Difference between revisions of "Aufgaben:Exercise 3.6: PM or FM? Or AM?"

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[[File:P_ID1102__Mod_A_3_6.png|right|frame|Zwei verschiedene Ortskurven für Winkelmodulation]]
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[[File:P_ID1102__Mod_A_3_6.png|right|frame|Two different locus curves for angle modulation]]
Zur Analyse eines Modulators wird an seinen Eingang das Signal
+
To analyze a modulator, apply the signal
 
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N})$$
 
:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N})$$
angelegt, wobei die Signalamplitude stets  $A_{\rm N}  = 2\ \rm  V$  beträgt.  
+
to its input, where the signal amplitude is always  $A_{\rm N}  = 2\ \rm  V$ .  
*Mit der Signalfrequenz  $f_{\rm N}  = f_1 = 5 \ \rm  kHz$  wird die Ortskurve  $\rm O_1$  ermittelt.  
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*Using the signal frequency  $f_{\rm N}  = f_1 = 5 \ \rm  kHz$  the locus  $\rm O_1$  is determined.  
*Verwendet man die Nachrichtenfrequenz  $f_{\rm N}  = f_2$, so stellt sich die Ortskurve  $\rm O_2$  ein.
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*If one uses the message frequency  $f_{\rm N}  = f_2$, then the  $\rm O_2$  locus is established.
  
  
Beachten Sie bei Ihrer Lösung, dass bei Winkelmodulation dies ist der Sammelbegriff für Phasen– und Frequenzmodulation der folgende Zusammenhang zwischen dem Modulationsindex  $η$  und der Modulatorkonstanten  $K_{\rm WM}$ besteht:
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In your solutions, note that for angle modulation the umbrella term for phase and frequency modulation the following relationship holds between the modulation index  $η$  and the modulator constant  $K_{\rm WM}$:
 
:$$\eta = \left\{ \begin{array}{c} K_{\rm WM} \cdot A_{\rm N} \\ {K_{\rm WM} \cdot A_{\rm N}}/({2 \pi \cdot f_{\rm N})} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{bei}} \\ {\rm{bei}} \\ \end{array}\begin{array}{*{20}c} {\rm PM} \hspace{0.05cm}, \\ {\rm FM}. \hspace{0.05cm} \\ \end{array}$$
 
:$$\eta = \left\{ \begin{array}{c} K_{\rm WM} \cdot A_{\rm N} \\ {K_{\rm WM} \cdot A_{\rm N}}/({2 \pi \cdot f_{\rm N})} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{bei}} \\ {\rm{bei}} \\ \end{array}\begin{array}{*{20}c} {\rm PM} \hspace{0.05cm}, \\ {\rm FM}. \hspace{0.05cm} \\ \end{array}$$
  
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''Hinweise:''
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*Die Aufgabe gehört zum  Kapitel  [[Modulationsverfahren/Frequenzmodulation_(FM)|Frequenzmodulation]].
+
 
*Bezug genommen wird aber auch auf das Kapitel   [[Modulationsverfahren/Phasenmodulation_(PM)|Phasenmodulation]].
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 +
''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]].
 
   
 
   
  
  
  
===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Um welchen Modulator handelt es sich?
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{Which modulator do we have here?
|type="[]"}
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|type="()"}
- AM–Modulator.
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- AM modulator.
- PM–Modulator.
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- PM modulator.
+ FM–Modulator.
+
+ FM modulator.
  
  
{Wie groß ist der Modulationsindex mit der Nachrichtenfrequenz &nbsp;$f_{\rm N}  = f_1 = 5 \ \rm  kHz$?
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{What is the modulation index when the message frequency is &nbsp;$f_{\rm N}  = f_1 = 5 \ \rm  kHz$?
 
|type="{}"}
 
|type="{}"}
 
$η_1 \ = \ $ { 1.3 3% }  
 
$η_1 \ = \ $ { 1.3 3% }  
  
{Welchen Wert besitzt die Modulatorkonstante? &nbsp; ''Hinweis:'' &nbsp; Die &bdquo;Einheit&rdquo; steht für $\rm V^{-1}$ (bei PM) oder $\rm (Vs)^{-1}$  (bei FM).
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{What is the value of the modulator constant? &nbsp; ''Hint:'' &nbsp; the "unit" stands for $\rm V^{-1}$ (for PM) or $\rm (Vs)^{-1}$  (for FM).
 
