Convolution of two impulse functions

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I was trying to solve fourier transform of $cos(2pi f_0 t)u(t)$ where $u(t)$ is a unit step function,



$$
u(t) = begincases
1 text for t$ge$ 0\
0 text for $t< 0$\
endcases
$$



As far i have done that
$$
mathscrFcos(2pi f_0 t)u(t) = mathscrFcos(2pi f_0 t) * mathscrFu(t)\
= frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]
$$
I know that $delta(f-f_0)*x(t) = x(f-f_0)$, so
$$
frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]\
= frac14jpi (f-f_0)+frac14jpi (f+f_0)+frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)]
$$
but what is the value of $frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)] = ?$



Thanks in advance.




fourier-analysis fourier-transform convolution




share|cite|improve this question
























    up vote
    0
    down vote

    favorite












    I was trying to solve fourier transform of $cos(2pi f_0 t)u(t)$ where $u(t)$ is a unit step function,



    $$
    u(t) = begincases
    1 text for t$ge$ 0\
    0 text for $t< 0$\
    endcases
    $$



    As far i have done that
    $$
    mathscrFcos(2pi f_0 t)u(t) = mathscrFcos(2pi f_0 t) * mathscrFu(t)\
    = frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]
    $$
    I know that $delta(f-f_0)*x(t) = x(f-f_0)$, so
    $$
    frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]\
    = frac14jpi (f-f_0)+frac14jpi (f+f_0)+frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)]
    $$
    but what is the value of $frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)] = ?$



    Thanks in advance.




    fourier-analysis fourier-transform convolution




    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I was trying to solve fourier transform of $cos(2pi f_0 t)u(t)$ where $u(t)$ is a unit step function,



      $$
      u(t) = begincases
      1 text for t$ge$ 0\
      0 text for $t< 0$\
      endcases
      $$



      As far i have done that
      $$
      mathscrFcos(2pi f_0 t)u(t) = mathscrFcos(2pi f_0 t) * mathscrFu(t)\
      = frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]
      $$
      I know that $delta(f-f_0)*x(t) = x(f-f_0)$, so
      $$
      frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]\
      = frac14jpi (f-f_0)+frac14jpi (f+f_0)+frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)]
      $$
      but what is the value of $frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)] = ?$



      Thanks in advance.




      fourier-analysis fourier-transform convolution




      share|cite|improve this question












      I was trying to solve fourier transform of $cos(2pi f_0 t)u(t)$ where $u(t)$ is a unit step function,



      $$
      u(t) = begincases
      1 text for t$ge$ 0\
      0 text for $t< 0$\
      endcases
      $$



      As far i have done that
      $$
      mathscrFcos(2pi f_0 t)u(t) = mathscrFcos(2pi f_0 t) * mathscrFu(t)\
      = frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]
      $$
      I know that $delta(f-f_0)*x(t) = x(f-f_0)$, so
      $$
      frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)+frac12jpi f]\
      = frac14jpi (f-f_0)+frac14jpi (f+f_0)+frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)]
      $$
      but what is the value of $frac12[delta(f-f_0)+delta(f+f_0)]*[frac12delta(f)] = ?$



      Thanks in advance.





      fourier-analysis fourier-transform convolution





      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 26 at 6:54









      Leolime

      11




      11




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote













          For every distribution $u$ one has $u(f) * delta(f) = u(f)$ so,
          $$delta(f pm f_0) * delta(f) = delta(f pm f_0).$$






          share|cite|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: "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: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            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%2f2894777%2fconvolution-of-two-impulse-functions%23new-answer', 'question_page');

            );

            Post as a guest

















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote













            For every distribution $u$ one has $u(f) * delta(f) = u(f)$ so,
            $$delta(f pm f_0) * delta(f) = delta(f pm f_0).$$






            share|cite|improve this answer
























              up vote
              0
              down vote













              For every distribution $u$ one has $u(f) * delta(f) = u(f)$ so,
              $$delta(f pm f_0) * delta(f) = delta(f pm f_0).$$






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                For every distribution $u$ one has $u(f) * delta(f) = u(f)$ so,
                $$delta(f pm f_0) * delta(f) = delta(f pm f_0).$$






                share|cite|improve this answer












                For every distribution $u$ one has $u(f) * delta(f) = u(f)$ so,
                $$delta(f pm f_0) * delta(f) = delta(f pm f_0).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 26 at 7:54









                md2perpe

                6,51311023




                6,51311023



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2894777%2fconvolution-of-two-impulse-functions%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    -

                    這個網誌中的熱門文章

                    tkz-euclide: tkzDrawCircle[R] not working

                    圖片
                    Clash Royale CLAN TAG #URR8PPP up vote 4 down vote favorite tkzDrawArc[R,arc](B,1.1*dBD)(20,83) works well. But tkzDrawCircle[R] is not working... documentclass[border=2pt]standalone usepackage[usenames,dvipsnames,svgnames]xcolor usepackagetkz-euclide usetkzobjall definecolorfondpaillecmyk0,0,0.1,0 pagecolorfondpaille colorMaroon begindocument begintikzpicture[scale=1.0] tkzDefPoint(0,0)B tkzDefPoint(0,6)A tkzDefPoint(8,0)C tkzLabelPoints[below](B,C) tkzLabelPoints[left](A) tkzDrawSegment[thick](A,B) tkzDrawSegment[thick](C,B) tkzDrawSegment[thick](A,C) tkzDefMidPoint(B,C)tkzGetPointM tkzInterLC(A,C)(M,C) tkzGetPointsDE tkzLabelPoints[above](D) tkzDrawSegment[thick](B,D) tkzCalcLength[cm](B,D) tkzGetLengthdBD tkzDrawArc[R,arc](B,1.1*dBD)(20,83) tkzDrawCircle[R](A,1.1*dBD) %tkzInterLC[R](A,C)(B,1.1*dBD) tkzGetPointsD_1D_2 endtikzpicture enddocument tkz-euclide share | improve this question edited Aug 13 at 7:07 Max 5,023 1 15 25 asked Aug 13 at 5:...
                    閱讀完整內容

                    How to combine Bézier curves to a surface?

                    圖片
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
                    閱讀完整內容

                    1st Magritte Awards

                    圖片
                    1st Magritte Awards Official poster Date 5 February 2011  ( 2011-02-05 ) Site Square Mont des Arts, Brussels, Belgium Hosted by Helena Noguerra Produced by José Bouquiaux Directed by Vincent J. Gustin Highlights Best Film Mr. Nobody Most awards Mr. Nobody (6) Most nominations Illegal (8) Television coverage Network BeTV Magritte Awards 2nd → The 1st Magritte Awards ceremony, presented by the Académie André Delvaux, honored the best films of 2010 in Belgium and took place on 5 February 2011 at the Square in the historic site of Mont des Arts, Brussels, beginning at 7:30 p.m. CET. During the ceremony, the Académie André Delvaux presented Magritte Awards in twenty categories. The ceremony, televised in Belgium by BeTV, was produced by José Bouquiaux and directed by Vincent J. Gustin. [1] Film director Jaco Van Dormael presided the ceremony, while actress Helena Noguerra hosted the evening. [2] The pre-show ceremony was hoste...
                    閱讀完整內容