Exercise 3.8: Triple Convolution?

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 "direct 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 rectangular 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 interactive applet Graphical Convolution.

Question

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

$y_1(t=2T)\ = \ $

${\rm V}$

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

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

$y_2(t=2T)\ = \ $

${\rm V}$

$y_2(t=3T)\ = \ $

${\rm V}$

$y_3(t=0)\ = \ $

${\rm V}$

Solution

(1) The correct answer is 2:

Both $x_1(t)$ and $x_3(t)$ contain only one frequency, namely $f = 0$ or $f = f_0$. In each case, the diversions via the spectrum is preferable.
With the rectangular pulse $x_2(t)$ the calculation via convolution is more favourable, since the inverse Fourier transform of $Y_2(f)$ is complicated.

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

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

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

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:

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

"Rectangle" $\star$ "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.

(4a) For $T \leq t \leq 2T$ the lower $($German: "untere" ⇒ "u"$)$ limit is $τ_u = 0$, and the upper $($German: "obere" ⇒ "o"$)$ limit is $τ_0 = t - T$:

Convolution for $T \leq t \leq 2T$

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

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 $2T \leq t \leq 3T$, the lower limit $τ_0 = t - T$ 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 mirror can now be dispensed 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.