Check if the system is linear












4














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)










share|improve this question
























  • Could you please share a document saying it is not linear? Maybe from the context we could get it
    – Laurent Duval
    Nov 28 at 18:31






  • 1




    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 at 19:35










  • Maybe you could talk to the source to double check there is no misunderstanding
    – Laurent Duval
    Nov 28 at 20:53
















4














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)










share|improve this question
























  • Could you please share a document saying it is not linear? Maybe from the context we could get it
    – Laurent Duval
    Nov 28 at 18:31






  • 1




    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 at 19:35










  • Maybe you could talk to the source to double check there is no misunderstanding
    – Laurent Duval
    Nov 28 at 20:53














4












4








4


1





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)










share|improve this question















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)







linear-systems system-identification






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 28 at 16:52









Peter K.

16.8k83059




16.8k83059










asked Nov 28 at 16:04









sswwqqaa

1305




1305












  • Could you please share a document saying it is not linear? Maybe from the context we could get it
    – Laurent Duval
    Nov 28 at 18:31






  • 1




    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 at 19:35










  • Maybe you could talk to the source to double check there is no misunderstanding
    – Laurent Duval
    Nov 28 at 20:53


















  • Could you please share a document saying it is not linear? Maybe from the context we could get it
    – Laurent Duval
    Nov 28 at 18:31






  • 1




    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 at 19:35










  • Maybe you could talk to the source to double check there is no misunderstanding
    – Laurent Duval
    Nov 28 at 20:53
















Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31




Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31




1




1




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 at 19:35




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 at 19:35












Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53




Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53










4 Answers
4






active

oldest

votes


















5














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.






share|improve this answer

















  • 4




    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 at 17:26






  • 1




    Indeed, I have never heard about "being affine" in the ordinal variable.
    – Laurent Duval
    Nov 28 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 at 18:43












  • I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
    – Laurent Duval
    Nov 28 at 20:17






  • 1




    @LaurentDuval 😅😂🤣😜
    – Peter K.
    Nov 28 at 23:43



















4














[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?).






share|improve this answer



















  • 2




    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 at 20:42






  • 1




    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 at 9:12








  • 1




    Go ahead and I'd suggest you to add the non-additive magma notion
    – Laurent Duval
    Nov 29 at 9:31



















2














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:



$$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$



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:



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



It is easily shown that the given system



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



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



$$h[n]=adelta[n]+bdelta[n-3]tag{4}$$



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 $c$ ($cneq 0$) to its output:



$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$



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.






share|improve this answer





























    -1














    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?).






    share|improve this answer





















    • Maybe you need another folk copying your answer to get the first vote
      – Laurent Duval
      Nov 29 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 at 22:02










    • @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
      – Fat32
      Dec 1 at 0:19










    • Sometimes, less is more. Lo siento
      – Laurent Duval
      Dec 1 at 12:24











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "295"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f53727%2fcheck-if-the-system-is-linear%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    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.






    share|improve this answer

















    • 4




      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 at 17:26






    • 1




      Indeed, I have never heard about "being affine" in the ordinal variable.
      – Laurent Duval
      Nov 28 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 at 18:43












    • I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
      – Laurent Duval
      Nov 28 at 20:17






    • 1




      @LaurentDuval 😅😂🤣😜
      – Peter K.
      Nov 28 at 23:43
















    5














    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.






    share|improve this answer

















    • 4




      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 at 17:26






    • 1




      Indeed, I have never heard about "being affine" in the ordinal variable.
      – Laurent Duval
      Nov 28 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 at 18:43












    • I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
      – Laurent Duval
      Nov 28 at 20:17






    • 1




      @LaurentDuval 😅😂🤣😜
      – Peter K.
      Nov 28 at 23:43














    5












    5








    5






    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.






    share|improve this answer












    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.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 28 at 16:55









    Peter K.

    16.8k83059




    16.8k83059








    • 4




      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 at 17:26






    • 1




      Indeed, I have never heard about "being affine" in the ordinal variable.
      – Laurent Duval
      Nov 28 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 at 18:43












    • I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
      – Laurent Duval
      Nov 28 at 20:17






    • 1




      @LaurentDuval 😅😂🤣😜
      – Peter K.
      Nov 28 at 23:43














    • 4




      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 at 17:26






    • 1




      Indeed, I have never heard about "being affine" in the ordinal variable.
      – Laurent Duval
      Nov 28 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 at 18:43












    • I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
      – Laurent Duval
      Nov 28 at 20:17






    • 1




      @LaurentDuval 😅😂🤣😜
      – Peter K.
      Nov 28 at 23:43








    4




    4




    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 at 17:26




    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 at 17:26




    1




    1




    Indeed, I have never heard about "being affine" in the ordinal variable.
    – Laurent Duval
    Nov 28 at 18:29




    Indeed, I have never heard about "being affine" in the ordinal variable.
    – Laurent Duval
    Nov 28 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 at 18:43






    @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 at 18:43














    I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
    – Laurent Duval
    Nov 28 at 20:17




