Derive Fourier transform for summation of shifted Dirac-delta function












0














How to derive this Fourier transform:



$$ F{ sum_{n=-infty}^{infty} delta (t- nT) text{ }} =omega_o sum_{n=-infty}^{infty} delta (omega - n omega_o)$$



where:



$$ omega_o = 2 pi / T $$






fourier-transform







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  • en.wikipedia.org/wiki/Dirac_comb
    – caverac
    Nov 24 at 15:32










  • dspillustrations.com/pages/posts/misc/…
    – Bill Moore
    Nov 28 at 12:46
















0














How to derive this Fourier transform:



$$ F{ sum_{n=-infty}^{infty} delta (t- nT) text{ }} =omega_o sum_{n=-infty}^{infty} delta (omega - n omega_o)$$



where:



$$ omega_o = 2 pi / T $$






fourier-transform







share|cite|improve this question
























  • en.wikipedia.org/wiki/Dirac_comb
    – caverac
    Nov 24 at 15:32










  • dspillustrations.com/pages/posts/misc/…
    – Bill Moore
    Nov 28 at 12:46














0












0








0







How to derive this Fourier transform:



$$ F{ sum_{n=-infty}^{infty} delta (t- nT) text{ }} =omega_o sum_{n=-infty}^{infty} delta (omega - n omega_o)$$



where:



$$ omega_o = 2 pi / T $$






fourier-transform







share|cite|improve this question















How to derive this Fourier transform:



$$ F{ sum_{n=-infty}^{infty} delta (t- nT) text{ }} =omega_o sum_{n=-infty}^{infty} delta (omega - n omega_o)$$



where:



$$ omega_o = 2 pi / T $$






fourier-transform




fourier-transform






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edited Nov 24 at 15:22









Bernard

117k637110




117k637110










asked Nov 24 at 14:33









Bill Moore

1176




1176












  • en.wikipedia.org/wiki/Dirac_comb
    – caverac
    Nov 24 at 15:32










  • dspillustrations.com/pages/posts/misc/…
    – Bill Moore
    Nov 28 at 12:46


















  • en.wikipedia.org/wiki/Dirac_comb
    – caverac
    Nov 24 at 15:32










  • dspillustrations.com/pages/posts/misc/…
    – Bill Moore
    Nov 28 at 12:46
















en.wikipedia.org/wiki/Dirac_comb
– caverac
Nov 24 at 15:32




en.wikipedia.org/wiki/Dirac_comb
– caverac
Nov 24 at 15:32












dspillustrations.com/pages/posts/misc/…
– Bill Moore
Nov 28 at 12:46




dspillustrations.com/pages/posts/misc/…
– Bill Moore
Nov 28 at 12:46










2 Answers
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oldest

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1














First note that



$$f(t) = sum_{n=-infty}^{+infty}delta(t-nT)$$



is a periodic tempered distribution with period $T$. Therefore you can do a Fourier series expansion:



$$f(t) = sum_{k=-infty}^{+infty}C_k,e^{ik2pi t/T},;mathrm{with};C_k=frac{1}{T}int_a^{a+T}f(t);e^{-ik2pi t/T},dt = frac{1}{T},.$$



In other words:



$$f(t) = sum_{n=-infty}^{+infty}delta(t-nT) = frac{1}{T}sum_{k=-infty}^{+infty}e^{ik2pi t/T},.$$



Now, taking the Fourier transform of the last series we get the desired result:



$$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
=frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






share|cite|improve this answer





























    0














    We start with the Dirac Comb Function:



