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The system is as follows: $y[n] = x[2n]$

Shouldnt the system be time invariant because

$y[n-n_0] = x[2n-2n_0]$

and

$T(x[n-n_0]) = x[2n-2n_0]$

These are both equal, therefore why is the system not time invariant?

Jason
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1 Answers1

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$T$ is time invariant if and only if $$ T(x[n-n_0]) = x[2n-n_0] $$ A shift of $n_0$ on $x$ reflects to a shift of $2n_0$ on $y$, therefore the system is not invariant.

You can also check that complex exponentials are not eigenfunctions of this system. (It is linear)

Henricus V.
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