# Thread: Limits of convolution integral

1. ## Limits of convolution integral

Hello, I've once again run into a small hurdle in my quest to learn maths on my own, with no one to talk to about it except for you fine folk! So If you would, please lend a hand again, I really do appreciate it!

The question is:

A linear, time invariant system has the impulse response $h(t) = e^{-et}u(t)$ find the system response to the input $x(t) = u(t) - u(t-3)$

Now, I know the answer is
$\frac{e^{-3t}}{3}\left(e^{9}-1\right)$

$\frac{e^{-3t}}{3}\left(e^{3t}-1\right)$

Now I can see how to get to that answer. I'm taking the limits of the integral as follows:

$\int_0^t x(\tau)h(t-\tau) \, d\tau$

But I guess it should be

$\int_0^3 x(\tau)h(t-\tau) \, d\tau$

Can anyone help to explain why you need the upper limit to be three? Surely when $t$ is less than 3, you don't want $\tau$ to go beyond 3? But then in my example, you don't want $\tau$ to go beyond 3 when $t$ is more than 3... So that doesn't make sense really either...

(I do grasp the logic behind convolution, can visualise what is happening, and can implement it in some basic signal processing. It's just parts of the mathematical proof that get to me!)

2. ## Re: Limits of convolution integral

Now, once again, just the process of writing it down and explaining it on this forum has helped me have a brainwave I think! I've been musing on this all day and it's just hit me... So here is my working:

$\int_0^t x(\tau)h(t-\tau) \, d\tau$

$\int_0^t [u(\tau)-u(\tau-3)]e^{-3(t-\tau)}u(t-\tau) \, d\tau$

because $u(\tau) = 1$ for $0 \leq \tau \leq t$
and $u(t - \tau) = 1$ for $0 \leq \tau \leq t$

$\int_0^t [1-u(\tau-3)]e^{-3(t-\tau)} \, d\tau$

take out the factors of t

$e^{-3t}\int_0^t [1-u(\tau-3)]e^{3\tau} \, d\tau$

$e^{-3t}\int_0^t e^{3\tau} - e^{3\tau}u(\tau-3) \, d\tau$

--------
Now here, I went to this:

$e^{-3t}\int_0^t e^{3\tau} \, d\tau$

$e^{-3t} \left [\frac{e^{3\tau}}{3}\right ]_0^t$

$\frac{e^{-3t}}{3} \left(e^{-3t} - 1\right)$

--------
But I think now, because

$u(\tau - 3) = 1 \, for \, 3\leq \tau$

then for $3 \leq t$ the integral is 0. So I should change the upper limit at this point to 3, and get:

$e^{-3t}\int_0^3 e^{3\tau} \, d\tau$

$e^{-3t} \left [\frac{e^{3\tau}}{3}\right ]_0^3$

$\frac{e^{-3t}}{3} \left(e^{-9} - 1\right)$

------------

Does that all seem good?!

3. ## Re: Limits of convolution integral

Originally Posted by halfnormalled
Hello, I've once again run into a small hurdle in my quest to learn maths on my own, with no one to talk to about it except for you fine folk! So If you would, please lend a hand again, I really do appreciate it!

The question is:

A linear, time invariant system has the impulse response $h(t) = e^{-et}u(t)$ find the system response to the input $x(t) = u(t) - u(t-3)$

Now, I know the answer is
$\frac{e^{-3t}}{3}\left(e^{9}-1\right)$

$\frac{e^{-3t}}{3}\left(e^{3t}-1\right)$

Now I can see how to get to that answer. I'm taking the limits of the integral as follows:

$\int_0^t x(\tau)h(t-\tau) \, d\tau$

But I guess it should be

$\int_0^3 x(\tau)h(t-\tau) \, d\tau$

Can anyone help to explain why you need the upper limit to be three? Surely when $t$ is less than 3, you don't want $\tau$ to go beyond 3? But then in my example, you don't want $\tau$ to go beyond 3 when $t$ is more than 3... So that doesn't make sense really either...

(I do grasp the logic behind convolution, can visualise what is happening, and can implement it in some basic signal processing. It's just parts of the mathematical proof that get to me!)
The limits on the convolution in general should be $-\infty$ and $\infty$. In this case the support of $x(t)$ is the interval $[0,3]$ (that is $x(t)$ is zero outside this interval) so the limits of integration can be reduced to $0$ and $3$.

CB