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Calculating Nyquist sampling rate and interval

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine the Nyquist sampling rate and the Nyquist sampling interval for this signal.

    sinc(2100[itex]\pi[/itex]t)
    2. Relevant equations

    N/A

    3. The attempt at a solution

    Ok I know that the Nyquist sampling rate is double or 2 times the bandwidth of a bandlimited signal. So I would assume the procedure for solving is find the bandwidth and multiply by 2. However I dont know where to start with finding the bandwidth of this signal. Also im unsure of how to treat the "sinc" function. Any help would be greatly appreciated!

    Thanks,
     
  2. jcsd
  3. Feb 13, 2013 #2

    I like Serena

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    Hey Evo8! Long time no see! :smile:

    More specifically the Nyquist sampling rate is double the highest relevant frequency ##f##.

    Your highest relevant frequency is the ##f## in ##\text{sinc}(2\pi f t) = {\sin(2\pi f t) \over 2\pi f t}##.
    The Nyquist sampling interval is the inverse of this doubled frequency.
     
  4. Feb 13, 2013 #3

    rude man

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    I think you need to look at the spectrum of sinc(2πf0t).
    That spectrum is derived by taking the Fourier integral of sinc(2πf0t) → |1/2πf0| rect(f/2πf0)
    (this is a rectangular function between f = -πf0 and +πf0 and height = 1).

    So the spectrum is flat from zero to f = πf0
    and here 2πf0 = 2100π or f0 = 1050 Hz. So the spectrum goes from zero Hz to 1050π Hz and the Nyquist sampling rate is 2100π Hz.

    So that makes the Nyquist rate 3.14... times as high as if the sampled wave were a simple sine wave of f0 = 1050 Hz.

    If I did it right ... comments welcomed.
     
  5. Feb 14, 2013 #4
    Hi I Like Serena! good to hear from you!

    Im a little confused now...

    I understand what ILS is saying. So if the[itex]f[/itex] is the highest relevant frequency then my Nyquist sampling rate would then be [itex]2f[/itex].

    And if my sampling rate is simply the inverse then that would turn out [itex]\frac{1}{2f}[/itex] right? So this means that the 2100 in the sinc function literally has no effect on the nyquist sampling rate or interval..

    I have a hard time seeing where rude man gets [itex]f_{0}=1050Hz[/itex] Further more how its determined that the rate is [itex]2100n Hz....[/itex]

    Thanks again for the help. It will take a few min until the mathematical portion of my brain comes out of hibernation. I simply dont use this type of math in my actual work practice so its gets uh lets say dusty....

    EDIT:
    I took another look at the question and im even more confused. Where did the [itex]f[/itex] come from? The original signal was [itex]sinc(2100\pi t)[/itex] So is the 2100 my highest relevant frequency?

    Thanks again for the help
     
    Last edited: Feb 14, 2013
  6. Feb 14, 2013 #5

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    After reading rude man's comment, I agree that the Nyquist rate will be ##2100\pi \text{ Hz} \approx 6597 \text{ Hz}##.

    To get the proper sampling, Nyquist stated that you need to sample at a frequency that is double the highest frequency that can occur, which is ##1050 \cdot \pi## in your case.
    (A typical trick is to pass the signal through a low pass filter to remove unwanted high frequencies.)

    He got 1050 Hz, since ##2\pi f t = 2100\pi t##.
    Dividing left and right by ##2\pi t## yields ##f = 1050 Hz##.
     
    Last edited: Feb 14, 2013
  7. Feb 14, 2013 #6
    Ok i think i understand.

    Heres one more related question then. Say I have the following signal ##5*sinc^{2}(200 \pi t)##

    Would the Nyquist sampling rate then be simply be ##2*100 \pi = 200 \pi Hz##?

    Thanks for braking this down for me. I do appreciate it!
     
  8. Feb 14, 2013 #7

    rude man

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    Once again, I would take the Fourier integral of that function of time to determine the spectrum.

