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Given is a system that can be described as

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

with

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

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?

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1 Answer 1

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

$$y(t)=\int_{-\infty}^{\infty}\delta(t-T-\tau)h(t,\tau)d\tau=h(t,t-T)\tag{1}$$

For the given system we obtain

$$\delta(t-T)\sigma(t)=h(t,t-T)\tag{2}$$

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

$$h(t,\tau)=\delta(\tau)\sigma(t)\tag{4}$$

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.

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  • $\begingroup$ 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

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