View source for Aufgaben:Exercise 2.7: Is the Modulation Depth Too High?


{{quiz-Header|Buchseite=Modulation_Methods/Envelope_Demodulation
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[[File:P_ID1032__Mod_A_2_7.png|right|frame|Signal waveforms for&nbsp; "DSB&ndash;AM with carrier"]]
The cosine-shaped source signal &nbsp;$q(t)$&nbsp; with amplitude &nbsp;$A_{\rm N} = 5\ \rm  V$&nbsp; and frequency &nbsp;$f_{\rm N} = 1 \ \rm kHz$&nbsp; is &nbsp; $\rm DSB$&nbsp; amplitude modulated.&nbsp; Assuming an ideal channel,&nbsp; the received signal is given by:
:$$r(t) = s(t) =\left(q(t) + A_{\rm T}\right) \cdot \cos (2\pi \cdot f_{\rm T}\cdot t )\hspace{0.05cm}.$$ 
Thus,&nbsp; we are dealing with&nbsp; "DSB–AM with carrier”.

In the graph,&nbsp; besides the source signal &nbsp;$q(t)$,&nbsp; and the received signal &nbsp;$r(t)$ including its envelope &nbsp;$a(t)$,&nbsp; we can also see the sink signal &nbsp;$v(t)$&nbsp; and the error signal
:$$ \varepsilon(t) = v(t) - q(t).$$
The sink signal drawn in red,
:$$v_{\rm A}(t) = a(t) - A_{\rm T}$$
belongs to an envelope demodulator,&nbsp; where exactly the carrier added at the transmitter &nbsp;$(A_{\rm T})$&nbsp; is subtracted from the envelope &nbsp;$a(t)$.&nbsp; 

This signal &nbsp;$v_{\rm A}(t)$&nbsp; contains a DC component just like the corresponding error signal &nbsp;$ε_{\rm A}(t)$&nbsp;.&nbsp; Due to the periodicity,&nbsp; it can be approximated by the following Fourier series:
:$$v_{\rm A}(t) = A_0 + \sum_{n=1}^{6} A_i \cdot \cos (n \cdot \omega_{\rm N}\cdot t ), \hspace{0.3cm}{\rm with}$$
::$$A_0  = 0.272\,{\rm V},\hspace{0.3cm}A_1 = 4.480\,{\rm V},\hspace{0.3cm}A_2 = 0.458\,{\rm V},\hspace{0.3cm}A_3 = -0.367\,{\rm V},\hspace{0.3cm}$$
::$$A_4  =  0.260\,{\rm V},\hspace{0.3cm}A_5 = -0.155\,{\rm V},\hspace{0.3cm}A_6 = 0.066\,{\rm V}\hspace{0.05cm}.$$
If the DC component is instead removed using an ideal high-pass filter,&nbsp; this would result in the DC-free signals 
:$$ v_{\rm B}(t) = \sum_{n=1}^{6} A_i \cdot \cos (n \cdot \omega_{\rm N}\cdot t ),$$
:$$ \varepsilon_{\rm B}(t) = v_{\rm B}(t) - q(t) = a(t) - A_{\rm T} - A_0 \hspace{0.05cm}.$$






Hints: 
*This exercise belongs to the chapter [[Modulation_Methods/Envelope_Demodulation|Envelope Demodulation]].
*Particular reference is made to the chapter &nbsp;[[Modulation_Methods/Quality_Criteria|Quality Criteria]]&nbsp; in this book,&nbsp; as well as the chapter &nbsp;[[Linear_and_Time_Invariant_Systems/Nonlinear_Distortions|Nonlinear Distortions]]&nbsp; in the book&nbsp; "Linear and Time Invariant Systems". 
*To solve these problems,&nbsp; the following indefinite integrals are given:
:$$ \int { \cos (a x )}\hspace{0.1cm}{\rm d}x =\frac{1}{a} \cdot \sin (a x ), \hspace{0.5cm} \int { \cos^2 (a x )}\hspace{0.1cm}{\rm d}x = \frac{x}{2} +\frac{1}{4a} \cdot \sin (2a x ).$$
*The distortion factors are calculated according to the formulas:
:$$K_2 = {A_2}/{A_1}, \hspace{0.3cm} K_3 = {A_3}/{A_1}, \hspace{0.1cm} \text{...} \hspace{0.5cm} \Rightarrow \hspace{0.5cm}K = \sqrt{K_2^2 +K_3^2 + \text{...}}\hspace{0.1cm} .$$


