Why does decimation make a system time variant?












2














On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."



Why does decimation makes system time variant?






wavelet linear-systems







share|improve this question




















  • 1




    On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
    – Olli Niemitalo
    1 hour ago












  • That link doesnt work.
    – Sweeper
    1 hour ago






  • 1




    try again please.
    – Olli Niemitalo
    1 hour ago
















2














On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."



Why does decimation makes system time variant?






wavelet linear-systems







share|improve this question




















  • 1




    On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
    – Olli Niemitalo
    1 hour ago












  • That link doesnt work.
    – Sweeper
    1 hour ago






  • 1




    try again please.
    – Olli Niemitalo
    1 hour ago














2












2








2







On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."



Why does decimation makes system time variant?






wavelet linear-systems







share|improve this question















On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."



Why does decimation makes system time variant?






wavelet linear-systems




wavelet linear-systems






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 36 mins ago









Matt L.

49.2k13684




49.2k13684










asked 2 hours ago









Sweeper

1497




1497








  • 1




    On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
    – Olli Niemitalo
    1 hour ago












  • That link doesnt work.
    – Sweeper
    1 hour ago






  • 1




    try again please.
    – Olli Niemitalo
    1 hour ago














  • 1




    On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
    – Olli Niemitalo
    1 hour ago












  • That link doesnt work.
    – Sweeper
    1 hour ago






  • 1




    try again please.
    – Olli Niemitalo
    1 hour ago








1




1




On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
1 hour ago






On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
1 hour ago














That link doesnt work.
– Sweeper
1 hour ago




That link doesnt work.
– Sweeper
1 hour ago




1




1




try again please.
– Olli Niemitalo
1 hour ago




try again please.
– Olli Niemitalo
1 hour ago










2 Answers
2






active

oldest

votes


















2














HINT:
Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.






share|improve this answer





















  • I dont understand.
    – Sweeper
    1 hour ago






  • 1




    @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
    – Matt L.
    1 hour ago










  • Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
    – Sweeper
    1 hour ago



















1














Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.



However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance.



Instead of aliasing cancellation, usually with quadrature mirror filters, it would also work to have perfect anti-alias filters, but this is typically not practical.






share|improve this answer























    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%2f54500%2fwhy-does-decimation-make-a-system-time-variant%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    HINT:
    Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.






    share|improve this answer





















    • I dont understand.
      – Sweeper
      1 hour ago






    • 1




      @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
      – Matt L.
      1 hour ago










    • Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
      – Sweeper
      1 hour ago
















    2














    HINT:
    Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.






    share|improve this answer





















    • I dont understand.
      – Sweeper
      1 hour ago






    • 1




      @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
      – Matt L.
      1 hour ago










    • Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
      – Sweeper
      1 hour ago














    2












    2








    2






    HINT:
    Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.






    share|improve this answer












    HINT:
    Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 1 hour ago









    Matt L.

    49.2k13684




    49.2k13684












    • I dont understand.
      – Sweeper
      1 hour ago






    • 1




      @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
      – Matt L.
      1 hour ago










    • Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
      – Sweeper
      1 hour ago


















    • I dont understand.
      – Sweeper
      1 hour ago






    • 1




      @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
      – Matt L.
      1 hour ago










    • Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
      – Sweeper
      1 hour ago
















    I dont understand.
    – Sweeper
    1 hour ago




    I dont understand.
    – Sweeper
    1 hour ago




    1




    1




    @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
    – Matt L.
    1 hour ago




    @Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
    – Matt L.
    1 hour ago












    Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
    – Sweeper
    1 hour ago




    Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
    – Sweeper
    1 hour ago











    1














    Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.



    However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance.



    Instead of aliasing cancellation, usually with quadrature mirror filters, it would also work to have perfect anti-alias filters, but this is typically not practical.






    share|improve this answer




























      1














      Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.



      However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance.



      Instead of aliasing cancellation, usually with quadrature mirror filters, it would also work to have perfect anti-alias filters, but this is typically not practical.






      share|improve this answer


























        1












        1








        1






        Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.



        However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance.



        Instead of aliasing cancellation, usually with quadrature mirror filters, it would also work to have perfect anti-alias filters, but this is typically not practical.






        share|improve this answer














        Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.



        However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance.



        Instead of aliasing cancellation, usually with quadrature mirror filters, it would also work to have perfect anti-alias filters, but this is typically not practical.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 49 mins ago

























        answered 1 hour ago









        Olli Niemitalo

        7,7121233




        7,7121233






























            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%2f54500%2fwhy-does-decimation-make-a-system-time-variant%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

            Morgemoulin

            Image
            Morgemoulin — miejscowość i gmina — Państwo   Francja Region Grand Est Departament Moza Okręg Verdun Kod INSEE 55357 Powierzchnia 6,82 km² Populacja  (1990) • liczba ludności 75 • gęstość 11 os./km² Kod pocztowy 55400 Położenie na mapie Francji Morgemoulin 49°14′N   5°35′E / 49,233333   5,583333 Portal Francja Morgemoulin – miejscowość i gmina we Francji, w regionie Grand Est, w departamencie Moza. Według danych na rok 1990 gminę zamieszkiwało 75 osób, a gęstość zaludnienia wynosiła 11 osób/km² (wśród 2335 gmin Lotaryngii Morgemoulin plasuje się na 970. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 848.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Verdun Abaucourt-Hautecourt • Aincreville • Ambly-sur-Meuse • Amel-sur-l'Étang • Ancemont • Arrancy-sur-Crusne • Aubréville • Autré...
            Read more

            Scott Moir

            Image
            Scott Moir Tessa Virtue i Scott Moir w 2016 r. Reprezentacja   Kanada Data i miejsce urodzenia 2 września 1987 London Wzrost 180 cm Konkurencja Pary taneczne Partner sportowy Tessa Virtue Trener Marie-France Dubreuil, Patrice Lauzon, Romain Haguenauer Poprzednio: Marina Zujewa, Oleg Epstein, Igor Szpilband, Johnny Johns, Carol Moir, Paul MacIntosh, Suzanne Killing Klub Montreal International School of Skating Poprzednio: Arctic Edge FSC Ilderton Skating Club Lokalizacja treningowa Montreal Poprzednio: Canton Kitchener Rekordy życiowe ISU przed sezonem 2018/2019 Nota łączna 206,07 Igrzyska Olimpijskie 2018 Taniec krótki 83,67 Igrzyska Olimpijskie 2018 Taniec dowolny 122,40 Igrzyska Olimpijskie 2018 Dorobek medalowy Reprezentacja   Kanada Igrzyska olimpijskie Złoto Vancouver 2010 pary taneczne Złoto Pjongczang 2018 pary taneczne Złoto Pjongczang 2018 drużynowo Srebro S...
            Read more

            Souastre

            Image
            Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe...
            Read more