Aufgaben:Exercise 3.1: Phase Modulation Locus Curve

• Return to book

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  Phase Modulation.
• Particular reference is made to the page  Equivalent low-pass signal in phase modulation.

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

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, one moves counterclockwise along the arc.