1. Commutative:#

2. Associative:#

3. Distributive:#

Graphical Evaluation of the Convolution Integral#

The convolution integral is most conveniently evaluated by a graphical evaluation. We give three examples (5.4—5.6) which we will demonstrate in class using a graphical visualization tool developed by Teja Muppirala of the Mathworks and updated by Rory Adams.

The tool: convolutiondemo.m (see license.txt).

We will then work through the examples again in the examples class.

cd /Users/eechris/code/src/github.com/cpjobling/eg-150-textbook/lti_systems/matlab/convolution_demo
pwd

ans = '/Users/eechris/code/src/github.com/cpjobling/eg-150-textbook/lti_systems/matlab/convolution_demo'

convolutiondemo % ignore warnings

Warning: The EraseMode property is no longer supported and will error in a future release.

Summary of Steps#

1. The inpulse response $h(\tau )$ is time reversed (that is, reflected about the origin) to obtain $h(-\tau )$ and then shifted by $t$ to form $h(t-\tau )=h[-(\tau -t)]$, which is a function of $\tau$ with parameter $t$.

2. The signal $x(\tau )$ and $h(t-\tau )$ are multiplied together for all values of $\tau$ with $t$ fixed at some value.

3. The product $x(\tau )h(t-\tau )$ is integrated over all $\tau$ to produce a single output value $y(t)$.

4. Steps 1 to 3 are repeated as $t$ varies over $-\mathrm{\infty }$ to $\mathrm{\infty }$ to produce the entire output $y(t)$.

Examples of the above convolution integral operation are given in Examples 5.4 to 5.6.

Example#

We will do Exercise 5.4 in class

Impulse response from step response#

Differentiating the step response with respect to $t$, we get

Thus the impulse response $h(t)$ can be determined by differentiating the step response $s(t)$.

Exercise 5.1#

Verify the following properties of the convolution integral; that is,

(a) $x(t)\ast h(t)=h(t)\ast x(t)$

(b) $\{x(t)\ast h_1(t)\}\ast h_2(t)=x(t)\ast \{h_1(t)\ast h_2(t)\}$

For the answer, refer to the lecture recording or see solved problem 2.1 in in [Hsu, 2020].

Exercise 5.2#

Show that

(a) $x(t)\ast \delta (t)=x(t)$

(b) $x(t)\ast \delta (t-t_0)=x(t-t_0)$

(c) $x(t)\ast u_0(t)={\int }_{-\mathrm{\infty }}^{t}x(\tau )d\tau$

(d) $x(t)\ast u_0(t-t_0)={\int }_{-\mathrm{\infty }}^{t_0}x(\tau )d\tau$

For the answer, refer to the lecture recording or see solved problem 2.2 in in [Hsu, 2020].

Exercise 5.3#

Let $y(t)=x(t)\ast h(t)$. Then show that

For the answer, refer to the lecture recording or see solved problem 2.3 in in [Hsu, 2020].

Exercise 5.4#

The input $x(t)$ and the impulse response $h(t)$ of a continuous-time LTI system are given by

(a) Compute the output $y(t)$ by using the convolution integral

(b) Compute the output $y(t)$ by using the convolution integral

Solutions#

(a) Graphical#

Using the convolutiondemo tool chose a value for $\alpha$. I will use $\alpha =1$.

Then set f(t), which represents $x(t)$, to heaviside(t) and g(t). which represents $h(t)$ to exp(-1*t)

Manual solution#

For the manual solution, refer to the lecture recording or see solved problem 2.3 in in [Hsu, 2020].

MATLAB Solution#

We can also use the Symbolic Math Toolbox to solve the problem directly:

syms t tau alpha
assume(alpha > 0)

x(t) = heaviside(t); % unit step function
subplot(211)
fplot(x(t)),title('x(t)'),ylim([0,1.25])
h(t) = exp(-alpha*t)*heaviside(t);
subplot(212)
fplot(subs(h(t),alpha,1)),title('h(t)')

Compute $y(t)$ using the MATLAB function int to compute the convolution integral symbolically.

y(t) = int(x(tau)*h(t - tau),tau,0,t)

$$y(t)=-\frac{2{\mathrm{e}}^{-\frac{\alpha t}{2}}(\frac{{\mathrm{e}}^{-\frac{\alpha t}{2}}}{2}-\frac{{\mathrm{e}}^{\frac{\alpha t}{2}}}{2})(\frac{\text{sign}\left(t\right)}{2}+\frac{1}{2})}{\alpha }$$

Plot the result for $\alpha =1$

clf
ya(t) = subs(y(t),alpha,1)
fplot(ya(t))

$$ya(t)=-2{\mathrm{e}}^{-\frac{t}{2}}(\frac{{\mathrm{e}}^{-\frac{t}{2}}}{2}-\frac{{\mathrm{e}}^{t/2}}{2})(\frac{\text{sign}\left(t\right)}{2}+\frac{1}{2})$$

(b) Graphical#

Reverse the settings for f(t)and g(t) in the convolutiondemo tool.

Manual solution#

For the manual solution, refer to the lecture recording or see solved problem 2.3 in in [Hsu, 2020].

