Sampling Time Calculation for Bode Plots: Plant Comparison & ZOH Preceding

  • #1
jami8337
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0

Homework Statement


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


Homework Equations



I figure it has something to do with f= 1/T, but other than that I am really not sure how to use it with the bode plot, any pointers would be great!
 
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  • #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?
 
  • #3
Ok, is the highest frequency then represented by fs/2 ?
 
  • #4
jami8337 said:
Ok, is the highest frequency then represented by fs/2 ?
Yes, the Nyquist frequency of the sampling system.
 
  • #5
So if the Nyquist frequency is fs/2 and fs= 1/Ts,

Ts = 1/2*Nyquist frequency ?
 
  • #6
jami8337 said:
So if the Nyquist frequency is fs/2 and fs= 1/Ts,

Ts = 1/2*Nyquist frequency ?
Well, to be more precise, Ts = 1/(2*nyquist freq.).
 
  • #7
Ah yeah that's 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?
 
  • #8
jami8337 said:
Ah yeah that's 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?
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.
 
  • #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!
 
  • #10
jami8337 said:
Ah ok, so (1/2Π)*16rad/sec = 2.55 Hz

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

Thats great, thanks for all your help!
Right. 0.2s.
 

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