Difference between revisions of "Aufgaben:Exercise 3.9: Circular Arc and Parabola"

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{{quiz-Header|Buchseite=Modulationsverfahren/Frequenzmodulation (FM)
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{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
 
}}
 
}}
  
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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:
+
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:
 
:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
 
:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
 
In this case, the result is a parabolic locus curve
 
In this case, the result is a parabolic locus curve
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''Hints:''
 
''Hints:''
*Die Aufgabe gehört zum  Kapitel  [[Modulation_Methods/Frequenzmodulation_(FM)|Frequenzmodulation]].
+
*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]].
*Bezug genommen wird aber auch auf das Kapitel   [[Modulation_Methods/Phasenmodulation_(PM)|Phasenmodulation]]  und insbesondere auf den Abschnitt  [[Modulation_Methods/Frequenzmodulation_(FM)#Signalverl.C3.A4ufe_bei_Frequenzmodulation|Signalverläufe bei Frequenzmodulation]].
+
*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:
*Gehen Sie bei der Berechnung von folgenden Werten der Besselfunktion aus:
 
 
:$${\rm J}_0 (2.4) \approx 0, \hspace{0.2cm}{\rm J}_1 (2.4) = -{\rm J}_{-1} (2.4)\approx 0.52, \hspace{0.2cm}{\rm J}_2 (2.4) = {\rm J}_{-2} (2.4)\approx 0.43.$$
 
:$${\rm J}_0 (2.4) \approx 0, \hspace{0.2cm}{\rm J}_1 (2.4) = -{\rm J}_{-1} (2.4)\approx 0.52, \hspace{0.2cm}{\rm J}_2 (2.4) = {\rm J}_{-2} (2.4)\approx 0.43.$$
  
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===Fragebogen===
+
===Question===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß ist die Modulatorkonstante &nbsp;$K_{\rm FM}$?
+
{What is the modulator constant &nbsp;$K_{\rm FM}$?
 
|type="{}"}
 
|type="{}"}
 
$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$
 
$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$
  
  
{Berechnen Sie den Realteil &nbsp;$x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$&nbsp; des äquivalenten Tiefpass-Signals und geben Sie dessen Maximum und Minimum an.
+
{Calculate the real part&nbsp;$x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$&nbsp; of the equivalent low-pass signal and give its maximum and minimum.
 
|type="{}"}
 
|type="{}"}
 
$x_{\rm max} \ = \ $ { 0.86 3% }
 
$x_{\rm max} \ = \ $ { 0.86 3% }
 
$x_{\rm min} \ = \ $ { -0.89--0.83 }  
 
$x_{\rm min} \ = \ $ { -0.89--0.83 }  
  
{Wie groß ist das Maximum und Minimum des Imaginärteils &nbsp;$y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?
+
{What is the maximum and minimum of the imaginary part &nbsp;$y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?
 
|type="{}"}
 
|type="{}"}
 
$y_{\rm max} \ = \ $ { 1.04 3% }  
 
$y_{\rm max} \ = \ $ { 1.04 3% }  
 
$y_{\rm min} \ = \ $ { -1.07--1.01 }  
 
$y_{\rm min} \ = \ $ { -1.07--1.01 }  
  
{Welche Phasenwerte ergeben sich bei allen Vielfachen von &nbsp;$T_{\rm N}/2$?
+
{What are the phase values for all multiples of &nbsp;$T_{\rm N}/2$?
 
|type="{}"}
 
|type="{}"}
$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm Grad$  
+
$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm degrees$  
  
{Wie groß ist der maximale Phasenwinkel &nbsp;$ϕ_{\rm max}$?&nbsp; Interpretieren Sie das Ergebnis.
+
{What is the maximum phase angle &nbsp;$ϕ_{\rm max}$?&nbsp; Interpret the result.
 
|type="{}"}
 
|type="{}"}
$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm Grad$  
+
$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm degrees$  
  
{Zeigen Sie, dass man die Ortskurve in der Form &nbsp;$y^2 + a · x + b = 0$&nbsp; angeben kann.&nbsp; Bestimmen Sie die Parabelparameter &nbsp;$a$&nbsp; und &nbsp;$b$.  
+
{Show that the locus curve can be given in the form &nbsp;$y^2 + a · x + b = 0$&nbsp;.&nbsp; Determine the parabolic parameters  &nbsp;$a$&nbsp; and &nbsp;$b$.  
 
|type="{}"}
 
|type="{}"}
 
$a\ = \ $ { 0.629 3%  }  
 
$a\ = \ $ { 0.629 3%  }  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Bei Frequenzmodulation eines Cosinussignals gilt für den Modulationsindex:
+
'''(1)'''&nbsp; In the frequency modulation of a cosine signal, the modulation index is:
 
:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} =  \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$
 
:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} =  \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$
  
  
  
'''(2)'''&nbsp; Die angegebene Gleichung für das äquivalente TP–Signal lautet in ausgeschriebener Form mit&nbsp; $γ = ω_{\rm N} · t$&nbsp; unter Berücksichtigung von&nbsp; ${\rm J}_{–1} = –{\rm J}_1$&nbsp; und&nbsp; ${\rm J}_{–2} = {\rm J}_2$:
+
'''(2)'''&nbsp; The equation given for the equivalent low-pass signal when&nbsp; $γ = ω_{\rm N} · t$&nbsp; and considering &nbsp; ${\rm J}_{–1} = –{\rm J}_1$&nbsp; and&nbsp; ${\rm J}_{–2} = {\rm J}_2$, is written out as:
 
:$$r_{\rm TP}(t)  =  {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 =  {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
 
