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We know that for an LTI system, if $y(t)$ is the output for $x(t)$ then the response for $x(t-2)$ will be $y(t-2)$ and so on.

But my question is what will be the system response for the input $x(-2t)$ ?Is it possible to determine the output without knowing $h(t)$?

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  • $\begingroup$ Are $x(t)$ and $y(t)$ known for all time $t$? $\endgroup$ – fibonatic Aug 19 at 14:22
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I think it's instructive to look at this in the time domain. For LTI systems, the output $y(t)$ is given by the convolution of the input $x(t)$ with the impulse response $h(t)$:

$$y(t)=(x\star h)(t)=\int_{-\infty}^{\infty}x(t-\tau)h(\tau)d\tau\tag{1}$$

From $(1)$, the response to $x_a(t)=x(t-a)$, $a\in\mathbb{R}$, is given by $y(t-a)$:

$$\begin{align}(x_a\star h)(t)&=\int_{-\infty}^{\infty}x_a(t-\tau)h(\tau)d\tau\\&=\int_{-\infty}^{\infty}x(t-a-\tau)h(\tau)d\tau\\&=y(t-a)\end{align}\tag{2}$$

which proves time-invariance.

However, the response to $x(at)$, $x\in\mathbb{R}$, is

$$\int_{-\infty}^{\infty}x(a(t-\tau))h(\tau)d\tau\neq y(at)\tag{3}$$

Consequently, in general we cannot deduce from the response $y(t)$ to an input $x(t)$ the response to a time-scaled input $x(at)$.

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In an LTI system the two things you can ascertain about your output without looking at your impulse response are time invariance and linearity, as they are the intrinsic properties held by the input-output of the system. However, you cannot compare/predict between a time-scaled input, and a scalar multiplied input (or even a delayed input for that matter) because the frequency composition of the two cases are different (scaling time by a factor more than 1 increases frequency and less than 1 decreases frequency) which necessitates the use of the impulse response (the system may behave differently at different frequencies, which is why we have H(s), the frequency response of the system).

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Let us start with some observations. There are systems that have an known behavior to scale changes in the signal. Two natural examples are the constant system, and the Fourier transform.

First, a system $\mathcal{S}$ with constant $h(t)=C$ will be somehow insensitive to a positive scale factor $a>0$. Then you should get: $\mathcal{S}(s(at))= \frac{1}{a}\mathcal{S(s(t))}$, because the system is itself invariant to scaling.

Second, for the Fourier transform, you have the famous "scaling" property for arbitrary $a$: if $s(t) \to S(\omega)$, then $s(at) \to \frac{1}{|a|}S\left(\frac{\omega}{a}\right)$.

The sines and cosines are the prototypical invariant functions of LTI systems, i.e. those that keep their shapes, and are only modified in amplitude and phase. So what would this mean for $t\to \cos(\omega t)$? If one could knew the output of $\mathcal{S}$ to $t\to \cos(-2\omega t)=\cos(2\omega t)$ from $t\to \cos(\omega t)$, than would be too good to be true: you could know the answer of a twice faster signal from a slow one! That would have revolutionized all the sensor industry. And that would be the first tool to teach to DSP students.

So, as we know that observations may come at different scales (like a scene at different distances), image processing and signal analysis people have ad to develop of lot of more or less precise tools to work at different scales, like multiscale transformations, scale-invariant descriptors, etc.

However, some processes studied in signal processing possess self-similarity and scale-invariant behavior, and some works have been trying to devise scale-invariant systems (instead of being time-invariant) even in discrete time (LSI for Linear Scale-Invariant system), for instance in Continuous-dilation discrete-time self-similar signals and linear scale-invariant systems, ICASSP 1998:

In this paper we present a novel model for purely discrete-time self-similar processes and scale-invariant systems. The results developed are based on a new interpretation of the discrete-time scaling (equivalently dilation or contraction) operation which is defined through a mapping between discrete and continuous time. It is shown that it is possible to have continuous scaling factors through this operation even though the signal itself is discrete-time. We study both deterministic and stochastic discrete-time self-similar signals. We then derive the existence conditions of discrete-time deterministically self-similar signals with respect to some specific mappings. Finally, we discuss the construction of discrete-time linear scale-invariant system and present results related to white noise driven system models of stochastic self-similar signals. Unlike continuous-time self-similar signals, it is possible to construct a wide class of non-trivial discrete-time self-similar signals.

I could however not find a lot of contributions in that direction.

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    $\begingroup$ :-)) after the edit I saw that it was already $a>0$ ;-)) $\endgroup$ – Fat32 Aug 19 at 23:01
  • $\begingroup$ My mistake, I left a trap in the $|a|$ for the scaling property $\endgroup$ – Laurent Duval Aug 20 at 6:46
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Not without knowing $h(t)$.

You could see this by the following simple argumentation. Let your compressed signal be $x(2t)$ whose CTFT is $0.5 X(\omega/2)$ and denote its output as $y_2(t)$ and $Y_2(\omega)$. Then you will have the following relationships:

$$ Y(\omega) = H(\omega) X(\omega) \tag{1} $$

$$ Y_2(\omega) = H(\omega) X_2(\omega) = 0.5 H(\omega) X(\omega/2) = \tag{2} $$

Use Eq.(1) to get rid of $X(\omega/2)$ in Eq.(2)

$$ Y_2(\omega) =0.5 H(\omega) \frac{Y(\omega/2)}{ H(\omega/2)} = \frac{ 0.5 H(\omega)}{ H(\omega/2)} Y(\omega/2) \tag{3} $$

$$ Y_2(\omega) = G(\omega) Y(\omega/2) \tag{4} $$

Now if this multiplier $G(\omega)$ is known without knowing $H(\omega)$ then you would also know $Y_2(\omega)$ in terms of $Y(\omega)$. But it's not the case.

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