Aufgaben:Exercise 3.4: Simple Phase Modulator

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"Approximate phase modulator"

The adjacent circuit allows the approximate realization of a phase-modulated signal. 

From the cosinusoidal carrier,  $z(t)$ , the  $90^\circ$ phase shifter forms a sinusoidal signal of the same frequency, such that the modulated signal can be written as:

$$ s(t) = z(t) + q(t) \cdot \frac{z(t- T_0/4)}{A_{\rm T}} = A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t) + q(t) \cdot \sin (\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

The second term describes a "DSB–AM without carrier".  Additionally, the carrier, phase-shifted by $90^\circ$ , is added.  Thus, with a cosine source signal $q(t) = A_{\rm N} \cdot \cos (\omega_{\rm N} \cdot t)$ , we get:

$$s(t) = A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t) + A_{\rm N} \cdot \cos (\omega_{\rm N} \cdot t) \cdot \sin (\omega_{\rm T} \cdot t) $$

$$\Rightarrow \hspace{0.3cm}s(t) = A_{\rm T} \cdot \big[\cos (\omega_{\rm T} \cdot t) + \eta \cdot \cos (\omega_{\rm N} \cdot t) \cdot \sin (\omega_{\rm T} \cdot t) \big] \hspace{0.05cm}.$$

We refer to the ratio  $η = A_{\rm N}/A_{\rm T}$  as the modulation index;  in the following, the carrier amplitude is set to   $A_{\rm T} = 1$  for simplicity.

• In contrast to  ideal phase modulation  the modulation index  $η$  and the phase deviation  $ϕ_{\rm max}$ may differ in this "approximate phase modulation".
• Additionally, we can see that the envelope  $a(t) ≠ 1$ .  This means that an unwanted amplitude modulation is superimposed on the phase modulation.

From the representation of the equivalent low-pass signal  $s_{\rm TP}(t)$  in the complex plane (locus curve), the following are to be calculated in this task:

• the envelope  $a(t)$  and
• the phase function $ϕ(t)$.

Then, you are to analyse the distortions arising when an ideal PM demodulator, which sets the sink signal  $v(t)$  proportional to the phase  $ϕ(t)$ , is used on the receiving side of this nonideal PM modulator.

Hints:

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

• You can use the following equations to approximate the distortion factor:

$$\arctan(\gamma) \approx \gamma - {\gamma^3}/{3} \hspace{0.05cm}, \hspace{0.3cm} \cos^3(\gamma) ={3}/{4} \cdot \cos(\gamma) +{1}/{4} \cdot \cos(3 \cdot \gamma) \hspace{0.05cm}.$$

Questions

1

Calculate the equivalent low-pass signal. Which statement is true?

	The locus curve  $s_{\rm TP}(t)$  is a circular arc.
	The locus curve  $s_{\rm TP}(t)$  is a horizontal straight line.
	The locus curve  $s_{\rm TP}(t)$  is a vertical straight line.

2

Calculate the (normalized) envelope $a(t)$  for  $A_{\rm T} = 1$.  What are its minimum and maximum values when  $η = 1$?

$a_{\rm min} \ = \ $

$a_{\rm max} \ = \ $

3

Calculate the maximum value of the phase $ϕ(t)$  for  $η = 1$  and  $η = 0.5$.

$η = 1.0\text{:} \ \ \ ϕ_{\rm max} \ = \ $ $\ \rm Grad$

$η = 0.5\text{:} \ \ \ ϕ_{\rm max} \ = \ $ $\ \rm Grad$

4

What are distortions result after ideal phase demodulation of  $s(t)$?

	No distortions occur.
	Linear distortions occur.
	Nonlinear distortions occur.

5

Calculate the distortion factor  $K$  considering the trigonometric relationships given on the exercise page.