|type="{}"}
 
|type="{}"}
$K_{\rm WM} \ = \ $ { 2.04 3% } $\ \cdot 10^4 $ &bdquo;Einheit&rdquo;
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$K_{\rm WM} \ = \ $ { 2.04 3% } $\ \cdot 10^4 $ "unit"
  
{Welchen Winkel &nbsp;$ϕ_0$&nbsp; (gegenüber der reellen Achse) weist die Ortskurve &nbsp;$\rm O_1$&nbsp; mit &nbsp;$ϕ_{\rm N} = 30^\circ$&nbsp; zum Zeitpunkt &nbsp;$t = 0$&nbsp; auf?
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{What is the angle &nbsp;$ϕ_0$&nbsp; (with respect to the real axis) of the locus curve &nbsp;$\rm O_1$&nbsp; when &nbsp;$ϕ_{\rm N} = 30^\circ$&nbsp; at time &nbsp;$t = 0$&nbsp;?
 
|type="{}"}
 
|type="{}"}
$ϕ_0 \ = \ $ { 37.5 3% } $\ \rm Grad$  
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$ϕ_0 \ = \ $ { 37.5 3% } $\ \rm degrees$  
  
{Mit welcher Nachrichtenfrequenz &nbsp;$f_{\rm N} = f_2$&nbsp; wurde die Ortskurve &nbsp;$\rm O_2$&nbsp; ermittelt?
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{What message frequency &nbsp;$f_{\rm N} = f_2$&nbsp; was used to determine the locus &nbsp;$\rm O_2$&nbsp;?
 
|type="{}"}
 
|type="{}"}
 
$f_2 \ = \ ${ 3 3% } $\ \rm kHz$  
 
$f_2 \ = \ ${ 3 3% } $\ \rm kHz$  
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist die <u>Antwort 3</u>:
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'''(1)'''&nbsp; <u>Answer 3</u> is correct:
*Da die Ortskurve einen Kreisbogen beschreibt, handelt es sich um einen Winkelmodulator (PM oder FM) mit dem Modulationsindex $η$. *Da aber hier $η$ offensichtlich von der Nachrichtenfrequenz $f_{\rm N}$ abhängt, kann eine Phasenmodulation ausgeschlossen werden.  
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*Since the locus curve describes an arc, we have an angle modulator (PM or FM) with modulation index&nbsp; $η$.
 +
*But since&nbsp; $η$&nbsp; here clearly depends on the message frequency&nbsp; $f_{\rm N}$&nbsp;, we can rule out phase modulation.
 +
 
 +
 
  
 +
'''(2)'''&nbsp; The modulation index can be read off the graph. It is &nbsp; $η_1 = 75^\circ/180^\circ · π\hspace{0.15cm}\underline { ≈ 1.3}$.
  
'''(2)'''&nbsp; Der Modulationsindex kann aus der Grafik abgelesen werden. Es gilt $η_1 = 75°/180° · π\hspace{0.15cm}\underline { ≈ 1.3}$.
 
  
  
'''(3)'''&nbsp; Bei Frequenzmodulation gilt:
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'''(3)'''&nbsp; For frequency modulation, it holds that:
 
:$$ K_{\rm WM} = K_{\rm FM} = \frac{ 2 \pi \cdot f_{\rm N} \cdot \eta}{A_{\rm N}} = \frac{ 2 \pi \cdot 5 \cdot 10^3 \,\,{\rm Hz}\cdot 1.3}{2\,{\rm V}} \hspace{0.15cm}\underline {\approx 2.04 \cdot 10^4 \hspace{0.1cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$
 
:$$ K_{\rm WM} = K_{\rm FM} = \frac{ 2 \pi \cdot f_{\rm N} \cdot \eta}{A_{\rm N}} = \frac{ 2 \pi \cdot 5 \cdot 10^3 \,\,{\rm Hz}\cdot 1.3}{2\,{\rm V}} \hspace{0.15cm}\underline {\approx 2.04 \cdot 10^4 \hspace{0.1cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$
  
'''(4)'''&nbsp; Der Frequenzmodulator kann als Phasenmodulator realisiert werden, wenn vorher das Quellensignal integriert wird. Dieses lautet:
+
 
 +
 
 +
'''(4)'''&nbsp; The frequency modulator can be realized as a phase modulator if the source signal is integrated beforehand.&nbsp; This is:
 
