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Say input-output of a system is defined as:

$$ x[n] \longrightarrow x[nM] $$

then what will be the output of $x[n-1]$?

will it be:

\begin{align} x[n-1] \longrightarrow &x\left[(n-1)M\right] = x\left[nM - M\right]\\ &\textrm{or}\\ x[n-1] \longrightarrow &x\left[nM - 1\right] \end{align}

And if someone says shift a signal by 1, if the signal is $x[2n-1]$ then the result will be

$$x[2n-1-1]\quad \textrm{or}\quad x[2(n-1)-1]\quad ?$$

I'm really confused between these. Can anyone make me understand it, please?

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Perhaps the easiest way to understand it is by seeing what happens as we plot graphs of an arbitrary sequence $x[n]$:

enter image description here

As it goes through the system and take one every second sample (i.e. $x[0]$, $x[2]$, $x[4]$, $x[6]$, $x[8]$ and $x[10]$ in the graph above), you get $y[n] = x[nM]$ which can be depicted as:

enter image description here

Now when you shift the input you get $x'[n] = x[n-1]$:

enter image description here

If we again take one out of every second sample we would then get $y'[n] = x'[nM]$:

enter image description here

We then note that this corresponds to the sequence $x[1]$, $x[3]$, $x[5]$, $x[7]$ and $x[9]$ from the original graph. For this simple case where $M=2$ you can see that the result is $$ \begin{align*} y'[n] &= x'[nM] \\ &= x[2n-1] \\ & =x[nM-1] &\mbox{given $M=2$} \end{align*} $$

A similar process could be followed to arrive at the general result whereby the output of that system given an input of $x[n-1]$ would be $x[nM-1]$.

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  • $\begingroup$ thank you. now I think I know why in multi-rate systems we do analysis in the frequency domain instead of the time domain. $\endgroup$ – magneto Aug 3 '16 at 4:19
  • $\begingroup$ It just occurred to me that the 3rd and 4th figures are incorrectly shifted to the left. I'll refresh the figures as soon as I get the chance. $\endgroup$ – SleuthEye Aug 3 '16 at 12:41
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  • Define the relation: $$y[n]=x[nM], \quad -\infty<n<\infty\tag{1}$$

    Let $x_1[n]=x[n-1]$, then from $(1)$ the output $y_1[n]$ from $x_1[n]$ (i.e. your $x[n-1]$) can be computed as follows:
    $$y_1[n]=x_1[nM]=x[nM-1], \quad -\infty<n<\infty\tag{2}$$

  • If the signal is $x[2n−1]$, let $x_2[n]=x[2n-1]$. Then let's have the relation $$y_2[n]=x_2[n]=x[2n−1], \quad -\infty<n<\infty\tag{3}$$ Shifting the output by 1, you get: $$y_2[n-1]=x_2[n-1]=x[2(n-1)−1], \quad -\infty<n<\infty\tag{4}$$

N.B. With $n\in\mathbb Z \textrm{ and } M\in \mathbb N$.

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