Show if signal is time variant or not

Question

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?

Answer 1

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.