Convolution exercise with block diagram and cases

[coutnoob]

1. First of all, a block diagram is given where x(t) is the input and y(t) the output.The y(t) is asked when given the h(t) and x(t).We have in parallel h(t) and δ(t-1)*h(t) and those 2 go through a summer (the second one with minus and the first one with a plus) and give y(t).

2. I know from a previous question that x(t) is a rectangular signal with length from -1 to 2.
And also, h(t)=e^{2-t}u(t-2).
3. i find that y(t)=x(t)*h(t)-x(t)*h(t-1) , cause δ(t-1)*h(t)=h(t-1) (time shift).
When i have to find first the x(t)*h(t) i get 3 cases and get :
1.0 for t<1
2.1-e^{1-t} for t>=1 and t<4
3.e^{4-t}-e^{1-t} for t>4.

Of course i could do the same for x(t)*h(t-1) and get another 3 cases.
Here's my main problem : the limits of the 2 convolutions are totally diferrent : should i get a lot of cases for y(t) with all the limits from the 2 convolutions or is a smarter-more accurate way of calculating y(t) ?
I've been working this for hours and tomorrow i must have it solved with another 2 exercises :(
Any help would be much appreciated.Thanks in advance !

 

[Adrmay]

The best way to solve this problem is to use the convolution theorem. The convolution theorem states that the convolution of two functions, f and g, is equal to the product of the Fourier transforms of the two functions. By doing this, you can simplify the problem and avoid having to consider multiple cases with different limits.