# Is applying a linear operation to the input and output of a continuous linear operator OK?

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.

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

$$Tr\{ax_1[n]+bx_2[n]\}=n(ax_1[n]+bx_2[n])=aTr\{x_1[n]\}+bTr\{x_2[n]\}$$

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.

• 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 } – user45664 Apr 24 '17 at 15:52
• 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? :) – user45664 Apr 24 '17 at 16:57
• 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? – Maximilian Matthé Apr 24 '17 at 18:01
• 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. – user45664 Apr 24 '17 at 23:17