Using dirac delta functions to get $h(t)$ that satisfies $[u(t+1/2)-u(t-1/2)] \ast h(t) = [u(t+6)-u(t+2)]$ where $u(t)$ is unit-step function

Question

Using dirac delta functions, how does one get $h(t)$ that satisfies $[u(t+1/2)-u(t-1/2)] \ast h(t) = [u(t+6)-u(t+2)]$ where $u(t)$ is unit(heaviside)-step function and $\ast$ is convolution?

Answer 1

Since this looks like a homework problem, I'll give you some hints instead of the complete solution. If you follow these hints you should be able to solve the problem yourself. First of all, it helps a lot to sketch the two functions $f_1(t)$ and $f_2(t)$ as defined in

$$f_1(t)*h(t)=f_2(t)$$

Note that $f_2(t)$ can be written as a sum of several shifted versions of $f_1(t)$:

$$f_2(t)=f_1(t+a_1)+f_1(t+a_2)+f_1(t+a_3)+f_1(t+a_4)\tag{1}$$

with appropriate constants $a_i$. If you know the identity

$$f_1(t)*\delta(t+a)=f_1(t+a)$$

then you can write (1) as a convolution of $f_1(t)$ with several shifted delta impulses, which will give you the required $h(t)$.