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Homework Help: Filtering problem. Help with the output

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the output if this signal is processed by a filter with the following transfer
    [itex]u= \sum_{k=-\infty}^\infty c_n*exp(j*\pi*n*x)= \sum_{k=-\infty}^\infty rect(x-2k-.5)[/itex]

    c_n = 0 for even
    = 4/(j*[itex]\pi[/itex]*n) for odd
    Determine the output if this signal is processed by a filter with the following transfer
    a) H([itex]\xi[/itex])=exp{[itex]\phi[/itex]([itex]\xi[/itex])}
    [itex]\phi[/itex]([itex]\xi[/itex])= [itex]\pi[/itex]/2 ; [itex]\xi[/itex]>0
    0; [itex]\xi[/itex]=0
    -[itex]\pi[/itex]/2 ; [itex]\xi[/itex]<0

    2. Relevant equations

    f(x)= 2*rect(x-.5) [itex]\otimes[/itex] .5 comb(x/2)
    [itex]\otimes[/itex]- convolution.

    3. The attempt at a solution

    f(x) is a series of rect functions of width 1 and centered at .5, 2.5,-1.5 etc
    In the frequency domain F([itex]\xi[/itex]) is a shifted since envelope with delta functions at .5 intervals.

    F([itex]\xi[/itex])=2* sinc([itex]\xi[/itex])* exp(-i*pi*[itex]\xi[/itex]).comb( 2*[itex]\xi[/itex])

    The filter H([itex]\xi[/itex]) is a phase filter which is equal to 1 at [itex]\xi[/itex]=0, i for [itex]\xi[/itex] >0 and -i for [itex]\xi[/itex]<0

    G([itex]\xi[/itex])= H([itex]\xi[/itex])* F([itex]\xi[/itex])
    I cant figure out what G([itex]\xi[/itex]) and g(x) {inverse fft} looks like. Please help me
    Last edited: Nov 2, 2012
  2. jcsd
  3. Nov 2, 2012 #2


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    Before you start you need to correct some mistakes in the question. The given Fourier series is incorrect for given rectangular pulse train (I'm assuming that it is a rectangular pulse train and that the sum from k=infinity to infinity was a typo - which I corrected in the quote).


    1. If you're going to use the complex exponential form of the series on that real time function then you'll need to sum from a_n from -infinity to infinity, not from zero to infinity.

    2. [itex]a_n = 0[/itex] for n even, except for a_0 which equals 1/2.

    3. [itex]a_n = 1/(j n \pi)[/itex] for n odd.

    4. We usually denote the coefficients of the complex series as c_n (a_n is usually reserved for the real cosine series).
    Last edited: Nov 2, 2012
  4. Nov 2, 2012 #3

    Thank you for correcting the typo. The series can alternatively be expressed in terms of rect and that is what I have been working on. Could you help me with the output of the filter. I know the phase change should the change the sines into cosines. I can understand what is happening but I cant figure out the math behind it
  5. Nov 2, 2012 #4


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    Yes I agree that the series can alternatively be represented as a single rectangular pulse convolved with an impulse train. However since the problem gave the Fourier series I thought perhaps that was how they wanted you do deal with it.

    You realize that with the Fourier series your "phase filter" (also known as a Hilbert transform filter) simply turns each sine component to an equivalent cosine component.
    Last edited: Nov 2, 2012
  6. Nov 2, 2012 #5


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    BTW. There is also a typo in your filter definition.


    In my previous response I've assumed that you meant to write:

    H([itex]\xi[/itex])=exp{[itex]j \phi[/itex]([itex]\xi[/itex])}
  7. Nov 4, 2012 #6

    Yes you are right. Sorry about having so many errors. This was the first time I was using latex.
    A gentle bounce. Can anyone help me?
  8. Nov 4, 2012 #7


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    Look at how your filter behaves, term (pair) by term (pair), on the Fourier series.

    Input term: [itex]\frac{\exp(j k \pi x) - \exp(-j k \pi x)}{j k} = \frac{2 \sin(k \pi x)} {k \pi}[/itex]

    Output term: [itex]\frac{\exp(j k \pi x + j \pi/2) - \exp(-j k \pi x - j \pi/2)}{j k \pi} = \frac{2 \cos(k \pi x)} {k \pi}[/itex]

    The easiest way to express the output of that filter is in terms of its Fourier series.

    Here's a plot of the first 50 terms of the FS of both input and output.

    Attached Files:

    Last edited: Nov 4, 2012
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