    $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) $$



    We know the formula for Complex Fourier Series Expansion is:



    $$x(t)=sum_{n=-a+e}^{a+e}c_n exp( j 2 pi n t / T)$$
    Where:



    $$ c_n = frac{1}{T_o}int_{-infty}^{infty}x(t)exp(-j 2 pi n t / T_o) dt$$



    First, we calculate the coefficients for the series expansion of f(t):



    $$ c_n = frac{1}{T}intlimits_{-T/2}^{T/2} sumlimits_{k=-infty}^{+infty} delta(t - kT) e^{-j 2 pi n t /T} dt \
    $$
    Since the period of f(t) is T, and we are integrating between -T/2 and T/2, it follow that only one value of k applies to the integral range, namely k=0. Thus, the equation simplifies to:



    $$c_n = frac{1}{T}intlimits_{-T/2}^{T/2} delta(t) e^{-j 2 pi n t/T} dt quad quad\$$



    Now we use this property of dirac delta:



    $$int_{-a+e}^{a+e}f(t)delta(t - a)dt = f(a)$$



    The equation further simplifies to:



    $$ begin{align}
    & c_n = frac{1}{T}e^{-j 2 pi n 0/T} \
    & c_n = frac{1}{T} quad quad text{(for all integers of n)} \
    end{align} $$



    Substituting the coefficient value, $c_n$, into the formula for Fourier Series Expansion yields:



    $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) = sumlimits_{n=-infty}^{+infty} frac{1}{T} e^{j 2 pi n t/T} $$



    Next we find the Fourier Transform of the Fourier Series Expansion of f(t):



    we know that:
    $$ F { e^{i omega_o t} } = 2 pi delta(omega - omega_o) $$



    Thus:



    $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
    =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






    share|cite|improve this answer























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      2 Answers
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      2 Answers
      2






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      active

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      active

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      1














      First note that



      $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT)$$



      is a periodic tempered distribution with period $T$. Therefore you can do a Fourier series expansion:



      $$f(t) = sum_{k=-infty}^{+infty}C_k,e^{ik2pi t/T},;mathrm{with};C_k=frac{1}{T}int_a^{a+T}f(t);e^{-ik2pi t/T},dt = frac{1}{T},.$$



      In other words:



      $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT) = frac{1}{T}sum_{k=-infty}^{+infty}e^{ik2pi t/T},.$$



      Now, taking the Fourier transform of the last series we get the desired result:



      $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
      =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






      share|cite|improve this answer


























        1














        First note that



        $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT)$$



        is a periodic tempered distribution with period $T$. Therefore you can do a Fourier series expansion:



        $$f(t) = sum_{k=-infty}^{+infty}C_k,e^{ik2pi t/T},;mathrm{with};C_k=frac{1}{T}int_a^{a+T}f(t);e^{-ik2pi t/T},dt = frac{1}{T},.$$



        In other words:



        $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT) = frac{1}{T}sum_{k=-infty}^{+infty}e^{ik2pi t/T},.$$



        Now, taking the Fourier transform of the last series we get the desired result:



        $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
        =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






        share|cite|improve this answer
























          1












          1








          1






          First note that



          $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT)$$



          is a periodic tempered distribution with period $T$. Therefore you can do a Fourier series expansion:



          $$f(t) = sum_{k=-infty}^{+infty}C_k,e^{ik2pi t/T},;mathrm{with};C_k=frac{1}{T}int_a^{a+T}f(t);e^{-ik2pi t/T},dt = frac{1}{T},.$$



          In other words:



          $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT) = frac{1}{T}sum_{k=-infty}^{+infty}e^{ik2pi t/T},.$$



          Now, taking the Fourier transform of the last series we get the desired result:



          $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
          =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






          share|cite|improve this answer












          First note that



          $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT)$$



          is a periodic tempered distribution with period $T$. Therefore you can do a Fourier series expansion:



          $$f(t) = sum_{k=-infty}^{+infty}C_k,e^{ik2pi t/T},;mathrm{with};C_k=frac{1}{T}int_a^{a+T}f(t);e^{-ik2pi t/T},dt = frac{1}{T},.$$



          In other words:



          $$f(t) = sum_{n=-infty}^{+infty}delta(t-nT) = frac{1}{T}sum_{k=-infty}^{+infty}e^{ik2pi t/T},.$$



          Now, taking the Fourier transform of the last series we get the desired result:



