# Find convolution sum of continuous signal [Signals&Systems]

1. Sep 20, 2010

### ColdStart

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 $$\tau$$. Now i have:
x($$\tau$$) = u($$\tau$$) - 2*u($$\tau$$-2) + u($$\tau$$-5)
h($$\tau$$) = exp(2*$$\tau$$) * u(1-$$\tau$$)

Now, i must derive h(t-$$\tau$$), and i have:

h($$\tau$$) = exp(2*(t-$$\tau$$)) * u(t + $$\tau$$ + 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($$\tau$$) and h(t-$$\tau$$) in general term of convolution sum, and got:

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

Now, solving inequalilties inside above expression gives:
$$\tau$$ >= 0, $$\tau$$ >=2, $$\tau$$ >= 5, $$\tau$$ <= t, $$\tau$$ <= 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!