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\begin{equation} G_c \qquad y[n] = \begin{cases} u[n] & \text{if} & n<0\\ 2u[n] & \text{if} & n\ge 0 \end{cases} \end{equation}

I found this question in web: Whether or not is this time-invariant. The short answer is no. But I want to make sure I know how to prove this.

Let's take a certain $n<0$ then $y[n] = 2u[n]$. Now, let's put instead of $n \rightarrow (n + n_0)>0$. Since, it is time-invariant then $y[n+n_0] = u[n+n_0]$.
But $(n+n_0)>0$, so $y[n+n_0] = 2u[n+n_0]$. So the only chance is: $u[n+n_0] = 0$, but since $n$ and $n_0$ where randomly chosen then, this system can be time-varying as long as always $u[n+n_0]=0 \rightarrow u[n] =0$ and $y[n] =0$ which doesn't hold as a system. So it is not time-varying. Does this makes sense?

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    $\begingroup$ It looks like y[n] is a signal, not a system. Time invariance is a property of systems and makes no sense for signals. $\endgroup$
    – Hilmar
    Dec 20 '20 at 20:08
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    $\begingroup$ Usually $u[n]$ is notation for the unit step function, is this what you meant? Just by the definition, I suspect not and you should edit you question to clarify what is $u[n]$ and what is $y[n]$. Is $u[n]$ the input signal and $y[n]$ the output signal? $\endgroup$
    – Engineer
    Dec 20 '20 at 20:46
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    $\begingroup$ @Engineer Some texts use $u_n$ as a signal. But yes -- it can be confusing, and should be clarified. $\endgroup$
    – TimWescott
    Dec 21 '20 at 3:54
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Let us suppose that the system with input $x$ is defined by:

\begin{equation} y[n] = \begin{cases} x[n] & \text{if} & n<0\\ 2x[n] & \text{if} & n\ge 0 \end{cases} \end{equation}

As answered by @TimWescott, the two-part definition suggests a lack of time invariance. Finding a counter-example could be a good start. For instance a two-part signal whose behavior when multiplied by 2 is visible. Indeed a modification of the Heavide jump is interesting:

\begin{equation} x_0[n] = \begin{cases} -1& \text{if} & n<0\\ \phantom{-}1 & \text{if} & n\ge 0 \end{cases} \end{equation} along with the right-shifted version: \begin{equation} x_1[n] = \begin{cases} -1 & \text{if} & n<1\\ \phantom{-}1 & \text{if} & n\ge 1\,. \end{cases} \end{equation}

You get:

\begin{equation} y_0[n] = \begin{cases} -1& \text{if} & n<0\\ \phantom{-}2 & \text{if} & n\ge 0 \end{cases} \end{equation} along with the output for the right-shifted version: \begin{equation} y_1[n] = \begin{cases} -1 & \text{if} & n<0\\ -2 & \text{if} & n=0\\ \phantom{-}2 & \text{if} & n\ge 1\,. \end{cases} \end{equation} which are not time-shifted versions of each other.

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Your proof errs in two places. The first is probably just a typo, where you say "$n<0$ then $y[n] = 2u[n]$". If $n < 0$, then $y[n] = 0$, not the other way around.

The second error is the deeper one:

but since $n$ and $n_0$ where randomly chosen then, this system can be time-varying as long as always $u[n+n_0]=0 \rightarrow u[n] =0$ and $y[n] =0$ which doesn't hold as a system. So it is not time-varying.

The system is time invariant if for *absolutely any choice of $n$ and $n_0$, the results are the same, only time shifted. So you can't put constraints on $n$ and $n_0$ -- the system has to behave the same way no matter what time it is.

All I have to do to show time invariance is to choose $n$ and $n_0$ such that either $n \ge 0$ and $n + n_0 < 0$, or $n < 0$ and $n + n_0 \ge 0$. That's it -- the system clearly and obviously behaves differently in those two cases.

Basically, at $n = 0$, the behavior of the system changes markedly -- it's time-varying.

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