Time shifted Fourier series












0












$begingroup$


The periodic pulse function can be represented as a Fourier series as,



$$f_f(t) = a_0 + sum_{i=1}^inf (a_n cos(nomega_0t))$$



where



$$a_0 = Afrac{T_p}{T}$$



$$a_n = 2frac{A}{npi}sin(npifrac{T_p}{T})$$



with period $T$, amplitude $A$ and pulse width $T_p$.



Two periodic pulse functions with different pulse widths/duty cycles represented as Fourier series can be summed as shown graphically here. In these functions, $omega_2 = omega_1N$ where $N$ is an arbitrary number. The duty cycles are given as $frac{T_{p1}}{T_{1}} = frac{1}{N}$, and $frac{T_{p2}}{T_{2}} = 0.5$. Function 1 is negative.



However, I would like the second Fourier series representation to be time shifted, so that the summed function is as shown graphically here.



Is this possible to do with a Fourier series representation, if so, how?










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    The periodic pulse function can be represented as a Fourier series as,



    $$f_f(t) = a_0 + sum_{i=1}^inf (a_n cos(nomega_0t))$$



    where



    $$a_0 = Afrac{T_p}{T}$$



    $$a_n = 2frac{A}{npi}sin(npifrac{T_p}{T})$$



    with period $T$, amplitude $A$ and pulse width $T_p$.



    Two periodic pulse functions with different pulse widths/duty cycles represented as Fourier series can be summed as shown graphically here. In these functions, $omega_2 = omega_1N$ where $N$ is an arbitrary number. The duty cycles are given as $frac{T_{p1}}{T_{1}} = frac{1}{N}$, and $frac{T_{p2}}{T_{2}} = 0.5$. Function 1 is negative.



    However, I would like the second Fourier series representation to be time shifted, so that the summed function is as shown graphically here.



    Is this possible to do with a Fourier series representation, if so, how?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The periodic pulse function can be represented as a Fourier series as,



      $$f_f(t) = a_0 + sum_{i=1}^inf (a_n cos(nomega_0t))$$



      where



      $$a_0 = Afrac{T_p}{T}$$



      $$a_n = 2frac{A}{npi}sin(npifrac{T_p}{T})$$



      with period $T$, amplitude $A$ and pulse width $T_p$.



      Two periodic pulse functions with different pulse widths/duty cycles represented as Fourier series can be summed as shown graphically here. In these functions, $omega_2 = omega_1N$ where $N$ is an arbitrary number. The duty cycles are given as $frac{T_{p1}}{T_{1}} = frac{1}{N}$, and $frac{T_{p2}}{T_{2}} = 0.5$. Function 1 is negative.



      However, I would like the second Fourier series representation to be time shifted, so that the summed function is as shown graphically here.



      Is this possible to do with a Fourier series representation, if so, how?










      share|cite|improve this question











      $endgroup$




      The periodic pulse function can be represented as a Fourier series as,



      $$f_f(t) = a_0 + sum_{i=1}^inf (a_n cos(nomega_0t))$$



      where



      $$a_0 = Afrac{T_p}{T}$$



      $$a_n = 2frac{A}{npi}sin(npifrac{T_p}{T})$$



      with period $T$, amplitude $A$ and pulse width $T_p$.



      Two periodic pulse functions with different pulse widths/duty cycles represented as Fourier series can be summed as shown graphically here. In these functions, $omega_2 = omega_1N$ where $N$ is an arbitrary number. The duty cycles are given as $frac{T_{p1}}{T_{1}} = frac{1}{N}$, and $frac{T_{p2}}{T_{2}} = 0.5$. Function 1 is negative.



      However, I would like the second Fourier series representation to be time shifted, so that the summed function is as shown graphically here.



