace{0.03cm}\cdot \hspace{0.03cm}\mu /{N} } } =S( 64 \cdot f_0 ) \cdot {\rm{e} }^{ {\rm{j \hspace{0.03cm}\cdot \hspace{0.03cm}2 \pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} \nu \hspace{0.03cm}\cdot \hspace{0.03cm}64 /512} + S( 448 \cdot f_0 ) \cdot {\rm{e} }^{ {\rm{j \hspace{0.03cm}\cdot \hspace{0.03cm}2 \pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} \nu \hspace{0.03cm}\cdot \hspace{0.03cm}448 /512} = \\ & = {1}/ {2} \cdot (a + {\rm{j} } \cdot b ) \cdot {\rm{e} }^{ {\rm{j \hspace{0.03cm}\cdot \hspace{0.03cm}2 \pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} \nu \hspace{0.03cm}\cdot \hspace{0.03cm}64 /512} + {1}/{2} \cdot (a - {\rm{j} } \cdot b ) \cdot {\rm{e} }^{ {\rm{j \hspace{0.03cm}\cdot \hspace{0.03cm}2 \pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} \nu \hspace{0.03cm}\cdot \hspace{0.03cm}448 /512}.\end{align*}$$

'''(2)'''   Applying  [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Euler's theorem]]  to this,  we get:
:$$s_\nu = \frac{a + {\rm{j} } \cdot b } {2} \cdot \left[ {\rm{cos} } ( 2\pi \cdot \frac{ {64} } { {512} }\cdot \nu ) + {\rm{j} } \cdot {\rm{sin} } ( 2\pi \cdot \frac{ {64} } { {512} }\cdot \nu ) \right] + \frac{a - {\rm{j} } \cdot b } {2} \cdot \left[ {\rm{cos} } ( 2\pi \cdot \frac{ {448} } { {512} }\cdot \nu ) + {\rm{j} } \cdot {\rm{sin} } ( 2\pi \cdot \frac{ {448} } { {512} }\cdot \nu )\right],$$
: where:
:$${\rm{cos} } ( 2\pi \cdot \frac{ {448} } { {512} }\cdot \nu ) = {\rm{cos} } ( 2\pi \cdot \frac{ {64} } { {512} }\cdot \nu ), \qquad {\rm{sin} } ( 2\pi \cdot \frac{ {448} } { {512} }\cdot \nu ) = - {\rm{sin} } ( 2\pi \cdot \frac{ {64} } { {512} }\cdot \nu ).$$

'''(3)'''   Summarizing here yields purely real time signal coefficients:
:$$s_\nu = a \cdot {\rm{cos} } ( 2\pi \cdot \frac{ {64} } { {512} }\cdot \nu ) - b \cdot {\rm{sin} } ( 2\pi \cdot \frac{ {64} } { {512} }\cdot \nu ).$$

:*The parameter  $a$  represents the  "in-phase component"  of the QAM modulation of the respective carrier (bins).
:* The parameter  $b$  indicates the  "quadrature component".

The same procedure can be followed for the other "carrier pairs".}}

==Exercise for the chapter==
 
[[Aufgaben:Exercise_5.10:_DMT_Process_for_DSL|Exercise 5.10: DMT Process for DSL]]

==References==
<references/>

{{Display}}

Modulation Methods/OFDM for 4G Networks

2022-04-09T14:02:15Z

Reed:

{{Header
|Untermenü=Multiple Access Methods
|Vorherige Seite=Implementation of OFDM Systems
|Nächste Seite=Further OFDM Applications
}}
==Multiplexing vs. multiple access methods==
 
As a first example,  let's now take a look at the fourth-generation  $\rm (4G)$  mobile communications networks,  which are the successors to the earlier mobile communications networks based on
*$\rm TDMA/FDMA$, see  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_GSM|$\rm GSM$]]  ("Global System for Mobile Communications"),  and
*$\rm CDMA$, see  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_UMTS|$\rm UMTS$]]  ("Universal Mobile Telecommunications System")

have replaced them.

*Another name often used synonymously with  "4G"  is  [[Mobile_Communications/General_Information_on_the_LTE_Mobile_Communications_Standard|$\rm LTE$]]  ("Long Term Evolution").  Like  [https://en.wikipedia.org/wiki/WiMAX WiMAX]  ("Worldwide Interoperability for Microwave Access"),  LTE uses  [https://www.techtarget.com/searchnetworking/definition/orthogonal-frequency-division-multiple-access-OFDMA  $\rm OFDMA$]  ("Orthogonal Frequency Division Multiple Access")  as its multiple access method.

*The main difference between a pure  "$\rm multiplexing \; method$"  $($TDM,  FDM,  CDM,  OFDM$)$  and a  "$\rm multiple \; access \; method$"  $($TDMA,  FDMA,  CDMA,  OFDMA$)$  is the user separation realized by resource allocation.  In the case of OFDMA,  this means that not only time slots  (TDMA)  or spreading codes  (CDMA)  are allocated to the individual subscribers,  but also different and  "preferably"  orthogonal subcarriers.

*The realization of a suitable allocation mechanism is a non-trivial problem which can only be dealt with very superficially here.  It makes sense that a method is not limited to only one  (physical)  layer,  but works across layers.  The term  "layer"  is to be understood here in the sense of the  [https://en.wikipedia.org/wiki/OSI_model OSI reference model]. 

*This is also necessitated by the changing requirements of a mobile communications network.  Whereas  '''connection-oriented'''  services such as voice telephony were in the foreground at the beginning of mobile communications,  today  '''packet-oriented'''  applications such as  "Voice over IP" (VoIP),  "video telephony"  or  "mobile data services"  represent the main traffic load and are the cause of the increased demand on the available data rates.

==Some characteristics of mobile radio systems==
 
First of all,  the special features of the mobile radio channel will be discussed very briefly.  The figure shows a typical mobile radio scenario.  More detailed information on this subject can be found in the book  [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading|Mobile Communications]].

[[File:EN_Mob_T1_1_S1.png |right|frame| Characteristics of the mobile communications channel]]
The main characteristics of the mobile communications channel are:
*the distance-dependent attenuation  ("path loss"),
*refraction,  scattering and reflection effects and thus multipath propagation,
*possible shadowing by mountains,  trees and houses,
*the Doppler effect due to the relative speed between transmitter and receiver.
 
Regarding  "attenuation"  one distinguishes
*between  "time-dependent attenuation"  (  "time selective fading")
*and  "frequency-dependent attenuation"  (  "frequency selective fading").
 
[[File:Mod_T_5_7_S2_2neu2.png |left|frame| Description of time-dependent and frequency-dependent attenuation]]
 The table on the left shows the respective descriptive quantities.  Please note:  

*The coherence bandwidth  $f_{\rm coh}$  is not the reciprocal of the coherence time  $T_{\rm coh}$,  although one might assume this due to comparable naming.

* $f_{\rm coh}$  is a parameter for the frequency-dependent attenuation and results as the reciprocal of the  [https://en.wikipedia.org/wiki/Delay_spread delay spread].

* $T_{\rm coh}$  on the other hand,  describes the time-dependent attenuation and is the reciprocal of the  [https://en.wikipedia.org/wiki/Fading Doppler spread].
 
==Determination of some OFDM parameters==
 
Now we will try to adapt the OFDM system to the channel in such a way that neither time nor frequency dependent fading occurs.  For this,  the following must apply:
*Time-dependent fading is avoided  (i.e.:  the channel is time-invariant)  if the frame duration  ("Time of Interest")  $T_{\rm R} = T + T_{\rm G})$  is significantly smaller than the coherence time  $T_{\rm coh}$: 
:$$T_{\rm R} \ll T_{\rm coh}.$$
*Frequency selective fading  (within a subcarrier)  is avoided if the bandwidth of all carriers  $(≈ f_0)$  is significantly smaller than the coherence bandwidth  $f_{\rm coh}$: 
:$$f_0 = {1}/{T} \ll f_{{\rm{coh}}} = {1}/{{T_{\rm{D}} }} \approx {1}/{{T_{\rm{G}} }}
\hspace{1.0cm}({\rm e.g.\hspace{0.12cm} for }\hspace{0.15cm} \tau_{\rm min}= 0,\hspace{0.12cm}\tau_{\rm max}= T_{\rm G}\hspace{0.12cm}{\rm applies}).$$
*Both requirements can be summarized by the following inequality with respect to the symbol duration  $T$: 
:$$T_{\rm{G}} \ll T \ll T_{{\rm{coh}}} - T_{\rm{G}}.$$
*However,  the two requirements cannot be satisfied simultaneously:   A larger  $T$  decreases the frequency selectivity,  but at the same time makes the transmission more susceptible to Doppler spreading  $($smaller ratio  $T_{\rm coh}/T)$.

{{GraueBox|TEXT=
$\text{Example 1:}$  Assuming a given coherence time  $T_{\rm coh}$  and maximum delay  $τ_{\rm max}$  through the channel,  one could proceed as follows:
#Determination of the preliminary guard interval to  ${T_{\rm G} }' ≥ τ_{\rm max}$,
#Calculating the upper and lower bounds:  ${T_{\rm G} }' \ll T \ll T_{\rm coh} – {T_{\rm G} }'$,
#Calculating the optimal symbol duration as a geometric mean:
::$$T_{\rm opt} = \sqrt { {T_{\rm G} }' \cdot (T_{\rm coh} - {T_{\rm G} }') }.$$
*The required number  $N_{\rm user}$  of subcarriers  – and thus the minimum FFT support number –  is obtained from this together with the data rate  $R$  and the number  $M$  of signal space points of the mapping used after rounding up:
:$$N_{ {\rm{user} } } = \left\lceil {\frac{ {R \cdot (T + {T_{\rm G} }') } } { { {\rm{log}_2}(M)} } } \right\rceil \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
N_{ {\rm{FFT} } } = 2^{\left\lceil { {\rm{log}_2} (N_{ {\rm{use} } } )} \right\rceil }.$$

*The value  $N_{\rm FFT}$  takes into account that the number of interpolation points of the FFT must be a power of two.  The unused carriers due to the FFT adjustment are used as an additional guard band at the edges of the spectrum.
*The resulting guard interval must now be adjusted to the new parameters:
:$$T_{\rm{G} } = N_{\rm{G} } \cdot \frac{{T_{{\rm{opt} } } } } { {N_{ {\rm{FFT} } } }} \quad {\rm{with}}
\quad N_{\rm{G} } = \left\lceil {\frac{ { {T_{\rm G} }' } } { {T_{ {\rm{opt} } } } } \cdot N_{ {\rm{FFT} } } } \right\rceil.$$

*The frame duration is given by  $T_{\rm R} = T + T_{\rm G}$  and the total number of samples in a frame is given by  $N_{\rm total} = N_{\rm FFT} + N_{\rm G}$. 
*Finally,  the number  $N_{\rm user}$  of the useful carriers must be calculated again with the above equation. }}

==Resource management in 4G networks==
 
A design parameter not considered so far is the choice of the respective modulation scheme of the individual subcarriers,  which has a decisive influence on the error probability during transmission. 
[[File: P_ID1652__Mod_T_5_7_S4a_ganz_neu.png |right|frame| Multi-user diversity gain from [Vie17]<ref>Viering, I.:  System Aspects in Communications.  Lecture manuscript, Lehrstuhl für Nachrichtentechnik, TU München, 2017.</ref>.     We thank Ingo Viering for permission to use the diagram.]]

[[File:EN_Mod_T_5_7_S5.png |right|frame| Cross-layer scheduler]]

*In principle,  it can be said that the robustness decreases with increasing  ${\rm log_2}(M)$,  since this reduces the spacing of the possible signal space points.

*However,  a large symbol range  $M$  is the prerequisite for the desired high data rates,  which is only possible in good channel states.  The characteristics of the channel often differ greatly for the participants and change over time.

The aim of a suitable  '''scheduling procedure'''  is now,
*transmit only when the channel is good,  and
*to serve other subscribers during the dips in attenuation.

The gain in system throughput that can be achieved in this way is often referred to as  [https://en.wikipedia.org/wiki/Diversity_scheme "multi-user diversity gain"]. 

The diagram on the right shows the procedure.  The blue line represents the desired  "riding on the peaks"  for the case of eight participants  (eight fading processes). 

 When implementing a channel-adaptive scheduler,  however,  information from higher layers should also be included,  since long waiting times   (large delays)  must be avoided depending on the  "Quality of Service"  $\rm (QoS)$  requirements and traffic type.  The diagram on the right shows a schematic representation of such a   "cross–layer approach". 
*The scheduler is located in the MAC layer  ("Medium Access Control Layer")  and communicates with the other  (not necessarily neighboring)  layers.  It should be noted that this approach contradicts the modularity principle envisaged in the OSI model.
*A similar approach has already been implemented as part of the extension of the UMTS network to include the [[Examples_of_Communication_Systems/Weiterentwicklungen_von_UMTS#High.E2.80.93Speed_Downlink_Packet_Access|$\rm HSDPA$]]  ("High Speed Downlink Packet Access") standard,  which has increased the maximum data rate in the 3G network  (downstream)  from  $\text{384 kbit/s}$  to the theoretical value  $\text{14.4 Mbit/s}$.

This section was only intended to provide a rough outlook on possible future mobile communications networks.  It should be noted that the improved adaptivity required for very high data rates is only made possible by the use of a multi-carrier system such as OFDM/OFDMA.

