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
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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









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    1 Answer
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    1 Answer
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    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






























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