Aufgaben:Exercise 3.8: Triple Convolution?

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Impulse response  $h(t)$  and
three input signals  $x(t)$

The impulse response of an LTI system has the following course in the time range between  $0$  and  $2T$ :

$$h( t ) = \frac{1}{T}\cdot \left( {1 - \frac{t}{ {2T}}} \right).$$

Outside this interval,  $h(t) = 0$. The corresponding spectral function is:

$$H( f ) = \frac{1}{ {8\left( {{\rm{\pi }}fT} \right)^2 }} \cdot \left( {1 - {\rm{j \cdot 4\pi }}fT - {\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.05cm} 4\pi }}\hspace{0.05cm}f\hspace{0.05cm}T} } \right).$$

This equation is not suitable for calculating the so-called „equal signal transfer factor”   ⇒   $H(f = 0)$ , since both the bracket expression and the denominator become zero. However, it is also valid:

$$H( {f = 0} ) = \int_0^{2T} {h( t )\hspace{0.1cm}{\rm d}t = 1.}$$

Three different time signals are applied to the input of this filter (see sketch):

• $x_1(t)$  is a DC signal with height  $x_0 = 1 \hspace{0.05cm}{\rm V}$.
• $x_2(t)$  is a square-wave pulse with duration  $T$  and height  $x_0 = 1\hspace{0.05cm} {\rm V}$, starting at  $t = T$.
• $x_3(t)$  is a cosine signal with frequency  $f_0 = 3/T$  and amplitude  $x_0 = 1 \hspace{0.05cm}{\rm V}$.

Hints:

• This exercise belongs to the chapter  The Convolution Theorem and Operation.
• The topic of this section is also illustrated in the  Zur Verdeutlichung der grafischen Faltung  veranschaulicht.

Question

1

For which of the three signals is it more appropriate to calculate the output signal directly in the time domain?

	$y_1(t) = x_1(t) \ast h(t)$.
	$y_2(t) = x_2(t) \ast h(t)$.
	$y_3(t) = x_3(t) \ast h(t)$.

2

What is the signal  $y_1(t)$  at the filter output when the DC signal  $x_1(t) = 1 \hspace{0.03cm}{\rm V}$  is applied to the input? Give the signal value at  $t = 2T$ .

$y_1(t=2T)\ = \ $  ${\rm V}$

3

To which time range between  $t_{\text{min}}$  and  $t_{\text{max}}$  is the output signal  $y_2(t) = x_2(t) \ast h(t)$  limited, i.e. not equal to zero?

$t_{\text{min}}/T\ = \ $

$t_{\text{max}}/T \ = \ $

4

Calculate the values of the signal  $y_2(t)$  at times  $t = 2T$  and  $t = 3T$.

$y_2(t=2T)\ = \ $  ${\rm V}$

$y_2(t=3T)\ = \ $  ${\rm V}$

5

What is the output signal  $y_3(t)$, when the cosine signal  $x_3(t)$  is applied to the input? Give the signal value at  $t =0$  an.

$y_3(t=0)\ = \ $  ${\rm V}$

Solution

Solution

(1)  The correct answer is 2:

• $x_1(t)$  and  $x_3(t)$  each contain only one frequency, namely  $f = 0$  or  $f = f_0$ . In each case, the diversions via the spectrum is preferable.
• Beim Rechtecksignal  $x_2(t)$  the calculation via convolution is more favourable, since the Fourier back transformation of  $Y_2(f)$  is complicated.

(2)  For the square wave signal  $x_1(t)$  the output signal is also a DC signal, since the following equations apply:

$$Y_1 (f) = X_1 (f) \cdot H(f)\quad {\rm{mit}}\quad X_1 (f) = 1\;{\rm{V}} \cdot \delta (f)$$

$$ \Rightarrow Y_1 (f) = 1\;{\rm{V}} \cdot \delta (f) \cdot H( {f = 0} ) = 1\;{\rm{V}} \cdot \delta (f) \; \Rightarrow \; y_1 (t) = 1\;{\rm{V}} \cdot H( {f = 0} ) \hspace{0.15 cm}\underline{= 1\;{\rm{V}}}.$$

• The calculation via convolution leads to the same result if one takes into account that the integral over the impulse response is equal to  $1$  in the present case.

(3)  The mirrored signal  $x_2(-t)$  has signal components between  $-2T$  and  $-T$.

• Only a shift by  $T \hspace{-0.1cm}+ \hspace{-0.1cm}\varepsilon$  leads to an overlap with  $h(t)$. Here  $\varepsilon$  denotes an arbitrarily small but positive time.

• However, if the shift is greater than  $4T\hspace{-0.1cm} - \hspace{-0.1cm}\varepsilon$, the integration over the product also yields the value zero. From this follows:

$$t_{\text{min}} \;\underline{= T}, \ \ \ t_{\text{max}} \;\underline{= 4T}.$$

Convolution rectangle and triangle

(4)  The result of the graphical convolution for the times  $t = 2T$  and  $t = 3T$  can be seen in the adjacent sketch.