:$$q_{\rm I}(t)  =  \int q(t)\hspace{0.15cm}{\rm d}t = A_{\rm N} \cdot\int \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.15cm}{\rm d}t =\frac{A_{\rm N}}{\omega_{\rm N}} \cdot \sin(\omega_{\rm N} \cdot t + \phi_{\rm N}) = \frac{A_{\rm N}}{\omega_{\rm N}} \cdot \cos(\omega_{\rm N} \cdot t + \phi_{\rm N} - 90^\circ)\hspace{0.05cm}.$$
 
:$$q_{\rm I}(t)  =  \int q(t)\hspace{0.15cm}{\rm d}t = A_{\rm N} \cdot\int \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.15cm}{\rm d}t =\frac{A_{\rm N}}{\omega_{\rm N}} \cdot \sin(\omega_{\rm N} \cdot t + \phi_{\rm N}) = \frac{A_{\rm N}}{\omega_{\rm N}} \cdot \cos(\omega_{\rm N} \cdot t + \phi_{\rm N} - 90^\circ)\hspace{0.05cm}.$$
Somit ergibt sich für das äquivalente TP-Signal mit $ϕ_N = 30°$:
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*Thus, for the equivalent low-pass signal with&nbsp; $ϕ_{\rm N} = 30^\circ$:
 
:$$s_{\rm TP}(t) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\omega_{\rm N} \hspace{0.03cm}\cdot \hspace{0.03cm}t \hspace{0.03cm} - \hspace{0.03cm}60^\circ)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{\rm TP}(t = 0) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\hspace{0.03cm}60^\circ)} = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta /2}\hspace{0.05cm}.$$
 
:$$s_{\rm TP}(t) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\omega_{\rm N} \hspace{0.03cm}\cdot \hspace{0.03cm}t \hspace{0.03cm} - \hspace{0.03cm}60^\circ)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{\rm TP}(t = 0) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\hspace{0.03cm}60^\circ)} = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta /2}\hspace{0.05cm}.$$
Der Nullphasenwinkel ist somit gleich $η/2$ entsprechend $ϕ_0\hspace{0.15cm}\underline {\approx 37.5^\circ}$.
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*Thus, the zero phase angle is equal to&nbsp; $η/2$&nbsp; corresponding to &nbsp; $ϕ_0\hspace{0.15cm}\underline {\approx 37.5^\circ}$.
 +
 
  
  
'''(5)'''&nbsp; Aus der Definition des Modulationsindex bei Frequenzmodulation  folgt:
+
'''(5)'''&nbsp; From the definition of modulation index in frequency modulation, it follows that:
 
:$$\eta_1 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 1}}\hspace{0.05cm},\hspace{0.3cm} \eta_2 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 2}} \hspace{0.3cm}
 
:$$\eta_1 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 1}}\hspace{0.05cm},\hspace{0.3cm} \eta_2 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 2}} \hspace{0.3cm}
 
\Rightarrow \hspace{0.3cm}\frac{\eta_1}{\eta_2} = \frac{f_2}{f_1}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} f_2 = \frac{\eta_1}{\eta_2} \cdot f_1 = \frac{75^\circ}{125^\circ} \cdot 5\,{\rm kHz} \hspace{0.15cm}\underline {= 3\,{\rm kHz}}\hspace{0.05cm}.$$
 
\Rightarrow \hspace{0.3cm}\frac{\eta_1}{\eta_2} = \frac{f_2}{f_1}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} f_2 = \frac{\eta_1}{\eta_2} \cdot f_1 = \frac{75^\circ}{125^\circ} \cdot 5\,{\rm kHz} \hspace{0.15cm}\underline {= 3\,{\rm kHz}}\hspace{0.05cm}.$$
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[[Category:Aufgaben zu Modulationsverfahren|^3.2 Frequenzmodulation (FM)^]]
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[[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]]

Latest revision as of 14:13, 17 March 2022

Two different locus curves for angle modulation

To analyze a modulator, apply the signal

$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N})$$

to its input, where the signal amplitude is always  $A_{\rm N} = 2\ \rm V$ .

  • Using the signal frequency  $f_{\rm N} = f_1 = 5 \ \rm kHz$  the locus  $\rm O_1$  is determined.
  • If one uses the message frequency  $f_{\rm N} = f_2$, then the  $\rm O_2$  locus is established.


