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Homework Help: Maximum frequency of a sinc

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Since the Fourier Transform of a rectangular Pulse with a half period of Tp contains an infinite number of frequencies, how can you find the max one to check if Nyquist's theorem holds?

    2. Relevant equations

    3. The attempt at a solution
    I don't suppose Ωmax = 2*pi / Tp ?
  2. jcsd
  3. Jan 30, 2012 #2
    There is no maximum frequency component of a rectangular pulse in the continuous time domain, as you wrote. So don't attempt to find one! What is the problem statement?

    Also, typically we'd use ω to represent angular frequency, not Ω.
  4. Jan 31, 2012 #3
    I just have a rectangular pulse, with a half period T0, and i'm asked to justify :
    "if we take samples with sampling period Ts < 2*T0 , no aliasing occurs"

    Don't i need to find the maximum frequency of the signal to do that?
  5. Feb 1, 2012 #4

    rude man

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    ".. a half-period of T0 ..." suggests you have not one pulse but a pulse TRAIN, alternating between 0 and V at a 50% duty cycle ...

    Assume your first sample occurs just before the rising edge of the pulse, at a sample rate of 2T0 - ε where ε → 0, satisfying Ts < 2T0, then obviously your samples would all be zero forever!
    Last edited: Feb 1, 2012
  6. Feb 1, 2012 #5
    half period T0, meaning that the singal is P(t) = u(t+T0) - u(T-T0) , where u is the heavside step.
  7. Feb 1, 2012 #6

    rude man

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    You can only talk about a "period" in reference to a pulse train, not a single pulse. A single pulse is not "periodic".

    OK, so we have a single rectangular pulse of duration 2T0, centered at t = 0. Is that right?

    OK, now we're guaranteed that one of the first two samples will be 1, then the rest of course will all be zero. We have to know ahead of time that the pulse is rectangular. Only in that sense can we claim some kind of valid sampling.

    In general, a sampler cannot ever reconstruct a single pulse faithfully unless Ts → 0.

    Suppose for example your pulse is u(t+T0)sinωt - U(t-T0)sin(ωt), ω = π/2T0. That pulse is zero for t < - T0, max'es out to 1 at t = 0, then goes back to zero for t > +T0. The sampler might sample at t = -T0 and then again at t = +T0 - ε where ε is an arbitrarily small quantity. This meets the criterion that Ts < 2T0. But your samples would be 0, ε, 0, 0, .... , totally losing the peak at t = 0.
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