1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    functions:
    [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
    functions:
    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

    uart

    User Avatar
    Science Advisor

    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).

    Also

    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
    Hi,

    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

    uart

    User Avatar
    Science Advisor

    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

    uart

    User Avatar
    Science Advisor

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

    H([itex]\xi[/itex])=exp{[itex]\phi[/itex]([itex]\xi[/itex])}

    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
    H(ξ)=exp{jϕ(ξ)}

    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

    uart

    User Avatar
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Filtering problem. Help with the output
Loading...