In your solutions, note that for angle modulation – the umbrella term for phase and frequency modulation – the following relationship holds between the modulation index  $η$  and the modulator constant  $K_{\rm WM}$:

$$\eta = \left\{ \begin{array}{c} K_{\rm WM} \cdot A_{\rm N} \\ {K_{\rm WM} \cdot A_{\rm N}}/({2 \pi \cdot f_{\rm N})} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{bei}} \\ {\rm{bei}} \\ \end{array}\begin{array}{*{20}c} {\rm PM} \hspace{0.05cm}, \\ {\rm FM}. \hspace{0.05cm} \\ \end{array}$$





Hints:



Questions

1

Which modulator do we have here?

AM modulator.
PM modulator.
FM modulator.

2

What is the modulation index when the message frequency is  $f_{\rm N} = f_1 = 5 \ \rm kHz$?

$η_1 \ = \ $

3

What is the value of the modulator constant?   Hint:   the "unit" stands for $\rm V^{-1}$ (for PM) or $\rm (Vs)^{-1}$ (for FM).

$K_{\rm WM} \ = \ $

$\ \cdot 10^4 $ "unit"

4

What is the angle  $ϕ_0$  (with respect to the real axis) of the locus curve  $\rm O_1$  when  $ϕ_{\rm N} = 30^\circ$  at time  $t = 0$ ?

$ϕ_0 \ = \ $

$\ \rm degrees$

5

What message frequency  $f_{\rm N} = f_2$  was used to determine the locus  $\rm O_2$ ?

$f_2 \ = \ $

$\ \rm kHz$


Solution

(1)  Answer 3 is correct:

  • Since the locus curve describes an arc, we have an angle modulator (PM or FM) with modulation index  $η$.
  • But since  $η$  here clearly depends on the message frequency  $f_{\rm N}$ , we can rule out phase modulation.


(2)  The modulation index can be read off the graph. It is   $η_1 = 75^\circ/180^\circ · π\hspace{0.15cm}\underline { ≈ 1.3}$.


(3)  For frequency modulation, it holds that:

$$ K_{\rm WM} = K_{\rm FM} = \frac{ 2 \pi \cdot f_{\rm N} \cdot \eta}{A_{\rm N}} = \frac{ 2 \pi \cdot 5 \cdot 10^3 \,\,{\rm Hz}\cdot 1.3}{2\,{\rm V}} \hspace{0.15cm}\underline {\approx 2.04 \cdot 10^4 \hspace{0.1cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$


(4)  The frequency modulator can be realized as a phase modulator if the source signal is integrated beforehand.  This is:

$$q_{\rm I}(t) = \int q(t)\hspace{0.15cm}{\rm d}t = A_{\rm N} \cdot\int \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.15cm}{\rm d}t =\frac{A_{\rm N}}{\omega_{\rm N}} \cdot \sin(\omega_{\rm N} \cdot t + \phi_{\rm N}) = \frac{A_{\rm N}}{\omega_{\rm N}} \cdot \cos(\omega_{\rm N} \cdot t + \phi_{\rm N} - 90^\circ)\hspace{0.05cm}.$$
  • Thus, for the equivalent low-pass signal with  $ϕ_{\rm N} = 30^\circ$:
$$s_{\rm TP}(t) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\omega_{\rm N} \hspace{0.03cm}\cdot \hspace{0.03cm}t \hspace{0.03cm} - \hspace{0.03cm}60^\circ)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{\rm TP}(t = 0) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\hspace{0.03cm}60^\circ)} = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta /2}\hspace{0.05cm}.$$
  • Thus, the zero phase angle is equal to  $η/2$  corresponding to   $ϕ_0\hspace{0.15cm}\underline {\approx 37.5^\circ}$.


(5)  From the definition of modulation index in frequency modulation, it follows that:

$$\eta_1 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 1}}\hspace{0.05cm},\hspace{0.3cm} \eta_2 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 2}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{\eta_1}{\eta_2} = \frac{f_2}{f_1}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} f_2 = \frac{\eta_1}{\eta_2} \cdot f_1 = \frac{75^\circ}{125^\circ} \cdot 5\,{\rm kHz} \hspace{0.15cm}\underline {= 3\,{\rm kHz}}\hspace{0.05cm}.$$