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Bode Plot Sampling time

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure, the Bode plots of a continuous-time plant (thin line) and of its discrete-time counterpart, representing the discrete-time operation of the plant preceded by a Zero Order Hold (ZOH) (bold line), are displayed. What is the Sampling time used?
    Capture.PNG

    2. Relevant equations

    I figure it has something to do with f= 1/T, but other than that im really not sure how to use it with the bode plot, any pointers would be great!
     
  2. jcsd
  3. May 24, 2015 #2
    That vertical line tells you something about the highest frequency component you can represent in a discrete-time system, which is very intimately related to the sampling frequency of the system.

    Does that help?
     
  4. May 24, 2015 #3
    Ok, is the highest frequency then represented by fs/2 ?
     
  5. May 24, 2015 #4
    Yes, the Nyquist frequency of the sampling system.
     
  6. May 24, 2015 #5
    So if the Nyquist frequency is fs/2 and fs= 1/Ts,

    Ts = 1/2*Nyquist frequency ?
     
  7. May 25, 2015 #6

    rude man

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    Well, to be more precise, Ts = 1/(2*nyquist freq.).
     
  8. May 25, 2015 #7
    Ah yeah thats what I meant. So the highest frequency here would be about 17 rad/sec? Do i need to convert this into Hz in order to use it in the equation?
     
  9. May 25, 2015 #8

    rude man

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    Yes. That's because sampling rate is always given in samples/sec. and so the sampling time is 1/sampling rate and fs = 1/2 sampling rate. Avoid rad/sec. in sampling questions, in general.
     
  10. May 25, 2015 #9
    Ah ok, so (1/2Π)*16rad/sec = 2.55 Hz

    Then Ts= 1/(2*fs) = 0.2

    Thats great, thanks for all your help!
     
  11. May 25, 2015 #10

    rude man

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    Right. 0.2s.
     
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