Solve Convolution Problem: Find y(t) Given h(t) and f(t)

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Homework Statement



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).

Homework 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)

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.
 
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Hint: Use ##t = (t-3)+3## in the exponential in f(t).
 
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.
 
You can pull the negative sign right out of the integral, right? So you should be able to see that
$$[-\delta(t)]*e^t u(-t) = -[\delta(t)*e^t u(-t)]$$
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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