Given is a system that can be described as

$y(t) = x(t)\cdot \sigma(t)$


$\sigma(t) = \left\{\begin{array}{ll} 1, & t \geq 0 \\ 0, & t<0\end{array}\right. .$

The output of a time-varying linear system can be written as:

$\int_{-\infty}^{\infty}x(t-\hat{\tau})\cdot h(t,\hat{\tau})d\hat{\tau}$

where t is the absolute time and $\hat{\tau}$ is the time lag. I need to define a time-varying impulse response $h(t,\hat{\tau})$ to obtain a system behavior as given above and verify your output.

My idea so far was to use the relation:

$y(t) = x(t)\cdot \sigma(t) = \int_{-\infty}^{\infty}x(t-\hat{\tau})\cdot h(t,\hat{\tau})d\hat{\tau}$

and try to do some sort of comparison of coefficients either in the time-domain or in the Frequency domain, but since x(t) itself is not given and can be arbitrary I could not get anywhere.

My second idea is to somehow get $h(t,\tau)$ by using the delta-dirac as an input signal, but i am not really sure how to insert it in this relation, since we never really did much with time-varying systems.

Is there a general way to solve examples like this?


1 Answer 1


If you use a shifted Dirac impulse $\delta(t-T)$ as an input, the corresponding output is


For the given system we obtain


or, using $\tau=t-T$,


This is also intuitively clear because the operation $x(t)\sigma(t)$ is memoryless, i.e., $h(t,\tau)$ must be zero for $\tau\neq 0$, and it must depend on $t$ because the system is time-varying.

Also take a look at this related question and its answer.

  • $\begingroup$ thanks for the help. Seeing the solution makes me wonder how i couldn't see it before in all my tries. The same idea also works for the other question so this was a big help $\endgroup$
    – Kaiser F
    Jan 9 at 13:07

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.