    I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
    – Laurent Duval
    Nov 28 at 20:17




    1




    1




    @LaurentDuval 😅😂🤣😜
    – Peter K.
    Nov 28 at 23:43




    @LaurentDuval 😅😂🤣😜
    – Peter K.
    Nov 28 at 23:43











    4














    [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?).






    share|improve this answer



















    • 2




      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 at 20:42






    • 1




      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 at 9:12








    • 1




      Go ahead and I'd suggest you to add the non-additive magma notion
      – Laurent Duval
      Nov 29 at 9:31
















    4














    [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?).






    share|improve this answer



















    • 2




      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 at 20:42






    • 1




      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 at 9:12








    • 1




      Go ahead and I'd suggest you to add the non-additive magma notion
      – Laurent Duval
      Nov 29 at 9:31














    4












    4








    4






    [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?).






    share|improve this answer














    [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?).







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 1 at 12:16

























    answered Nov 28 at 19:44









    Laurent Duval

    16.2k32058




    16.2k32058








    • 2




      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 at 20:42






    • 1




      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 at 9:12








    • 1




      Go ahead and I'd suggest you to add the non-additive magma notion
      – Laurent Duval
      Nov 29 at 9:31














    • 2




      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 at 20:42






    • 1




      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 at 9:12








    • 1




      Go ahead and I'd suggest you to add the non-additive magma notion
      – Laurent Duval
      Nov 29 at 9:31








    2




    2




    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 at 20:42




    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 at 20:42




    1




    1




    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 at 9:12






    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 at 9:12






    1




    1




    Go ahead and I'd suggest you to add the non-additive magma notion
    – Laurent Duval
    Nov 29 at 9:31




    Go ahead and I'd suggest you to add the non-additive magma notion
    – Laurent Duval
    Nov 29 at 9:31











    2














    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:



    $$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$



    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:



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



    It is easily shown that the given system



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



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



    $$h[n]=adelta[n]+bdelta[n-3]tag{4}$$



    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 $c$ ($cneq 0$) to its output:



    $$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$



    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.






    share|improve this answer


























      2














      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:



      $$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$



      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:



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



      It is easily shown that the given system



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



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



      $$h[n]=adelta[n]+bdelta[n-3]tag{4}$$



      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 $c$ ($cneq 0$) to its output:



      $$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$



      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.






      share|improve this answer
























        2












        2








        2






        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:



        $$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$



        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:



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



        It is easily shown that the given system



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



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



        $$h[n]=adelta[n]+bdelta[n-3]tag{4}$$



        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 $c$ ($cneq 0$) to its output:



        $$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$



        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.






        share|improve this answer












        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:



        $$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$



        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:



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



        It is easily shown that the given system



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



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



        $$h[n]=adelta[n]+bdelta[n-3]tag{4}$$



        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 $c$ ($cneq 0$) to its output:



        $$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$



        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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 28 at 21:32









        Matt L.

        49k13683




        49k13683























            -1














            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?).






            share|improve this answer





















            • Maybe you need another folk copying your answer to get the first vote
              – Laurent Duval
              Nov 29 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 at 22:02










            • @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
              – Fat32
              Dec 1 at 0:19










            • Sometimes, less is more. Lo siento
              – Laurent Duval
              Dec 1 at 12:24
















            -1














            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?).






            share|improve this answer





















            • Maybe you need another folk copying your answer to get the first vote
              – Laurent Duval
              Nov 29 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 at 22:02










            • @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
              – Fat32
              Dec 1 at 0:19










            • Sometimes, less is more. Lo siento
              – Laurent Duval
              Dec 1 at 12:24














            -1












            -1








            -1






            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?).






            share|improve this answer












            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?).







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 29 at 9:09









            Fat32

            14.2k31129




            14.2k31129












            • Maybe you need another folk copying your answer to get the first vote
              – Laurent Duval
              Nov 29 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 at 22:02










            • @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
              – Fat32
              Dec 1 at 0:19










            • Sometimes, less is more. Lo siento
              – Laurent Duval
              Dec 1 at 12:24


















            • Maybe you need another folk copying your answer to get the first vote
              – Laurent Duval
              Nov 29 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 at 22:02










            • @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
              – Fat32
              Dec 1 at 0:19










            • Sometimes, less is more. Lo siento
              – Laurent Duval
              Dec 1 at 12:24
















            Maybe you need another folk copying your answer to get the first vote
            – Laurent Duval
            Nov 29 at 19:37




            Maybe you need another folk copying your answer to get the first vote
            – Laurent Duval
            Nov 29 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 at 22:02




            @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 at 22:02












            @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
            – Fat32
            Dec 1 at 0:19




            @LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
            – Fat32
            Dec 1 at 0:19












            Sometimes, less is more. Lo siento
            – Laurent Duval
            Dec 1 at 12:24




            Sometimes, less is more. Lo siento
            – Laurent Duval
            Dec 1 at 12:24


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Signal Processing Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f53727%2fcheck-if-the-system-is-linear%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Wiesbaden

            Marschland

            Dieringhausen