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Difference between revisions of "Aufgaben:Exercise 3.1: Phase Modulation Locus Curve"

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'''(5)'''  One moves clockwise along the circular arc:
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'''(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.  
 
*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.  
 
*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.  

Revision as of 19:48, 10 March 2022

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:


Questions

1

Which modulation method is used by modulator  \rm M_1?

Double-sideband amplitude modulation.
Single sideband amplitude modulation.
Phase modulation.

2

Which modulation method is used by modulator  \rm M_2?

Double-sideband amplitude modulation.
Single sideband amplitude modulation.
Phase modulation.

3

What is the carrier amplitude  A_{\rm T}  for the phase modulator?  Note the normalization to  1 \ \rm V.

A_{\rm T} \ = \

\ \rm V

4

What are the values of the modulation index  η  and the modulator constant  K_{\rm PM}?

η\ = \

K_{\rm PM}\ = \

\ \rm 1/V

5

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?

t_1\ = \

\ \rm µ s


Solution

(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.