$η = 1.0\text{:} \ \ \ K \ = \ $ $\ \text{%}$

$η = 0.5\text{:} \ \ \ K \ = \ $ $\ \text{%}$

Solution

Solution

Konstruktion der "vertikalen" Ortskurve aus den Zeigern

(1)  Answer 3 is correct:

• The equivalent low-pass signal is::$$s_{\rm TP}(t) = A_{\rm T} \cdot \left ( 1 + {\rm j}\cdot \frac {\eta}{2}\cdot \left ({\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}\right) \right) = A_{\rm T} \cdot \big ( 1 + {\rm j}\cdot {\eta}\cdot \cos (\omega_{\rm N} \cdot t) \big)\hspace{0.05cm}.$$
• The graph illustrates that the locus curve  $s_{\rm TP}(t)$  is now a a vertical straight line in contrast to the ideal PM (circular arc) and DSB–AM  (horizontal straight line). 
• In the following, we set   $A_{\rm T} = 1$ .

(2)  The envelope is obtained from the time-dependent pointer length as

$$a(t) = \sqrt{1 + \eta^2 \cdot \cos^2 (\omega_{\rm N} \cdot t)} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_{\rm min} \hspace{0.15cm}\underline { = 1}, \hspace{0.3cm}a_{\rm max} = \sqrt{1 + \eta^2 }\hspace{0.05cm}.$$

• For  $η = 1$  the maximum value becomes  $a_{\rm max} = \sqrt{2}\hspace{0.15cm}\underline { ≈ 1.414}$.

(3)  The phase function of this simple phase demodulator is given by:

$$\phi(t) = \arctan \frac{{\rm Im}[s_{\rm TP}(t)]}{{\rm Re}[s_{\rm TP}(t)]} = \arctan (\eta \cdot \cos (\omega_{\rm N} \cdot t)) \hspace{0.05cm}.$$

• The maximum value occurs at time  $t = 0$ , for example, and is   $ϕ_{\rm max} = \arctan(η)$.

• When  $η = 1$ , one obtains   $ϕ_{\rm max}\hspace{0.15cm}\underline { = 45^\circ}$  $($to compare:  for ideal PM  $57.3^\circ)$,
• When  $η = 0.5$  one gets   $ϕ_{\rm max}\hspace{0.15cm}\underline { \approx 26.6^\circ}$  $($for ideal PM  $28.7^\circ)$.

(4) Answer 3 is correct:

• It is  not true that:      $\arctan\big [η · \cos(γ)\big ] = η · \cos(γ)$.
• This means that the sink signal is not cosine, in contrast to the source signal.
• This points to nonlinear distortions.

(5)  Using  $γ = η · \cos(ω_N · t)$  and  $\arctan(γ) ≈ γ – γ^3/3$ , we get:

$$ \phi(t) = \eta \cdot \cos (\omega_{\rm N} \cdot t) - \frac{\eta^3}{3}\cdot \cos^3 (\omega_{\rm N} \cdot t))= \eta \cdot \cos (\omega_{\rm N} \cdot t) - \frac{\eta^3}{3}\cdot \left [ {3}/{4}\cdot \cos (\omega_{\rm N} \cdot t) + {1}/{4}\cdot \cos (3 \omega_{\rm N} \cdot t)\right ] $$

$$\Rightarrow \hspace{0.3cm} \phi(t) = \left(\eta - {\eta^3}/{4} \right) \cdot \cos (\omega_{\rm N} \cdot t) - {\eta^3}/{12}\cdot \cos (3\omega_{\rm N} \cdot t) \hspace{0.05cm}.$$

• This means:  using the given series expansion (where 5th and higher order terms are ignored), only the third-order harmonic distortion is non-zero. Thus:

$$K = K_3 = \frac{\eta^3/12}{\eta-\eta^3/4}= \frac{1}{12/\eta^2 -3} \hspace{0.05cm}.$$

• When  $η = 1$  the numerical value is  $K = 1/9 \hspace{0.15cm}\underline { ≈ 11.1\%}$.
• When  $η = 0.5$  the distortion factor is  $K = 1/45 \hspace{0.15cm}\underline {≈ 2.2\%}$.

A simulation shows that by stopping the series after the third order term, we have made the error of over-estimating the distortion factor:

• The values obtained by simulation are   $K ≈ 6%$  $($for  $η = 1)$  and  $K ≈ 2%$  $($for  $η = 0.5)$.
• Thus, the error increases more than proportionally with increasing  $η$ .