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Fourier Transfrom of scaled periodic impulse train

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Fourier transform of scaled pulse train from -inf to +inf. Starting at 0 the first impulse is scaled at 2, the second impulse at 1 is scaled as 1, third at three scaled at 2, etc....


    2. Relevant equations
    Does my solution look correct?


    3. The attempt at a solution

    X(jw) = 2[tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex]k) + [tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex](k + 1))
    the sigma notation is from k = -inf to inf. My thinking behind this is to represent a pulse for each scaled delta funciton in freq domain
     
    Last edited: Dec 8, 2009
  2. jcsd
  3. Dec 9, 2009 #2

    Redbelly98

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    I'm having trouble seeing any pattern in the pulse train you describe. Is there no pulse at t=2? What about the pulses at negative times?

    The transform of a simple pulse train might come in useful here.

    If you can describe the input pulse train better, it will be easier for us to see what is going on.
     
  4. Dec 9, 2009 #3
    The pulse train is periodic so it runs from plus -inf to +inf.
    At t = -2 the impulse = 2
    At t = -1 the impulse = 1
    At t = 0 the impulse = 2
    At t = 1 the impulse = 1
    At t = 2 the impulse = 2
    At t = 3 the impulse = 1
    At t = 4 the impulse = 2
    Etc.... from -inf to +inf
     
  5. Dec 9, 2009 #4

    Redbelly98

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    Thank you for clarifying that.
    No, it does not look correct. Can you show your work?
     
  6. Dec 9, 2009 #5
    The Fourier transform for a pulse train is 2*pi*sigma*delta(w - 2*pi/T). Here T = 2. the problem is that a regular pulse train has a constant value of 1 for the impulse while this one does not. This means the above equation needs to be modified to alternate between value of 2 and a value of 1. That is how I arrived at the above equation. The first part of the equation is for value of 2. When k = 0 the first part 2*pi. The second part = pi when w = pi, etc.... This was my thinking behind this X(jw). Because it's linear I could split the problem into two summations to represent each of the 2 values of the delta function.
     
  7. Dec 9, 2009 #6
    Oops! I am overwriting my impulses. It should be:

    X(jw) = 2*pi*Sigma[delta(w - pi*(2k))] + pi*Sigma[delta(w - pi*(2k + 1))]
     
  8. Dec 9, 2009 #7

    Redbelly98

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    Okay, I agree with your solution now. I had in mind expressing it a different way, but it is equivalent to what you have.

    What I had in mind:
    X(ω) = π(∑δ(ω-πk) + ∑δ(ω-2πk))
    where k runs from -∞ to +∞.
     
  9. Dec 9, 2009 #8
    Cool. Thanks a lot.
     
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