0
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\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

1
\$\begingroup\$

\$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.

\$\endgroup\$
1
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.