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Assume a bandlimited signal $X(t)$. Given that the output for this signal is $X(t/2)$, what will be the impulse response $h(t)$ of such a system?

\begin{array}{l} X( \omega ) \ =\ \int ^{\infty }_{-\infty } x( t) e ^{-j\omega t} dt \\ Y( \omega ) \ =2X( 2\omega )\\ \end{array}

As the system won't be LTI the output won't be the convolution of the input and the system response. How do I proceed with finding the response of such a system? Any guidance in the direction of the solution will be very helpful.

Edit: Thanks for all the answers, i had reframed this question which was on a similar line. The question goes like this -> Given an input pulse of duration T, the system scales it to a pulse of 2T, design such a system? It would be of great help if someone could point out the right direction to proceed.

user8206055
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    Note that this system is not time-invariant. You may well calculate the response to an impulse signal, but calling it H (as if it were a transfer function for an LTI system) is not correct. – Juancho Jun 02 '20 at 12:59
  • Yes i agree, but is there any way to calculate this impulse response, as convolution also won't be defined between the input and the impulse response. – user8206055 Jun 02 '20 at 13:10
  • I have edited the question, the earlier version was how I had reframed the question from what I understood, I have posted the original question in the Edit. If anybody could give any direction, it would be of great help. – user8206055 Jun 03 '20 at 02:40
  • You could convolve the impulse with itself, this will give you a new impulse twice as long. – Matt L. Jun 03 '20 at 07:33
  • Maintaining the shape of pulse is also needed. – user8206055 Jun 03 '20 at 07:38

2 Answers2

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The input-output relation of such a system can be written as

$$y(t)=\int_{-\infty}^{\infty}\delta(t/2-\tau)x(\tau)d\tau\tag{1}$$

Note that the system is linear but time-varying, and such systems can generally be described by a two-dimensional impulse response $h(t,\tau)$:

$$y(t)=\int_{-\infty}^{\infty}h(t,\tau)x(\tau)d\tau\tag{2}$$

In the given example we simply have

$$h(t,\tau)=\delta(t/2-\tau)\tag{3}$$

For linear time-invariant (LTI) systems, $h(t,\tau)$ only depends on the difference $t-\tau$.

Matt L.
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Because you have this information about the fourier transform of the output that $Y(\omega) = 2X(2\omega)$, so I will suggest you to use the time-frequency scaling property of Fourier Transform itself to compute response of the system to any input $x(t)$.

Time-Frequency scaling of the Fourier Transform says that: $$x(at) \leftrightarrow \frac{1}{|a|}X(\frac{\omega}{a})$$

Notice that the arrow is double sided and so we know that the system's output for any input $x(t)$ in time domain will be inverse fourier transform of $Y(\omega)$. Here $a=\frac{1}{2}$ and so the time-domain output $y(t)$ will be given by: $$y(t) = \mathcal F^{-1}\{Y(\omega)\} = x(\frac{t}{2})$$.

You can use this relation between input and output to find the response of this system.

You can not call it impulse response of the system. It is just input-output relationship of the given system. But if you could give an impulse as input to the system then you would have gotten $y(t) = \delta(\frac{t}{2}) = 2\delta(t)$ as output.

DSP Rookie
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