# Convolution Problem

1. Jul 29, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

The unit impulse response of an LTIC system is h(t) = $e^{-t}u(t)$. Find the system's (zero-state) response y(t) if the input f(t) is $e^{-2t}u(t-3)$.

2. Relevant equations

y(t) = f(t) * h(t) = $∫^{∞}_{-∞}f(t)h(t-\tau)d\tau$
$f_{1}(t) * f_{2}(t ) = c(t)$
$f_{1}(t) * f_{2}(t - T) = c(t - T)$

3. The attempt at a solution

I'm not sure how to apply the shifting property because here in f(t) I have the unit step function only which is shifted and not the exponential. Is it possible to apply the shifting property above for this problem? I don't see how I can apply it for the reason mentioned above.

Thanks for any help.

Last edited: Jul 29, 2013
2. Jul 29, 2013

### vela

Staff Emeritus
Hint: Use $t = (t-3)+3$ in the exponential in f(t).

3. Jul 30, 2013

### GreenPrint

Thanks. I was able to do it. I'm however now stuck on a some what more challenging problem.

$-\delta(t) * e^{t}u(-t) \stackrel{Δ}{=} ∫^{\tau = ∞}_{\tau = -∞}e^{t}u(-\tau)(-\delta(t-\tau))d\tau$

I know that

$f(t) * \delta(t-T) = f(t - T)$

But I'm unsure how to apply this here. I guess in my case $T = 0$ but how do I tell with the negative sign of the dirac delta function? Is my solution

$-e^{t}u(-t)$?

Thanks for help.

4. Jul 30, 2013

### vela

Staff Emeritus
You can pull the negative sign right out of the integral, right? So you should be able to see that
$$[-\delta(t)]*e^t u(-t) = -[\delta(t)*e^t u(-t)]$$