    I don't have a table of transforms that includes that function. I supose the choice would be to either do a convolution of F{sinc(2πf0t)} with itself, or try to perform the actual integral. In the latter case I would try sin(x) = (1/2j)[exp(jx) - exp(-jx)] but it still looks like a bear to perform with the x2 in the denominator ... and with the infinite limits it might be necessary to invoke the delta function .. blah blah .. think I'll sit this one out. :frown:
     
  9. Feb 14, 2013 #8

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    It turns out not to be so bad.

    According to wiki the Fourier transform of ##\text{sinc}^2(ax)## is ##{1 \over \sqrt{2\pi a^2}} \text{tri}({\omega \over 2\pi a})##, where ##\text{tri}## is the triangular function that has its last non-zero value at 1.

    In our case we have ##~~a = 2\pi f_0## and ##\omega_{highest} = 2\pi f_{highest}##.
    So ##f_0 = 100 \text{ Hz}##.

    The highest frequency is given by:
    $${\omega_{highest} \over 2\pi a} = 1$$
    $${2\pi f_{highest} \over 2\pi \cdot 2\pi f_0} = 1$$
    $$f_{highest} = 2\pi f_0 = 2\pi \cdot 100 \text{ Hz}$$
    Therefore the Nyquist sample rate is:
    $$f_{Nyquist} = 2 f_{highest} = 4\pi\cdot 100 \approx 1257 \text{ Hz}$$
     
  10. Feb 14, 2013 #9

    rude man

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    Thanks ILS.
     
  11. Feb 14, 2013 #10
    Ok im hanging on here... barely.

    I understand how you applied transform 203 from the table and the basic algebra to get the ##f_{highest}=2 \pi f_{0}## then finding the sample rate is even more streight forward.

    The next signal I need to evaluate is something like this. ##sinc(2100 \pi t)+sinc^{2}(200 \pi t)##

    Soo looking at the tables on the wiki I don't see a transform that matches. Or a theorem to add two transforms. I cant image its as simple as adding the two highest frequencies together right?

    Giving me ##(2100 \pi)+(4 \pi 100)=7854Hz## for a nyquist rate?
     
  12. Feb 14, 2013 #11

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    I hope you intended [itex]T_{Nyquist} = \frac{1}{2f_{highest}}[/itex].
     
  13. Feb 14, 2013 #12

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    Transform 101 tells you what happens if you add: you just add the two transforms.
    So no, you don't add the two highest frequencies together.
    The frequency spectrums are simply added together.
    The highest resulting frequency is the maximum of the 2 frequencies.
     
  14. Feb 14, 2013 #13
    Yes! Thanks for catching that. I wrote one thing down in my notebook and typed another!
     
  15. Feb 14, 2013 #14
    Ok for some reason I thought 101 looked like it wasn't exactly what I had. On second inspection i see it.

    Now If i add them I get something like ##(\frac{1}{2100 \pi}* rect(\frac{f_{highest}}{2100 \pi})+(\frac{1}{200 \pi}*tri \frac{f_{highest}}{200 \pi})##

    However im nor sure I understand this comment.
    Soo I have the two highest frequencies of ##100 \pi## and ##1050 \pi## So the highest frequency would be ##1050 \pi## therefore the sampling rate is ##2100 \pi##? I have a feeling im not understanding fully.. right?
     
  16. Feb 14, 2013 #15

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    You'd get something like ##\text{rect}(\frac{f}{2100 \pi})+\text{tri}(\frac{f}{200 \pi})##.
    I'm leaving out the amplitudes for convenience.
    In particular this is a function of the frequency f and not of fhighest.
    See W|A.
    Their frequency spectrums overlap.
    The highest frequency where this function is non-zero is determined by the rectangle.
     
  17. Feb 14, 2013 #16

    rude man

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    Better than that, you just pick the higher of the two sampling rates!
     
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