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Fourier Series of Dirac Comb! Please help

  1. Oct 15, 2013 #1
    http://en.wikipedia.org/wiki/Dirac_comb

    Please have a look at the Fourier Series section, and its last equation.

    Let T = 1.

    After expanding the Equation

    x(t) = 1 + 2cos(2∏t) + 2cos(4∏t) + 2cos(6∏t) .......

    Now this does not give the original Dirac Comb.
    Eg: at t = 1/2

    x(1/2) = 0

    But RHS

    = 1 + 2cos(2∏*1/2) + 2cos(4∏*1/2) + 2cos(6∏*1/2) .......
    = 1 + 2cos(∏) + 2cos(2∏) + 2cos(3∏) .......
    = 1 -2 + 2 -2 + 2..........
    ≠ LHS


    What is the problem?
     
  2. jcsd
  3. Oct 15, 2013 #2

    mathman

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    The mathematics is tricky here. You need to set it up as a function of t in closed form and take limits for t = multiples of 1/2.
     
  4. Oct 15, 2013 #3
    Agreed Mathman, after doing that you will get the LHS ≠ RHS. Where LHS is the dirac comb graph, and RHS is the computation via fourier series. Can you please elaborate?
     
  5. Oct 16, 2013 #4

    mathman

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    What I was trying to say is that you need to see what the Fourier series sums to first.

    I'll give a simple example of the procees I was talking about.

    1 + x + x2 + ..... = 1/(1-x) for |x| < 1. for x = -1, the l.h.s is 1 - 1 + 1 ..., while the r.h.s = 1/2. This makes sense only if you get the sum first where it converges and then extend it.

    What I am suggesting is the closed form for values where it converges and then extend it.
     
  6. Oct 17, 2013 #5
    Mathman if we just look at the second equation on the wikipedia page, which has absolutely no convergence issues because we can choose T as per our convenience, and if we let T=1, and expand it, it will not give LHS = RHS for values other than t = nT. Where n is an integer.
     
  7. Oct 17, 2013 #6
    I had taken the limits for T=1 between t=1/2 and t=-1/2.
     
    Last edited: Oct 17, 2013
  8. Feb 18, 2014 #7

    I think i found the answer , it is unusual.

    Where 1-1+1-1+1..... = 1/2

    http://en.wikipedia.org/wiki/Grandi's_series#Heuristics

    So the original RHS = 1 -2 + 2 - 2 + 2....

    = 1 -2( 1 - 1 + 1 - 1....)

    =1 -2( 1/2 )............Using Grandi Series

    =1 -1

    =0

    =LHS.
     
  9. Feb 18, 2014 #8

    jasonRF

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    I am am an engineer not a mathematician, so my answer will not pretend to be rigorous.

    Anyway, when you are dealing with animals like delta functions it does not make sense to try to evaluate what they "equal to" point-wise. They only make any sense when multiplied by a "nice" function (continuous, differentiable, falls of fast as t-> infinity) and integrated. Thus, we would expect that the Fourier series doesn't make sense evaluated at individual points either. Also, since the Dirac comb has nasty discontinuities (in some sense), we expect Gibbs phenomenon to be present in the Fourier series.

    Going back to your problem, for finite [itex]N[/itex], if I did my math correctly (just using sum of geometric series here and basic trig identities, nothing fancy),
    [tex]
    S_N(t) = \frac{1}{T} \sum_{n=-N}^{N} \exp\left( i n \frac{2 \pi}{T} t \right) = \frac{\sin\left( \frac{2 \pi}{T} (N + \frac{1}{2}) t \right)}{T \sin\left( \frac{\pi}{T}t\right)}
    [/tex]
    With T=1 and t=0.5 for example, this is either 1 or -1, depending upon the value of N and no matter how big N gets. So pointwise the sequence [itex] S_N (t) [/itex] does not converge at all as N goes to infinity. However, if you look at the oscillations in the series as a function of t, as N gets bigger and bigger, the oscillations get more closely spaced in time, so when you multiply by a continuous, smooth function and integrate there is no contribution to the integral in the neighborhood of t=0.5 as N goes to infinity. Likewise for every point in between the locations of the delta functions.

    It can be instructive to actually plot the fourier series [itex] S_N(t)[/itex] over some time interval for a few values of N (say 10, 30, 100) just to see how it behaves.

    jason
     
    Last edited: Feb 18, 2014
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