          $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
          =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 17:23









          Anders Beta

          68319




          68319























              0














              We start with the Dirac Comb Function:



              $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) $$



              We know the formula for Complex Fourier Series Expansion is:



              $$x(t)=sum_{n=-a+e}^{a+e}c_n exp( j 2 pi n t / T)$$
              Where:



              $$ c_n = frac{1}{T_o}int_{-infty}^{infty}x(t)exp(-j 2 pi n t / T_o) dt$$



              First, we calculate the coefficients for the series expansion of f(t):



              $$ c_n = frac{1}{T}intlimits_{-T/2}^{T/2} sumlimits_{k=-infty}^{+infty} delta(t - kT) e^{-j 2 pi n t /T} dt \
              $$
              Since the period of f(t) is T, and we are integrating between -T/2 and T/2, it follow that only one value of k applies to the integral range, namely k=0. Thus, the equation simplifies to:



              $$c_n = frac{1}{T}intlimits_{-T/2}^{T/2} delta(t) e^{-j 2 pi n t/T} dt quad quad\$$



              Now we use this property of dirac delta:



              $$int_{-a+e}^{a+e}f(t)delta(t - a)dt = f(a)$$



              The equation further simplifies to:



              $$ begin{align}
              & c_n = frac{1}{T}e^{-j 2 pi n 0/T} \
              & c_n = frac{1}{T} quad quad text{(for all integers of n)} \
              end{align} $$



              Substituting the coefficient value, $c_n$, into the formula for Fourier Series Expansion yields:



              $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) = sumlimits_{n=-infty}^{+infty} frac{1}{T} e^{j 2 pi n t/T} $$



              Next we find the Fourier Transform of the Fourier Series Expansion of f(t):



              we know that:
              $$ F { e^{i omega_o t} } = 2 pi delta(omega - omega_o) $$



              Thus:



              $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
              =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






              share|cite|improve this answer




























                0














                We start with the Dirac Comb Function:



                $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) $$



                We know the formula for Complex Fourier Series Expansion is:



                $$x(t)=sum_{n=-a+e}^{a+e}c_n exp( j 2 pi n t / T)$$
                Where:



                $$ c_n = frac{1}{T_o}int_{-infty}^{infty}x(t)exp(-j 2 pi n t / T_o) dt$$



                First, we calculate the coefficients for the series expansion of f(t):



                $$ c_n = frac{1}{T}intlimits_{-T/2}^{T/2} sumlimits_{k=-infty}^{+infty} delta(t - kT) e^{-j 2 pi n t /T} dt \
                $$
                Since the period of f(t) is T, and we are integrating between -T/2 and T/2, it follow that only one value of k applies to the integral range, namely k=0. Thus, the equation simplifies to:



                $$c_n = frac{1}{T}intlimits_{-T/2}^{T/2} delta(t) e^{-j 2 pi n t/T} dt quad quad\$$



                Now we use this property of dirac delta:



                $$int_{-a+e}^{a+e}f(t)delta(t - a)dt = f(a)$$



                The equation further simplifies to:



                $$ begin{align}
                & c_n = frac{1}{T}e^{-j 2 pi n 0/T} \
                & c_n = frac{1}{T} quad quad text{(for all integers of n)} \
                end{align} $$



                Substituting the coefficient value, $c_n$, into the formula for Fourier Series Expansion yields:



                $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) = sumlimits_{n=-infty}^{+infty} frac{1}{T} e^{j 2 pi n t/T} $$



                Next we find the Fourier Transform of the Fourier Series Expansion of f(t):



                we know that:
                $$ F { e^{i omega_o t} } = 2 pi delta(omega - omega_o) $$



                Thus:



                $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
                =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






                share|cite|improve this answer


























                  0












                  0








                  0






                  We start with the Dirac Comb Function:



