# Time-shift confusion

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?

Perhaps the easiest way to understand it is by seeing what happens as we plot graphs of an arbitrary sequence $x[n]$: As it goes through the system and take one every second sample (i.e. $x$, $x$, $x$, $x$, $x$ and $x$ in the graph above), you get $y[n] = x[nM]$ which can be depicted as: Now when you shift the input you get $x'[n] = x[n-1]$: If we again take one out of every second sample we would then get $y'[n] = x'[nM]$: We then note that this corresponds to the sequence $x$, $x$, $x$, $x$ and $x$ 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 $x[nM-1]$.

• thank you. now I think I know why in multi-rate systems we do analysis in the frequency domain instead of the time domain. Aug 3 '16 at 4:19
• 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. Aug 3 '16 at 12:41
• 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 $n\in\mathbb Z \textrm{ and } M\in \mathbb N$.