Exercise 4.2Z: Multiplication with a Sine Signal

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Spectral functions  $Q(f)$  and  $Z(f)$

A periodic message signal  $q(t)$  is considered, whose spectral function  $Q(f)$  can be seen in the upper graph.

A multiplication with the dimensionless carrier  $z(t)$, whose spectrum  $Z(f)$  is also shown, leads to the signal  $s(t) = q(t) \cdot z(t).$

In this task, the spectral function  $S(f)$  of this signal is to be determined, whereby the solution can be either in the time or frequency domain.




Hint:



Questions

1

Give the source signal  $q(t)$  in analytical form.  Which values result for  $t = 0$  und  $t = 0.125\, \text{ms}$?

$q(t = 0)\ = \ $

 $\text{V}$
$q(t = 0.125 \,\text{ms})\ = \ $

$\text{V}$

2

What is the (dimensionless) carrier signal  $z(t)$?  What is its maximum value?

$z_{\rm max}\ = \ $

3

Calculate the spectrum  $S(f)$  separately for real and imaginary parts.  At which frequencies are there lines with a non-zero real part?

$3\ \text{kHz},$
$4\ \text{kHz},$
$5\ \text{kHz},$
$6\ \text{kHz},$
$7\ \text{kHz}.$

4

At which frequencies do purely imaginary spectral lines occur?

$3\ \text{kHz},$
$4\ \text{kHz},$
$5\ \text{kHz},$
$6\ \text{kHz},$
$7\ \text{kHz}.$


Solution

(1)  The source signal can be represented with the abbreviations  $f_1 = 1\ \text{kHz}$  and  $T_1 = 1/f_1 = 1 \ \text{ms}$  as follows  $($  $f_2 = 2f_1 applies)$:

$$q(t ) = 4\hspace{0.05cm}{\rm V} \cdot {\cos} ( 2 \pi f_1 t) - 2\hspace{0.05cm}{\rm V} \cdot {\sin} ( 4 \pi f_1 t)= 4\hspace{0.05cm}{\rm V} \cdot {\cos} ( 2 \pi {t}/{T_1}) - 2\hspace{0.05cm}{\rm V} \cdot {\sin} ( 4 \pi {t}/{T_1}) .$$
  • At time  $t = 0$ , the second component disappears and  $q(t = 0)\; \underline{= 4 \ \text{V}}$.
  • On the other hand, for  $t = 0.125 \ \text{ms} = T_1/8$ is obtained:
$$q(t = 0.125{\rm ms}) = 4\hspace{0.05cm}{\rm V} \cdot {\cos} ( {\pi}/{4}) - 2\hspace{0.05cm}{\rm V} \cdot {\sin} ( {\pi}/{2}) = \frac {4\hspace{0.05cm}{\rm V}}{\sqrt{2}} - 2\hspace{0.05cm}{\rm V} \hspace{0.15 cm}\underline{= 0.828 \hspace{0.05cm}{\rm V}}.$$


(2)  According to the purely imaginary spectrum  $Z(f)$  and the impulse weights  $\pm 3$  must hold:

$$z(t ) = 6 \cdot {\sin} ( 2 \pi \cdot 5\hspace{0.05cm}{\rm kHz})\hspace{0.5cm}\Rightarrow \hspace{0.5cm} z_{\rm max}\hspace{0.15 cm}\underline{ = 6} .$$


Discrete band-pass spectrum

(3)  The spectral function  $S(f)$  results from the convolution between  $Q(f)$  and  $Z(f)$. One obtains:

$$S(f) = - 3{\rm j} \cdot Q(f- f_{\rm T}) + 3{\rm j} \cdot Q(f+ f_{\rm T}).$$

This results in spectral lines at

  • $3\ \text{kHz}\ (–3\ {\rm V})$,
  • $4\ \text{kHz} (–{\rm j} \cdot 6\ {\rm V})$,
  • $6\ \text{kHz} (–{\rm j} \cdot 6\ {\rm V})$,
  • $7\ \text{kHz}\ (–3\ {\rm V})$.


Plus the conjugate-complex components at negative frequencies.

Lines with real weights at  $\underline{\pm 3 \ \text{kHz}}$  and  $\underline{\pm 7 \ \text{kHz}}$.


(4)  Imaginary lines appear at  $\underline{\pm 4 \ \text{kHz}}$  and  $\underline{\pm 6 \ \text{kHz}}$.

An alternative way to solve this problem is to use trigonometric equations.

In the following, for example,  $f_5 = 5 \text{ kHz}$.  Then it applies:

$$4\hspace{0.05cm}{\rm V} \cdot {\cos} ( 2 \pi f_1 \hspace{0.03cm}t) \cdot 3 \cdot {\sin} ( 2 \pi f_5 \hspace{0.03cm} t)= \frac{12\hspace{0.05cm}{\rm V}}{2}\cdot \big[{\sin} ( 2 \pi f_4 \hspace{0.03cm} t)+ {\sin} ( 2 \pi f_6 \hspace{0.03cm} t)\big],$$
$$-2\hspace{0.05cm}{\rm V} \cdot {\sin} ( 2 \pi f_2 \hspace{0.03cm}t) \cdot 3 \cdot {\sin} ( 2 \pi f_5 \hspace{0.03cm} t)= \frac{-6\hspace{0.05cm}{\rm V}}{2}\cdot \big[{\cos} ( 2 \pi f_3 \hspace{0.03cm} t)+ {\cos} ( 2 \pi f_7 \hspace{0.03cm} t)\big].$$
  • From the first equation, the following spectral lines are obtained:
  • at  $+f_4$  and  $-f_4$  with weights  $–{\rm j} \cdot 3\ {\rm V}$  bzw.  $+{\rm j}\cdot 3 \ {\rm V}$ respectively,
  • at  $+f_6$  and  $-f_6$  with weights  $–{\rm j} \cdot 3 \ {\rm V}$  bzw.  $+{\rm j} \cdot 3 \ {\rm V}$ respectively.
  • The second equation gives a total of four Dirac lines  (all  $6 \ {\rm V}$, real and negative) at  $\pm f_3$  and  $\pm f_7$.


A comparison with the sketch above shows that both solutions lead to the same result.