===Questions===



<quiz display=simple>

{What is the modulation depth &nbsp;$m$&nbsp; of the DSB–AM?
|type="{}"} 
$m \ = \ $ { 1.25 3%  }

{At which times&nbsp;$t_1$&nbsp; and &nbsp;$t_2$&nbsp; (see graph) does the envelope first &nbsp;$a(t)$&nbsp;become  zero?
|type="{}"}
$t_1 \ = \ $ { 0.4 3% } $\ \rm ms$ 
$t_2 \ = \ $ { 0.6 3% } $\ \rm ms$ 

{Calculate the distortion factors &nbsp;$K_2$, ... , &nbsp;$K_6$&nbsp; as well as the total distortion factor &nbsp;$K$.
|type="{}"}
$K \ = \ $ { 14.8 3% } $\text{%}$ 

{Calculate the power &nbsp;$P_{ε{\rm A}} = Ε\big [ε_{\rm A}^2(t)\big ]$&nbsp; for the red error signal &nbsp;$ε_{\rm A}(t)$.
|type="{}"}
$P_{ε{\rm A}} \ = \ $ { 0.432 3% } $\ \rm V^2$

{Calculate the power&nbsp;$P_{ε{\rm B}} = Ε\big [ε_{\rm B}^2(t)\big ]$&nbsp; for the green error signal &nbsp;$ε_{\rm B}(t)$.
|type="{}"}
$P_{ε{\rm B}} \ = \ $ { 0.358 3% }  $\ \rm V^2$

{Calculate the corresponding sink SNR &nbsp;$ρ_v = P_q/P_ε$&nbsp; for both demodulators.
|type="{}"}
$ρ_{v{\rm A}} \ = \ $ { 28.94 3% } 
$ρ_{v{\rm B}} \ = \ $ { 34.92 3% } 
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; From the graph we can see  that &nbsp; $A_{\rm T} = 4\ \rm  V$.&nbsp; With $A_{\rm N} = 5\ \rm  V$&nbsp; this results in the modulation depth
:$$m = A_{\rm N}/A_{\rm T}\hspace{0.15cm}\underline { = 1.25}.$$



'''(2)'''&nbsp; From the condition&nbsp; $a(t) = q(t) + A_{\rm T} = 0$&nbsp; it follows directly for the first zero crossing:
:$$ \cos (2\pi \cdot f_{\rm N}\cdot t_1 ) = \frac{-A_{\rm T}}{A_{\rm N}}= -0.8 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}t_1 = \frac{\rm arccos(-0.8)}{2\pi \cdot f_{\rm N}}\approx \frac{0.795 \cdot \pi}{2\pi \cdot f_{\rm N}} \hspace{0.05cm}.$$
*When&nbsp; $f_{\rm N} = 1 \ \rm  kHz$&nbsp; this gives &nbsp; $t_1\hspace{0.15cm}\underline {  ≈ 0.4 \ \rm  ms}$. 
*Accordingly,&nbsp; the second zero crossing is at &nbsp; $t_2\hspace{0.15cm}\underline {  ≈ 0.6 \ \rm  ms}$.




'''(3)'''&nbsp; The second order distortion factor is&nbsp; $K_2 = 0.458/4.48 ≈ 0.102$.&nbsp; 
*Correspondingly,&nbsp; the third order distortion factor is:&nbsp; $K_3 = 0.367/4.48 ≈ 0.082$. 
*The further distortion factors are &nbsp; $K_4 ≈ 0.058$,&nbsp; $K_5 ≈ 0.035$&nbsp; and&nbsp; $K_6 ≈ 0.015$. 
*Thus,&nbsp; for the total distortion factor we get:
:$$ K = \sqrt{K_2^2 + K_3^2 + K_4^2 + K_5^2 + K_6^2 }\hspace{0.15cm}\underline { \approx 14.8 \%}.$$



