Exercise 2.11Z: Once again SSB-AM and Envelope Demodulator

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Equivalent low-pass signal in
single-sideband AM

The adjacent graph shows the locus curve – i.e., the representation of the equivalent low-pass signal  (German:  "äquivalentes Tiefpass-Signal"   ⇒   subscript:  "TP")  in the complex plane – for a  single-sideband amplitude modulation  $\text{(SSB-AM)}$  system.

It is further given that the carrier frequency is  $f_{\rm T} = 100 \ \rm kHz$  and the channel is ideal:

$$ r(t) = s(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} r_{\rm TP}(t) = s_{\rm TP}(t) \hspace{0.05cm}.$$

An ideal envelope demodulator is used at the receiver.

The following values are used in these exercises:

• the sideband-to-carrier ratio

$$\mu = \frac{A_{\rm N}/2}{A_{\rm T}}\hspace{0.05cm},$$

• the envelope

$$a(t) = |s_{\rm TP}(t)| \hspace{0.05cm},$$

• the maximum deviation  $τ_{\rm max}$  of the zero crossings between the transmitted signal  $s(t)$  and the carrier signal  $z(t)$.

Hints:

• This exercise belongs to the chapter  Single-sideband Modulation.
• Particular reference is made to the page  Sideband-to-carrier ratio.
• In this exercise,  apply the same assumptions as in  Exercise 2.11.

Questions

1

Find the equivalent low-pass signal  $s_{\rm TP}(t)$  in analytical form and choose which statements apply.

	We are dealing with an  "upper-sideband amplitude modulation"  $\text{(USB-AM)}$.
	We are dealing with a  "lower-sideband amplitude modulation"  $\text{(LSB-AM)}$.
	The source signal  $q(t)$  is cosine-shaped.
	The source signal  $q(t)$  is sine-shaped.

2

Find the amplitude  $A_{\rm N}$  and the frequency  $f_{\rm N}$  of the source signal.  Take into account that we are dealing with a SSB-AM.

$A_{\rm N} \ = \ $ $\ \rm V$

$f_{\rm N} \ = \ $ $\ \rm kHz$

3

Which value results for the sideband-to-carrier ratio   $μ$?  Use this value to describe  $s_{\rm TP}(t)$.

$μ \ = \ $

4

Calculate the time course of the envelope  $a(t)$.  Which values arise for  $t = 50 \ \rm µ s$,  $t = 100 \ \rm µ s$  and  $t = 150 \ \rm µ s$?

$a(t = 50 \ \rm µ s) \hspace{0.32cm} = \ $ $\ \rm V$

$a(t = 100 \ \rm µ s) \ = \ $ $\ \rm V$

$a(t = 150 \ \rm µ s) \ = \ $ $\ \rm V$

5

By what time difference   $τ_{\rm max}$  (in magnitude) are the zero crossings of  $s(t)$  maximally shifted with respect to  $z(t)$ ?

$τ_{\rm max} \ = \ $ $\ \rm µ s$

Solution

Solution

(1)  Answers 2 and 4 are correct:

• The equivalent low-pass signal is:

$$ s_{\rm TP}(t) = 1\,{\rm V} + {\rm j}\cdot 1\,{\rm V}\cdot {\rm e}^{-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \hspace{0.05cm}.$$

• The locus curve is a circle with its center at   $A_{\rm T} = 1 \ \rm V$.
• Since the rotation is clockwise, we are dealing with a LSB-AM.
• The rotating (green) pointer points towards the start time  $t = 0$  in the direction of the imaginary axis.
• It follows that the source signal is characterized by:   $q(t) = A_{\rm N} \cdot \sin(\omega_{\rm N} \cdot t).$

(2)  For the LSB, only the lower sideband is transmitted, with pointer length   $A_{\rm N}/2 = 1 \ \rm V$ .

• This results in   $A_{\rm N}\hspace{0.15cm}\underline { = 2 \ \rm V}$.
• For one revolution in the locus, the pointer needs the time   $200 \ \rm µ s$.
• The reciprocal of this is the frequency   $f_{\rm N}\hspace{0.15cm}\underline { = 5 \ \rm kHz}$.

(3)  According to the definition on the exercise page and the results in subtasks  (1)  and  (2) , the following holds:

$$ \mu = \frac{A_{\rm N}/2}{A_{\rm T}}\hspace{0.15cm}\underline {= 1}\hspace{0.05cm}.$$

• Thus, the equivalent low-pass signal can also be written as:

$$s_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + {\rm j} \cdot \mu \cdot {\rm e}^{-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right),\hspace{0.3cm}{\rm hier}\hspace{0.15cm}\mu = 1 \hspace{0.05cm}.$$

(4)  Splitting the complex exponential function into real and imaginary parts using Euler's theorem, we get:

$$s_{\rm TP}(t) = A_{\rm T} \cdot \big( 1 + \sin(\omega_{\rm N}\cdot t) + {\rm j} \cos(\omega_{\rm N}\cdot t)\big) \hspace{0.05cm}.$$

• By applying the "Pythagorean Theorem", this can also be written as:

$$a(t) = |s_{\rm TP}(t)| = A_{\rm T} \cdot \sqrt{ (1 + \sin(\omega_{\rm N}\cdot t))^2 + \cos^2(\omega_{\rm N}\cdot t)} = A_{\rm T} \cdot \sqrt{ 2 + 2 \cdot \sin(2\omega_{\rm N}\cdot t)} \hspace{0.05cm}.$$

• The retrieved values when  $A_{\rm T} = 1\ \rm V$ are:

$$ a(t = 50\,{\rm µ s}) \hspace{0.15cm}\underline {= 2\,{\rm V}},\hspace{0.3cm}a(t = 100\,{\rm µ s}) \hspace{0.15cm}\underline {= 1.414\,{\rm V}},\hspace{0.3cm}a(t = 150\,{\rm µ s}) \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$$

• These results can be directly read off the graph on the exercise page.

(5)  A hint for the location of the zero crossings of   $s(t)$  with respect to the grid given by the carrier signal   $z(t)$  is provided by the phase function  $ϕ(t)$.

• For the given locus, these can take on values between  $±π/2\ (±90^\circ)$ .
• For example, these maximum values arise in the region around   $t ≈ 150 \ \rm µ s$ , since a phase jump occurs there.
• The relationship between  $τ_{\rm max}$  and  $\Delta ϕ_{\rm max}$  is:

$$ \tau_{\rm max} = \frac {\Delta \phi_{\rm max}}{2 \pi }\cdot \frac{1 }{f_{\rm T}} = \frac {1}{4}\cdot 10\,{\rm µ s} \hspace{0.15cm}\underline {= 2.5\,{\rmµ s}} \hspace{0.05cm}.$$