      Is this possible to do with a Fourier series representation, if so, how?







      functions trigonometry fourier-analysis fourier-series generating-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 14 '18 at 10:14







      dagfinae

















      asked Mar 13 '18 at 16:51









      dagfinaedagfinae

      65




      65






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          We can get a general answer for a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = a_v + sum_{n = 1}^{infty}left[a_n cos(nomega_c (t-tau)) + b_n sin(nomega_c (t - tau))right]$$



          Let $alpha = cos(nomega_c (t-tau))$ and let $beta= sin(nomega_c (t-tau))$. Then



          begin{align*}
          alpha &= cos(nomega_c t - nomega_c tau) = cos(nomega_c t)cos(nomega_c tau) + sin(nomega_c t)sin(nomega_c tau) \
          beta &= sin(nomega_c t - nomega_c tau) = sin(nomega_c t)cos(nomega_c tau) - cos(nomega_c t)sin(nomega_c tau)end{align*}



          We can then combine these results as follows



          $$a_nalpha + b_nbeta = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }cos(nomega_c t) + { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }sin(nomega_c t). $$



          We can define the following to make things simpler



          $$Av_f = { a_v }_f = { frac{a_0}{2} }_f\
          A_f(n,tau) = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }_f \
          B_f(n,tau) = { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }_f.$$



          Using the above definitions, we can define a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = Av_f + sum_{n = 1}^{infty}left[A_f(n,tau) cos(nomega_c t) + B_f(n,tau) sin(nomega_c t)right].$$



          This will allow us to add two trigonometric Fourier series, each with their own distinct time-shifts as follows



          $$ g(t - tau) = Av_g + sum_{n = 1}^{infty}left[A_g(n,tau) cos(nomega_c t) + B_g(n,tau) sin(nomega_c t)right] \
          h(t - lambda) = Av_h + sum_{n = 1}^{infty}left[A_h(n,lambda) cos(nomega_c t) + B_h(n,lambda) sin(nomega_c t)right] $$



          $$ f(t) = g(t - tau) + h(t - lambda) $$



          We can setup some more definition to make things simpler as follows



          begin{align*}
          Av_f &= Av_g + Av_h \
          A_f(n, tau, lambda) &= { A_g(n, tau) + A_h(n, lambda)} \
          B_f(n, tau, lambda) &= { B_g(n, tau) + B_h(n, lambda)}. end{align*}



          Then



          $$
          f(t) = Av_f + sum_{n = 1}^{infty}left[A_f(n, tau, lambda) cos(nomega_c t) + B_f(n, tau, lambda) sin(nomega_c t)right].
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Xander Thanks for cleaning it up a bit.
            $endgroup$
            – Gustav
            Jan 16 at 0:18











          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',
          autoActivateHeartbeat: false,
          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%2f2689559%2ftime-shifted-fourier-series%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          We can get a general answer for a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = a_v + sum_{n = 1}^{infty}left[a_n cos(nomega_c (t-tau)) + b_n sin(nomega_c (t - tau))right]$$



          Let $alpha = cos(nomega_c (t-tau))$ and let $beta= sin(nomega_c (t-tau))$. Then



          begin{align*}
          alpha &= cos(nomega_c t - nomega_c tau) = cos(nomega_c t)cos(nomega_c tau) + sin(nomega_c t)sin(nomega_c tau) \
          beta &= sin(nomega_c t - nomega_c tau) = sin(nomega_c t)cos(nomega_c tau) - cos(nomega_c t)sin(nomega_c tau)end{align*}



          We can then combine these results as follows



          $$a_nalpha + b_nbeta = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }cos(nomega_c t) + { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }sin(nomega_c t). $$



          We can define the following to make things simpler



          $$Av_f = { a_v }_f = { frac{a_0}{2} }_f\
          A_f(n,tau) = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }_f \
          B_f(n,tau) = { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }_f.$$



          Using the above definitions, we can define a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = Av_f + sum_{n = 1}^{infty}left[A_f(n,tau) cos(nomega_c t) + B_f(n,tau) sin(nomega_c t)right].$$



          This will allow us to add two trigonometric Fourier series, each with their own distinct time-shifts as follows



          $$ g(t - tau) = Av_g + sum_{n = 1}^{infty}left[A_g(n,tau) cos(nomega_c t) + B_g(n,tau) sin(nomega_c t)right] \
          h(t - lambda) = Av_h + sum_{n = 1}^{infty}left[A_h(n,lambda) cos(nomega_c t) + B_h(n,lambda) sin(nomega_c t)right] $$



          $$ f(t) = g(t - tau) + h(t - lambda) $$



          We can setup some more definition to make things simpler as follows



          begin{align*}
          Av_f &= Av_g + Av_h \
          A_f(n, tau, lambda) &= { A_g(n, tau) + A_h(n, lambda)} \
          B_f(n, tau, lambda) &= { B_g(n, tau) + B_h(n, lambda)}. end{align*}



