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Fourier Transform of Dirac Comb/Impulse Train

  1. Aug 28, 2015 #1
    With Dirac Comb is defined as follow:
    $$III(t)=\sum_{n=-\infty}^\infty\delta(t-nT)$$
    Fourier Transform from t domain to frequency domain can be obtained by:
    $$F(f)=\int_{-\infty}^{\infty}f(t)\cdot e^{-i2\pi ft}dt$$
    I wonder why directly apply the above equation does not work for the Dirac Comb:
    $$F(III(t))=\sum_{n=-\infty}^\infty\int_{-\infty}^{\infty}\delta(t-nT)\cdot e^{i2\pi ft}dt$$
    $$III(f)=\sum_{n=-\infty}^\infty e^{-i2\pi fnT}dt\cdots\cdots [1]$$
    Where the correct way to obtain the FT of Dirac Comb is to first find the Fourier series, and then do the Fourier Transform for each term in the summation.
    I copied the upper section from a open lecture slide, and I don't even understand how it goes from [2] to [3], not to mention [1] and [3] are totally not the same thing. Any hints guys?
     
  2. jcsd
  3. Aug 28, 2015 #2

    blue_leaf77

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    The ##dt## shouldn't be there.
    They are equivalent. To see this, consider a function ##f(x) = \sum_n \delta (x-na)##. This is a periodic function with periodicity ##a##, which means it can be expanded in Fourier series. So
    $$
    f(x) = \sum_{n=-\infty}^{\infty} \delta (x-na) = \sum_{k=-\infty}^{\infty} c_k \hspace{0.5mm}e^{-i2\pi x/a}
    $$
    Find ##c_k## and you will see that [1] and [3] are equivalent.
    That's the Fourier definition of delta function
    $$
    \delta(x-a) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-a)} \hspace{1mm} dk
    $$
     
  4. Aug 28, 2015 #3
    Oh I get it, with III(f) having a period of 1/T, the Fourier series should write:
    $$\frac{1}{T}\sum_{n=-\infty}^\infty\delta(f-\frac{n}{T})=\sum_{k=-\infty}^\infty c_k\cdot e^{i2\pi kfT}\cdots\cdots[4]$$
    $$c_k= T\cdot\int_{0}^{1/T} \sum_{n=-\infty}^\infty\delta(f-\frac{n}{T}) \cdot e^{-i2\pi kfT}=1$$

    Putting c_k=1 back to [4] results in:
    $$\sum_{k=-\infty}^\infty e^{i2\pi kfT} = \text{(1)}$$

    Just to clear my concept, s this equivalent to (1) since the exponent missed out a -ve sign, but k ranged from -inf to inf so its just the same?
     
  5. Aug 28, 2015 #4

    blue_leaf77

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    The used definition for Fourier series in complex form is to use negative exponent.
     
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