:$$r_{\rm TP}(t)  =  {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 =  {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
*Somit ergibt sich für den Realteil allgemein bzw. für&nbsp; $η = 2.4$, das heißt&nbsp; ${\rm J}_0 = 0$&nbsp; und&nbsp; ${\rm J}_2 = 0.43$:
+
*Thus, for the real part in general, including for &nbsp; $η = 2.4$, that is,&nbsp; ${\rm J}_0 = 0$&nbsp; and&nbsp; ${\rm J}_2 = 0.43$:
 
:$$ 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}  
 
:$$ 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}.$$
 
\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}.$$
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'''(3)'''&nbsp; Entsprechend dem Ergebnis der Teilaufgabe&nbsp; '''(2)'''&nbsp; erhält man für den Imaginärteil&nbsp; $({\rm J}_1 = 0.52)$:
+
'''(3)'''&nbsp; According to the result of question&nbsp; '''(2)'''&nbsp; we get an imginary part &nbsp; $({\rm J}_1 = 0.52)$ of:
 
:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$
 
:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$
  
  
  
'''(4)'''&nbsp; Der Imaginärteil ist zu diesen Zeitpunkten jeweils Null und damit auch die Phasenfunktion: &nbsp;
+
'''(4)'''&nbsp; The imaginary part is zero at each of these time points and so is the phase function:
 
:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$   
 
:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$   
*Diesen Sachverhalt erkennt man auch aus der Skizze auf der Angabenseite.
+
*This fact can also be seen from the sketch on the exercise page.
  
  
  
  
'''(5)'''&nbsp; Aus der Skizze ist bereits zu erkennen, dass der Phasenwinkel beispielsweise für&nbsp; $t = T_{\rm N}/4$&nbsp; seinen Maximalwert erreicht.  
+
'''(5)'''&nbsp; From the sketch it can already be seen that the phase angle reaches its maximum value at&nbsp; $t = T_{\rm N}/4$&nbsp;, for example.  
*Dieser kann mit&nbsp; $y_{\rm max} = 1.04$&nbsp; und&nbsp; $x_{\rm min} = -0.86$&nbsp; wie folgt berechnet werden:
+
*This can be calculated with&nbsp; $y_{\rm max} = 1.04$&nbsp; and&nbsp; $x_{\rm min} = -0.86$&nbsp; as follows:
 
:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$
 
:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$
*Ohne Bandbegrenzung würde sich hier der Phasenwinkel&nbsp; $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$&nbsp; ergeben.  
+
*Without band-limiting, the phase angle here would be $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$&nbsp;.  
*Die maximale Abweichung des Sinkensignals vom Quellensignal tritt somit beispielsweise zur Zeit&nbsp; $t = T_{\rm N}/4$&nbsp; auf.
+
*Thus, the maximum deviation of the sink signal from the source signal occurs at time&nbsp; $t = T_{\rm N}/4$&nbsp;, for example.
  
  
  
[[File:P_ID1113__Mod_A_3_8_f.png|right|frame|Parabelverlauf]]
+
[[File:P_ID1113__Mod_A_3_8_f.png|right|frame|Parabolic progression]]
'''(6)'''&nbsp; Mit&nbsp; $γ = ω_{\rm N} · t$&nbsp; und&nbsp; $\cos(2γ) = 1 - 2 · \cos^2(γ)$&nbsp; kann für Real– und Imaginärteil geschrieben werden:
+
'''(6)'''&nbsp; Using&nbsp; $γ = ω_{\rm N} · t$&nbsp; and&nbsp; $\cos(2γ) = 1 - 2 · \cos^2(γ)$&nbsp; the real and imaginary parts can be written as:
  
 
:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$
 
:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$
  
*Diese Gleichungen können wie folgt umgeformt werden:
+
*These equations can be rearranged:
 
:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$
 
:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$
  
 
$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$
 
$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$
  
*Damit lauten die Parabelparameter für&nbsp; $ {\rm J}_0 = 0$:
+
*Thus, the parabolic parameters for&nbsp; $ {\rm J}_0 = 0$ are:
 
:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$
 
:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$
*Zur Kontrolle verwenden wir&nbsp; $y = 0$ &nbsp; &rArr; &nbsp; $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$
+
*To double-check, we use &nbsp; $y = 0$ &nbsp; &rArr; &nbsp; $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$
*Die Werte bei&nbsp; $x = 0$&nbsp; sind somit: &nbsp; &nbsp; $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$
+
*Thus, the values at&nbsp; $x = 0$&nbsp; are &nbsp; &nbsp; $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Latest revision as of 16:17, 9 April 2022

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:

$${\rm J}_0 (2.4) \approx 0, \hspace{0.2cm}{\rm J}_1 (2.4) = -{\rm J}_{-1} (2.4)\approx 0.52, \hspace{0.2cm}{\rm J}_2 (2.4) = {\rm J}_{-2} (2.4)\approx 0.43.$$



Question

1

What is the modulator constant  $K_{\rm FM}$?

$K_{\rm FM} \ = \ $

$\ \cdot 10^4 \ \rm (Vs)^{-1}$

2

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.

$x_{\rm max} \ = \ $

$x_{\rm min} \ = \ $

3

What is the maximum and minimum of the imaginary part  $y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?

$y_{\rm max} \ = \ $

$y_{\rm min} \ = \ $

4

What are the phase values for all multiples of  $T_{\rm N}/2$?

$ϕ(t = n · T_{\rm N}/2) \ = \ $

$\ \rm degrees$

5

What is the maximum phase angle  $ϕ_{\rm max}$?  Interpret the result.

$ϕ_{\rm max} \ = \ $

$\ \rm degrees$

6

Show that the locus curve can be given in the form  $y^2 + a · x + b = 0$ .  Determine the parabolic parameters  $a$  and  $b$.

$a\ = \ $

$b\ = \ $


Solution

(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.


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     $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$