==Exercise for the chapter==
 
[[Aufgaben:Exercise_5.9:_Selection_of_OFDM_Parameters|Exercise 5.9:   Selection of OFDM Parameters]]

==References==

{{Display}}

Modulation Methods/Implementation of OFDM Systems

2022-04-09T14:01:45Z

Reed:

{{Header
|Untermenü=Multiple Access Methods
|Vorherige Seite=General Description of OFDM
|Nächste Seite=OFDM for 4G Networks
}}
==OFDM using discrete Fourier transform (DFT)==
 
We now consider again the temporally non-overlapping transmitted signal frames
:$$s_k (t) = \sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot g_\mu (t - k \cdot T_{\rm{R}} )},$$
where  $k$  indicates the frame number.  At sampling times  $k · T_{\rm R} + ν · T_{\rm A}$  with  $0 ≤ ν < N$  and  $T_{\rm A} = T/N$,  these frames have the sampling values
:$$s_{\nu ,\hspace{0.08cm}k} = \sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot {\rm{e}}^{ {\kern 1pt} {\rm{j\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi}} {\kern 1pt}\cdot \hspace{0.03cm}\nu \hspace{0.03cm}\cdot \hspace{0.03cm}{\mu}/{N}} }.$$
Here  $T_{\rm R}$  denote the  "frame duration"  (German:  "Rahmendauer"   ⇒   subscript  "R")  and  $T_{\rm A}$  the  "sampling distance"  (German:  "Abtastabstand"   ⇒   subscript  "A").
*With the renaming  $s_{ν,\hspace{0.08cm}k} = d_{ν,\hspace{0.08cm}k}$  and  $a_{\mu,\hspace{0.08cm}k} = D_{\mu,\hspace{0.08cm}k}$  the equation corresponds exactly to the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)#Inverse_discrete_Fourier_transform|Inverse Discrete Fourier Transform]]  $\rm (IDFT)$  in the  $k$–th interval:
:$$d_{\nu ,\hspace{0.08cm}k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu ,\hspace{0.08cm}k} \cdot w^{ - \nu \hspace{0.03cm}\cdot \hspace{0.03cm} \mu } } \quad {\rm{with}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi}}/N}.$$
:Here,   $d_{ν,\hspace{0.08cm}k}$  are the time samples and  $D_{ν,\hspace{0.08cm}k}$  are the discrete spectral coefficients.

*The equation for the transition from the discrete time function to the discrete spectral function   ⇒    [[Signal_Representation/Discrete_Fourier_Transform_(DFT)#From_the_continuous_to_the_discrete_Fourier_transform|Discrete Fourier Transform]]  $\rm (DFT)$  is:
:$$D_{\mu ,\hspace{0.08cm}k} = \frac{1}{N}\cdot \sum\limits_{\nu = 0}^{N - 1} {d_{\nu ,\hspace{0.08cm}k} \cdot w^{\hspace{0.05cm}\nu \hspace{0.03cm}\cdot \hspace{0.03cm}\mu } }.$$

*Furthermore:
#The coefficients  $d_{ν,\hspace{0.08cm}k}$  and  $D_{μ,\hspace{0.08cm}k}$  are periodic with the grid number  $N$.  Moreover,  they are in general complex-valued.
#In principle,  DFT and IDFT have the same structure.
#They only differ by the sign in the exponent of the complex rotation factor  $w$  and the normalization factor  $1/N$  in the case of DFT.

{{BlaueBox|TEXT=
$\text{Notes:}$ 
*The applet  [[Applets:Discrete_Fouriertransform_and_Inverse|"Discrete Fourier Transform"]]  clarifies the properties of DFT and IDFT.
*The possibility of an efficient realization of the multicarrier system results with the  [[Signal_Representation/Fast_Fourier_Transform_(FFT)|Fast Fourier Transform]]. 
*For the use of  FFT/IFFT,  the number of interpolation points  (or samples)  in the time and frequency domain must be a power of two in each case.
*Under this condition,  an implementation with the complexity  $\mathcal{O}(N · {\rm log_2} \ N)$  is possible with the different known algorithms for the implementation of the FFT.}}

==Realization of the OFDM transmitter==
 
The diagram shows the block diagram for the realization of the OFDM transmitter using the "Inverse Discrete Fourier Transform"  $\rm (IDFT)$.
[[File:EN_Mod_T_5_6_S2.png|right|frame |Block diagram of the OFDM transmitter ]]
*In the  [[Modulation_Methods/General_Description_of_OFDM#The_principle_of_OFDM_-_system_consideration_in_the_time_domain|general model]]  at the beginning of the last chapter,  this replaces the very complex parallel demodulation of the  $N$  orthogonal carriers.
*The implementation of the  $\rm IDFT$  as  $\rm IFFT$  (Inverse Fast Fourier Transform) results in a further reduction in effort.

One recognizes from this diagram:
*In the input buffer,  the source signal  $q(t)$  is implicitly serial/parallel  $\rm (S/P)$  converted.  After that,  a signal space mapping to the  $N$  spectral coefficients  $D_{\mu,\hspace{0.08cm}k}$  is performed.  The index  $k$  again denotes the time frame.
*In  $\rm 4–QAM$  mapping,  each two source symbols together yield a complex coefficient  $D_{\mu,\hspace{0.08cm}k}$,  which can take four different values.
*The spectral coefficients  $D_{\mu,\hspace{0.08cm}k}$  generated in this way are then fed to the  $\rm IDFT$ block,  which generates the time domain values $d_{ν,\hspace{0.08cm}k}$  from them.  These are again parallel/serial  $\rm (P/S)$  converted. 

After the subsequent  $\rm (D/A)$ conversion and low-pass filtering the  $\rm OFDM$ transmitted signal  $s(t)$  is finally obtained in the equivalent low-pass range.

==Realization of the OFDM receiver==
 
The diagram shows the block diagram for the realization of the OFDM receiver using the  "Discrete Fourier Transform"  $\rm (DFT)$.
This replaces in the  [[Modulation_Methods/General_Description_of_OFDM#The_principle_of_OFDM_-_system_consideration_in_the_time_domain|general model]]  (see last chapter)  the very complex parallel demodulation of the  $N$  orthogonal carriers.

The realization of the  $\rm DFT$  as  $\rm FFT$  ("Fast Fourier Transform")  results in a further reduction of effort.  The essential steps are:
[[File:EN_Mod_T_5_6_S3.png |right|frame|Block diagram of the OFDM receiver]]

*The input signal  $r(t)$  of the receiver is first digitalized  $(\rm A/D$ conversion$)$.  This is followed by a pre-equalization in the time domain  (optional),  e.g. with  [[Digital_Signal_Transmission/Entscheidungsrückkopplung|Decision Feedback Equalization]]  $($ $\rm DFE)$  or the  [[Digital_Signal_Transmission/Viterbi–Empfänger|Viterbi algorithm]].
*It should be noted,  that the decisive equalization happens in the frequency domain.   This is explained in section  [[Modulation_Methods/Realisierung_von_OFDM-Systemen#OFDM.E2.80.93Entzerrung_im_Frequenzbereich|OFDM equalization in the frequency domain]]  at the end of the chapter and is not included in the diagram above.
*After serial/parallel  $\rm (S/P)$  conversion,  the discrete time values  $d_{ν,\hspace{0.08cm}k}$  are fed to the DFT block.  The generated spectral samples  $D_{\mu,\hspace{0.08cm}k}$  are decoded by the QAM detector and implicitly parallel/serial converted in the output buffer,  resulting in the sink signal  $v(t)$. 
*Note,  that the receiver-side coefficients $d_{ν,\hspace{0.08cm}k}$  and  $D_{\mu,\hspace{0.08cm}k}$  may well differ from the corresponding quantities of the OFDM transmitter due to channel distortion and noise,  which is not reflected in the chosen nomenclature.
*Only in the case of error-free detection,  the coefficients  $\hat{a}_{\mu,\hspace{0.08cm}k}$  of the sink signal $v(t)$  are identical to the coefficients  $a_{\mu,\hspace{0.08cm}k}$  of the source signal  $q(t)$.  In general,  they differ,  which is captured by the  '''symbol error rate'''.

==Intercarrier interference and intersymbol interference==
 
{{BlaueBox|TEXT=
$\text{Definitions:}$  Orthogonality of OFDM carriers is lost during transmission over a frequency-selective channel.
*The resulting interference between the individual carriers is called  "'''intercarrier interference'''"  $\rm (ICI)$.
*However,  transmission over a multipath channel ultimately also causes superimposition of successive symbols and thus  [[Digital_Signal_Transmission/Ursachen_und_Auswirkungen_von_Impulsinterferenzen|"'''intersymbol interference'''"]]. }}

{{GraueBox|TEXT=
$\text{Example 1:}$  The diagram shows the real part of a received OFDM  (equivalent low-pass)  signal after transmission via a noise-free multipath channel with parameters:
*for the path  "0":   Attenuation  $h_0 = 0.5$;   delay  $τ_0 = 0$,
:for the path  "1":   Attenuation  $h_1 = 0.5$;   delay  $τ_1 = T/4$.
[[File:EN_Mod_T_5_6_S4b.png|right|frame|Received OFDM signal via multipath channel in the equivalent low-pass range]]
*The carrier of frequency  $1 · f_0$  of the interval  $k$  assigned with the coefficient  "$+1$"  is drawn in black.
*The carrier weighted with  "$-1$"  with frequency  $3 · f_0$  in the previous interval  $(k-\hspace{-0.08cm}1)$  is shown in red. 
*Other intervals and carriers are not considered.

One can see from this diagram:
#Transient events at the symbol beginning lead to  "intercarrier interference"  $\rm (ICI)$  in the spectrum. 
#In the time domain,  $\rm ICI$  can be recognized by the jumps that occur  (marked yellow in the diagram). 
# As a result, orthogonality is lost with respect to the frequency grid points.
#Further one recognizes  "intersymbol interference"  $\rm (ISI)$  in the green framed time interval  $0 ≤ t < τ_1$:         The red predecessor symbol  $k-\hspace{-0.08cm}1$   $($frequency  $3 · f_0)$  interferes with the black symbol  $k$   $($frequency $1 · f_0)$. }}

==Guard gap to reduce intersymbol interference==
 
A first possible solution for the second problem  $\rm (ISI)$  is the introduction of a guard gap of length  $T_{\rm G}$:
[[File: P_ID1643__Mod_T_5_6_S4b_1_neu.png|right|frame|Principle of the  "guard gap"]]
*Here,  the signal between two symbols is set to zero for the duration of the protection time  $T_{\rm G}$. 

*As a result,  possible pulse trailers of symbol  $k-\hspace{-0.08cm}1$  no longer extend into the following symbol  $(k)$,  provided that the guard gap is selected  "wider"  than the maximum channel delay.

*The new frame duration  $T_{\rm R}$ – i.e. the distance between successive transmitted symbols – is thus given by
:$$T_{\rm R} = T + T_{\rm G}.$$
 
{{GraueBox|TEXT=
$\text{Example 2:}$ 
This diagram again shows the real part of the received OFDM signal,  but now with  "guard gap".  The assumptions of  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference|$\text{Example 1}$]]  have been kept.
[[File:EN_Mod_T_5_6_S5a.png |right|frame| OFDM reception signal over multipath channel with guard gap '''KORREKTUR''': duration]]

The diagram shows:
# In addition,  $T_{\rm G} = T/4$  is set,  which corresponds to the limiting case  $T_{\rm G} = τ_{\rm max}$  for the present channel.
# By using a guard gap of corresponding width,   intersymbol interference  $\rm (ISI)$  can be avoided   ⇒   in interval  $k$  only one frequency occurs.
# '''But''':  Intercarrier interference  $\rm (ICI)$  cannot be prevented by this,  because the symbols still have a transient phase and thus jumps.

The  "guard gap"  approach will not be considered further.  Rather,  a better alternative is presented in the next section.}}

==Cyclic Prefix==
 
A better solution for the described problem is the introduction of a  '''cyclic extension of the transmitted symbols'''  in the so-called  "guard interval"  of length  $T_{\rm G}$.
[[File:EN_Mod_T_5_6_S5a_neu.png|right|frame| Principle of the cyclic prefix]]
*For this,  the end of a symbol in the time interval  $T \ – \ T_{\rm G} ≤ t < T$  is prefixed again to the actual symbol.
*This procedure thus generates a  '''cyclic prefix'''.
*As with the  "guard gap",  the interval duration increases from symbol duration  $T$  to the new frame duration  $T_{\rm R} = T + T_{\rm G}$. 
*The new number of samples of the extended discrete-time signal in the  $k$–th interval is then:
:$$N_{\rm{total}} = N + N_{\rm{G}} = N \cdot (1 + T_{\rm{G}} /T) .$$
*The number of carriers and the number of useful IDFT values is still  $N$.  Here,  the expansion is only achieved by repeating the end of the symbol  $N\hspace{-0.03cm}-\hspace{-0.08cm}N_0$, ... , $N\hspace{-0.08cm}-\hspace{-0.08cm}1$  in the guard interval  (highlighted in red).
*The use of the  "cyclic prefix"  seems to be particularly useful if the  $\rm ISI$  are mainly caused by tracking. This applies in particular to the copper twisted pairs used in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_DSL|DSL systems]]. 
 
{{GraueBox|TEXT=
$\text{Example 3:}$  The diagram shows the operation of the guard interval in the continuous-time case. 
The parameters from the consideration of the guard gap in  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference|$\text{Example 1}$]]  still apply,  although only one symbol  $($with frequency  $f_0)$ is now considered. 
[[File:EN_Mod_T_5_6_S5b_neu.png |right|frame| Received OFDM signal over multipath channel with cyclic prefix]]

Further system parameters are again  $T_{\rm G} = T/4$  and for path  "0"  or path  "1":
*Attenuation  $h_0 = 0.5$;   delay  $τ_0 = 0$,
*Attenuation  $h_1 = 0.5$;   delay  $τ_1 = T/4$.