          Then



          $$
          f(t) = Av_f + sum_{n = 1}^{infty}left[A_f(n, tau, lambda) cos(nomega_c t) + B_f(n, tau, lambda) sin(nomega_c t)right].
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Xander Thanks for cleaning it up a bit.
            $endgroup$
            – Gustav
            Jan 16 at 0:18
















          1












          $begingroup$

          We can get a general answer for a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = a_v + sum_{n = 1}^{infty}left[a_n cos(nomega_c (t-tau)) + b_n sin(nomega_c (t - tau))right]$$



          Let $alpha = cos(nomega_c (t-tau))$ and let $beta= sin(nomega_c (t-tau))$. Then



          begin{align*}
          alpha &= cos(nomega_c t - nomega_c tau) = cos(nomega_c t)cos(nomega_c tau) + sin(nomega_c t)sin(nomega_c tau) \
          beta &= sin(nomega_c t - nomega_c tau) = sin(nomega_c t)cos(nomega_c tau) - cos(nomega_c t)sin(nomega_c tau)end{align*}



          We can then combine these results as follows



          $$a_nalpha + b_nbeta = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }cos(nomega_c t) + { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }sin(nomega_c t). $$



          We can define the following to make things simpler



          $$Av_f = { a_v }_f = { frac{a_0}{2} }_f\
          A_f(n,tau) = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }_f \
          B_f(n,tau) = { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }_f.$$



          Using the above definitions, we can define a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = Av_f + sum_{n = 1}^{infty}left[A_f(n,tau) cos(nomega_c t) + B_f(n,tau) sin(nomega_c t)right].$$



          This will allow us to add two trigonometric Fourier series, each with their own distinct time-shifts as follows



          $$ g(t - tau) = Av_g + sum_{n = 1}^{infty}left[A_g(n,tau) cos(nomega_c t) + B_g(n,tau) sin(nomega_c t)right] \
          h(t - lambda) = Av_h + sum_{n = 1}^{infty}left[A_h(n,lambda) cos(nomega_c t) + B_h(n,lambda) sin(nomega_c t)right] $$



          $$ f(t) = g(t - tau) + h(t - lambda) $$



          We can setup some more definition to make things simpler as follows



          begin{align*}
          Av_f &= Av_g + Av_h \
          A_f(n, tau, lambda) &= { A_g(n, tau) + A_h(n, lambda)} \
          B_f(n, tau, lambda) &= { B_g(n, tau) + B_h(n, lambda)}. end{align*}



          Then



          $$
          f(t) = Av_f + sum_{n = 1}^{infty}left[A_f(n, tau, lambda) cos(nomega_c t) + B_f(n, tau, lambda) sin(nomega_c t)right].
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Xander Thanks for cleaning it up a bit.
            $endgroup$
            – Gustav
            Jan 16 at 0:18














          1












          1








          1





          $begingroup$

          We can get a general answer for a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = a_v + sum_{n = 1}^{infty}left[a_n cos(nomega_c (t-tau)) + b_n sin(nomega_c (t - tau))right]$$



          Let $alpha = cos(nomega_c (t-tau))$ and let $beta= sin(nomega_c (t-tau))$. Then



          begin{align*}
          alpha &= cos(nomega_c t - nomega_c tau) = cos(nomega_c t)cos(nomega_c tau) + sin(nomega_c t)sin(nomega_c tau) \
          beta &= sin(nomega_c t - nomega_c tau) = sin(nomega_c t)cos(nomega_c tau) - cos(nomega_c t)sin(nomega_c tau)end{align*}



          We can then combine these results as follows



          $$a_nalpha + b_nbeta = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }cos(nomega_c t) + { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }sin(nomega_c t). $$



          We can define the following to make things simpler



          $$Av_f = { a_v }_f = { frac{a_0}{2} }_f\
          A_f(n,tau) = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }_f \
          B_f(n,tau) = { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }_f.$$



          Using the above definitions, we can define a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = Av_f + sum_{n = 1}^{infty}left[A_f(n,tau) cos(nomega_c t) + B_f(n,tau) sin(nomega_c t)right].$$



