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Find convolution sum of continuous signal [Signals&Systems]

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ColdStart
#1
Sep20-10, 08:23 PM
P: 19
1. The problem statement, all variables and given/known data
x(t) = u(t) - 2*u(t-2) + u(t-5)
h(t) = exp(2*t) * u(1-t)


2. Relevant equations
Need to find convolution integrals for all possible intervals


3. The attempt at a solution
Ok, the first thing to do is to rewrite these equations in terms of [tex]\tau[/tex]. Now i have:
x([tex]\tau[/tex]) = u([tex]\tau[/tex]) - 2*u([tex]\tau[/tex]-2) + u([tex]\tau[/tex]-5)
h([tex]\tau[/tex]) = exp(2*[tex]\tau[/tex]) * u(1-[tex]\tau[/tex])

Now, i must derive h(t-[tex]\tau[/tex]), and i have:

h([tex]\tau[/tex]) = exp(2*(t-[tex]\tau[/tex])) * u(t + [tex]\tau[/tex] + 1)

Ok, now i am stuck, how to derive the needed intervals? And form the appropriate integral for each interval without graphing?

The answer says, that the intervals are: t <=1, 1 <= t <= 3, 3 <= t <= 6, 6 <= t

I have no clear idea now how did they derive these intervals WITHOUT GRAPHING!

What else i did is i rewrote x([tex]\tau[/tex]) and h(t-[tex]\tau[/tex]) in general term of convolution sum, and got:

y(t) = Integral[-Inf, Inf] ( (u([tex]\tau[/tex]) - 2*u([tex]\tau[/tex]-2) + u([tex]\tau[/tex]-5)) * exp(2*[tex]\tau[/tex]) * u(1-[tex]\tau[/tex]) ) d[tex]\tau[/tex]

Now, solving inequalilties inside above expression gives:
[tex]\tau[/tex] >= 0, [tex]\tau[/tex] >=2, [tex]\tau[/tex] >= 5, [tex]\tau[/tex] <= t, [tex]\tau[/tex] <= t+1

And now im kind of stuck... how to derive now the correct integrals for bolded intervals above?

what is the clear logic to do this, and similar problems? I have several other problems too like this, and i think once i get the robust logic here, i will be able to solve others too.

thanks!
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