'''(4)'''&nbsp; The distortion power is obtained by averaging &nbsp; $ε_{\rm A}(t)^2$&nbsp; over the period &nbsp; $T_0 = 1\ \rm  ms$:
[[File:EN_Mod_A_2_7_d.png|right|frame|Calculating the distortion power]]
:$$P_{\varepsilon \rm A} = \frac{1}{T_{\rm 0}}\hspace{0.05cm} \cdot \int_{t_1}^{\hspace{0.1cm} t_2} {\varepsilon_{\rm A}^2(t) }\hspace{0.1cm}{\rm d}t = \frac{I_{\varepsilon}}{T_{\rm 0}}\hspace{0.05cm}.$$
*This takes into account that the error signal &nbsp; $ε_{\rm A}(t)$&nbsp; is equal to zero outside of the interval from&nbsp; $t_1$&nbsp; to&nbsp; $t_2$.&nbsp;

*As seen from the graph, &nbsp; $I_ε$&nbsp; is twice as large as the integral &nbsp; $I_γ$&nbsp; of the auxiliary quantity &nbsp; $γ$&nbsp; in the interval from zero to &nbsp; $t_3 = (t_2 – t_1)/2 ≈ 0.1 \ \rm ms$:
:$$I_{\gamma} = \int_{0}^{\hspace{0.1cm} t_3} {\gamma^2(t) }\hspace{0.1cm}{\rm d}t  \hspace{0.5cm}{\rm with}\hspace{0.5cm} \gamma(t) = 2 \cdot \left( A_{\rm N} \cdot \cos (\omega_{\rm N}\cdot t ) - A_{\rm T}\right)\hspace{0.05cm}.$$
*An additional calculation yields &nbsp; $I_{\gamma} = 4 \cdot \left( I_1 + I_2 + I_3 \right)$&nbsp; with
:$$I_1  =  A_{\rm N}^2 \cdot \int_{0}^{\hspace{0.1cm} t_3} {\cos^2 (\omega_{\rm N}\cdot t ) }\hspace{0.1cm}{\rm d}t = A_{\rm N}^2 \cdot \left[ \frac{t_3}{2} + \frac{\sin (2 \omega_{\rm N}\cdot t_3 )}{4 \omega_{\rm N}} \right] =  25\,{\rm V}^2 \cdot \left[ 0.05\,{\rm ms} + 0.0378\,{\rm ms} \right] = 2.196 \cdot 10^{-3}\,{\rm V^2 s} \hspace{0.05cm},$$
:$$ I_2  =  - 2 \cdot A_{\rm N}\cdot A_{\rm T} \cdot \int_{0}^{\hspace{0.1cm} t_3} \hspace{-0.2cm}{\cos (\omega_{\rm N}\cdot t ) }\hspace{0.1cm}{\rm d}t = - 2 \cdot A_{\rm N}\cdot A_{\rm T} \cdot \frac{\sin (\omega_{\rm N}\cdot t_3 )}{\omega_{\rm N}} =  - 2 \cdot 5\,{\rm V} \cdot 4\,{\rm V}\cdot 0.0935\,{\rm ms} = -3.742 \cdot 10^{-3}\,{\rm V^2 s} \hspace{0.05cm},$$
:$$ I_3 = A_{\rm T}^2 \cdot \int_{0}^{\hspace{0.1cm} t_3} {}\hspace{0.1cm}{\rm d}t = A_{\rm T}^2 \cdot {t_3} = 1.6 \cdot 10^{-3}\,{\rm V^2 s}$$
:$$\Rightarrow \hspace{0.3cm}I_{\gamma} = 0.216 \cdot 10^{-3}\,{\rm V^2 s} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}I_{\varepsilon} = 2 \cdot I_{\gamma} =0.432 \cdot 10^{-3}\,{\rm V^2 s} \hspace{0.05cm}.$$
*Thus,&nbsp; the final result is obtained as: &nbsp;
:$$P_{\varepsilon \rm A} = {I_{\varepsilon}}/{T_{\rm 0}}\hspace{0.15cm}\underline {=0.432 \,{\rm V^2 }}\hspace{0.05cm}.$$