In the frame  $k$  of duration  $T_{\rm R}$,  there is now no interference at all:
# Since the preceding symbols completely fade away during the guard interval, there is no  "intersymbol interference"  $\rm (ISI)$.
# Since the respective transients do not extend into the useful symbols,  no  "intercarrier interference"  $\rm (ICI)$  occurs either. }}
 
{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
#  By using a cyclic prefix alone, both "intercarrier interference"  $\rm (ICI)$  and  "intersymbol interference"  $\rm (ISI)$ can be completely avoided.
#  This requires that the length of the guard interval  $(T_{\rm G})$  is at least equal to the maximum duration  $τ_{\rm max}$  of the channel impulse response:   $T_{\rm G} \ge τ_{\rm max}$. 
#  In the example considered,  $T_{\rm G} = τ_{\rm max} = \tau_1$ .
#  The quantity  $τ_{\rm max}$  generally limits the ISI– and ICI–free section within the guard interval to the range  $ \ –T_{\rm G} + τ_{\rm max} ≤ t < T$.}}

==OFDM system with cyclic prefix==
 
The   "Cyclic prefix"  block must therefore be added to the  [[Modulation_Methods/Implementation_of_OFDM_Systems#Realization_of_the_OFDM_transmitter|transmitter structure]]  already shown at the beginning.  At the  [[Modulation_Methods/Implementation_of_OFDM_Systems#Realization_of_the_OFDM_receiver|receiver]]  this prefix must be removed again.

[[File:EN_Mod_T_5_6_S6a.png |right|frame| OFDM transmitter  $($subscript  $\rm S)$  and receiver  $($subscript  $\rm E)$  with cyclic prefix]]

*The definition of a suitable guard interval is an important design criterion for OFDM-based transmission systems.  A possible approach to this is presented as an example in the section  [[Modulation_Methods/OFDM_für_4G–Netze|OFDM for 4G Networks]]. 
*However,  the use of a cyclic prefix degradesthe   "bandwidth efficiency".  The degradation increases with increasing duration  $T_{\rm G}$  of the guard interval  (hereafter abbreviated as  "GI").
*Under the simplifying assumption of a transmission spectrum  $S(f)$  hard limited to $1/T$,  the bandwidth efficiency – see [Kam04]<ref>Kammeyer, K.D.:  Nachrichtenübertragung.  Stuttgart: B.G. Teubner, 4. Auflage, 2004.</ref>:
:$$\beta = \frac{ \text{symbol rate} }{ {\rm bandwidth} } = \frac{1/(T + T_{\rm G})}{1/T} = \frac{1}{{1 + T_{\rm{G}} /T}}.$$
*However,  in a system using the so-called  "matched-filter approach",  increasing the frame duration from  $T$  to  $T_{\rm G} + T$  leads to a decrease in the signal-to-noise ratio if the impulse responses  $g_{\rm S}(t)$  and  $g_{\rm E}(t)$  of the transmission and reception filters are matched to the symbol duration  $T$. 
*The resulting  signal–to–noise ratio  $\rm (SNR)$  of the overall system  (in dB)  can be calculated as follows,  taking into account the guard interval:
:$${\rm{SNR}}_{\hspace{0.08cm}{\rm{ {\rm{with} }\hspace{0.08cm} GI} } } = {\rm{SNR}}_{\hspace{0.08cm}{\rm{{\rm{without}}\hspace{0.08cm} GI}}} + 10 \cdot \lg (\beta ), \quad {\rm{where}}$$
:$$\beta = \frac{{\left[ {\int\limits_0^T {g_{\rm{S}} (\tau ) \cdot g_{\rm{E}} ( - \tau )d\tau } } \right]^2 }}{{\int\limits_{ - T_{\rm{G}} }^T {g_{\rm{S}}^2 (\tau )} \,d\tau \cdot \int\limits_{\rm{0}}^T {g_{\rm{E}}^2 (\tau )} \,d\tau }} = \frac{ {T^2 } } { {(T + T_{\rm{G} } ) \cdot T} } = \frac{1}{ {1 + T_{\rm{G} } /T} }.$$

{{GraueBox|TEXT=
$\text{Example 4:}$ 
We assume a guard interval of length  $T_{\rm G} = T/3$.  Then the bandwidth efficiency is given by:
:$$\beta = \frac{1}{ {1 + 1/3} } = 3/4.$$
*The share of the cyclic prefix in the frame duration  $T_{\rm R}$  is  $25\%$,  and
*the  (logarithmic)  SNR loss is then  $10 · \lg \ (4/3) ≈ 1.25 \ \rm dB$. }}

The  (German language)  SWF applet  [[Applets:OFDM|"OFDM-Spektrum und Signale"]]   ⇒   "OFDM Spectrum and Signals"  illustrates the operation of a cyclic prefix in the continuous-time case with respect to  "intercarrier interference"  $\rm (ICI)$.

==OFDM equalization in the frequency domain==
 
We continue to consider the  [[Modulation_Methods/Realisierung_von_OFDM-Systemen#OFDM.E2.80.93System_mit_zyklischem_Pr.C3.A4fix|OFDM system]]  in the noise-free case and assume a time-invariant channel impulse response whose length is smaller than the duration  $T_{\rm G}$  of the cyclic prefix added at the transmit end.
*The observation is made in the  $k$–th interval,  and indexing is omitted.
*The discrete-time channel impulse response can be written as   $h_ν = h(ν · T_{\rm A})$  with the abbreviation  $T_{\rm A} = T/N$. 
*The discrete-time reception signal is thus obtained by linear  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Faltung_im_Zeitbereich|convolution]]  to:
:$$r_\nu = s_\nu * h_\nu = d_\nu * h_\nu.$$
This takes into account that the time samples  $s_ν$  of the transmitted signal coincide with the IDFT coefficients  $d_ν$. 

{{BlaueBox|TEXT=
$\text{To be noted:}$  In general,  for the conventional linear convolution:
:$${\rm{DFT} } \{ d_\nu * h_\nu \} \ne {\rm{DFT} } \{d_\nu \} \cdot {\rm{DFT} } \{ h_\nu \}.$$
*Nevertheless,  in order to specify the discrete spectrum of the received signal by the discrete Fourier transform  $\rm (DFT)$,  one needs the  [https://en.wikipedia.org/wiki/Circular_convolution "cyclic convolution"]: 
:$$r_\nu = d_\nu * _{\rm (circ)} h_\nu \quad \circ\hspace{0.01cm}\!\!\!-\!\!\!-\!\!\!-\!\!\bullet \quad R_\mu = {\rm{DFT} } \{ d_\nu * _{\rm (circ)} h_\nu \}.$$
:The terms  "circular convolution"  and  "periodic convolution"  are also used synonymously for this purpose.
*Using the convolution theorem for linear time-invariant systems,  one can then also write the spectrum as a product of two discrete Fourier transforms:
:$$R_\mu = {\rm{DFT} }\{ d_\nu \} \cdot {\rm{DFT} }\{ h_\nu \} = D_\mu \cdot H_\mu.$$
*To compensate for the influence of the channel on the reception sequence,  it is convenient to multiply the spectrum by the inverse transfer function  $1/H_{\mu}$. 
*This  "zero forcing"  approach leads to the ideal signal reconstruction in the noise-free case.  The equalization can be done point by point:
:$$\hat {D}_\mu = \frac{1}{ {H_\mu } } \cdot R_\mu.$$}}

{{BlaueBox|TEXT=
$\text{Conclusion:}$  
*In the  '''OFDM system''',   '''channel equalization can be realized with a single multiplication per subcarrier'''  if the channel frequency response is known.
*In contrast, a  '''classic single-carrier system'''  would require  '''equalization of the entire frequency range used'''. }}

==OFDM equalization in matrix-vector notation==
 
In the following,  a renewed but more in-depth consideration of OFDM equalization will be given,  where we use a  [https://en.wikipedia.org/wiki/Matrix_multiplication matrix-vector notation].    The consideration still refers to the  $k$–th interval,  without any special note:
*The vector of a channel with  $L$  echoes is  $\mathbf h = (h_0$, ... , $h_L)$.  The transmission matrix with  $N$  rows and  $N + L$  columns is:
:$${\rm\bf{H}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & \cdots & {h_L } & {} & {} & {} \\ {} & {h_0 } & {h_1 } & \cdots & {h_L } & {} & {} \\ {} & {} & \ddots & \ddots & {} & \ddots & {} \\ {} & {} & {} & {h_0 } & {h_1 } & \cdots & {h_L } \\ \end{array}} \right).$$
*Here,  $N$  indicates the number of carriers and hence the number of time samples of the IDFT.  With the transmit vector  ${\bf d} = (d_0$,  ...  , $d_{N–1})$  the receive vector is:
:$$\bf r = d · H.$$

*Considering the cyclic prefix,  the extended transmit vector is obtained:
:$${\rm\bf{d}}_{{\rm{ext}}} = (d_{N - N_G } , \ \ldots \ ,d_{N - 1} ,d_0 , \ \ldots \ ,d_{N - 1} ).$$
*Now,  one could extend the above transmission matrix  $\bf H$  likewise accordingly on   $(N + N_{\rm G})$  rows   and   $(N + L + N_{\rm G})$  columns   as well as remove the prefix at the receiver again,  which is not to be pursued here further.

Alternatively,  one can use the  "cyclic matrix"  $\rm \bf H_C$  with  $N$  rows and  $N$  columns as well as the  "Fourier transform  $\rm \bf F$  in matrix–vector notation": 
:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c}
{h_0 } & {h_1 } & \cdots & \cdots & {h_L } & {} & {} & {} \\
{} & {h_0 } & {h_1 } & \cdots & \cdots & {h_L } & {} & {} \\
{} & {} & \ddots & \ddots & {} & {} & \ddots & {} \\
{} & {} & {} & {h_0 } & {h_1 } & \cdots & \cdots & {h_L } \\
\hline
{h_L } & {} & {} & {} & {h_0 } & {h_1 } & \cdots & {h_{L - 1} } \\
\vdots & \ddots & {} & {} & {} & \ddots & {} & \vdots \\
\vdots & {} & \ddots & {} & {} & {} & \ddots & \vdots \\
{h_1 } & \cdots & \cdots & {h_L } & {} & {} & {} & {h_0 } \\
\end{array}} \right), \hspace{1cm} {\rm\bf{F}} = \left( {\begin{array}{*{20}c}
1 & 1 & \cdots & 1 \\
1 & {} & {} & {} \\
\vdots & {} & {{\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi }}{\kern 1pt}
\cdot \hspace{0.02cm}\nu {\kern 1pt} \cdot\mu /N} } & {} \\
1 & {} & {} & {} \\
\end{array}} \right) .$$

*The Discrete Fourier Transform  $\rm (DFT)$  can be represented by  $1/N · \bf F$  and its inverse  $\rm (IDFT)$  by  $\rm \bf F^{\star}$ such that for the transmit vector:  $\rm {\bf d} = {\bf D} · {\bf F}^{\star}$.

*The  $N$  spectral coefficients are described by the vector  ${\bf D} = 1/N · {\bf d} · {\bf F}$  and the reception vector is  ${\bf r} = {\bf d} · {\bf H}_{\rm C} = {\bf D} · {\bf F}^{\star} · {\bf H}_{\rm C}$.

*The (discrete) Fourier transform  $\rm \bf R$  of the reception vector  $\rm \bf r$  can then be written in the following way:
:$${\rm\bf{R}} = \frac{1}{N} \cdot {\rm\bf{r}} \cdot {\rm\bf{F}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
{H_0 } & {} & {} & {} \\
{} & {H_1 } & {} & {} \\
{} & {} & \ddots & {} \\
{} & {} & {} & {H_{N - 1} } \\
\end{array}} \right),\hspace{0.25cm} {\rm with}\hspace{0.25cm} H_\mu = \sum\limits_{l = 0}^L {h_l \cdot
{\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi }}\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm}\mu /N} }.$$

[[File:EN_Mod_T_5_6_S8b.png|right|frame| Block diagram of the OFDM receiver]]
{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
*The reception symbol on the  $\mu$–th carrier is:  
:$$R_{\mu} = D_{\mu} · H_{\mu} \ \ (\mu = 0, \text{...}\ ,\ N–1).$$
*This can thus be equalized using the  "Zero Forcing"  approach:
:$$\hat {D}_\mu = \frac{1}{ {H_\mu } } \cdot R_\mu = e_\mu \cdot R_\mu .$$
*Equalization   ⇒   multiplication with  $e_{\mu} = 1/H_{\mu} \ (\mu = 0,$ ... , $N–1)$.
* The complete block diagram of OFDM receiver is shown on the right.
}}

{{GraueBox|TEXT=
$\text{Example 5:}$ 
We assume a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes,
*so that for the transmit vector  ${\bf d} = (d_0, d_1, d_2, d_3)$,  and
*for the channel impulse response  ${\bf h} = (h_0, h_1, h_2)$. 

