Control System Design Exam Prep: Analytical Solutions & MATLAB

[danago]

I am studying for an exam coming up and came across a question that i am having a little trouble with.

[PLAIN]http://img442.imageshack.us/img442/825/screenshot20101112at110.png

I think i could do this in MATLAB, however i will obviously not have MATLAB in this exam. I have a set of equations that relate phase margin, damping ratio, percentage overshoot and peak time to each other analytically, however these were derived for a standard second order system. My initial thought was to use these equations to obtain a first approximation for the phase lead parameters and the gain K, followed by any necessary adjustments, however i am not sure how i would check if the specifications have been met (as requested in part b).

Any suggestions are greatly appreciated

Thanks,
Dan.

 

[CEL]

I am studying for an exam coming up and came across a question that i am having a little trouble with.

[PLAIN]http://img442.imageshack.us/img442/825/screenshot20101112at110.png

I think i could do this in MATLAB, however i will obviously not have MATLAB in this exam. I have a set of equations that relate phase margin, damping ratio, percentage overshoot and peak time to each other analytically, however these were derived for a standard second order system. My initial thought was to use these equations to obtain a first approximation for the phase lead parameters and the gain K, followed by any necessary adjustments, however i am not sure how i would check if the specifications have been met (as requested in part b).

Any suggestions are greatly appreciated

Thanks,
Dan.

From your equations, you can calculate the positions of the dominant poles of your system (the ones you calculated for a second order system).
You know that the sum of the angle of one of the poles of your closed loop system (CLS) to the poles of the open loop system (OLS) minus the sum of the angles to the zeros of the OLS, should be an odd multiple of 180 degrees.
There is more unknowns than equations, so you can choose the value of one unknown. For example, use the zero of C(s) to cancel the pole s = 0.1 of G(s).
Now you can determine the pole of C(s).
The gain K times b/a is the product of the distances of the chosen pole of the CLS to all of the poles of the OLS, divide by the distance to the zero of the OLS.

 

[danago]

Thanks for the reply CEL.

I am not completely sure what you mean when you say to calculate the position of the dominant poles.

Anyway, your statement "For example, use the zero of C(s) to cancel the pole s = 0.1 of G(s)." helped me i think. The closed loop transfer function i obtained was:

<br /> \frac{{Kb(s + a)}}{{as(10s + 1)(s + b) + Kb(s + a)}}<br />

If i choose a=0.1 then the TF becomes:

<br /> \frac{{Kb}}{{s(s + b) + Kb}}<br />

Which is a second order system. I then found that if b=K=0.7255 then the system would have the desired transient response (i checked in MATLAB).

Is this along the lines of what you were suggesting?

 

[CEL]

Thanks for the reply CEL.

I am not completely sure what you mean when you say to calculate the position of the dominant poles.

Anyway, your statement "For example, use the zero of C(s) to cancel the pole s = 0.1 of G(s)." helped me i think. The closed loop transfer function i obtained was:

<br /> \frac{{Kb(s + a)}}{{as(10s + 1)(s + b) + Kb(s + a)}}<br />

If i choose a=0.1 then the TF becomes:

<br /> \frac{{Kb}}{{s(s + b) + Kb}}<br />

Which is a second order system. I then found that if b=K=0.7255 then the system would have the desired transient response (i checked in MATLAB).

Is this along the lines of what you were suggesting?

Yes!

 

[danago]

Alright, thank you very much!

 

[CEL]

Alright, thank you very much!

You have dropped the constant a in the formula you posted. Have tou taken it in consideration in your calculations?

 

[danago]

You have dropped the constant a in the formula you posted. Have tou taken it in consideration in your calculations?

Do you mean in the denominator of this one?

<br /> <br /> \frac{{Kb}}{{s(s + b) + Kb}}<br /> <br />

If so, when i took out the factor of 10 from the brackets i had 0.1*10=1, which is why the 'a' was removed. Is this what you mean?

 

[CEL]

Do you mean in the denominator of this one?

<br /> <br /> \frac{{Kb}}{{s(s + b) + Kb}}<br /> <br />

If so, when i took out the factor of 10 from the brackets i had 0.1*10=1, which is why the 'a' was removed. Is this what you mean?

You're right!

 

[danago]

Ok, thanks again