Exercise 2.2: Modulation Depth

From LNTwww

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 the message signal  $q(t)$  and the 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,   ⇒   $A_{\rm max} = 2 \ \rm V$.
  • 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:


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 carrier".
The modulation depth is  $m = 0$.
The modulation depth  $m$  is infinite.

3

Let  $A_{\rm T} = A_{\rm N} = 1\ \rm V$   ⇒   $m = 1$.  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 statements seem reasonable to you?

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


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 carrier".
  • The modulation depth  $m = A_{\rm N}/A_{\rm T}$  is infinitely large.


Analytical signal's spectrum

(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 (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:
$$P_{\rm LSB} = P_{\rm USB} =({A_{\rm N}}/{2})^2/2 = 0.125\,{\rm V}^2 \hspace{0.05cm}.$$
  • 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  $m = 3)$,     $0$  $($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.