I have an LTI system with input and output related as per below:
$$
y(t) = \int_{-\infty}^t \! x(T-2)e^{-(t-T)} \, \mathrm{d}T
$$
and I need to find \$h(t)\$.
I am familiar with two methods of finding \$h(t)\$, namely, comparing the form to the traditional convolution integral and knowing that \$ h(t) = L[\Delta(t)] \$ and relating those forms, but each time, the \$(T-2)\$ bit trips me up.
For the first comparison method, if I set \$\lambda = T-2\$, then \$T = \lambda + 2\$. That puts the x function in an expected form, but turns \$ e^{-(t-T)}\$ into \$e^{-(t – \lambda + 2)}\$ and then I'm not sure how to proceed, given that the added \$+2\$ doesn't give the expected form of \$t – \lambda\$ alone.
Best Answer
\$h(t) = \int_{-\infty}^t \! \delta(T-2)e^{-(t-T)} \, \mathrm{d}T = e^{-(t-2)}u(t-2)\$
The delta "function" is zero except where the argument is zero, i.e., when T=2, where it has an area of 1.
So, if \$ t < 2\$, the integral is zero.
If \$ t \ge 2\$, the integral equals the area of the delta function multiplied by the value of the exponential when T = 2.