I'm stuck at exercise 1.18 in Understanding Digital Signal Processing by Richard G. Lyons:

There is an often-used process in DSP called decimation, and in that process we retain some samples of an $x(n)$ input sequence and discard other $x(n)$ samples. Decimation by a factor of two can be described algebraically by $$y(m) = x(2n) \tag{P1-3}$$ where index $m = 0,1,2,3,. . .$ The decimation defined by Eq. (P1-3) means that $y(m)$ is equal to alternate samples (every other sample) of $x(n)$. For example: $y(0) = x(0)$, $y(1) = x(2)$, $y(2) = x(4)$, $y(3) = x(6)$, $. . .$ and so on. Here is the question: Is that decimation process time invariant?

Illustrate your answer by decimating a simple sinusoidal $x(n)$ time-domain sequence by a factor of two to obtain $y(m)$. Next,create a shifted-by-one-sample version of $x(n)$ and call it $x_{shift}(n)$. That new sequence is defined by (P1-4):

$$x_{shift}(n) = x(n+1) \tag{P1-4}$$ Finally, decimate $x_{shift}(n)$ according to Eq. (P1-3) to obtain $y_{shift}(m)$. The decimation process is time invariant if $y_{shift}(m)$ is equal to a time-shifted version of $y(m)$. That is, decimation is time invariant if $$y_{shift}(m) = y(m+1)$$

I'm somewhat confused by the notation $y(m) = x(2n)$--according to the definition of the decimation, I would hve written $y(n) = x(2n)$. Anyway, if I try to resolve the problem as it is stated:

  1. "Illustrate your answer by decimating a simple sinusoidal 𝑥(𝑛) time-domain sequence by a factor of two to obtain 𝑦(𝑚)"

    Let say $x(n)=sin(\omega n) \text{ with } \omega=2\pi f_0 t_s$ $\implies y(m) = x(2n)=sin(2\omega n)$

  2. "create a shifted-by-one-sample version of $x(n)$ and call it $x_{shift}(n)$"

    $x_{shift}(n) = x(n+1) = sin(\omega n+\omega)$

  3. "decimate $x_{shift}(n)$ according to Eq. (P1-3) to obtain $y_{shift}(m)$"

    $y_{shift}(m) = x_{shift}(2n) = x(2n+1) = sin(2\omega n +\omega)$

  4. "That is, decimation is time invariant if $y_{shift}(m) = y(m+1)$"

    $y(m+1) = x(\text{???})$

Eq (P1-3) doesn't really give a relation between $m$ and $n$. So what is $m+1$? According to the decimation definition given in the exercice, I would be tempted to say that $m+1=2n+2$. But that leads to $y(m+1)=sin(2\omega n + 2\omega) \neq y_{shift}(m)$

Could you pinpoint my mistake or clarify my misunderstanding in this exercise?

  • $\begingroup$ Simple answer: when you shift $x[n]$ by $M$, $y[n]$ = shifts by $M/2$. That's not the same shift, so is violates the invariance condition $\endgroup$
    – Hilmar
    Dec 7, 2019 at 15:10
  • $\begingroup$ Thank you @Hilmar. As you see, I do my homework ;) Joke aside, that was mostly what I was writing in a comment below: $y_{shift}(m) \neq y(m+1)$ is the right conclusion. I don't know why I was persuaded we had to prove it was otherwise. $\endgroup$ Dec 7, 2019 at 15:15
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    $\begingroup$ You should indeed be confused by the notation, because the notation is wrong. Decimation by a factor $M$ is described by $y[n]=x[Mn]$. Please browse this site for book recommendations. In my opinion there are much better (and free) books available, such as the one by Orfanidis. $\endgroup$
    – Matt L.
    Dec 7, 2019 at 15:15
  • $\begingroup$ Thanks, @Matt. I will do that. That being said, the Lyons book is very accessible at my level. It refreshes concepts I studied a long time ago and introduce new ones progressively. I saw the post you mentioned and already downloaded a couple of free books from the list. I will double-check to see if the one by Orfanidis was part of them. $\endgroup$ Dec 7, 2019 at 15:20

1 Answer 1


Sample rate conversion systems (expansion or decimation) are time-varying operators.

For your example system of decimation by $M$ :

$$ y[n] = T\{x[n]\} = x[Mn] $$

you can easily see that results of shift in the input and output are not the same; i.e.,

$$ y_1[n] = T\{ x[n-d] \} = x[Mn - d] $$


$$ y_d[n] = y[n-d] = x[M(n - d)] \neq y_1[n]$$

  • $\begingroup$ The books I'm reading right now aren't using (yet?) the $T\{...\}$ notation> But, as a conclusion, and to use that particular exercise wordings, I was right to find that $y_{shift}(m) \neq y(m+1)$--so I don't need to spend more time trying to prove decimation is time-invariant, since this is the exact opposite... $\endgroup$ Dec 7, 2019 at 15:09
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    $\begingroup$ The system notation $y[n] = T\{x[n]\}$ is just a depiction that an input $x[n]$ is mapped (or transformed) into an output $y[n]$, nothing special ;-). Also your conflict was correct as the notation $y[m] = x[Mn]$ is misleading (or may be just a typo)... Time invariance is proved if you can show that the shifted output $y[n-d]$ equals the response for the equally shifted input $T\{x[n-d]\}$. $\endgroup$
    – Fat32
    Dec 7, 2019 at 19:35

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