'''(1)'''   To represent the cyclic prefix,  we use the cyclic transmission matrix  ${\rm\bf{H} }_{\rm{C} }$,  instead of the extended transmission vector with the corresponding transmission matrix,  resulting in the reception vector  ${\rm \bf r}= {\rm \bf d} \cdot {\rm \bf H}_{\rm{C} }$: 
:$${\rm\bf{H} }_{\rm{C} } = \left( {\begin{array}{*{20}c}
{h_0 } & {h_1 } & {h_2 } & { } \\
{} & {h_0 } & {h_1 } & {h_2 } \\
\hline
{h_2 } & {} & {h_0 } & {h_1 } \\
{h_1 } & {h_2 } & {} & {h_0 } \\
\end{array} } \right), \hspace{1cm}
{\rm\bf{r} } = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {d_0 ,d_1 ,d_2 ,d_3 } \right) \cdot \left( {\begin{array}{*{20}c}
{h_0 } & {h_1 } & {h_2 } & {} \\
{} & {h_0 } & {h_1 } & {h_2 } \\
\hline
{h_2 } & {} & {h_0 } & {h_1 } \\
{h_1 } & {h_2 } & {} & {h_0 } \\
\end{array} } \right) $$
:$$\Rightarrow \hspace{0.3cm} r_0 = d_0 \cdot h_0 + d_2 \cdot h_2 + d_3 \cdot h_1, \hspace{0.5cm}
r_1 = d_0 \cdot h_1 + d_1 \cdot h_0 + d_3 \cdot h_2,$$
:$$\Rightarrow \hspace{0.3cm} r_2 = d_0 \cdot h_2 + d_1 \cdot h_1 + d_2 \cdot h_0, \hspace{0.5cm}
r_3 = d_1 \cdot h_2 + d_2 \cdot h_1 + d_3 \cdot h_0.$$

'''(2)'''   The  (discrete)  Fourier transform of the receive vector is calculated to be
:$${\rm\bf{R} } = \frac{1}{N} \cdot {\rm\bf{r} } \cdot {\rm\bf{F} } = {\rm\bf{D} } \cdot \left( {\begin{array}{*{20}c}
{H_0 } & {} & {} & {} \\
{} & {H_1 } & {} & {} \\
{} & {} & {H_2 } & {} \\
{} & {} & {} & {H_3 } \\
\end{array} } \right) ,\hspace{0.25cm} {\rm mit}\hspace{0.25cm} H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot
{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$

'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded transmit sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$: 
:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =
\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$
:$$
{\rm\bf{h} }= \left( h_0, h_1,h_2\right) = \left(
0.5,\ 0.3,\ 0.2\right).$$

'''(4)'''   First,  we determine the elements  $H_{\mu}$  of the diagonal matrix:
:$$\begin{array}{l}
H_0 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^0 = 0.5 + 0.3 + 0.2 = 1,} \\
H_1 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {1}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } /2 } +
0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } } = 0.3 - {\rm{j} } \cdot 0.3, \\
H_2 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {2}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } }
+ 0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}2\pi } } } = 0.4, \\
H_3 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{2} \pi } } }
+ 0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}3\pi } } } = 0.3 + {\rm{j} } \cdot 0.3. \\
\end{array}$$
'''(5)'''   Thus,  the vector of frequency support points at the receiver is given by
:$$\begin{align*}{\rm\bf{R} } &= \left( {\rm{1, -1, \; \; 1, -1} } \right) \cdot \left( {\begin{array}{*{20}c}
1 & {} & {} & {} \\
{} & {0.3 - {\rm{j} } \cdot 0.3} & {} & {} \\
{} & {} & {0.4} & {} \\
{} & {} & {} & {0.3 + {\rm{j} } \cdot 0.3} \\
\end{array} } \right) \ = \ {\rm{ (1, -0.3 + j \cdot 0.3, \; \; 0.4, -0.3 - j \cdot 0.3) } }.\end{align*}$$

'''(6)'''   One chooses the equalizer coefficients according to  $e_{\mu} = 1/H_{\mu}$,  where  $\mu = 0$, ... , $3$   holds:
:$$e_0 = 1, \quad e_1 = \frac{1}{ {0.3 - {\rm{j} } \cdot 0.3} }, \quad e_2 = \frac{1}{ {0.4} }, \quad e_3 = \frac{1}{{0.3 + {\rm{j} } \cdot 0.3} }.$$

'''(7)'''   The rectified symbol sequence with  ${\bf e} = (e_0, e_1, e_2, e_3)$  finally results in
:$$\hat {\rm\bf{D} } = {\rm\bf{R} } \cdot {\rm\bf{e} }^{\rm{T} } = (R_0 ,R_1 ,R_2 ,R_3) \cdot \left( {\begin{array}{*{20}c}
{e_0 } \\
{e_1 } \\
{e_2 } \\
{e_3 } \\
\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$
⇒   This corresponds exactly to the transmit symbol sequence  $\bf D$.  That is: 

:'''Knowing the channel,  the signal can be completely equalized,  requiring only a single multiplication per symbol  (carrier)'''. }}

==Advantages and disadvantages of OFDM==
 
Major  '''advantages'''  of OFDM over single-carrier or other multi-carrier systems are:
*flexible with respect to adaptation to bad channel conditions,
*simple channel organization,
*very easy to realize equalization,
*very robust against multipath propagation due to guard interval technique,
*high spectral efficiency,
*simple implementation using  $\rm IFFT/FFT$  (Fast Fourier Transform),
*relatively insensitive to inaccurate time synchronization.

Major  '''disadvantages'''  of OFDM are:
*susceptible to Doppler spreading due to a relatively long symbol duration,
*sensitive to oscillator fluctuations,
*an unfavorable crest factor.

'''Note''':   The so-called  "crest factor"  describes the ratio of peak value to rms value of an alternating quantity.  In an OFDM system,  this can be very large.  As a result,  the demands on the amplifier circuits used are very high  (linearity over a wide range),  if efficiency  (energy consumption,  waste heat)  is not to be ignored.

{{BlaueBox|TEXT=
$\text{Conclusions:}$ 
*The advantages of OFDM far outweigh the disadvantages.
*Although the principle has been known at least since the publication [Wei71]<ref>Weinstein, S. B.:  Data Transmission by Frequency Division Multiplexing Using the Discrete Fourier Transform.  IEEE Transactions on Communications, COM-19, S. 628-634, 1971.</ref>,  OFDM systems have,  however,  only been used since the 1990s.
*The main reason for this is among other things,  that the powerful signal processors required for IFFT or FFT have only been available for a few years. }}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_5.7:_OFDM_Transmitter_using_IDFT|Exercise 5.7: OFDM Transmitter using IDFT]]

[[Aufgaben:Exercise_5.7Z:_Application_of_the_IDFT|Exercise 5.7Z: Application of the IDFT]]

[[Aufgaben:Exercise_5.8:_Equalization_in_Matrix_Vector_Notation|Exercise 5.8: Equalization in Matrix Vector Notation]]

[[Aufgaben:Exercise_5.8Z:_Cyclic_Prefix_and_Guard_Interval|Exercise 5.8Z: Cyclic Prefix and Guard Interval]]

==References==
<references/>

{{Display}}

Modulation Methods/General Description of OFDM

2022-04-09T14:01:13Z

Reed:

{{Header
|Untermenü= Multiple Access Methods
|Vorherige Seite=Error Probability of Direct-Sequence Spread Spectrum Modulation
|Nächste Seite=Implementation of OFDM Systems
}}
==The principle of OFDM - system consideration in the time domain==
 
Orthogonal Frequency Division Multiplex  $\rm (OFDM)$  is a digital multicarrier modulation method with the following characteristics:
[[File:P_ID1635__Mod_T_5_5_S1a_neu.png|right|frame| Principle of an  $\rm OFDM$ transmitter based on  $\text{4-QAM}$]]

*Instead of a broadband,  strongly modulated signal,  a large number of narrowband,  mutually orthogonal subcarriers are used for data transmission.  Among other things,  this enables adaptation to a frequency-selective channel.

*In OFDM,  the subcarriers themselves are usually modulated by conventional  [[Modulation_Methods/Quadratur–Amplitudenmodulation|quadrature amplitude modulation]]  $\rm (QAM)$  or by  [[Modulation_Methods/Lineare_digitale_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|binary phase modulation]]  $\rm (BPSK)$,  although the individual carriers may well differ in terms of the modulation type.

*Differences in the degree of modulation result in different data rates for the subcarriers.  This means that a high-rate source signal must be split into several signals of significantly lower symbol rate for transmission.
 
The diagram shows the principle of an OFDM transmitter based on the modulation method  $\text{4-QAM}$. 
*The representation of the  "zeroth"  branch  $(\mu = 0)$,  which represents the DC part,  has been deliberately omitted here because it is often set to zero   ⇒   for all frames  $k$   ⇒   $a_{0,\hspace{0.08cm} k} =0 $  holds.
*The  $N–1$  parts of the data stream  $〈q_{\mu,k}〉$  present at time $k$  are first 4-QAM encoded by combining two bits at a time. 
*Then the generally complex amplitude $a_{\mu,\hspace{0.08cm}k}$  $($with control variables  $\mu = 1$, ... , $N–1)$  is pulse-shaped and modulated with the  $\mu$–th multiple of the basic frequency  $f_0$. 
*The transmitted signal is now the additive superposition of the individual partial signals.  The consideration takes place here and also in the following in the [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function|equivalent low-pass range]],  whereby the index  "TP"  (German:  "Tiefpassbereich")  is renounced.
*The pulse shape filter  $g_s(t)$  is a rectangle of height  $s_0$  limited to the range  $0 ≤ t < T$. 
*We call  $T$  the  '''symbol duration'''  and denote the reciprocal  $f_0 = 1/T$  as the  '''basic frequency'''.

If we now summarize this filter with the respective modulation to
:$$g_\mu (t) = \left\{ \begin{array}{l} s_0 \cdot {\rm{e}}^{ {\kern 1pt} {\rm{j2 \pi}} {\kern 1pt} \mu f_0 t} \quad 0 \le t < T, \\ 0 \quad \quad \quad \quad \quad {\rm otherwise} \\ \end{array} \right.$$
with  $\mu ∈ \{0, \ \text{...}\ , N–1\}$,  we obtain the  '''OFDM transmitted signal  $s_k(t)$  in the  $k$–th time interval''':
:$$s_k (t) = \sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot g_\mu (t - k \cdot T_{\rm{R}} )}.$$

{{BlaueBox|TEXT=
The total  '''OFDM transmitted signal considering all time intervals'''  is then:
:$$s(t) = \sum\limits_{k = - \infty }^{+\infty} {\sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot g_\mu (t - k \cdot T_{\rm{R} } )} }.$$
*$T_{\rm R}$  denotes the frame  (German:  "Rahmen"   ⇒   subscript "R")  duration.  Within this time the same data is present at the input and after  $T_{\rm R}$  the next frame follows.

*For a multicarrier system with  $M$  QAM signal space points and the bit duration  $T_{\rm B}$  of the binary source symbols,  the symbol duration  $T$  is generally given by
:$$T = N \cdot {\rm{log}_2}(M) \cdot T_{\rm{B} } ,$$
:where  $N$  is again the number of subcarriers. 
*For the frame duration,   $T_{\rm R} \ge T$  must hold.  Initially,  let  $T_{\rm R} = T$.}}

{{GraueBox|TEXT=
$\text{Example 1:}$ 
*We first assume a single-carrier system with data rate  $R_{\rm B} = 768 \ \rm kbit/s$   ⇒   $T_{\rm B} ≈ 1.3 \ \rm µ s$   and a mapping with  $M = 4$  signal space points  $\text{(4–QAM)}$.  
*The symbol duration in the single carrier  $($ $\rm SC)$  case is then:
:$$T_\text{SC} = 1 \cdot {\rm{log}_2}(4) \cdot 1.3 \,{\rm{µ s} } \approx 2.6 \,{\rm{µ s} }.$$
*Assuming that for a multi carrier  $($ $\rm MC)$  system with  $N = 32$  carriers the modulation method  $\text{16–QAM}$  is used,  the symbol duration is larger by a factor of  $64$: 
:$$T_\text{MC} = 32 \cdot {\rm{log}_2}(16) \cdot 1.3 \,{\rm{µ s} } \approx 0.167\, {\rm{ms} }.$$ }}

{{BlaueBox|TEXT=
$\text{Conclusions:}$ 
*The duration of a symbol increases significantly for a multicarrier system compared to a single-carrier system,  reducing the spurious influence of the channel impulse response and decreasing intersymbol interference.

*The ability to use different robust modulation schemes for different subbands is one of the  '''major advantages of OFDM'''.  This will be discussed in more detail in the following sections  [[Modulation_Methods/OFDM_für_4G–Netze|OFDM for 4G networks]]  and  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_DSL|Digital Subscriber Line]]  $\rm (DSL)$.  }}

==System consideration in the frequency domain with acausal basic pulse==
 
We consider again the OFDM transmitted signal in the  $k$–th time interval with setting  $T_{\rm R} = T$:
:$$s_k (t) = \sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot g_\mu (t - k \cdot T)}.$$
[[File:EN_Mod_T_5_5_S2_a.png|right|frame| Spectrum of a non-causal basic pulse ]]
For simplicity,  we assume the basic pulse  $g_{\mu}(t)$  to be symmetric at  $t = 0$.   Then it is valid with  $f_0 = 1/T$:
:$$g_\mu (t) = \left\{ \begin{array}{l} s_0 \cdot {\rm{e}}^{ {\kern 1pt} {\rm{j2 \pi}} {\kern 1pt} \mu f_0 t} \quad \quad - T/2 < t < T/2, \\ 0 \quad \quad \quad \quad \quad \quad \; {\rm otherwise.} \\ \end{array} \right.$$
In the spectral domain,  such an acausal rectangular function modulated by a  (complex)  exponential function of frequency  $\mu · f_0$  corresponds to an  $\rm sinc$ function shifted by $\mu · f_0$:
:$$G_\mu (f) = s_0 \cdot T \cdot {\rm{sinc}} \big((f - \mu f_0 )\cdot T \big ).$$
On the right,  this spectral function  $($normalized to the maximum  $s_0 · T)$  is shown for  $\mu = 5$.   –   The arrow is to suggest that if the basic pulse is unconstrained in time,  the  $\rm sinc$ function shown would have to be replaced by a Dirac delta function at  $\mu · f_0$.

