Find convolution sum of continuous signal [Signals&Systems]by ColdStart Tags: continuous, convolution, signal, signalsandsystems 

#1
Sep2010, 08:23 PM

P: 19

1. The problem statement, all variables and given/known data
x(t) = u(t)  2*u(t2) + u(t5) h(t) = exp(2*t) * u(1t) 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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