Aufgaben:Exercise 2.2: Modulation Depth

• Return to book

Definition of modulation depth for DSB–AM

The graph shows DSB amplitude-modulated signals  $s_1(t)$  to  $s_4(t)$  with differing modulation depth  $m$. Let message signal  $q(t)$  and carrier signal  $z(t)$  each be cosine:

$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$

$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$

The modulated signal (transmitted signal) with the DC component added in the modulator is  $A_{\rm T}$:

$$s(t ) = A(t) \cdot z(t), \hspace{0.2cm} A(t) = q(t) + A_{\rm T}\hspace{0.05cm}.$$

In the graphs, the chosen normalization was:

$$A_{\rm T}+ A_{\rm N} = 2\,{\rm V}\hspace{0.05cm}.$$

• If the modulation depth is  $m ≤ 1$, then  $A(t)= q(t) + A_{\rm T}$  is equal to the envelope  $a(t)$.
• In contrast, for a modulation depth  $m > 1$:

$$a(t ) = |A(t)|\hspace{0.05cm}.$$

• The cosine curve  $A(t)$  varies between  $A_{\rm max}$  and  $A_{\rm min}$; because of normalization,  $A_{\rm max} = 2 \ \rm V$ is always the case.
• The minimum values of  $A(t)$  occur at half the period of the source signal $($i.e., for  $t = 125 \ \rm µ s)$ :

$$A_{\rm min} = q(T_0/2)+ A_{\rm T} = A_{\rm T}-A_{\rm N}.$$

• The numerical values are given in the graph.

Hints:

• This exercise belongs to the chapter   Double-Sideband Amplitude Modulation.
• Particular reference is made to the page  DSB-AM with carrier.

Questions

1

Determine the modulation depth for each of the signals  $s_1(t)$,  $s_2(t)$,  $s_3(t)$ .

$m_1 \ = \ $

$m_2 \ = \ $

$m_3 \ = \ $

2

Which statements are true for the signal  $s_4(t)$ ?

	This is a case of "DSB–AM without a carrier".
	The modulation depth is  $m = 0$.
	The modulation depth  $m$  is infinite.

3

Let  $A_{\rm T} = A_{\rm N} = 1\ \rm V$, so  $m = 1$.  What is the spectrum  $S_+(f)$  of the analytical signal?
Which Dirac weights occur at  $f_{\rm T}$  as well as at  $f_{\rm T}± f_{\rm N}$ ?

$S_+(f_{\rm T}) \ = \ $ $\ \text{V}$

$S_+(f_{\rm T} ± f_{\rm N}) \ = \ $ $\ \text{V}$

4

Now let  $m = 1$.  Which fraction  $P_{\rm T}/P_{\rm S}$  of the total transmission power  $P_{\rm S}$  is due to the carrier alone, and thus cannot be used for demodulation??

$P_{\rm T}/P_{\rm S} \ = \ $

5

Generalize the result from   (4)  for an arbitrary modulation depth $m$.
What are the power ratios for  $m = 0.5$,  $m = 3$  and  $m → ∞$ ?

$m = 0.5\text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $

$m = 3.0\text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $

$m → ∞ \text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $

6

Based on the calculations so far, which of the following assessments seem reasonable to you?

	$m ≈ 1$  s more favorable than a small  $m$ for energy reasons.
	The carrier is only useful for envelope demodulation.

Solution

Solution

(1)  From the two equations

$$ A_{\rm max} = A_{\rm T}+A_{\rm N}=2\,\,{\rm V},\hspace{0.3cm} A_{\rm min} = A_{\rm T}-A_{\rm N}\hspace{0.05cm}$$

directly follows:

$$A_{\rm N} = (A_{\rm max} - A_{\rm min})/2,\hspace{0.3cm} A_{\rm T} = (A_{\rm max} + A_{\rm min})/2\hspace{0.05cm}.$$

• Thus, the modulation depth is

$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$

• With the given numerical values, one obtains:

$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm} m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$

(2)  Answers 1 and 3 are correct:

• In this case,  $A_{\rm T} = 0$, which means it is indeed "DSB-AM without a carrier".
• The modulation depth  $m = A_{\rm N}/A_{\rm T}$  is infinitely large.

Spectrum: analytical signal

(3)  The spectrum  $S_+(f)$  is composed of three Dirac lines for each modulation depth   $m$  with the following weights:

• $A_{\rm T}$  $($at  $f = f_{\rm T})$,
• $m/2 · A_{\rm T}$  $($at  $f = f_{\rm T} ± f_{\rm N})$.

For $m = 1$, the weights are obtained according to the graph:

• $S_+(f_{\rm T}) = 1\ \rm V$,
• $S_+(f_{\rm T} ± f_{\rm T}) = 0.5\ \rm V$.

(4)  The power (root mean square) of a harmonic oscillation with amplitude   $A_{\rm T} = 1 \ \rm V$  referenced to the   $1 \ Ω$  resistor is:

$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$

• In the same way, for the powers of the lower and the upper sideband we obtain:

$$P_{\rm USB} = P_{\rm OSB} =({A_{\rm N}}/{2})^2/2 = 0.125\,{\rm V}^2 \hspace{0.05cm}.$$

• Thus, for  $m=1$, the ratio we are looking for is:

$${P_{\rm T}}/{P_{\rm S}}= \frac{P_{\rm T}}{P_{\rm USB} + P_{\rm T}+ P_{\rm OSB}}= \frac{0.5\,{\rm V}^2}{0.125\,{\rm V}^2 + 0.5\,{\rm V}^2+ 0.125\,{\rm V}^2}= 2/3\hspace{0.15cm}\underline { = 0.667}\hspace{0.05cm}.$$

(5)  Using the Dirac weights  $m/2 · A_{\rm T}$  of the two sidebands corresponding to subtask  (3) , we get:

$${P_{\rm T}}/{P_{\rm S}}= \frac{A_{\rm T}^2/2}{A_{\rm T}^2/2 + 2 \cdot (m/2)^2 \cdot A_{\rm T}^2/2}= \frac{2}{2 + m^2}\hspace{0.05cm}.$$

• This leads to the numerical values   $8/9 = 0.889$  $($for  $m = 0.5)$,     $2/11 = 0.182$  $($for  $m = 3)$  und     $0$  $($for  $m \to ∞$).

(6)  Both statements are true:

• The addition of the carrier only makes sense in order to use the simpler envelope demodulator. This is only possible for  $m \le 1$.
• However, should the modulation depth be  $m > 1$  and the use of a synchronous demodulator therefore be required, the carrier should be (almost) completely omitted for energy reasons.
• Similarly due to energy concerns, if an envelope demodulator is used, the largest possible modulation depth  $m < 1$    ⇒   $m \to 1$  should be aimed for.
• However, a small residual carrier can facilitate carrier recovery, which is needed in a synchronous demodulator for frequency and phase synchronization. Thus, the second statement is only conditionally correct.