{{BlaueBox|TEXT=
$\text{Conclusions:}$   If all amplitude coefficients  $a_{μ,\hspace{0.08cm}k} ≠ 0$,  the spectrum  $S_k(f)$  of the transmitted signal in the  $k$–th time domain interval
*is composed of  $N$   $\rm sinc$  functions   ⇒   ${\rm sinc}(x) = \sin(\pi x)/(\pi x),$
*each shifted by a multiple of the basic frequency  $f_0$.}}

==System consideration in the frequency domain with causal basic pulse==
 
{{BlaueBox|TEXT=
$\text{Important result:}$  

*If one also takes into account that in reality a causal basic pulse
:$$g_\mu (t) = \left\{ \begin{array}{l} s_0 \cdot {\rm{e} }^{ {\kern 1pt} {\rm{j{\kern 1pt}\cdot {\kern 1pt}2 \pi} } {\kern 1pt}\cdot {\kern 1pt} \mu f_0 {\kern 1pt}\cdot {\kern 1pt}t} \quad 0 \le t < T, \\ 0\quad \quad \quad \quad \quad \quad {\rm otherwise}, \\ \end{array} \right.$$
:has to be assumed,  the spectrum is given by
:$$S_k (f) = s_0 \cdot T \cdot \sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot \,} {\rm{sinc} }\big( (f - \mu \cdot f_0 )\cdot T\big) \cdot {\rm{e} }^{ - {\rm{j{\kern 1pt}\cdot {\kern 1pt}2\pi} }\hspace{0.05cm}\cdot \hspace{0.05cm} {T}/{2} \hspace{0.05cm}\cdot \hspace{0.05cm} (f - \mu \hspace{0.05cm}\cdot \hspace{0.05cm}f_0 )} .$$
*The complex exponential function comes from the limits of the rectangle used here for pulse shaping in the time domain  $0$ ... $T$   $($shift by  $T/2)$.
*The purely real  $\rm sinc$  function shown before,  on the other hand,  would correspond to the non-causal rectangle of  $ -T/2$ ... $+T/2$.}}

The diagram shows an example of the magnitude spectrum of an OFDM signal with five carriers.
[[File:ENneu_Mod_T_5_5_S4.png|right|frame| Spectrum of an OFDM signal]]

*It is remarkable that the maximum of each subcarrier coincides with the zeros of all other carriers.  This corresponds to the first Nyquist condition in the frequency domain.
*This property allows  "ICI"–free sampling  (that is:   without intercarrier interference)  of the spectrum at multiples of  $f_0$.  Orthogonality is therefore guaranteed.
*If one were to dispense with the time limit in pulse shaping, the displayed  $\rm sinc$  functions would each become Dirac lines at the distance  $f_0$  (drawn in gray in the diagram).
*This idealizing simplification is unfortunately not realizable in practice.  Indeed,  the requirement  $T → ∞$  means at the same time that only one frame could be transmitted in an infinitely long time.
 

{{BlaueBox|TEXT=
$\text{Conclusion:}$  An OFDM signal under the condition of a rectangular pulse shaping and a subcarrier spacing of  $f_0$ 
*fulfills the  [[Digital_Signal_Transmission/Eigenschaften_von_Nyquistsystemen#Erstes_Nyquistkriterium_im_Zeitbereich|first Nyquist condition in the time domain]]  and thus, of course,
*also the  [[Digital_Signal_Transmission/Eigenschaften_von_Nyquistsystemen#Erstes_Nyquistkriterium_im_Frequenzbereich |first Nyquist condition in the frequency domain]].}}

==Orthogonality properties of the carriers==
 
The time limit of the basic pulse allows the separate consideration of the two sums in the equation of the OFDM transmitted signal:
:$$s(t) = \sum\limits_{k = - \infty }^{+\infty} {s_k (t)} \quad {\rm{with}} \quad s_k (t) = \sum\limits_{\mu = 0}^{N - 1} {a_{\mu ,\hspace{0.08cm}k} \cdot g_\mu (t - k \cdot T )}.$$
Here,  the  $k$–th transmitted pulse
*is the sum of the basic pulses  $g_{\mu}(t)$ shifted by $k · T$, 
*each of which is weighted by the  $\mu$–th amplitude coefficients of the QAM coder at time  $k$. 

{{BlaueBox|TEXT=
$\text{Another important result:}$ 

This gives for the spectrum  $S_{\mu,\hspace{0.08cm}k}(f)$  of the  $\mu$–th carrier in the  $k$–th interval:
:$$S_{\mu ,\hspace{0.08cm}k} (f) = s_0 \cdot a_{\mu ,\hspace{0.08cm}k} \cdot T \cdot {\rm{sinc} }\big( (f - \mu \cdot f_0 )\cdot T\big) \cdot {\rm{e} }^{ - {\rm{j \hspace{0.05cm} \cdot \hspace{0.05cm} \pi} } \hspace{0.05cm} \cdot \hspace{0.05cm}T \hspace{0.05cm} \cdot \hspace{0.05cm} (f - \mu \hspace{0.05cm} \cdot \hspace{0.05cm} f_0)}.$$
The following properties, which are important for the OFDM principle, apply:
*The  $s_k(t)$  are orthogonal to each other in time $($control variable  $k)$,  since they do not overlap in time due to the time limitation of the pulse shape filter  $g_s(t)$.
*Although the time limitation of the pulses leads to a spectral overlap,  there is nevertheless also orthogonality with respect to the carriers $($control variable $\mu)$,  since:
:$$S_k (\mu \cdot f_0 ) = S_{\mu ,\hspace{0.08cm}k} (\mu \cdot f_0 ) = s_0 \cdot a_{\mu ,\hspace{0.08cm}k} \cdot T.$$}}

{{BlaueBox|TEXT=
$\text{Proof:}$ 
For orthogonality at the frequency support points  $\mu · f_0$  must hold:
:$$S(\mu \cdot f_0 ) = S_0 (\mu \cdot f_0 ) + \ \text{...} \ + S_\mu (\mu \cdot f_0 ) + \ \text{...} \ + S_{N - 1} (\mu \cdot f_0 ) = S_\mu (\mu \cdot f_0 ).$$
*Here and in the following,  we omit the index  $k$  of the frame number. From
:$$s_\mu (t) = s_0 \cdot a_\mu \cdot {\rm{e} }^{{\rm j \hspace{0.03cm}\cdot\hspace{0.03cm}2\pi } \hspace{0.03cm}\cdot \hspace{0.03cm} \mu \hspace{0.03cm}\cdot \hspace{0.03cm} f_0 \hspace{0.03cm}\cdot \hspace{0.03cm} t} \cdot {\rm{rect} } \left( {\frac{ {t - T/2} }{T} } \right) \hspace{0.45cm} {\rm{and }} \hspace{0.45cm} S_\mu (f) = \int_{ - \infty }^{+\infty} {s_\mu (t) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } \hspace{0.03cm}\cdot \hspace{0.03cm} f \hspace{0.03cm}\cdot
\hspace{0.03cm} t} \hspace{0.06cm}{\rm d}t}$$

:the spectrum  $S(f)$  results in general to:
:$$S(f) = \left( {s_0 \cdot a_0 \cdot T \cdot {\rm{sinc} }(f T ) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}/{2}\hspace{0.03cm}\cdot \hspace{0.03cm} f} } \right) * \int_{ - \infty }^{+\infty} { {\rm{e} }^{\rm{0} } \cdot {\rm{e} }^{ - {\rm{j\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} f \hspace{0.03cm}\cdot \hspace{0.03cm} t} \hspace{0.06cm}{\rm d}t} \hspace{0.08cm}+ \text{...} $$
:$$\hspace{0.5cm}\text{...} + \left( {s_0 \cdot a_\mu \cdot T \cdot {\rm{sinc} } ( f T ) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi }\hspace{0.03cm}\cdot \hspace{0.03cm}{T}/{2}\hspace{0.03cm}\cdot \hspace{0.03cm} f} } \right) * \int_{ - \infty }^{+\infty} { {\rm{e} }^{ {\rm{j\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } } \hspace{0.03cm}\cdot \hspace{0.03cm}\mu \hspace{0.03cm}\cdot \hspace{0.03cm} f_0 \hspace{0.03cm}\cdot \hspace{0.03cm} t} \cdot {\rm{e} }^{ - {\rm{j\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} f \hspace{0.03cm}\cdot
\hspace{0.03cm} t} \hspace{0.06cm}{\rm d}t} \hspace{0.08cm}+ \text{...} $$
:$$ \hspace{0.5cm}\text{...} +\left( {s_0 \cdot a_{N - 1} \cdot T \cdot {\rm{sinc} } ( f T ) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}/{2}\hspace{0.03cm}\cdot \hspace{0.03cm} f} }\right)
* \int_{ - \infty }^{+\infty} { {\rm{e} }^{ {\rm{j\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } } \hspace{0.03cm}\cdot \hspace{0.03cm}(N-1) \hspace{0.03cm}\cdot \hspace{0.03cm} f_0 \hspace{0.03cm}\cdot \hspace{0.03cm} t} \cdot {\rm{e} }^{ - {\rm{j\hspace{0.03cm}\cdot \hspace{0.03cm}2\pi } } \hspace{0.03cm}\cdot \hspace{0.03cm} f \hspace{0.03cm}\cdot \hspace{0.03cm} t} \hspace{0.06cm}{\rm d}t} .$$

*With distributions, this equation can be expressed as follows:
:$$S(f) = \left( {s_0 \cdot a_0 \cdot T \cdot {\rm{sinc} }(f T ) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}\hspace{0.03cm}\cdot \hspace{0.03cm} f} } \right) * \delta (f) \hspace{0.08cm}+ \text{...} $$
:$$\hspace{0.5cm} \text{...} + \left( {s_0 \cdot a_\mu \cdot T \cdot {\rm{sinc} } ( fT )\cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot
\hspace{0.03cm}\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}\hspace{0.03cm}\cdot
\hspace{0.03cm} f} } \right) * \delta (f - \mu \cdot f_0 )\hspace{0.08cm}+ \text{...} $$
:$$\hspace{0.5cm} \text{...}
+ \left( {s_0 \cdot a_{N - 1} \cdot T \cdot {\rm{sinc} } ( f T ) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}\hspace{0.03cm}\cdot
\hspace{0.03cm} f} }\right) * \delta (f-(N - 1) \cdot f_0 ) .$$
:$$\Rightarrow \hspace{0.3cm}S(f) = {s_0 \cdot a_0 \cdot T \cdot {\rm{sinc} }( T \cdot f) \cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}\hspace{0.03cm}\cdot \hspace{0.03cm} f} } \hspace{0.08cm}+\hspace{0.08cm} \text{...} $$
:$$\hspace{1.4cm}\text{...} + {s_0 \cdot a_\mu \cdot T \cdot {\rm{sinc} } ( T \cdot (f - \mu \cdot f_0 ))\cdot {\rm{e} }^{ - {\rm j \hspace{0.03cm}\cdot \hspace{0.03cm}\pi }\hspace{0.03cm}\cdot
\hspace{0.03cm} {T}\hspace{0.03cm}\cdot \hspace{0.03cm} (f - \mu \hspace{0.03cm}\cdot \hspace{0.03cm}f_0 )} } \hspace{0.08cm}+ \hspace{0.08cm}\text{...}$$
:$$ \hspace{1.4cm}\text{...} + s_0 \cdot a_{N - 1} \cdot T \cdot {\rm sinc } ( T \cdot \big [f-(N - 1) \cdot f_0 ) \big ] \cdot {\rm e}^{ - {\rm j \hspace{0.03cm}\cdot
\hspace{0.03cm}\pi }\hspace{0.03cm}\cdot \hspace{0.03cm} {T}\hspace{0.03cm}\hspace{0.03cm}\cdot \hspace{0.03cm} \big [f-(N - 1) \hspace{0.03cm}\cdot \hspace{0.03cm}f_0 \big ]}.$$
*Now setting  $f = \mu · f_0$,  we obtain:
:$$S (\mu \cdot f_0) = 0 \hspace{0.08cm}+ \hspace{0.08cm} \text{...} \hspace{0.08cm}+\hspace{0.08cm} s_0 \cdot a_\mu \cdot T \cdot {\rm{sinc} } (0) \cdot {\rm{e} }^0 \hspace{0.08cm}+\hspace{0.08cm} \text{...}+ 0 = s_0 \cdot a_\mu \cdot T = S_\mu ( \mu \cdot f_0 ).$$

*Thus,  the spectrum at  $f = \mu · f_0$  is composed only of components of the  $\mu$–th carrier,  with all other carriers becoming identically zero.
*Orthogonality is guaranteed. <div align="right">'''q.e.d.'''</div> }}

{{BlaueBox|TEXT=
$\text{Conclusion:}$    The  '''orthogonality of the OFDM signal'''   $s(t)$ 
*is given for the control variable  $k$  $\rm (time)$ 
*as well as for the control variable  $\mu$  $\text{(carrier frequencies)}$ !}}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_5.6:_OFDM_Spectrum|Exercise 5.6: OFDM Spectrum]]

[[Aufgaben:Exercise_5.6Z:_Single_and_Multiple_Beam_System|Exercise 5.6Z: Single and Multiple Beam System]]

{{Display}}

Modulation Methods/Error Probability of Direct-Sequence Spread Spectrum Modulation

2022-04-09T14:00:40Z

Reed:

{{Header
|Untermenü=Multiple Access Methods
|Vorherige Seite=Spreading Sequences for CDMA
|Nächste Seite=General Description of OFDM
}}
==The IS-95 CDMA system==
 
The properties of PN modulation are now given using the example of the American mobile radio standard  [https://en.wikipedia.org/wiki/CdmaOne IS–95],  which has resulted from the work of  [https://en.wikipedia.org/wiki/Qualcomm Qualcomm Inc.]  and in particular of  [https://en.wikipedia.org/wiki/Andrew_Viterbi Andrew J. Viterbi].   In a somewhat simplified representation – without convolutional encoder,  interleaver and de-interleaver as well as the Viterbi decoder at the receiver - the following block diagram results.  The following statements are valid:

[[File:EN_Mod_T_5_4_S1.png |right|frame| Considered block diagram for this chapter]]

*The spreading signal  $c(t)$  causes  "band spreading"  by the spreading factor  $J$,  with [[Modulation_Methods/Spreading_Sequences_for_CDMA#Walsh_functions|Walsh functions]]  or  [[Modulation_Methods/Spreading_Sequences_for_CDMA#Pseudo-noise_sequences_of_maximum_length|M–Sequences]].   

