# Finding Impulse Response for System?

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.

• It is unclear from your question what the integral should look like. Is this what you are trying to show? $$y(t) = \int_{-\infty}^t \! x(T-2)e^{-(t-T)} \, \mathrm{d}T$$ Or this? $$y(t) = \int_{-\infty}^t \! e^{-(t-T)x(T-2)} \, \mathrm{d}T$$ Jun 3, 2013 at 23:04

$h(t) = \int_{-\infty}^t \! \delta(T-2)e^{-(t-T)} \, \mathrm{d}T = e^{-(t-2)}u(t-2)$
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.