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Sampling with aliasing

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Let the signal [itex]x(t) = \left(\displaystyle\frac{\sin(50\pi t)}{\pi t}\right)^2[/itex], which we want to sample with sampling frequency [itex]\omega_s = 150\pi[/itex] in order to obtain a signal, [itex]g(t)[/itex] whose Fourier transform is [itex]G(\omega)[/itex]. Determine the maximun value for [itex]\omega_0[/itex] which guarantees that [itex]G(\omega) = 75X(\omega)[/itex] for [itex]\left|\omega\right| \leq{\omega_0}[/itex]



    2. Relevant equations
    Sampling Nyquist theoreme: [itex]\omega_s > 2B[/itex], where [itex]B[/itex] is the signal band-with.


    3. The attempt at a solution
    [itex]X(\omega) = FT\{x(t)\}[/itex] is a triangular signal with [itex]B = 100\pi[/itex] and amplitude [itex]X(0) = 25[/itex].

    From sampling Nyquist theoreme, [itex]150\pi > 200\pi[/itex] is false, so there is aliasing.

    I don't know how to finish the problem.

    Thank you.
     
  2. jcsd
  3. Aug 23, 2011 #2

    marcusl

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    There is no aliasing. You should not define signal bandwidth to include both positive and negative frequencies if you plan to use a relation with 2B. (That is, 2B assumes that B was defined as one-sided, whereupon the factor of 2 accounts for the negative frequency portion.) With this definition, you should write
    [tex]B=50\pi[/tex] and 2B<ws, so you are sampled properly.
     
  4. Aug 23, 2011 #3
    I've defined bandwith to include only positive frequencies, so my problem must be an error in Fourier transform.

    I think [itex]X(\omega) = FT\{x(t)\}[/itex] is a triangular signal whose bandwith is [itex]B = 100\pi[/itex] and [itex]X(0) = 25[/itex]

    Is this wrong?

    Thank you.
     
  5. Aug 23, 2011 #4

    marcusl

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    Hmm. Now that I work it out, I also get B=100pi but X(0)=50 (although the value of X(0) is not needed to solve the problem). Aliasing changes the shape of G so that the spectrum becomes horizontal between w=50pi and 75pi. You can compare the shape of G to that of X; I don't see that G exceeds X by 75, however.
     
    Last edited: Aug 23, 2011
  6. Aug 23, 2011 #5
    You're right: B=100pi and X(0)=50.

    Intuitively I understand that a horizontal line appears in the spectrum, but how can you know that this occurs between w=50pi and 75pi?

    Regarding the problem question (which I've checked that it is correctly copied), I know [itex]G(\omega) = \displaystyle\frac{1}{T_s}\sum_{k=-\infty}^{\infty}X(\omega-k\omega_s)[/itex]. As [itex]\omega_s = 150\pi[/itex], then [itex]T_s = 2\pi/150\pi = 1/75[/itex], this expression can be written as [itex]G(\omega) = 75\displaystyle\sum_{k=-\infty}^{\infty}X(\omega-k\omega_s)[/itex], where 75 appears, but I don't know how to continue.

    Thank you.
     
  7. Aug 24, 2011 #6

    marcusl

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    You can do it graphically. I'll discuss only the positive frequencies, but everything to follow is mirrored for w<0. The spectrum X is a triangle X(0)=50 and X(100*pi)=0, and G is X replicated at every integer multiple of ws. The first folding frequency is 75*pi, so fold the bit of triangle from 75 to 100 pi back and add it to X to give G. The flat part of G extends from 50 to 75 pi.

    Excellent, you've shown that G=75X in the absence of aliasing. Aliasing raises the value of G between 50 and 75 pi, so 50pi is the maximum frequency where G=75X.
     
  8. Aug 25, 2011 #7
    Thank you.

    I've just understood this problem.
     
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