decimation in signal processing Recently I was struck with the following question: !enter image description here And this is what I think about it: "Downsampling is one of the rare processes that are NOT time invariant. From the very nature of it's operation, we know if we delay the input sequence by one sample, a downsampler will generate an entirely different output sequence. For example, if we apply an input sequence x(n) = x(0), x(1), x(2), x(3), x(4), etc., to a downsampler and m = 3, the output y(m) will be the sequence x(0), x(3), x(6), etc., Should we delay the input sequence by one our delayed xd(n) input would be x(1),x(2),x(3),x(4),x(5), etc., In this case the downsampled output sequence yd(m) would be x(1),x(4),x(7),etc., which is NOT a delayed version of y(m). Thus downsampling is not time invariant. Is that an adequate answer, in your opinion?

A simple proof for showing time-variancy of a system is as follows:

Let $\vec{x}_1(n) = \vec{x}(n)$ and $\vec{x}_2(n) = \vec{x}(n-n_0)$ so that $\vec{x}_2 (n)$ is a shifted version of $\vec{x}_1(n)$ by $n_0$. Then calculate system output (downsample the input signal in this case):

$\vec{y}_1(n) = \vec{x}_1(2n) = \vec{x}(2n)$

$\vec{y}_2(n) = \vec{x}_2(2n) = \vec{x}(2n - n_0)$

here we can see that : $\vec{y}_1(n-n_0) = \vec{x}(2(n-n_o)) \
e \vec{x}(2n-n_0) = \vec{y}_2(n) $

The output of $\vec{x}(n-n_0)$ is not a shifted version of the output to $\vec{x}(n)$. So the system isn't time-invariant. $\surd$