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I know that I have to show that

\begin{align*} y[n-n_0] &= f \Big( \{x[n - n_0]\} \Big) \end{align*}

in order to tell if a signal is time-varying of not.

Having a signal $y[n] = \frac{1}{2}\delta[n] - 2x[n-1]x[n-2]$ is it sufficient to just show that:

$ \begin{align*} y[n - n_0] &= \frac{1}{2}\delta[n-n_0] - 2x[n-n_0-1]x[n-n_0-2] \\ f \Big( \{ x[n-n_0]\} \Big) &= \frac{1}{2}\delta[n] - 2x[n-n_0-1]x[n-n_0-2] \\ \\ &\Rightarrow y[n] \text{ is time variant} \end{align*} $

and if that's correct: Could somebody explain to me what I just did there? My question arises at the $\delta[n]$ as it is affected in the first equation but in the second it isn't. It might be possible that I just don't understand the notation very well.

$f \Big( \{ x[n-n_0]\} \Big)$ is to me an input signal that gets shifted by $n_0$. So all the $x[n]$ of the function "receive" that shifting value.

$y[n - n_0]$ is the ouput signal that gets shifted by $n_0$. Why is it that the $\delta$-function gets shifted too here?

Stefan Falk
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  • I guess you mean if a system is time-varying or not. – Calculon Apr 09 '15 at 17:30
  • @Calculon Imho both should be fine for Time-variant systems, isn't it? Awesome nickname btw :D – Stefan Falk Apr 09 '15 at 17:31
  • the idea of time-variant signal sounds weird. if a signal is not time-variant, then it is just a constant. there is nothing to study there. anyway, in your question you have a system where $y$ is the output and $x$ is the input and what you are trying to show is whether or not $y$ shifts in time accordingly when the input shifts in time. – Calculon Apr 09 '15 at 17:37
  • @Calculon This makes sense to me but all I got is this receipt that you can see in my question. I am absolutely new to SP and at this point I don't get it why the $\delta$-function gets shifted in the first line but not in the second. Dr. MV the third line is the conclusion? do you mean the second line? – Stefan Falk Apr 09 '15 at 17:45
  • In the first line you are looking at the delayed version of the output. In the second line you are looking at what the output would be if you were to delay the input by the same amount. The two are not the same, which satisfies the definition of a time-variant system. – Calculon Apr 09 '15 at 17:50
  • @Calculon I've specified my question a little more. The thing that is bugging me is that the $\delta$-function does not get affected by the shifting and I don't seem to get why that is so. – Stefan Falk Apr 09 '15 at 17:52
  • That is because $\delta$ is not an input of your system. (It is actually an input in the sense that it affects the output but that is not relevant here.) You have a system that is driven by $x$ and outputs $y$. – Calculon Apr 09 '15 at 17:59

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The more common term is a "time invariant system." It can be aptly described as the following:

If the input signal x(t) produces an output y(t) then any time shifted input, x(t + \delta), results in a time-shifted output y(t + \delta)

If you look at $y$ shifted by $n_0$, you get

$$ y[n-n_0] = \delta[n-n_0] -2x[n-1-n_0] x[n-2-n_0] $$

Since $f(x[n]) = \delta[n]-2x[n-1]x[n-2]$, for $f(x[n-n_0])$ we get

$$ f(x[n-n_0]) = \delta[n]-2x[n-1-n_0]x[n-2-n_0] $$

Thus, $y[n-n_0] \neq f(x[n-n_0])$ and the the system is not time invariant.

AnonSubmitter85
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