Amplitude of derivatives approximated by continuous wavelet transform











up vote
0
down vote

favorite












I do have a question regarding the meaning of the amplitude of an approximated derivative derived from a continuous wavelet transform. From what is known, applying the CWT (Haar wavelet) to a signal results in an approximation of the first derivative. Applying it several times result in higher order derivatives.



Since the amplitude of the resulting derivative is altered by the scale factor, the real amplitude of the derivative is not known.



My goal is to calculate the highest value of a noisy signal's first derivative. However, the noise is hurting a lot. I found the Savitzky Golay algorithm that should work out for me.



However, I was wondering if there is any way to get the real amplitude of a derivative derived from a CWT.



Ideas are welcome? Thanks






signal-processing wavelets







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    I do have a question regarding the meaning of the amplitude of an approximated derivative derived from a continuous wavelet transform. From what is known, applying the CWT (Haar wavelet) to a signal results in an approximation of the first derivative. Applying it several times result in higher order derivatives.



    Since the amplitude of the resulting derivative is altered by the scale factor, the real amplitude of the derivative is not known.



    My goal is to calculate the highest value of a noisy signal's first derivative. However, the noise is hurting a lot. I found the Savitzky Golay algorithm that should work out for me.



    However, I was wondering if there is any way to get the real amplitude of a derivative derived from a CWT.



    Ideas are welcome? Thanks






    signal-processing wavelets







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I do have a question regarding the meaning of the amplitude of an approximated derivative derived from a continuous wavelet transform. From what is known, applying the CWT (Haar wavelet) to a signal results in an approximation of the first derivative. Applying it several times result in higher order derivatives.



      Since the amplitude of the resulting derivative is altered by the scale factor, the real amplitude of the derivative is not known.



      My goal is to calculate the highest value of a noisy signal's first derivative. However, the noise is hurting a lot. I found the Savitzky Golay algorithm that should work out for me.



      However, I was wondering if there is any way to get the real amplitude of a derivative derived from a CWT.



      Ideas are welcome? Thanks






      signal-processing wavelets







      share|cite|improve this question















      I do have a question regarding the meaning of the amplitude of an approximated derivative derived from a continuous wavelet transform. From what is known, applying the CWT (Haar wavelet) to a signal results in an approximation of the first derivative. Applying it several times result in higher order derivatives.



      Since the amplitude of the resulting derivative is altered by the scale factor, the real amplitude of the derivative is not known.



      My goal is to calculate the highest value of a noisy signal's first derivative. However, the noise is hurting a lot. I found the Savitzky Golay algorithm that should work out for me.



      However, I was wondering if there is any way to get the real amplitude of a derivative derived from a CWT.



      Ideas are welcome? Thanks






      signal-processing wavelets




      signal-processing wavelets






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 15 hours ago









      Henno Brandsma

      100k344107




      100k344107










      asked Nov 6 at 10:32









      user2799180

      1085




      1085



























          active

          oldest

          votes











          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: "69"
          };
          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',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          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%2fmath.stackexchange.com%2fquestions%2f2986993%2famplitude-of-derivatives-approximated-by-continuous-wavelet-transform%23new-answer', 'question_page');
          }
          );

          Post as a guest























          active

          oldest

          votes













          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2986993%2famplitude-of-derivatives-approximated-by-continuous-wavelet-transform%23new-answer', 'question_page');
          }
          );

          Post as a guest






































































          -

          Popular posts from this blog

          How do I know what Microsoft account the skydrive app is syncing to?

          Image
          up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
          Read more

          When does type information flow backwards in C++?

          Image
          up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
          Read more

          Grease: Live!

          Image
          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
          Read more