# Check if the system is linear

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)

• Could you please share a document saying it is not linear? Maybe from the context we could get it – Laurent Duval Nov 28 '18 at 18:31
• 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. – sswwqqaa Nov 28 '18 at 19:35
• Maybe you could talk to the source to double check there is no misunderstanding – Laurent Duval Nov 28 '18 at 20:53

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.

• 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. – Matt L. Nov 28 '18 at 17:26
• Indeed, I have never heard about "being affine" in the ordinal variable. – Laurent Duval Nov 28 '18 at 18:29
• @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. – Peter K. Nov 28 '18 at 18:43
• I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma? – Laurent Duval Nov 28 '18 at 20:17
• @LaurentDuval 😅😂🤣😜 – Peter K. Nov 28 '18 at 23:43

[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?).

• 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. – Matt L. Nov 28 '18 at 20:42
• 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 ;-) – Fat32 Nov 29 '18 at 9:12
• Go ahead and I'd suggest you to add the non-additive magma notion – Laurent Duval Nov 29 '18 at 9:31

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.

This system is, as far as Peter K., Laurent, and I know, nicely linear. It is also time-invariant, causal, and stable. Indeed it's LTI with impulse response $$h[n] = a \delta[n] + b \delta[n-3]$$.

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

• Maybe you need another folk copying your answer to get the first vote – Laurent Duval Nov 29 '18 at 19:37
• @LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-)) – Fat32 Nov 29 '18 at 22:02
• @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice ! – Fat32 Dec 1 '18 at 0:19
• Sometimes, less is more. Lo siento – Laurent Duval Dec 1 '18 at 12:24