How to derive the discrete delta function from geometric sum of complex sinusoids?












0












$begingroup$


Using this in the context of Fourier transforms. This should probably be an easy derivation for you guys, but I forget how to derive it.



$$
sum_{n=0}^{L-1}e^{-jfrac{2pi k}{L}n} = frac{1-e^{-j2pi k}}{1-e^{frac{-j2pi k}{L}}}
$$



which somehow can also be reduced to $$Ldelta [k] $$






exponential-sum







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Write the thing inside the sum as $u^n$, where $u = exp(-2jpi k/L)$. Then it's just a sum of a geometric series.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:53












  • $begingroup$
    Gah, yes I see it now. Any hints about the reduction to the discrete delta function?
    $endgroup$
    – user50420
    Dec 11 '18 at 22:56










  • $begingroup$
    Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 23:00












  • $begingroup$
    true, I made the update. Thanks for helping with the first half, though.
    $endgroup$
    – user50420
    Dec 11 '18 at 23:13










  • $begingroup$
    Is the term inside the summation $e^{-j frac{2 pi k n}{L}}$? (i.e. missing $n$?)
    $endgroup$
    – Aditya Dua
    Dec 11 '18 at 23:15
















0












$begingroup$


Using this in the context of Fourier transforms. This should probably be an easy derivation for you guys, but I forget how to derive it.



$$
sum_{n=0}^{L-1}e^{-jfrac{2pi k}{L}n} = frac{1-e^{-j2pi k}}{1-e^{frac{-j2pi k}{L}}}
$$



which somehow can also be reduced to $$Ldelta [k] $$






exponential-sum







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Write the thing inside the sum as $u^n$, where $u = exp(-2jpi k/L)$. Then it's just a sum of a geometric series.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:53












  • $begingroup$
    Gah, yes I see it now. Any hints about the reduction to the discrete delta function?
    $endgroup$
    – user50420
    Dec 11 '18 at 22:56










  • $begingroup$
    Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 23:00












  • $begingroup$
    true, I made the update. Thanks for helping with the first half, though.
    $endgroup$
    – user50420
    Dec 11 '18 at 23:13










  • $begingroup$
    Is the term inside the summation $e^{-j frac{2 pi k n}{L}}$? (i.e. missing $n$?)
    $endgroup$
    – Aditya Dua
    Dec 11 '18 at 23:15














0












0








0





$begingroup$


Using this in the context of Fourier transforms. This should probably be an easy derivation for you guys, but I forget how to derive it.



$$
sum_{n=0}^{L-1}e^{-jfrac{2pi k}{L}n} = frac{1-e^{-j2pi k}}{1-e^{frac{-j2pi k}{L}}}
$$



which somehow can also be reduced to $$Ldelta [k] $$






exponential-sum







share|cite|improve this question











$endgroup$




Using this in the context of Fourier transforms. This should probably be an easy derivation for you guys, but I forget how to derive it.



$$
sum_{n=0}^{L-1}e^{-jfrac{2pi k}{L}n} = frac{1-e^{-j2pi k}}{1-e^{frac{-j2pi k}{L}}}
$$



which somehow can also be reduced to $$Ldelta [k] $$






exponential-sum




exponential-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 3:12







user50420

















asked Dec 11 '18 at 22:51









user50420user50420

253




253








  • 2




    $begingroup$
    Write the thing inside the sum as $u^n$, where $u = exp(-2jpi k/L)$. Then it's just a sum of a geometric series.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:53












  • $begingroup$
    Gah, yes I see it now. Any hints about the reduction to the discrete delta function?
    $endgroup$
    – user50420
    Dec 11 '18 at 22:56










  • $begingroup$
    Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 23:00












  • $begingroup$
    true, I made the update. Thanks for helping with the first half, though.
    $endgroup$
    – user50420
    Dec 11 '18 at 23:13










  • $begingroup$
    Is the term inside the summation $e^{-j frac{2 pi k n}{L}}$? (i.e. missing $n$?)
    $endgroup$
    – Aditya Dua
    Dec 11 '18 at 23:15














  • 2




    $begingroup$
    Write the thing inside the sum as $u^n$, where $u = exp(-2jpi k/L)$. Then it's just a sum of a geometric series.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:53












  • $begingroup$
    Gah, yes I see it now. Any hints about the reduction to the discrete delta function?
    $endgroup$
    – user50420
    Dec 11 '18 at 22:56










  • $begingroup$
    Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note.
    $endgroup$
    – John Hughes
    Dec 11 '18 at 23:00












  • $begingroup$
    true, I made the update. Thanks for helping with the first half, though.
    $endgroup$
    – user50420
    Dec 11 '18 at 23:13










