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.

  • \$\begingroup\$ 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? $$ y(t) = \int_{-\infty}^t \! e^{-(t-T)x(T-2)} \, \mathrm{d}T $$ \$\endgroup\$
    – benjwy
    Jun 3, 2013 at 23:04

1 Answer 1


\$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.

  • \$\begingroup\$ Ahh, I always forget about the sifting property and how to apply it. I see now. Thank you. \$\endgroup\$ Jun 4, 2013 at 0:16

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