'''(5)'''&nbsp; The two sink signals &nbsp; $v_{\rm A}(t)$&nbsp; and&nbsp; $v_{\rm B}(t)$&nbsp; differ by the DC component &nbsp; $A_0$,&nbsp; as do the two error signals &nbsp; $ε_{\rm A}(t)$&nbsp; and&nbsp; $ε_{\rm B}(t)$.&nbsp; Therefore:
:$$ P_{\varepsilon \rm B} = \frac{1}{T_{\rm 0}}\hspace{0.05cm} \cdot \int_{0}^{T_0} {\varepsilon_{\rm B}^2(t) }\hspace{0.1cm}{\rm d}t = \frac{1}{T_{\rm 0}}\hspace{0.05cm} \cdot \int_{0}^{T_0} {\left[\varepsilon_{\rm A}(t) - A_0 \right]^2}\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
*This can also be written as:
:$$P_{1}  =  \frac{1}{T_{\rm 0}}\hspace{0.05cm} \cdot \int_{0}^{T_0} {\varepsilon_{\rm A}^2(t) }\hspace{0.1cm}{\rm d}t = P_{\varepsilon \rm A} \hspace{0.05cm},\hspace{0.3cm}
P_{2}  =  - 2 A_0 \cdot \frac{1}{T_{\rm 0}}\hspace{0.05cm} \cdot \int_{0}^{T_0} {\varepsilon_{\rm A}(t) }\hspace{0.1cm}{\rm d}t = - 2 A_0^2 \hspace{0.05cm},\hspace{0.3cm} 
P_{3}  =  A_0^2 \cdot \frac{1}{T_{\rm 0}}\hspace{0.05cm} \cdot \int_{0}^{T_0} { }\hspace{0.1cm}{\rm d}t = A_0^2 \hspace{0.05cm}.$$
*This gives the power of the error signal&nbsp; $ε_{\rm B}(t)$:
:$$P_{\varepsilon \rm B} = P_{\varepsilon \rm A}- A_0^2 = 0.432\,{\rm V}^2 - (0.272\,{\rm V})^2 \hspace{0.15cm}\underline {= 0.358\,{\rm V}^2} \hspace{0.05cm}.$$
*A similar result would have been obtained from the following calculation:
:$$ P_{\varepsilon \rm B} = \frac{1\,{\rm V}^2}{2} \cdot \left[ (5 - 4.48)^2 + 0.458^2 + 0.367^2 + ... + 0.066^2 \right] \approx 0.356\,{\rm V}^2 \hspace{0.05cm}.$$
*The small difference in both calculation processes is due to the fact that the Fourier coefficients &nbsp; $A_7$,&nbsp; $A_8$,&nbsp; ..., are very small,&nbsp; but not zero.



'''(6)'''&nbsp; The power of the source signal &nbsp; $q(t)$&nbsp; is &nbsp; $P_q = A_{\rm N}^2/2 = 12.5 V^2$.&nbsp; This gives the two S/N ratios:
:$$\rho_{v {\rm A}} = \frac{P_{q}}{P_{\varepsilon \rm A}} \hspace{0.15cm}\underline {= 28.94} \hspace{0.05cm}, \hspace{0.3cm} \rho_{v {\rm B}} = \frac{P_{q}}{P_{\varepsilon \rm B}}\hspace{0.15cm}\underline { = 34.92} \hspace{0.05cm}.$$
*This shows that the envelope demodulator &nbsp; $\rm B$&nbsp; (using a high-pass) is better than the demodulator &nbsp; $\rm A$&nbsp; (using subtraction)&nbsp; by about &nbsp; $6 \ \rm dB$.&nbsp;
*It should also be noted that the approximation &nbsp; $ρ_v = α_2 · P_q/K^2$&nbsp; would here lead to the distorted numerical value &nbsp; $ρ_v = 36.66$.&nbsp;
*This different result is justified in detail on the page&nbsp; "The distortion factor"&nbsp; in the book&nbsp; "Linear and Time Invariant systems",&nbsp; based exactly on the same assumptions made for this exercise.


{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^2.3 Envelope Demodulation^]]