Convolution of a linear and rectangular function

1. Dec 12, 2013

Lindsayyyy

Hi everyone

1. The problem statement, all variables and given/known data
I want to to calculate the convolution of the following two functions
$$h(t)=\left\{\begin{array}{ll} t, & 0 \leq t \leq 10 \\ 0, & otherwise\end{array}\right.$$ and the function
$$x(t)=\left\{\begin{array}{ll} A, & 0 \leq t \leq 10 \\ 0, & otherwise\end{array}\right.$$

2. Relevant equations
Convolution theorem

3. The attempt at a solution
I tried to solve it, but I have problems with the constant A.

I wrote
$$h(t) * x(t) = \int_{0}^{10} h(\tau) *x(t-\tau) d\tau =\int_{0}^{10} \tau \cdot A d\tau$$

I'm not sure about the last step, can anyone help me ?

edit: actually I think the limits are wrong aswell? Because If I'd do it this way I get a number as a result and not a function. But I don't know what the limits should be then?!

2. Dec 12, 2013

jbunniii

Please note that the above is correct only for $t=10$. To deal with $A$, recognize that it simply stands for some number which does not vary with $t$ or $\tau$, so you can slide it out of the integral just as if it were a number:
$$\int_{0}^{10} \tau \cdot A d\tau = A \int_{0}^{10} \tau d\tau$$
Yes, you're right. The integral you wrote gives you $h*x(10)$. In general, you should expect one or both of the integration limits to depend on $t$. You won't be able to write down a single integral which is correct for all values of $t$. Instead you will have to consider several cases.

I highly recommend sketching plots of $h(\tau)$ and $x(t - \tau)$ as functions of $\tau$. This will be very helpful to determine the endpoints of the integral. You should find that it makes sense to consider the following four cases:

(1) $t < 0$
(2) $0 \leq t \leq 10$
(3) $10 \leq t \leq 20$
(4) $t > 20$

3. Dec 12, 2013

Lindsayyyy

Thank you for your quick reply. But I don't understand why you differ the fours different cases. I'd take t<0 0<t<10 and t>10 but why do you mention the one between 10 and 20 ?

4. Dec 12, 2013

jbunniii

I'll assume you have sketched the graphs of $h(\tau)$ and $x(t - \tau)$ as functions of $\tau$. I'm too lazy to scan and attach an image, but your graphs should have the following characteristics:

• $h(\tau)$ is nonzero between $\tau = 0$ and $\tau = 10$
• $x(t - \tau)$ is nonzero between $\tau = t-10$ and $\tau = t$
Now, case (1) is when the graph of $x(t-\tau)$ is entirely to the left of $h(\tau)$, and the nonzero portions do not overlap. This happens when the right edge of $x(t-\tau)$ comes before the left edge of $h(\tau)$, i.e., when $t < 0$.

Case (2) is when the left edge of $h(\tau)$ is between the left and right edges of $x(t - \tau)$. This corresponds to the case $0 \leq \tau \leq 10$.

Case (3) is when the left edge of $x(t - \tau)$ is between the left and right edges of $h(\tau)$. This corresponds to the case $10 \leq \tau \leq 20$.

Case (4) is when the left edge of $x(t - \tau)$ comes after the right edge of $h(\tau)$. This corresponds to the case $\tau > 20$.

5. Dec 12, 2013

Lindsayyyy

Thank you very much again. I will try it tomorrow and if I have further problems I post again in this thread. Much appreciated your help.