• The value at  $t = 2T$  corresponds to the area highlighted in red:

$$y_2( {t = 2T} ) = \frac{1}{2}\cdot ( {\frac{1}{T} + \frac{1}{ {2T}}} ) \cdot T \cdot x_0 \hspace{0.15 cm}\underline{= 0.75 {\rm V}} .$$

• The green highlighted area indicates the value at  $t = 3T$:

$$y_2( {t = 3T} ) = \frac{1}{2}\cdot ( {\frac{1}{2T} + 0} ) \cdot T \cdot x_0 \hspace{0.15 cm}\underline{= 0.25 {\rm V}} .$$

To calculate the entire signal curve between  $t = T$  and  $t = 4T$  three areas must be considered separately. To simplify the representation,  $x_0 = 1$  is set in the following   ⇒   amplitude normalisation.

Convolution for  $T \leq t \leq 2T$

(4a)  For  $T \leq t \leq 2T$  the lower limit is  $τ_u = 0$, the upper limit is  $τ_0 = t - T$:

$$y_2 (t) = \int_{\tau _u }^{\tau _0 } {h(\tau )\,{\rm{d}}\tau = \int_0^{t - T} {\frac{1}{T}} }\cdot \left( {1 - \frac{\tau }{ {2T}}} \right)\,{\rm{d}}\tau .$$

• With the indefinite integral   $I(\tau ) = {\tau }/{T} - 0.25 \cdot \left( {{\tau }/{T}} \right)^2$   we obtain

$$y_2 (t) = I(t - T) - I(0) = \frac{ {t - T}}{T} - 0.25 \cdot \left( {\frac{ {t - T}}{T}} \right)^2 $$

$$\Rightarrow \hspace{0.3cm}y_2 (t) = 1.5 \cdot {t}/{T} - 0.25\cdot \left( {{t}/{T}} \right)^2 - 1.25.$$

• For verification we consider the two limits. We get the values  $y_2(T) = 0$  and  $y_2(2T) = 0.75$.

Convolution for  $2T \leq t \leq 3T$

(4b)  In the interval  $2T \leq t \leq 3T$  ,  $τ_0 = t - T$, still holds, while now  $τ_u = t - 2T$ :

$$y_2 (t) = I(t - T) - I(t - 2T) = 1.75 - 0.5 \cdot {t}/{T}.$$

This corresponds to a linear decrease with the two limit values 

$$y_2(2T) = 0.75,$$

$$y_2(3T) = 0.25.$$

Convolution for  $3T \leq t \leq 4T$

(4c)  For the interval  $3T \leq t \leq 4T$  and  $τ_0 = 2T$  und   $τ_u = t - 2T$:

$$y_2 (t) = I(2T) - I(t - 2T) = - 2 \cdot {t}/{T} + 0.25\left( {c{t}/{T}} \right)^2 + 4.$$

Here, too, the correct limit values result:

$$y_2 (3T) = 0.25,$$

$$y_2 (4T) = 0.$$

(5)  In principle, this sub-task could also be solved directly with the convolution.

• However, since  $x_3(t)$  is an even function, the reflection can now be dispensed with here and we obtain:

$$y_3 (t) = \int_{ - \infty }^{ + \infty } {h(\tau ) \cdot x_3 (t + \tau )\hspace{0.1cm}{\rm{d}}\tau = x_0 }\cdot \int_0^{2T} {h(\tau ) \cdot \cos (2{\rm{\pi }}f_0 (t + \tau )\hspace{0.1cm}{\rm{d}}\tau .}$$

• The simpler way here is via the spectra. $X(f)$  consists of two diraclines at  $\pm 3f_0$. Thus, the frequency response must also only be calculated for this frequency:

$$H( {f = 3f_0 } ) = \frac{1}{ {72{\rm{\pi }}^{\rm{2}} }}\big[ {1 - {\rm{j}}\cdot 12{\rm{\pi }} - {\rm{cos}}( {{\rm{12\pi }}} ) + {\rm{j}}\cdot \sin ( {{\rm{12\pi }}})} \big] = \frac{1}{ {72{\rm{\pi }}^{\rm{2}} }}\big[ 1 - {\rm j}\cdot 12{\rm \pi } - 1 + {\rm j}\cdot 0 \big]= { - {\rm{j}}} \cdot \frac{1}{ {6{\rm{\pi }}}}.$$

• Thus the spectrum of the output signal is:

$$Y(f) = - {\rm{j}} \cdot \frac{ {x_0 }}{{{\rm{12\pi }}}}\cdot \delta \left( {f - 3f_0 } \right) + {\rm{j}} \cdot \frac{ {x_0 }}{{{\rm{12\pi }}}}\cdot \delta \left( {f + 3f_0 } \right).$$

• The signal  $y_3(t)$  is thus sinusoidal with amplitude  $x_0/(6\pi )$.
• At  $t = 0$  the signal value  $y_3(t = 0)\; \underline{= 0}$ results.