The scientist and engineers guide to digital signal processing, Steven W. Smith, p. 134-135 states:

[...] imagine a linear system receiving an input signal, $x[n]$. and generating an output signal, $y[n]$. Now suppose the input signal is changed in some linear way, resulting in a new input signal, which we will call $x'[n]$. This results in a new output signal, $y'[n]$. The question is, how does the change in the input signal relate to the change in the output signal? The answer is: the output signal is changed in exactly the same linear way that the input signal was changed. [...] A linear change made to the input signal results in the same linear change to the output signal.'

I assume this can be written for the continuous case as:

If $g = L(f)$, then $M(g) = L(M(f))$

where $L, M$ are linear (LTI) operators, $f, g$ are functions corresponding to $x[n], y[n]$ and $M(f), M(g)$ correspond to $x'[n], y'[n]$.

My question is: is this true for the continuous case?

Any references, or a proof would be appreciated.

This question is a near duplicate of this question (answered) and also this one (now deleted). I am sorry about this-spent much time on the web but only found the reference I quoted-but really need an answer asap.


1 Answer 1


To be a bit picky, the claim does not hold for all linear systems. Here's a counter-example:

Let the system $Tr$ be defined by the input-output relation $$y[n]=Tr\{x[n]\}=nx[n].$$ This system is clearly linear, since


Now, let's apply the following linear operation $H$ to an input signal: $y[n]=H\{x[n]\}=x[n-1]$. This operation is also linear. Let's see what happens to the claim:

$$Tr\{H\{x[n]\}\}=nx[n-1]\neq H\{Tr\{x[n]\}=(n-1)x[n-1].$$

So, in general the claim does not hold for a linear system.

However: Assuming the claim is stated for linear time-invariant systems (LTI), then it is correct. (The claim was missing the constraint of time-invariance). The proof is simple: Any LTI system is described by its impulse response, and the output of a system is the convolution of the input signal with its impulse response. Given a system $L$ with impulse response $l[n]$, and a system $M$ with impulse response $m[n]$, we get

$$g=L\{f\}=l*f$$ $$M\{g\}=m*g = L\{M\{f\}\}=l*m*f=m*l*f=M\{L\{f\}\}$$

The reason why this is true is the commutativity of the convolution operation (i.e. $x*y=y*x$). Since the commutativity holds for both discrete and continuous time, the claim holds for both discrete and continuous time.

  • $\begingroup$ Do you have a reference for the proof? { I am trying to see the history of the 'claim', also don't have enough points for a +5 } $\endgroup$
    – user45664
    Commented Apr 24, 2017 at 15:52
  • $\begingroup$ Also, should the final result be L{M{f}} for the proof?? ( That easily follows from what you did. ) I think the associativity of convolution is also needed, is it not? :) $\endgroup$
    – user45664
    Commented Apr 24, 2017 at 16:57
  • 1
    $\begingroup$ Well, a proof that convolution is commutative and associative (what you also need) is e.g. given here: web.eecs.utk.edu/~roberts/WebAppendices/D-ConvProperties.pdf Is that what you were asking for? $\endgroup$ Commented Apr 24, 2017 at 18:01
  • $\begingroup$ My last question was bad--please ignore. Thanks for that reference. I was hoping to find a published paper or a book that contained the proof you gave. $\endgroup$
    – user45664
    Commented Apr 24, 2017 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.