*"Band compression"  at the receiver uses the same spreading sequence in a phase-synchronous manner.

*The additional  $±1$ signal  $w(t)$  provides additional  "scrambling",  but does not cause further band spreading. 

*The rectangular duration of  $w(t)$  is exactly as large  $(T_c)$  as the rectangular duration of  $c(t)$.   $T_c$  is called the  "chip duration".

*Without band spreading and scrambling  $($or with  $J = 1)$,  the transmission chain corresponds to  [[Modulation_Methods/Lineare_digitale_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|BPSK modulation]]. 

*The matched filter is implemented by the  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission#Optimal_binary_receiver_.E2.80.93_.22Integrate_.26_Dump.22_realization|"Integrate & Dump"]]  variant,  so it is an optimal system.

*With  $H_{\rm K}(f) = 1$,  the  [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|AWGN channel model]]  is obtained with the Gaussian distributed noise signal  $n(t)$  and the AWGN parameter  $E_{\rm B}/N_0$. 

*The additional interference component  $i(t)$  summarizes the  "interferences from other users".

*In the case of a  [[Examples_of_Communication_Systems/General_Description_of_UMTS#Properties_of_the_UMTS_radio_channel|multipath channel]]   (one main path and one or more secondary paths),  the resulting intersymbol interference can be reduced by using a [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Examinations_of_the_rake_receiver|"rake receiver"]]. 

==System configurations for minimum error probability==
 
[[File:EN_Mod_T_5_4_S2.png|right|frame| Bit error probability curves for the AWGN channel]]
The diagram shows the bit error probability at BPSK as a function of the logarithmized AWGN parameter  $E_{\rm B}/N_0$   (signal energy per bit related to the noise power density)   as a solid blue curve.  With the  [[Applets:Complementary_Gaussian_Error_Functions|complementary Gaussian error function]]  ${\rm Q}(x)$:
:$$p_{\rm B} = {\rm Q} \left ( \sqrt{{2\cdot E_{\rm B}}/{N_{\rm 0}}}\hspace{0.05cm}\right ) \hspace{0.05cm}.$$

For example,  with  $10 · \lg \ (E_{\rm B}/N_0) = 8 \ \rm dB$  the bit error probability is  $p_{\rm B} \approx \rm 2 · 10^{–4}$.  The same minimum value is provided by a system simulation  (red dots)  under the following conditions:
#  PN modulation,  whether with an  "M-sequence"  or with a  "Walsh function",  at any spreading factor  $J$,  if only one user is active.
#  Synchronous CDMA operation with Walsh functions,  even if other users  $($maximum  $J -1)$  are active in the same frequency band.
#  The rake receiver can be dispensed with,  since no intersymbol interference occurs with the AWGN channel.

==Two users with M-sequence spreading==
 
We consider the disturbing influence of a second  (clock-synchronous)  user on the error probability of user 1.

[[File:EN_Mod_T_5_4_S3a.png|right|frame| Bit error probability with band spreading  $(J = 15)$  with M-sequences]]

*Spreading is done using M-sequences which,  unlike Walsh functions,  are not orthogonal to each other.
*The spreading factor in each case is  $J = 15$.  Let the octal identifiers of the two PN spreading sequences involved be  $(23)$  and  $(31)$.

For example,  from this diagram can be read:

*With only one user,  it results in the blue solid curve. 
*The second user increases the error probability enormously:  At  $10 · \lg \ (E_{\rm B}/N_0) = 8 \ \rm dB$  from  $p_{\rm B} = 2· 10^{–4} = 0.02\%$  to  $p_{\rm B} =1.5\%$  $($brown markings,  $τ = 0)$.
*By phase-shifting the PN sequences with respect to each other by multiples of the chip duration,  one can achieve large improvements. 
*For example,  if the PN sequence  $(31)$  of the interfering user is shifted by  $τ = 2T_c$  to the right  (red markings),  instead of  $p_{\rm B} =1.5\%$  only  $p_{\rm B} = 0.034\%$  is obtained.
 
[[File:P_ID1896__Mod_T_5_4_S3b_ganz_neu.png |right|frame| PCCF of the PN sequences  $(23)$  and  $(31)$]]

 The results can be understood by considering the periodic cross-correlation function  $\rm (PCCF)$   $φ_\text{23, 31}(λ \cdot T_c)$  between the sequences  $(23)$  and  $(31)$. 
*The smaller the PCCF amount at  $\tau$,  the smaller  $p_{\rm B}$.
*Thus,  one could also shift the second PN sequence to the right by six or eight chip durations or to the left by five,  six,  or seven chip durations  (red dots).
*In all these cases,  the PCCF amount  $|φ_\text{23, 31}(2 T_c)| = 1/15$  is smaller compared to
:#$\ \ |φ_\text{23, 31}(0)| = 7/15$  (ochre dots),
:#$\ \ |φ_\text{23, 31}(3 T_c)| = 5/15$  (green dots), and
:#$\ \ |φ_\text{23, 31}(T_c)| = 3/15$  (purple dots).
 
==Asynchronous CDMA operation with Walsh functions==
 
On the page  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#System_configurations_for_minimum_error_probability|System configurations for minimum error probability]],  it was shown that when orthogonal Walsh functions are used,  the bit error probability  $p_{\rm B}$  is not changed compared with the BPSK comparison curve  (system without band spreading)  even in the presence of other CDMA subscribers,  as long as all subscribers operate synchronously.

[[File:EN_Mod_T_5_4_S4neu.png|right|frame| Bit error probability for asynchronous CDMA operation with Walsh spreading functions ]]

*In mobile communications, this condition is generally fulfilled for the downlink  (the transmitter is a single base station),
*but not in the uplink  (transmitters are many mobile terminals).

The diagram shows bit error probability curves for the most unfavorable phase offset of the spreading sequences of the considered user and the interfering user,  each with spreading factor  $J = 16$.  The first user always used Walsh function no.  $15$.

The result can be summarized as follows:
*If the second participant uses the Walsh function no.  $1$,  a phase shift does not have a negative effect,  since for all  $λ$–values the PCCF  $φ_\text{1, 15}(λ · T_c) = 0$. 
*On the other hand,  if the second participant uses Walsh function no.  $14$  $($or any other with identification number greater than/equal to  $8)$, a phase shift by one chip duration results in a huge degradation.
*Although  $φ_\text{14, 15}(\lambda=0) = 0$  holds,  for  $λ = 1$  this PCCF has a very large value with  $φ_\text{14, 15}(T_c) = 3/4$. 

==Bit error probability for the two-way channel==
 
For the rest of this chapter  "Error Probability of Direct-Sequence Spread Spectrum Modulation"  we assume:
*Walsh functions are used for band spreading.   The spreading factor in each case is  $J = 16$.  In particular, we consider the functions:
:$$ \langle w_\nu^{\hspace{0.12cm}(1)}\rangle =
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$ \langle w_\nu^{\hspace{0.12cm}(2)}\rangle =
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$ \langle w_\nu^{(12)}\rangle =
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{-\hspace{-0.05cm}1}\hspace{0.15cm}
{-\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.15cm}
{+\hspace{-0.05cm}1}\hspace{0.15cm}
{+\hspace{-0.05cm}1} \hspace{0.05cm}.$$

*Only one subscriber is active at a time   ⇒   interference from other users does not occur.
*The channel contains a multipath component in addition to the AWGN noise:   The main path is attenuated by a factor of  $0.8$. 
*In addition,  there is an echo at a distance of  $2T_c$  with a weight of  $0.6$.  The channel impulse response in this case is:
[[File:EN_Mod_T_5_4_S5a_neu.png |right|frame| CDMA error probability with two-way channel]]
:$$h_{\rm K}(t) = 0.8 \cdot \delta (t) + 0.6 \cdot \delta (t - 2 T_c).$$

The diagram shows bit error probability curves.  These show:
*The Walsh function no.  $2$  (brown dots)  is conceivably unsuitable for the two-way channel defined above,  since here the band spreading signal  $b(t)$  almost cancels itself out due to the echo at the distance  $2T_c$.   This can also be seen from the PACF value  $φ_{22}(τ = 2T_c) = -1$.
*The Walsh function no.  $1$  (yellow dots)  is very well suited for this channel.  The echo signal is constructively superimposed on the signal on the main path and  $b(t)$  is almost doubled.  The good result can be explained by the PACF value  $φ_{11}(τ = 2T_c) = +1$. 
*With Walsh function no.  $12$  (red dots),  the constructive and destructive superpositions almost balance out,  so that the error probability is approximately on the BPSK curve.  All other Walsh functions also lie between the brown and yellow boundary curves.

{{BlaueBox||TEXT=
$\text{Conclusion:}$ 
*The user with Walsh function no.  $1$  has significantly better conditions than the user with Walsh function no.  $2$  at the considered channel with echo delay  $2 \cdot T_c$. 
*For a channel with echo delay  $T_c$  the user with Walsh function no.  $2$  would have significantly better conditions than the user with Walsh function no.  $1$.
* However,  since a network operator must provide the same conditions for all users with any channel, this configuration is unsuitable for practical operation.
*In the following,  ungeeignet. it is shown how to provide approximately equal conditions for all subscribers.}}

==Influence of additional scrambling of the spreading sequence==
 
[[File:EN_Mod_T_5_4_S5b_neu.png|right|frame | CDMA error probability with two-way channel,  additional scrambling ]]
One way to equalize the quality for the individual subscribers even for the two-way channel is to add scrambling with  $w(t)$  according to the  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#The_IS-95_CDMA_system|block diagram]]  shown at the front.  For the diagram shown on the right it is assumed:
*The participating subscribers use different Walsh functions for band spreading,  all with spreading factor  $J = 16$.
*Each user is additionally scrambled by an M-sequence of period length  $P = 63$  $($register degree  $G = 6)$. 
*The plotted points apply to Walsh function no.  $12$  as spreading signal  $c(t)$  and scrambling  $w(t)$  by the M-sequence with octal identifier  $\rm (163)_{octal}$.

Compared with the pure AWGN channel,  the two-way channel considered with coefficients  $0.8$  and  $0.6$  gives a degradation of about  $2 \ \rm dB$ to $3 \ \rm dB$   ⇒   horizontal distance between the plotted points and the blue comparison curve.

'''Note:'''   The results for other Walsh functions, e.g. no.  $1$  or no.  $2$  differ only insignificantly from this diagram  $($valid for Walsh function no.  $12)$  within the character accuracy.
 
{{BlaueBox||TEXT=
$\text{Conclusion:}$ 
*The additional scrambling with  $w(t)$  serves only to create equal conditions for all participants in a channel with echo distortions.
*Thus,  $w(t)$  does not cause any additional band spreading.

All participants thus experience the same degradation due to an echo,  compared to the ideal channel.  Otherwise  (without this scrambling),  the customers with poor conditions would complain to the operator and possibly claim recourse,  while the other customers would accept the constructive echo overlay,  which is favorable to them,  perhaps delightedly but surely tacitly.}}

==Examinations of the rake receiver==
 
Finally,  let us consider the improvement by using a rake receiver,  which is illustrated by the following diagram.  For example,  for  $10 · \lg \ E_{\rm B}/N_0 = 8 \ \rm dB$   this reduces the bit error probability from  $\rm 6 · 10^{–3}$  to  $\rm 5 · 10^{-4}$.

[[File:EN_Mod_T_5_4_S6a_neu.png|right|frame |CDMA error probability curves with and without rake receiver]]

Note that the exact same assumptions apply to this diagram as to the  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Influence_of_additional_scrambling_of_the_spreading_sequence|diagram in the last section]]:
*Band spreading with  $J =16$  and Walsh function no.  $12$,
*additional scrambling by the M-sequence no.  $\rm (163)_{octal}$,
*two-way channel with   $h_{\rm K}(t) = 0.8 · δ(t) + 0.6 · δ(t - 2T_c)$,
*SNR:   $10 · \lg \ E_{\rm B}/N_0 = 8 \ \rm dB$.

The reason for this improvement is the smaller variance  ${σ_d}^2$  of the detection signal samples,  as can be seen from the probability density functions  $f_d(d)$  shown on the right.  Here,  ${σ_d}^2 = {σ_{\rm I}}^2 + {σ_{\rm N}}^2$  is composed of two components:
*The fraction  ${σ_{\rm N}}^2$  of the AWGN noise  $n(t)$  in the total variance  ${σ_d}^2$  depends solely on the abscissa  $(E_{\rm B}/N_0)$  and is the same with and without rake.
*The smaller  $σ_d^2$  with rake is solely due to the fact that the rake receiver combats the intersymbol interference.  Thus,  ${σ_{\rm I}}^2$  is not zero for the (red) receiver with rake, but it is significantly smaller than for the (black) receiver without rake.

