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Sampling a continuous-time signal, aliasing/Nyquist

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    The analog signal x(t) = cos(2pi f t) is sampled at a rate of 1 kHz, using ideal
    impulse sampling, to obtain the sampled signal x^(s)(t). The sampled signal is then sent through an ideal
    lowpass filter with transfer function H(2pi f ) = 0.001 rect (0.001 f ).
    (a) If f =1.01kHz, what is the output frequency from the filter? Is there aliasing or not?
    (b) If f = 0.99kHz, what is the output frequency from the filter? Is there aliasing or not?
    (c) If f = 0.49kHz, what is the output frequency from the filter? Is there aliasing or not?

    2. Relevant equations
    To perfectly reconstruct a signal, we need: sampling rate > 2*signal frequency

    3. The attempt at a solution
    I think these are correct, not 100% sure though:
    aliasing occurs, as 1.00 kHz !> 2 × 1.01 kHz
    fout = 1.01 kHz – n × 1.00 kHz = 1.01 kHz – 1 × 1.00 kHz = 0.01 kHz

    aliasing occurs, as 1.00 kHz !> 2 × 0.99 kHz
    fout = 0.99 kHz – n × 1.00 kHz = 0.99 kHz – 1.00 kHz = - 0.01 kHz

    aliasing does not occur, as 1.00 kHz > 2 × 0.49 kHz
    fout = 0.49 kHz
  2. jcsd
  3. Oct 26, 2016 #2

    rude man

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    Not sure I understand this terminology. What really is the ideal low-pass cutoff frequency? Is it really a function of input frequency f? Do you have to divide by 2π? Etc. ???

    I don't know your method, and as I say I'm not sure what your cutoff filter really looks like, but it basically seems to work.

    Just FYI I first determine the cutoff frequency of the low-pass filter = f0, the I take the sampling frequency fs and my input frequency f and find spectral components:

    f, |fs - f|, fs + f, |2fs - f|, 2fs + f, ...

    and then compare each component against the cutoff frequency f0.
    The only unaliased signal is at f so anything below f0 that is a mix of f and fs is an aliased component.
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