1
$\begingroup$

Suppose I have an system $$ y(t) = t^{2}x(t).$$

The impulse response of this system would be: $$h(t) = t^{2} \delta(t).$$

Since $\delta(t) = 0$ for $t \neq 0 , h(t) = 0$ for $t \neq 0$. And at $t=0, h(t) = 0$.

But since $$y(t) = \int_{-\infty}^{\infty}x(r)h(t-r)dr,$$ wouldn't $y(t) = 0$?

What am I missing here?

$\endgroup$
1
  • $\begingroup$ Your system's impulse response does not fully characterise your system, unless you take it at every point in time. The convolution integral does not apply here, instead you would have to use the more general Green's function integral which does not assume time-translation invariance of the impulse response. That's why your result is incorrect. $\endgroup$
    – Jazzmaniac
    Jan 9, 2022 at 19:39

1 Answer 1

7
$\begingroup$

This system $$ y(t) = t^2 x(t) $$

is not LTI and therefore does not have an impulse response of the form $h(t) = \mathcal{T}\{\delta(t)\}$.

So your statement $h(t) = t^2 \delta(t)$ is not correct... Hope this solves your confusion.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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