          This will allow us to add two trigonometric Fourier series, each with their own distinct time-shifts as follows



          $$ g(t - tau) = Av_g + sum_{n = 1}^{infty}left[A_g(n,tau) cos(nomega_c t) + B_g(n,tau) sin(nomega_c t)right] \
          h(t - lambda) = Av_h + sum_{n = 1}^{infty}left[A_h(n,lambda) cos(nomega_c t) + B_h(n,lambda) sin(nomega_c t)right] $$



          $$ f(t) = g(t - tau) + h(t - lambda) $$



          We can setup some more definition to make things simpler as follows



          begin{align*}
          Av_f &= Av_g + Av_h \
          A_f(n, tau, lambda) &= { A_g(n, tau) + A_h(n, lambda)} \
          B_f(n, tau, lambda) &= { B_g(n, tau) + B_h(n, lambda)}. end{align*}



          Then



          $$
          f(t) = Av_f + sum_{n = 1}^{infty}left[A_f(n, tau, lambda) cos(nomega_c t) + B_f(n, tau, lambda) sin(nomega_c t)right].
          $$






          share|cite|improve this answer











          $endgroup$



          We can get a general answer for a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = a_v + sum_{n = 1}^{infty}left[a_n cos(nomega_c (t-tau)) + b_n sin(nomega_c (t - tau))right]$$



          Let $alpha = cos(nomega_c (t-tau))$ and let $beta= sin(nomega_c (t-tau))$. Then



          begin{align*}
          alpha &= cos(nomega_c t - nomega_c tau) = cos(nomega_c t)cos(nomega_c tau) + sin(nomega_c t)sin(nomega_c tau) \
          beta &= sin(nomega_c t - nomega_c tau) = sin(nomega_c t)cos(nomega_c tau) - cos(nomega_c t)sin(nomega_c tau)end{align*}



          We can then combine these results as follows



          $$a_nalpha + b_nbeta = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }cos(nomega_c t) + { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }sin(nomega_c t). $$



          We can define the following to make things simpler



          $$Av_f = { a_v }_f = { frac{a_0}{2} }_f\
          A_f(n,tau) = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }_f \
          B_f(n,tau) = { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }_f.$$



          Using the above definitions, we can define a time-shifted trigonometric Fourier series as follows



          $$ f(t - tau) = Av_f + sum_{n = 1}^{infty}left[A_f(n,tau) cos(nomega_c t) + B_f(n,tau) sin(nomega_c t)right].$$



          This will allow us to add two trigonometric Fourier series, each with their own distinct time-shifts as follows



          $$ g(t - tau) = Av_g + sum_{n = 1}^{infty}left[A_g(n,tau) cos(nomega_c t) + B_g(n,tau) sin(nomega_c t)right] \
          h(t - lambda) = Av_h + sum_{n = 1}^{infty}left[A_h(n,lambda) cos(nomega_c t) + B_h(n,lambda) sin(nomega_c t)right] $$



          $$ f(t) = g(t - tau) + h(t - lambda) $$



          We can setup some more definition to make things simpler as follows



          begin{align*}
          Av_f &= Av_g + Av_h \
          A_f(n, tau, lambda) &= { A_g(n, tau) + A_h(n, lambda)} \
          B_f(n, tau, lambda) &= { B_g(n, tau) + B_h(n, lambda)}. end{align*}



          Then



          $$
          f(t) = Av_f + sum_{n = 1}^{infty}left[A_f(n, tau, lambda) cos(nomega_c t) + B_f(n, tau, lambda) sin(nomega_c t)right].
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 17 at 4:34

























          answered Jan 14 at 21:04









          GustavGustav

          1469




          1469












          • $begingroup$
            @Xander Thanks for cleaning it up a bit.
            $endgroup$
            – Gustav
            Jan 16 at 0:18


















          • $begingroup$
            @Xander Thanks for cleaning it up a bit.
            $endgroup$
            – Gustav
            Jan 16 at 0:18
















          $begingroup$
          @Xander Thanks for cleaning it up a bit.
          $endgroup$
          – Gustav
          Jan 16 at 0:18




          $begingroup$
          @Xander Thanks for cleaning it up a bit.
          $endgroup$
          – Gustav
          Jan 16 at 0:18


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics 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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2689559%2ftime-shifted-fourier-series%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

          Human spaceflight

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          張江高科駅