==Principle of the rake receiver==
 
[[File:EN_Mod_T_5_4_S6b.png |right|frame| Two-way channel and associated rake receiver]]
The rake principle is illustrated by the diagram.  The two-way channel consists of

*the direct path with delay time  $τ_0$  and weight  $h_0$,
*an echo with delay  $τ_1 > τ_0$  and weight  $h_1$.

Let both amplitude coefficients be real.  As a normalization condition is valid in the following:
:$${h_0}^2 + {h_1}^2 = 1.$$

The task of the rake receiver is to concentrate the signal energies of the two paths  (in general:   of all paths)  to a single point in time.  It therefore works like a  "rake"  for the garden.

If one applies a Dirac delta pulse at time  $t = 0$  to the channel input   ⇒   $s(t) = δ(t)$,  there are three Dirac delta pulses at the output of the rake receiver:
:$$b(t) = \big [ h_0 \cdot h_1\cdot \delta (t - 2 \tau_0)\big ] + \big [(h_0^2 + h_1^2) \cdot \delta (t - \tau_0- \tau_1) \big ]+
\big [h_0 \cdot h_1\cdot \delta (t - 2 \tau_1) \big ] \hspace{0.05cm}.$$
*The signal energy is concentrated in the output signal at time  $τ_0 + τ_1$,  and two of the total four paths contribute to this.
*The Dirac delta functions at  $2τ_0$  and  $2τ_1$  do cause intersymbol interference.
*However,  their weights  $(h_0 · h_1)$  are much smaller than the weight of the main path  $({h_0}^2 + {h_1}^2)$.

{{GraueBox|TEXT=
$\text{Example 1:}$ 
*With parameter values  $h_0 = 0.8$  and  $h_1 = 0.6$,  the main path  $($with weight  $h_0)$  contains only  $\rm 0.8^2/(0.8^2 + 0.6^2) = 64\%$  of the total signal energy.
*With rake receiver and the same weights the above equation is
:$$b(t) = \big [ 0.48 \cdot \delta (t - 2 \tau_0)\big ] + \big [1.0 \cdot \delta (t - \tau_0- \tau_1)\big ] +
\big [ 0.48 \cdot \delta (t - 2 \tau_1)\big ] \hspace{0.05cm}.$$
*The contribution of the main path to the total energy is now  $\rm 1^2/(1^2 + 0.48^2 + 0.48^2) ≈ 68\%$,  slightly larger than without rake. }}

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
*Rake receivers are preferred for implementation in mobile devices,  but have limited performance with many active subscribers.
*In a multipath channel with many  $(M)$  paths,  the rake also has  $M$  fingers.
*The main finger  (called the  "searcher")  is responsible for identifying and classifying the individual paths of multiple propagation in most mobile radio systems. }}

==Exercises for the chapter==
 
[[Aufgaben:Exercise_5.5:_Multi-User_Interference|Exercise 5.5: Multi-User Interference]]

[[Aufgaben:Exercise_5.5Z:_About_the_Rake_Receiver|Exercise 5.5Z: About the Rake Receiver]]

{{Display}}

Modulation Methods/Spreading Sequences for CDMA

2022-04-09T14:00:07Z

Reed:

{{Header
|Untermenü=Multiple Access Methods
|Vorherige Seite=Direct-Sequence Spread Spectrum Modulation
|Nächste Seite=Error Probability of Direct-Sequence Spread Spectrum Modulation
}}
==Properties of the correlation functions==
 
Important evaluation criteria for spreading sequences are the correlation functions.

{{BlaueBox|TEXT=
$\text{Definition:}$  Considering two ergodic processes with model functions  $x(t)$  and  $y(t)$,  the following applies to their  [[Theory_of_Stochastic_Signals/Kreuzkorrelationsfunktion_und_Kreuzleistungsdichte#Definition_der_Kreuzkorrelationsfunktion|'''cross-correlation function''']]  $\rm (CCF)$:
:$$\varphi_{xy}(\tau)=\overline{x(t)\cdot y(t+\tau)}=\lim_{T_{\rm M}\to\infty}\,\frac{1}{T_{\rm M} }\cdot\int^{T_{\rm M}/{\rm 2} }_{-T_{\rm M}/{\rm 2} }x(t)\cdot y(t+\tau)\,\,\rm d \it t.$$
The line crossing over indicates a  "time averaging."}}

$φ_{xy}(τ)$  is a quantitative measure of the linear statistical dependence of the instantaneous values of model functions  $x(t)$  and  $y(t + τ)$  of the two random processes, and thus serves to describe the statistical relationship between them.  It holds:
*If  $x(t)$  and  $y(t)$  are uncorrelated,  then  $φ_{xy}(τ) \equiv 0$  $($i.e.,  for all arbitrary values of  $τ)$.
*In general,  $φ_{xy}(τ)$  is not symmetric,  but the CCF maximum may well occur at  $τ_{\rm max} ≠ 0$. 
*Then the maximum correlation results from a mutual shift of the two considered signals by the time  $τ_{\rm max}$.

{{BlaueBox|TEXT=
$\text{Definition:}$ 
If we set  $y(t) = x(t)$ in the above equation,  we arrive at the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)|'''auto-correlation function''']]  $\rm (ACF)$
:$$ \varphi_{xx}(\tau)=\overline{x(t)\cdot x(t+\tau)}=\lim_{T_{\rm M}\to\infty}\,\frac{1}{T_{\rm M} }\cdot\int^{T_{\rm M}/{\rm 2} }_{-T_{\rm M}/{\rm 2} }x(t)\cdot x(t+\tau)\,\,\rm d \it t$$
with the following properties:
*The ACF is a measure of the internal statistical bindings of a stationary and ergodic process defined by the model function  $x(t)$. 
*If  $x(t)$  is real,  then  $φ_{xx}(τ)$  is a real even function:   $φ_{xx}(–τ) = φ_{xx}(τ)$.  Phase relations are thus lost in the ACF.
*If  $x(t)$  describes a complex process,  then the ACF is also complex.
*The maximum value of the ACF is at  $τ =0$.  It is always  $\vert φ_{xx}(τ)\vert ≤ φ_{xx}(τ = 0)$,  where  $φ_{xx}(τ =0)$  indicates the signal power  $P_x = {\rm E}\big[x^2(t)\big]$. 
*The DC component of  $x(t)$  can be determined from the limit value for  $(τ → ∞)$  as long as  $x(t)$  does not contain any periodic components:
:$$\overline{ x(t)} = {\rm E}\big[x(t)\big] = \sqrt{\lim_{\tau\to\infty}\,\varphi_{xx} (\tau)} \hspace{0.05cm}.$$}}

The auto-correlation function and cross-correlation function describe the internal bindings and the mutual statistical dependencies in the time domain.  The corresponding description functions in the frequency domain are
*the  [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|power-spectral density]]  ${\it Φ}_{xx}(f)$, as well as
*the  [[Theory_of_Stochastic_Signals/Kreuzkorrelationsfunktion_und_Kreuzleistungsdichte#Kreuzleistungsdichtespektrum|cross power-spectral density]]  ${\it Φ}_{xy}(f)$.

For ergodic processes,  these are obtained as the  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation|Fourier transforms]]  of ACF and CCF:
:$${\it \Phi}_{xx}(f) \hspace{0.2cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm}\varphi_{xx}(\tau)\hspace{0.05cm} ,\hspace{0.3cm} {\it \Phi}_{xy}(f) \hspace{0.2cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm}\varphi_{xy}(\tau)\hspace{0.05cm}.$$

'''Note:'''   In the following,  as in the book [[Theory of Stochastic Signals]],  we write simplifying for the ACF  $φ_x(τ)$  instead of  $φ_{xx}(τ)$  and for the PSD  ${\it Φ}_x(f)$  instead of  ${\it Φ}_{xx}(f)$.

==Periodic ACF and CCF==
 
For periodic signals,  the boundary transition can be omitted from the ACF and the CCF calculations,  and the period duration  $T_0$  (which must be the same for both signals) is used to obtain:
:$$\begin{align*}\varphi_{x}(\tau) & = \frac{1}{T_{\rm 0}}\cdot\int^{T_0}_{0}x(t)\cdot x(t+\tau)\,\,\rm d \it t\hspace{0.05cm} ,\\
\varphi_{xy}(\tau) & = \frac{1}{T_{\rm 0}}\cdot\int^{T_0}_{0}x(t)\cdot y(t+\tau)\,\,\rm d \it t\hspace{0.05cm}.\end{align*}$$
In this case,  the ACF is also a periodic function,  and it is called the  "periodic auto-correlation function"  $\rm (PACF)$.  This shows the following:
:$$\varphi_{x}(\pm T_0) = \varphi_{x}(\pm 2T_0) =\text{ ...} = \varphi_{x}(0)
\hspace{0.05cm}.$$
We now apply the above calculation rule to the spreading signal
:$$ c(t) = \sum\limits^{+\infty}_{\nu = -\infty}c_\nu\cdot g_c(t - \nu \cdot T_c)$$
assuming a rectangular pulse  $g_c(t)$  of width  $T_c$ ;  $T_c$  is called the  "chip duration".

Considering the periodicity  $(T_0 = P · T_c)$  of the amplitude coefficients  $c_ν ∈ \{±1\}$,  the discrete ACF values at multiples $($integer parameter $λ)$  of  $T_c$ are obtained:
:$$\varphi_{c}(\lambda \cdot T_c) = \frac{1}{P}\cdot\sum\limits^{P-1}_{\nu = 0} c_\nu \cdot c_{\nu+ {\it \lambda} }\hspace{0.05cm}.$$
*The maximum PACF value is obtained for  $λ = 0$  and for multiples of the period length  $P$.
*Due to the rectangular pulse  $g_c(t)$,  the PACF progression between two samples  $λ · T_c$  and  $(λ + 1) · T_c$  is always linear.
*Accordingly, the  "periodic cross-correlation function"  $\rm (PCCF)$  between two spreading sequences  $〈c_ν〉$  and  $〈c\hspace{0.04cm}'_ν〉$  of equal period length  $P$  is given as follows:
:$$\varphi_{cc\hspace{0.04cm}'}(\lambda \cdot T_c) = \frac{1}{P}\cdot\sum\limits^{P-1}_{\nu = 0} c_\nu \cdot c\hspace{0.04cm}'_{\nu+ \lambda }\hspace{0.05cm}.$$

==Evaluation criteria for PN spreading sequences==
 
The quality of a CDMA system based on PN modulation depends significantly on the PACF and PCCF properties of the spreading sequences used.  In summary:
*The PACF of the spreading code class used should be characterized by a pronounced peak at  ${\it λ} = 0$,  if possible,  in order to make synchronization at the receiver easy.  Moreover,  for multipath reception with an echo of delay difference  $λ · T_c$,  the smaller  $|φ_c(λ · T_c)|$  is,  the smaller is the degradation due to intersymbol interference.
*The disruptive influence of interfering CDMA subscribers can be estimated by the PCCF value  $φ_{cc\hspace{0.04cm}'} (λ = 0)$.  If this is equal to zero,  one speaks of   "orthogonal functions".  The error probability is not increased in this case.  If all spreading sequences are orthogonal,  then even if there are  $J$  participants,  the error probability is the same as if there were only one user.
*This last statement is of particular importance in synchronous systems with distortion-free channel  (for example,  in AWGN).   On the other hand,  in asynchronous operation or multipath reception,  "de-orthogonalization"  occurs and the stricter requirement that the PCCF between each sequence should take on the smallest possible values  (in terms of magnitude)  at all times.

When selecting code families for a CDMA system,  care must also be taken to ensure that as many code sequences as possible with favorable properties with respect to these three criteria can be found for a given spreading factor  $J = P$,  so that as many subscribers as possible can be supplied simultaneously in the same frequency band.

==Pseudo-noise sequences of maximum length==
 
A feedback shift register can be used to generate a sequence with favorable ACF properties if the feedback coefficients  $g_i$  with index  $i = 1, \text{...} \ , G-1$  are chosen appropriately.  
[[File:EN_Mod_T_5_3_S4.png|right|frame|PN generator  (realization with shift registers)]]
[[File:EN_Mod_T_5_3_S4b.png|right|frame| Polynomials of M-sequences with degree  $G$]]

*The sequence $〈c_ν〉$  is not random in the strict sense,  but periodic.  
*However,  due to the large period length  $P$  it appears stochastic to an uninitiated observer. 
*One speaks of a  '''pseudo–noise sequence'''  or  "PN sequence" for short.

 
$\text{PN generators}$  have the following properties,  see chapter  [[Theory_of_Stochastic_Signals/Erzeugung_von_diskreten_Zufallsgrößen|Generation of Discrete Random Variables]]  in the book "Theory of Stochastic Signals":
*The binary values  $c_{ν-1}, \text{...} \ , c_{ν-G}$  generated at earlier times are stored in the  $G$  memory cells of the shift register.  We refer to  $G$  as the  "degree of the shift register".

*The coefficients  $g_1, \ \text{...} \ , g_{G-1}$  are binary values,  where a  "1"  indicates a feedback at the corresponding location of the shift register and a  "0"  indicates no feedback.

*For the currently generated symbol,  with  $g_i ∈ \{0, 1\}$  and  $i = 1,\ \text{...}\ , G-1$:
:$$c_\nu = (g_1\cdot c_{\nu-1}+g_2\cdot c_{\nu-2}+\ \text{...}\ +g_i\cdot c_{\nu-i}+\ \text{...}\ +g_{G-1}\cdot c_{\nu-G+1}+ c_{\nu-G})\hspace{0.1cm} \rm mod \hspace{0.2cm}2.$$
*The modulo-2 addition can be realized for example by an  $\rm XOR$  operation:
:$$(x + y)\hspace{0.2cm} \rm mod\hspace{0.2cm}2 = {\it x}\hspace{0.2cm}\rm XOR\hspace{0.2cm} {\it y} = \left\{ \begin{array}{*{2}{c}} 0 & \rm if\hspace{0.1cm} {\it x}= {\it y},\\ 1 & \rm if\hspace{0.1cm} {\it x}\neq {\it y}. \\ \end{array} \right.$$
*If not all  $G$  memory cells are preallocated with zeros,  the result is a periodic random sequence  $〈c_ν〉$.  The period length  $P$  of this sequence depends strongly on the feedback coefficients  $g_i$. 