MATLAB Solution#

Reverse the arguments to the fplot and int functions.

subplot(211)
fplot(subs(h(t),alpha,1)),title('h(t)'),ylim([0,1.25])
subplot(212)
fplot(x(t)),title('x(t)'),ylim([0,1.25])

y(t) = int(h(tau)*x(t - tau),tau,-Inf,Inf)

$$y(t)=-\frac{2{\mathrm{e}}^{-\frac{\alpha t}{2}}(\frac{{\mathrm{e}}^{-\frac{\alpha t}{2}}}{2}-\frac{{\mathrm{e}}^{\frac{\alpha t}{2}}}{2})(\frac{\text{sign}\left(t\right)}{2}+\frac{1}{2})}{\alpha }$$

Plot the result for $\alpha =1$

clf
yb(t) = subs(y(t),alpha,1)
fplot(yb(t))

$$yb(t)=-2{\mathrm{e}}^{-\frac{t}{2}}(\frac{{\mathrm{e}}^{-\frac{t}{2}}}{2}-\frac{{\mathrm{e}}^{t/2}}{2})(\frac{\text{sign}\left(t\right)}{2}+\frac{1}{2})$$

Go back to F. Step Response

Exercise 5.5#

Compute the output $y(t)$ for a continuous-time LTI system whose impulse response $h(t)$ and the input $x(t)$ are given by

Solutions#

Manual solution#

For the manual solution, refer to the lecture recording or see solved problem 2.3 in in [Hsu, 2020].

MATLAB Solution#

We can also use the Symbolic Math Toolbox to solve the problem directly:

x(t) = exp(t)*heaviside(-t);
subplot(211)
fplot(x(t)),,title('x(t)')
h(t) = exp(-1*t)*heaviside(t);
subplot(212)
fplot(h(t)),title('h(t)')

Compute $y(t)$ using the convolution integral

y(t) = int(x(tau)*h(t - tau),tau,-Inf,Inf)

$$\begin{array}{r}y(t)=\{\begin{array}{cl}\frac{{\mathrm{e}}^{-t}}{2}& \text{if}0\le t\\ \frac{{\mathrm{e}}^t}{2}& \text{if}t\le 0\end{array}\end{array}$$

Plot the result for $\alpha =1$

fplot(y(t)),title('y(t) = x(t) * y(t)')

Exercise 5.6#

Evaluate $y(t)=x(t)\ast h(t)$, where $x(t)$ and $h(t)$ are shown in Fig. 34, by an alalytical technique, and (b) by a graphical method.

Solutions#

(a) Analytical#

We first express $x(t)$ and $h(t)$ in functional form using the unit step (or Heaviside function)

$$x(t)=u_0(t)-u_0(t-3)$$

$$h(t)=u_0(t)-u_0(t-2)$$

We will use the MATLAB Symbolic Math Toolbox:

x(t) = heaviside(t)-heaviside(t-3);
h(t) = heaviside(t)-heaviside(t-2);
subplot(121)
fplot(x(t),[-3,6]),grid,ylim([0,1.5]),ylabel('x(t)'),xlabel('t')
subplot(122)
fplot(h(t),[-3,6]),grid,ylim([0,1.5]),ylabel('h(t)'),xlabel('t')

Compute $y(t)$

y(t) = int(x(tau)*h(t - tau),tau,-Inf,Inf)

$$y(t)=\text{heaviside}(t-5)(t-5)-\text{heaviside}(t-3)(t-3)-\text{heaviside}(t-2)(t-2)+t\text{heaviside}\left(t\right)$$

Plot the result

clf
fplot(y(t),[-3,6]),grid,ylim([0,2.5]),ylabel('h(t)'),xlabel('t')

(b) Graphical#

Since both functions are unity between the limits set by the Heaviside function, graphical solution requires multiple applications of the definate integral

$${\int }_{t_0}^{t_1}1\times 1d\tau ={\int }_{t_0}^{t_1}1d\tau$$

with different values for the limits $t_0$ and $t_1$. The convolutiondemo tool can help us discover the limits for the piecewise continuous signal $y(t)$.

For the complete solution to Example 5.2 refer to the lecture recording or see solved problem 2.6 in in [Hsu, 2020].

Unit 3.1: Take Aways#

• Impulse response: $h(t)=\mathbf{T}\{\delta (t)\}$

• Step response: $s(t)=\mathbf{T}\{u_0(t)\}={\int }_0^{t}h(\tau )d\tau$

• Arbitrary system response: $y(t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)h(t-\tau )d\tau$

• Convolution itegral: $y(t)=x(t)\ast h(t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)h(t-\tau )d\tau ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t-\tau )h(\tau )d\tau$

• Properties of the convolution integral:

  • Communitative: $x(t)\ast h(t)=h(t)\ast x(t)$

  • Associative: $\{x(t)\ast h_1(t)\}\ast h_2(t)=x(t)\ast \{h_1(t)\ast h_2(t)\}$

  • Distributive: $x(t)\ast \{h_1(t)+h_2(t)\}=x(t)\ast h_1(t)+x(t)\ast h_2(t)$

• The convolution integral can be computed graphically or analytically.