Find convolution sum of continuous signal [Signals&Systems]

[ColdStart]

Homework Statement

x(t) = u(t) - 2*u(t-2) + u(t-5)
h(t) = exp(2*t) * u(1-t)

Homework Equations

Need to find convolution integrals for all possible intervals

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 I am 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!

 

[KataKoniK]

Thank you for your post. It seems like you are on the right track with your approach to solving this problem. The key to finding the intervals for the convolution integrals is to consider the different cases that can arise based on the given functions.

For the first interval, t <= 1, you can see that the function h(t-\tau) will only be non-zero when t <= 1 and \tau <= t. This gives you the following integral:

y(t) = Integral[0, t] ( (u(\tau) - 2*u(\tau-2) + u(\tau-5)) * exp(2*\tau) * u(1-\tau) ) d\tau

For the second interval, 1 <= t <= 3, you will have to consider two cases: when t <= 1 and t > 1. In the case when t <= 1, the integral will be the same as above. However, in the case when t > 1, the function h(t-\tau) will be non-zero for all values of \tau between 0 and t-1. This gives you the following integral:

y(t) = Integral[0, 1] ( (u(\tau) - 2*u(\tau-2) + u(\tau-5)) * exp(2*\tau) * u(1-\tau) ) d\tau + Integral[1, t] ( (u(\tau) - 2*u(\tau-2) + u(\tau-5)) * exp(2*\tau) * u(1-\tau) ) d\tau

For the third interval, 3 <= t <= 6, you will have to consider three cases: when t <= 1, 1 < t <= 3, and t > 3. In the case when t <= 1, the integral will be the same as above. In the case when 1 < t <= 3, the function h(t-\tau) will be non-zero for all values of \tau between t-3 and t-1. In the case when t > 3, the function h(t-\tau) will be non-zero for all values of \tau between 0 and t-1. This gives you the following integral:

y(t) = Integral[0, 1] ( (u(\tau) - 2