*For each degree  $G$  there are at least two configurations with the maximum period  $P_{\rm max} = 2^G – 1$  for this.

In the table on the right,  for some values of  $G$,  the generator polynomial  $G(D)$  of such an M-sequence and the corresponding  (maximum)  period length  $P$  are given,  where  "M"  stands for  "Maximum".  In the following the statements made here are clarified at the example  $G = 4$. 

{{GraueBox|TEXT=
$\text{Example 1:}$  The diagram shows two possible arrangements for generating a PN sequence of maximum length for  $G = 4$   ⇒   $P = 15$.
[[File:EN_Mod_T_5_3_S4c.png|right|frame| PN generators of degree  $G = 4$]]

[[File: P_ID1879__Mod_T_5_3_S4d_ganz_neu.png|right|frame|PACF of a PN sequence of maximum length  $P = 2^G – 1$]]

*The octal representation of the binary number  $(g_G, \ \text{...} \ ,g_2, g_1, g_0)$  is usually chosen as the abbreviation.  Basically  $g_0 = g_G = 1$  is to be set.

*For the left generator:  $\rm (11001)_{binary} = (31)_{octal}$.  The corresponding generator polynomial is:
:$$G_1(D) = D^4 + D^3 +1\hspace{0.05cm}.$$

*For the right generator:  $\rm (10011)_{binary} = (23)_{octal}$.  This generator can be described by the polynomial
:$$G_2(D) =D^4 + D +1 \hspace{0.05cm}.$$
*The polynomial  $G_2(D)$  is reciprocal to  $G_{\rm 1}(D)$:
:$$G_{\rm 2}(D) = D^4 \cdot (D^{-4} + D^{-3} +1) =D^4 + D +1 \hspace{0.05cm}.$$

Since both  $G_1(D)$  and  $G_{\rm 2}(D)$  are  [[Channel_Coding/Erweiterungskörper#Bin.C3.A4re_Erweiterungsk.C3.B6rper_.E2.80.93_Primitive_Polynome|primitive generator polynomials]]    (the proof for this is not easy),  both output sequences have the maximum period length  $P = 15$  for  $G = 4$.

As shown in  [[Aufgaben:Exercise_5.3:_PACF_of_PN_Sequences|Exercise 5.3]],  the PACF in unipolar representation   ⇒   $c_ν ∈ \{0, 1\}$  results in
:$${\it \varphi}_{c{\rm ,\hspace{0.15cm}unipolar} }(\lambda \cdot T_c) = \left\{ \begin{array}{c}(P+1)/(2P) \\ (P+1)/(4P) \\ \end{array} \right. \begin{array}{*{10}c} \ \ {\rm{for} }\ \lambda = 0, \pm P, \pm 2P, \text{...} \hspace{0.05cm} \\ {\rm{otherwise} } \hspace{0.05cm}. \\ \end{array}$$

After conversion to bipolar coefficients  $(0 \ ⇒ +1$,     $1 \ ⇒ -1)$,  we obtain:
:$${\it \varphi}_{c{\rm ,\hspace{0.15cm}bipolar} }(\lambda \cdot T_c) = \left\{ \begin{array}{c}1 \\ -P \\ \end{array} \right. \begin{array}{*{10}c} \ \ {\rm{for} }\ \lambda = 0, \pm P, \pm 2P, \text{...} \hspace{0.05cm} \\ {\rm{otherwise} } \hspace{0.05cm}. \\ \end{array}$$
*One can see from the bottom diagram the desired distinct PACF peaks at intervals of periode  $P$.
*The PCCF properties of PN sequences are less good,  as will be shown in chapter  [[Modulation_Methods/Fehlerwahrscheinlichkeit_der_PN–Modulation|Error Probability of Direct-Sequence Spread Spectrum Modulation]]. }}

==Code families with M-sequences==
 
In CDMA,  a specific spreading sequence of the same period length is required for each subscriber.  A  '''code family'''  is defined as a set  (as large as possible)  of spreading sequences of the same period length  $P$,  each valid for a register level  $G$.

The table shows that the number of PN sequences of maximum length is very small. For the degree  $G = 5$   ⇒   $P = 31$,  for example,  there are just six M-sequences,  namely the PN generators with the octal identifiers  $(45), (51), (57), (67), (73)$  and  $(75)$.

[[File:EN_Mod_T_5_3_S5neu.png|right|frame| M-sequence code family – powerfullness]]

Furthermore, the term  '''powerfulness'''  of the code family can also be found in the literature.
*This quantity indicates how many M-sequences and thus simultaneous CDMA subscribers are possible if it is required that all code pairs have  "favorable PCCF properties".
*For reasons of space,  it cannot be explicitly stated here what exactly is meant by  "favorable".  For this,  we refer to the original literature,  for example [ZP85]<ref>Ziemer, R.; Peterson, R. L.:  Digital Communication and Spread Spectrum Systems. New York: McMillon, 1985.</ref>.

The last row in the above table makes it clear that the powerfulness of M-sequence code families is extremely limited,  even if  $G$  is large and thus the period length  $P$ is large, too.  If  $G$  is a multiple of  $4$   ⇒   $P = 15$,  $P = 255$,  $P = 4095$ etc.,  there are essentially no favorable pairs.

==Gold Codes==
 
To obtain larger code families than with M-sequences,  the requirements for the periodic cross-correlation function  $\rm (PCCF)$  must be weakened.  With this restriction,  code families with a much greater powerfulness are obtained,  so that more CDMA subscribers can also be supplied.

Gold codes use this principle.  The rule for generating a  '''Gold code family'''   $\rm (GCF)$  is where a  „+”  denotes a modulo-2 addition:
:$${\rm GCF}(C_1,\ C_2) = \{ C_1,\ C_2,\ C_1 + C_2,\ C_1 + D \cdot C_2,\ C_1 + D^2 \cdot C_2, \ \text{...} \ ,\
C_1 + D^{P-1} \cdot C_2 \} \hspace{0.05cm}.$$
[[File:EN_Mod_T_5_3_S6.png|right|frame|Gold code generator  $(51, 75)$  of degree  $G = 5$]]

The principle is illustrated in the diagram by an example:
*The sequences  $C_1$  and  $C_2$  are a convenient pair of  "M-sequences"  of the same period length,  for example,  the PN generators with the octal identifiers  $(51)$  and  $(75)$  of degree  $G = 5$  and thus the period length  $P = 31$.
*The Gold code family consists,  on the one hand,  of the  "M-sequences"  $C_1$  and  $C_2$  and of some modulo-2 additions of these sequences.   $C_1$  is used with a fixed phase  (characterized by the preallocation of all memory cells with ones),  while all  $P$  possible initial phases are permissible for the sequence  $C_2$  (all preallocations except the zero allocation).
*If such a code family is used for CDMA,  a total of  $K_{\rm GCF} = P + 2 = 2^G + 1$  sequences are available.  However,  the PACF of these sequences is now no longer bivalent like the two PN sequences   $(+1$  and  $-1/31)$, but quadrivalent  $(+1$,  $+7/31$,  $-1/31$,  $-9/31)$.  The phase of  $C_2$  changes the specific progression,  but not the possible PACF values.

Gold codes are used for scrambling in UMTS,  for example,  as explained in the chapter  [[Examples_of_Communication_Systems/Nachrichtentechnische_Aspekte_von_UMTS#Spreizcodes_und_Verw.C3.BCrfelung_bei_UMTS|Spreading codes and scrambling in UMTS]]  in the book "Examples of Communication Systems".
*The two master code shift registers are each constructed with  $G = 18$  memory cells.
*This results in the period length  $P = 262\hspace{0.05cm} 143$.

==Walsh functions==
 
Spreading sequences with very favorable PCCF properties are the so-called  '''Walsh functions''',  whose construction is based on the  '''Hadamard matrix'''  and can be performed in a simple way by recursion.  Starting from the matrix  $\mathbf H_2$,  further Hadamard matrices  $\mathbf H_{2J}$  can be generated as follows:

[[File:P_ID1882__Mod_T_5_3_S7_neu.png|right|frame| Walsh spreading sequences  $(J = 8)$  and Hadamard matrix  $\mathbf H_8$ ]]

:$${\mathbf{H}_{2}} = \left[ \begin{array}{ccc} +1 & +1 \\ +1 & -1 \end{array} \right] \hspace{0.5cm} \Rightarrow \hspace{0.5cm}{\mathbf{H}_{2J}} = \left[ \begin{array}{ccc} \mathbf{H}_J & \mathbf{H}_J \\ \mathbf{H}_J & -\mathbf{H}_J \end{array} \right] $$
:$$\hspace{0.5cm} \Rightarrow \hspace{0.5cm}
{\mathbf{H}_{4}} = \left[ \begin{array}{cccc} +1 & +1 & +1 & +1 \\ +1 & -1 & +1 & -1 \\
+1 & +1 & -1 & -1 \\+1 & -1 & -1 & +1 \end{array} \right] .$$
The  $J$  rows of such a matrix describe the  $J$  possible spreading sequences  $($each of length  $J)$,  which are numbered  $w_0(t)$  to  $w_{J–1}(t)$. 

The diagram shows the Hadamard matrix  $\mathbf H_8$  (right) and the $J -1$  spreading sequences that can be constructed with it.
*$J - 1$ because the sequence  $w_0(t)$  without spreading effect is usually not used.
*Please note the color allocation between the rows of the Hadamard matrix and the spreading sequence  $w_j(t)$  in the graphic. The matrix  $\mathbf H_4$  is highlighted in yellow.
*The HTML5/JavaScript applet  [[Applets:Generation_of_Walsh_functions|Generation of Walsh functions]]  shows the construction algorithm of such sequences.

Further applies:
# If one takes any two lines and forms the correlation  (averaging over the products),  the PCCF value always results in zero.  Thus,  Walsh functions for a distortion-free channel and a synchronous CDMA system are optimal spreading sequences due to their orthogonality.
# In contrast,  for asynchronous operation  (example: uplink of a mobile radio system)  or de-orthogonalization due to multipath propagation,  Walsh functions alone are not necessarily suitable for band spreading - see  [[Aufgaben:Exercise_5.4:_Walsh_Functions_(PCCF,_PACF)|Exercise 5.4]].
# Regarding PACF  (periodic ACF),  these consequences are less good:   Each single Walsh function has a different PACF and each single PACF is less favorable than for a comparable PN sequence.  This means:   Synchronization is more difficult with Walsh functions than with PN sequences.

==Codes with variable spreading factor (OVSF codes)==
 
The 3G mobile communications system [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_UMTS|UMTS]]  provides various data rates.  For this purpose
*spreading sequences with different spreading factors  $J = 4$  to  $J = 512$  are required,
*which must all be orthogonal to each other.

[[File:P_ID1883__Mod_T_5_3_S8_neu.png|right|frame | Tree diagram for generating an OVSF code]]
 The so-called  '''OVSF codes'''  ("Orthogonal Variable Spreading Factor")  can be created with the help of a code tree.  Thereby two new codes  $(+C \ +\hspace{-0.05cm}C)$  and  $(+C \ -\hspace{-0.05cm}C)$  are created at each branching from a code  $C$,  as indicated in the graphic above right.

It should be noted that no predecessor and successor of a code may be used by other participants.  In this example,  the following selection could therefore be made  (among others):
*eight codes with the spreading factor  $J = 8$,  or
*the four highlighted codes – once with  $J = 2$, once with  $J = 4$  and twice with  $J = 8$.

In the second case, the other six  $J = 8$  codes cannot be used because they begin with  "$+1 \ +\hspace{-0.05cm}1$"  or with  "$+1 \ -\hspace{-0.05cm}1 \ +\hspace{-0.05cm}1 \ -\hspace{-0.05cm}1$". 

From the four spreading sequences in the graphic at the bottom right,  it can be seen that with a constant chip duration  $T_c$,  the user with spreading factor  $J = 2$  can transmit at a higher data rate than the users with  $J = 4$  or  $J = 8$ because his bit duration  $T_{\rm B}$  is smaller.

From the graph,  we can further see that the periodic cross-correlation function  $\rm (PCCF)$  is always zero at the point  $τ = 0$. 
*That means:   If one forms the product of any two of these sequences and integrates over the represented time range, the value  $0$  always results.
*This also means:   '''An OVSF code is orthogonal to all other OVSF codes of the same family as long as there are no shifts'''.

'''Note:'''   The  (German language)  SWF applet  [[Applets:OVSF-Codes_(Applet)|OVSF codes]]  shows the construction algorithm of these codes and the allowed selection of spreading sequences.

==Exercises for the chapter==
 

[[Aufgaben:Exercise_5.3:_PACF_of_PN_Sequences|Exercise 5.3: PACF of PN Sequences]]

[[Aufgaben:Exercise_5.3Z:_Realization_of_a_PN_Sequence|Exercise 5.3Z: Realization of a PN Sequence]]

[[Aufgaben:Exercise_5.4:_Walsh_Functions_(PCCF,_PACF)|Exercise 5.4: Walsh Functions (PCCF, PACF)]]

[[Aufgaben:Exercise_5.4Z:_OVSF_Codes|Exercise 5.4Z: OVSF Codes]]

==References==
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