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

[jami8337]

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?

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!

 

[milesyoung]

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?

 

[jami8337]

Ok, is the highest frequency then represented by fs/2 ?

 

[milesyoung]

Ok, is the highest frequency then represented by fs/2 ?

Yes, the Nyquist frequency of the sampling system.

 

[jami8337]

So if the Nyquist frequency is fs/2 and fs= 1/Ts,

Ts = 1/2*Nyquist frequency ?

 

[rude man]

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.).

 

[jami8337]

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?

 

[rude man]

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.

 

[jami8337]

Ah ok, so (1/2Π)*16rad/sec = 2.55 Hz

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

Thats great, thanks for all your help!

 

[rude man]

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.