# Exercise 4.2: Rectangular Spectra

Given low–pass and band-pass spectra

We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.

• It is obvious that
$$u(t) = u_0 \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$
is a low-pass signal whose two parameters  $u_0$  and  $T_u$  are to be determined in subtask  (1) .
• In contrast, the spectrum  $W(f)$ shows that  $w(t)$  describes a band-pass signal.

This task also refers to the band-pass signal

$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t) - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$

whose spectrum was determined in  Exercise 4.1Z . Let  $f_2 = 2 \ \rm kHz.$

Hints:

• Consider the following trigonometric relationship in the solution:
$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin (\alpha + \beta)+ \sin (\alpha - \beta)\big].$$

### Questions

1

What are the parameter values  $u_0$  and  $T_u$  of the low-pass signal?

 $u_0\ = \$  $\text{V}$ $T_u\ = \$  $\text{ms}$

2

Calculate the band-pass signal  $w(t)$.  What are the signal values at  $t = 0$  and  $t = 62.5 \, {\rm µ}\text{s}$?

 $w(t=0)\ = \$  $\text{V}$ $w(t=62.5 \,{\rm µ} \text{s})\ = \$  $\text{V}$

3

Which statements are true regarding the band-pass signals  $d(t)$  and  $w(t)$ ?  Justify your result in the time domain.

 The signals  $d(t)$  and  $w(t)$  are identical. $d(t)$  and  $w(t)$  differ by a constant factor. $d(t)$  und  $w(t)$  have different shapes.

### Solution

#### Solution

(1)  The time  $T_u$   ⇒   first zero of the low-pass signal  $u(t)$  – is equal to the reciprocal of the width of the rectangular spectrum, i.e.   $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$.

• The pulse amplitude is equal to the rectangular area as shown in the sample solution for  Exercise 4.1 .  From this follows  $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$.

Multiplication with a cosine function

(2)  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:

$$W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[ \delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$

According to the  Shifting Theorem,  the following then applies to the associated time signal:

$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0 \cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t).$$

The graph shows

• above the low–pass signal $u(t)$,
• then the oscillation $c(t) = 2 · \cos(2 \pi f_{\rm T}t$ ),
• below the band-pass signal  $w(t) = u(t) \cdot c(t)$.

In particular, at time  $t = 0$ one obtains:

$$w(t = 0) = 2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$

The time  $t=62.5 \,{\rm µ} \text{s}$  corresponds exactly to a quarter of the period of the signal  $c(t)$:

$$w(t = 62.5 \hspace{0.05cm}{\rm µ s}) = 2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm µ s}} {500 \hspace{0.05cm}{\rm µ s}}) \cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot 62.5 \hspace{0.05cm}{\rm µ s})$$
$$\Rightarrow \hspace{0.3cm}w(t = 4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$

(3)  Proposed solution 1 is correct:

• If we compare the spectral function  $W(f)$  of this task with the spectrum  $D(f)$  in the sample solution to  Exercise 4.1, we see that  $w(t)$  and  $d(t)$  are identical.
• This proof is somewhat more complex in the time domain.  With  $f_2 = 2 \,\text{kHz}$  can be written for the signal considered here:
$$w(t ) = 4\hspace{0.05cm}{\rm V} \cdot {\rm si} ( \pi f_2 t) \cdot {\cos} ( 4 \pi f_2 t) = ({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$
• Because of the trigonometric relationship
$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin (\alpha + \beta)+ \sin (\alpha - \beta)\big]$$
the above equation can be transformed:
$$w(t ) = \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ] = 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}- 6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$
• This shows that both signals are actually identical   ⇒   Proposed solution 1:
$$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t) - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$