Aufgaben:Exercise 3.7Z: Rectangular Signal with Echo

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Sendesignal $s(t)$ & Signal $r(t)$ mit Echo

We consider a periodic square wave signal  $s(t)$  with the possible amplitude values  $0\text{ V}$  and  $2\text{ V}$  and the period duration  $T_0 = T = 1 \text{ ms}$. At the jump points, for example at  $t = T/4$, the signal value is  $1\text{ V}$. The DC component $($also i.e. the Fourier coefficient  $A_0)$  of the signal is also  $1\text{ V}$.

Further applies:

• Due to symmetry (even function), all sine coefficients  $B_n = 0$.

• The coefficients  $A_n$  with even  $n$  are also zero.

• For odd values of  $n$  on the other hand, the following applies:

$$A_n = ( { - 1} )^{\left( {n - 1} \right)/2} \cdot \frac{{4\;{\rm{V}}}}{{n \cdot {\rm{\pi }}}}.$$

The signal  $s(t)$  reaches the receiver via two paths (see sketch below):

• Once on the direct path and secondly via a secondary path.
• The latter is characterised by the attenuation factor  $\alpha$  and the transit time  $\tau$ .
• Therefore, the following applies to the received signal:

$$r(t) = s(t) + \alpha \cdot s( {t - \tau } ).$$ The frequency response of the channel is  $H(f) = R(f)/S(f)$, the impulse response is denoted by  $h(t)$ .

Hints:

• This exercise belongs to the chapter  The Convolution Theorem and Operation.
• Important information can be found in particular on the page  Convolution of a Function With a Dirac Function.

Questions

1

Which statements are true regarding the impulse response  $h(t)$ ?

	For  $0 ≤ t < \tau$   $h(t) = 1$ is true, for  $t > \tau$    $h(t) = 1 + \alpha$.
	It holds that  $h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.
	$h(t)$  has a gaussian shape.

2

Calculate the signal  $r(t)$  for the channel parameters  $\alpha = -0.5$  and  $\tau = T/4$.
What values result at the given times?

$r(t = 0.2 \cdot T)\ = \ $  $\text{V}$

$r(t = 0.3 \cdot T)\ = \ $  $\text{V}$

3

Calculate the signal  $r(t)$  with  $\alpha = 1$  and  $\tau = T/2$. Interpret the result in the frequency domain.
What value results for  $t = T/2$?

$r(t = T/2)\ = \ $  $\text{V}$

Solution

Solution

(1)  The second suggested solution is correct:

• The impulse response is equal to the received signal  $r(t)$, if a single Dirac impulse is present at the input at time  $t = 0$ :

$$h(t) = \delta (t) + \alpha \cdot \delta( {t - \tau } ).$$

Faltung von Rechtecksignal  $s(t)$  und Impulsantwort  $h(t)$

(2)  It holds that  $r(t) = s(t) ∗ h(t)$. This convolution operation is most easily performed graphically:

The values of the received signal are generally:

• $0.00 < t/T < 0.25\text{:}\hspace{0.4cm} r(t) = +1\hspace{0.02cm}\text{ V}$,
• $0.25 < t/T < 0.50\text{:}\hspace{0.4cm} r(t) = -1 \hspace{0.02cm}\text{ V}$,
• $0.50 < t/T < 0.75\text{:}\hspace{0.4cm} r(t) = 0 \hspace{0.02cm}\text{ V}$,
• $0.75 < t/T < 1.00\text{:}\hspace{0.4cm} r(t) = +2 \hspace{0.02cm}\text{ V}$.

The values we are looking for are thus

$$r(t = 0.2 \cdot T) \hspace{0.15cm}\underline{= +1 \hspace{0.02cm}\text{ V}},$$

$$r(t = 0.3 · T) \hspace{0.15cm}\underline{= -1 \hspace{0.02cm}\text{ V}}.$$

(3)  Using a similar procedure as in (2) , a DC signal of $2\hspace{0.02cm}\text{ V}$ is obtained for $r(t)$ :

• The gaps in the signal $s(t)$ are completely filled by the echo $s(t - T/2)$.
• This result can also be derived in the frequency domain.
• The channel frequency response is with $\alpha = 1$ and $\tau = T/2$:

$$H( f ) = 1 + 1 \cdot {\rm{e}}^{ - {\rm{j\pi }}fT} = 1 + \cos ( {{\rm{\pi }}fT} ) - {\rm{j}} \cdot {\rm{sin}}( {{\rm{\pi }}fT} ).$$

• Apart from the DC component, the input signal ${s(t)}$ only has components at $f = f_0 = 1/T$, $f = 3 \cdot f_0$, $f = 5 \cdot f_0$ etc..
• At these frequencies, however, both the real– and the imaginary part of ${H(f)}$ are equal to zero.
• Thus, for the output spectrum with $A_0 = 1 \text{ V}$ and $H(f = 0) = 2$ we obtain:

$$R(f) = A_0 \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$

The Fourier inverse transformation thus also yields $r(t) \underline{= 2 \text{ V= const}}$.