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Spectrum of sampled signal

  1. Jun 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi, I need to find the spectrum of [tex]s(t)=sin(400\pi t) + 0.5cos(12000\pi t)
    [/tex] when sampled at 10KHz.

    2. Relevant equations
    Using the Discrete Fourier Transform:

    [tex]G_{k}=\sum_{n=0}^{N-1}s_{n}e^{\frac{-j2\pi kn}{N}}[/tex]
    where N is the amount of samples taken in the signal duration which is [tex]NT_{s}=\frac{N}{10000}[/tex] and [tex]s_{n}=s(nT_{s})[/tex]

    3. The attempt at a solution
    I decided to take 10 samples, so N=10, therefore the signal duration is 0.001 seconds
    Therefore: [tex]G_{k}=\sum_{n=0}^{9}s_{n}e^{\frac{-j2\pi kn}{10}}[/tex]
    [tex]G_{k}=\sum_{n=0}^{9}[sin(400\pi nT_{s}e^{\frac{-j2\pi kn}{10}}+0.5sin(12000\pi nT_{s})e^{\frac{-j2\pi kn}{10}}] [/tex]
    [tex]G_{k}=\sum_{n=0}^{9}[sin(\frac{4\pi n}{100})e^{\frac{-j2\pi kn}{10}}+0.5sin(\frac{12\pi n}{10})e^{\frac{-j2\pi kn}{10}}][/tex]
    [tex]G_{k}=\frac{1}{2}-0.28e^{\frac{-j\pi k}{5}}+0.403e^{\frac{-j\pi k2}{5}}+0.523e^{\frac{-j\pi k3}{5}}+0.077e^{\frac{-j\pi k4}{5}}+1.09e^{\frac{-j\pi k5}{5}}+0.28e^{\frac{-j\pi k6}{5}}+0.93e^{\frac{-j\pi k7}{5}}+0.999e^{\frac{-j\pi k8}{5}}+0.5e^{\frac{-j\pi k9}{5}}[/tex]
    I'm not sure what this all actually means, how can we find the frequency components based on all those exponentials? I know that exponentials in the time domian give a delta in the frequency domain, but these exponentials are already in the frequency domain.
    Any help would be greatly appreciated!
  2. jcsd
  3. Jun 6, 2010 #2
    Hey I think I worked it out, you don't need to use the DFT.
    The 200Hz component will be present after sampling because 200Hz is less than half the sampling frequency 10000Hz, however the 6000Hz component will not be present after sampling because 6000Hz is not less than half the sampling frequency, ie you need to sample at at least 12000Hz to recover the whole signal (Nyquist theorem). Instead the 6000Hz component will be shifted to 0.5Fs – (6000Hz-0.5Fs )=5000Hz-(1000Hz)=4000Hz.
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