                  $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) $$



                  We know the formula for Complex Fourier Series Expansion is:



                  $$x(t)=sum_{n=-a+e}^{a+e}c_n exp( j 2 pi n t / T)$$
                  Where:



                  $$ c_n = frac{1}{T_o}int_{-infty}^{infty}x(t)exp(-j 2 pi n t / T_o) dt$$



                  First, we calculate the coefficients for the series expansion of f(t):



                  $$ c_n = frac{1}{T}intlimits_{-T/2}^{T/2} sumlimits_{k=-infty}^{+infty} delta(t - kT) e^{-j 2 pi n t /T} dt \
                  $$
                  Since the period of f(t) is T, and we are integrating between -T/2 and T/2, it follow that only one value of k applies to the integral range, namely k=0. Thus, the equation simplifies to:



                  $$c_n = frac{1}{T}intlimits_{-T/2}^{T/2} delta(t) e^{-j 2 pi n t/T} dt quad quad\$$



                  Now we use this property of dirac delta:



                  $$int_{-a+e}^{a+e}f(t)delta(t - a)dt = f(a)$$



                  The equation further simplifies to:



                  $$ begin{align}
                  & c_n = frac{1}{T}e^{-j 2 pi n 0/T} \
                  & c_n = frac{1}{T} quad quad text{(for all integers of n)} \
                  end{align} $$



                  Substituting the coefficient value, $c_n$, into the formula for Fourier Series Expansion yields:



                  $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) = sumlimits_{n=-infty}^{+infty} frac{1}{T} e^{j 2 pi n t/T} $$



                  Next we find the Fourier Transform of the Fourier Series Expansion of f(t):



                  we know that:
                  $$ F { e^{i omega_o t} } = 2 pi delta(omega - omega_o) $$



                  Thus:



                  $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
                  =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$






                  share|cite|improve this answer














                  We start with the Dirac Comb Function:



                  $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) $$



                  We know the formula for Complex Fourier Series Expansion is:



                  $$x(t)=sum_{n=-a+e}^{a+e}c_n exp( j 2 pi n t / T)$$
                  Where:



                  $$ c_n = frac{1}{T_o}int_{-infty}^{infty}x(t)exp(-j 2 pi n t / T_o) dt$$



                  First, we calculate the coefficients for the series expansion of f(t):



                  $$ c_n = frac{1}{T}intlimits_{-T/2}^{T/2} sumlimits_{k=-infty}^{+infty} delta(t - kT) e^{-j 2 pi n t /T} dt \
                  $$
                  Since the period of f(t) is T, and we are integrating between -T/2 and T/2, it follow that only one value of k applies to the integral range, namely k=0. Thus, the equation simplifies to:



                  $$c_n = frac{1}{T}intlimits_{-T/2}^{T/2} delta(t) e^{-j 2 pi n t/T} dt quad quad\$$



                  Now we use this property of dirac delta:



                  $$int_{-a+e}^{a+e}f(t)delta(t - a)dt = f(a)$$



                  The equation further simplifies to:



                  $$ begin{align}
                  & c_n = frac{1}{T}e^{-j 2 pi n 0/T} \
                  & c_n = frac{1}{T} quad quad text{(for all integers of n)} \
                  end{align} $$



                  Substituting the coefficient value, $c_n$, into the formula for Fourier Series Expansion yields:



                  $$ f(t) = sumlimits_{k=-infty}^{+infty} delta(t - kT) = sumlimits_{n=-infty}^{+infty} frac{1}{T} e^{j 2 pi n t/T} $$



                  Next we find the Fourier Transform of the Fourier Series Expansion of f(t):



                  we know that:
                  $$ F { e^{i omega_o t} } = 2 pi delta(omega - omega_o) $$



                  Thus:



                  $$F{ f(t) } = frac{1}{T}F{ sum_{k=-infty}^{+infty} e^{ik2pi t/T} text{ }}
                  =frac{2pi}{T}sum_{k=-infty}^{+infty}delta(omega-kfrac{2pi}{T}).$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 at 15:53

























                  answered Nov 28 at 13:53









                  Bill Moore

                  1176




                  1176






























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