  • $begingroup$
    Is the term inside the summation $e^{-j frac{2 pi k n}{L}}$? (i.e. missing $n$?)
    $endgroup$
    – Aditya Dua
    Dec 11 '18 at 23:15








2




2




$begingroup$
Write the thing inside the sum as $u^n$, where $u = exp(-2jpi k/L)$. Then it's just a sum of a geometric series.
$endgroup$
– John Hughes
Dec 11 '18 at 22:53






$begingroup$
Write the thing inside the sum as $u^n$, where $u = exp(-2jpi k/L)$. Then it's just a sum of a geometric series.
$endgroup$
– John Hughes
Dec 11 '18 at 22:53














$begingroup$
Gah, yes I see it now. Any hints about the reduction to the discrete delta function?
$endgroup$
– user50420
Dec 11 '18 at 22:56




$begingroup$
Gah, yes I see it now. Any hints about the reduction to the discrete delta function?
$endgroup$
– user50420
Dec 11 '18 at 22:56












$begingroup$
Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note.
$endgroup$
– John Hughes
Dec 11 '18 at 23:00






$begingroup$
Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note.
$endgroup$
– John Hughes
Dec 11 '18 at 23:00














$begingroup$
true, I made the update. Thanks for helping with the first half, though.
$endgroup$
– user50420
Dec 11 '18 at 23:13




$begingroup$
true, I made the update. Thanks for helping with the first half, though.
$endgroup$
– user50420
Dec 11 '18 at 23:13












$begingroup$
Is the term inside the summation $e^{-j frac{2 pi k n}{L}}$? (i.e. missing $n$?)
$endgroup$
– Aditya Dua
Dec 11 '18 at 23:15




$begingroup$
Is the term inside the summation $e^{-j frac{2 pi k n}{L}}$? (i.e. missing $n$?)
$endgroup$
– Aditya Dua
Dec 11 '18 at 23:15










1 Answer
1






active

oldest

votes


















0












$begingroup$

When $k=0$, the summation is simply $L$.



For $k neq 0$, write the numerator as: $e^{-j pi k} left( e^{j pi k} - e^{-j pi k} right)$ = $e^{-j pi k} times 2j sin(k pi)$, which is $0$ for $k neq 0$, because $sin(k pi)=0$.



Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k neq 0$. This can be expressed as $L delta[k]$.



Also, the denominator should read $1- e^{-j frac{2 pi k}{L}}$.






share|cite|improve this answer









$endgroup$













    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%2f3035957%2fhow-to-derive-the-discrete-delta-function-from-geometric-sum-of-complex-sinusoid%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









    0












    $begingroup$

    When $k=0$, the summation is simply $L$.



    For $k neq 0$, write the numerator as: $e^{-j pi k} left( e^{j pi k} - e^{-j pi k} right)$ = $e^{-j pi k} times 2j sin(k pi)$, which is $0$ for $k neq 0$, because $sin(k pi)=0$.



    Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k neq 0$. This can be expressed as $L delta[k]$.



    Also, the denominator should read $1- e^{-j frac{2 pi k}{L}}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      When $k=0$, the summation is simply $L$.



      For $k neq 0$, write the numerator as: $e^{-j pi k} left( e^{j pi k} - e^{-j pi k} right)$ = $e^{-j pi k} times 2j sin(k pi)$, which is $0$ for $k neq 0$, because $sin(k pi)=0$.



      Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k neq 0$. This can be expressed as $L delta[k]$.



      Also, the denominator should read $1- e^{-j frac{2 pi k}{L}}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        When $k=0$, the summation is simply $L$.



        For $k neq 0$, write the numerator as: $e^{-j pi k} left( e^{j pi k} - e^{-j pi k} right)$ = $e^{-j pi k} times 2j sin(k pi)$, which is $0$ for $k neq 0$, because $sin(k pi)=0$.



        Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k neq 0$. This can be expressed as $L delta[k]$.



        Also, the denominator should read $1- e^{-j frac{2 pi k}{L}}$.






        share|cite|improve this answer









        $endgroup$



        When $k=0$, the summation is simply $L$.



        For $k neq 0$, write the numerator as: $e^{-j pi k} left( e^{j pi k} - e^{-j pi k} right)$ = $e^{-j pi k} times 2j sin(k pi)$, which is $0$ for $k neq 0$, because $sin(k pi)=0$.



        Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k neq 0$. This can be expressed as $L delta[k]$.



        Also, the denominator should read $1- e^{-j frac{2 pi k}{L}}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 23:21









        Aditya DuaAditya Dua

        1,18418




        1,18418






























            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%2f3035957%2fhow-to-derive-the-discrete-delta-function-from-geometric-sum-of-complex-sinusoid%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

            Le Mesnil-Réaume

            Image
            Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more

            web3.py web3.isConnected() returns false always

            Image
            1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
            Read more