4
$\begingroup$

The system: $$ T(x[n]) = ax[n] + bx[n-3] $$

For me it seems that the system is linear:

$$ \begin{align} T(\alpha_1x_1[n]+\alpha_2x_2[n]) & = a(\alpha_1x_1[n]+\alpha_2x_2[n]) + b(\alpha_1x_1[n-3]+\alpha_2x_2[n-3]) \\ & = \alpha_1(ax_1[n] + bx_1[n-3]) + \alpha_2(ax_2[n] + bx_2[n-3]) \\ & = \alpha_1T(x_1[n])+\alpha_1T(x_2[n]) \end{align}$$

Thus it's linear, however in the presentation I got it says it's not linear. (without reasoning) Where I'm making the mistake (or maybe there is a mistake in the presentation)

|improve this question
$\endgroup$
  • $\begingroup$ Could you please share a document saying it is not linear? Maybe from the context we could get it $\endgroup$ – Laurent Duval Nov 28 '18 at 18:31
  • 1
    $\begingroup$ It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability. $\endgroup$ – sswwqqaa Nov 28 '18 at 19:35
  • $\begingroup$ Maybe you could talk to the source to double check there is no misunderstanding $\endgroup$ – Laurent Duval Nov 28 '18 at 20:53
5
$\begingroup$

I believe there's either a mistake in the presentation or the presentation is using a different definition of linear.

For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset.

On this site, we tend to go with the system definition rather than split hairs about linearity versus affine-ness.

|improve this answer
$\endgroup$
  • 4
    $\begingroup$ The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context. $\endgroup$ – Matt L. Nov 28 '18 at 17:26
  • 1
    $\begingroup$ Indeed, I have never heard about "being affine" in the ordinal variable. $\endgroup$ – Laurent Duval Nov 28 '18 at 18:29
  • $\begingroup$ @MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it. $\endgroup$ – Peter K.♦ Nov 28 '18 at 18:43
  • $\begingroup$ I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma? $\endgroup$ – Laurent Duval Nov 28 '18 at 20:17
  • 1
    $\begingroup$ @LaurentDuval 😅😂🤣😜 $\endgroup$ – Peter K.♦ Nov 28 '18 at 23:43
4
$\begingroup$

[Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something]

This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable.

The only basic property it does not possess is "to be memoryless", unless one of the constant $a$ or $b$ (including both) is zero (see for instance an extended conception of a memoryless system from What is a memory less system?).

|improve this answer
$\endgroup$
  • 2
    $\begingroup$ It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter. $\endgroup$ – Matt L. Nov 28 '18 at 20:42
  • 1
    $\begingroup$ hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-) $\endgroup$ – Fat32 Nov 29 '18 at 9:12
  • 1
    $\begingroup$ Go ahead and I'd suggest you to add the non-additive magma notion $\endgroup$ – Laurent Duval Nov 29 '18 at 9:31
2
$\begingroup$

A system $\mathcal{T}$ is linear if its response to a weighted sum of two signals equals the weighted sum of its individual responses to those two input signals:

$$\mathcal{T}\big\{\alpha x_1+\beta x_2\big\}=\alpha\mathcal{T}\big\{x_1\big\}+\beta\mathcal{T}\big\{x_2\big\}\tag{1}$$

with arbitrary constants $\alpha$ and $\beta$, and arbitrary input sequences $x_1[n]$ and $x_2[n]$. A system $\mathcal{T}$ satisfying $(1)$ is completely characterized by its impulse response $h[n]$, and its input-output relation can be formulated as a convolution of the input sequence with its impulse response:

$$\mathcal{T}\big\{x\big\}=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\tag{2}$$

It is easily shown that the given system

$$\mathcal{T}\big\{x\big\}=y[n]=ax[n]+bx[n-3]\tag{3}$$

satisfies $(1)$, and that it is characterized by the impulse response

$$h[n]=a\delta[n]+b\delta[n-3]\tag{4}$$

Clearly, its input-output relation $(3)$ can be written as a convolution sum $(2)$. Equivalently, the $\mathcal{Z}$-transform of its response is given by the multiplication of the $\mathcal{Z}$-transform of its input sequence and its transfer function $H(z)=\mathcal{Z}\{h[n]\}$:

$$Y(z)=X(z)H(z)\tag{5}$$

By contrast, an affine system does not satisfy $(1)$, and it cannot be characterized by an impulse response or, equivalently, by a transfer function. The given linear system $(3)$ could be made affine by adding a constant $c$ ($c\neq 0$) to its output:

$$\mathcal{T}\big\{x\big\}=y[n]=ax[n]+bx[n-3]+c\tag{6}$$

This input-output relation cannot be formulated in terms of a convolution $(2)$, and it can easily be checked that the system $(6)$ doesn't satisfy $(1)$. Such a system is not linear, since part of the output (the constant $c$) does not depend on the input signal $x[n]$.

There is no reasonable definition of linearity according to which the given system $(3)$ could be classified as being non-linear.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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