# Control System Problem

1. Sep 24, 2013

1. The problem statement, all variables and given/known data
Since this is a third order system and there will be a zero in its transfer function i am confused that how the natural frequency and K will be linked ? Please do help i really get confused in these type of problems.

2. Relevant equations

3. The attempt at a solution
1 st image in the attachments is the attempt, and second image is the question.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Sep 24, 2013

### rude man

Assume Gc = (s+a)/(s+b) as you have done.

Then the open-loop gain is GGc = (k/s2)[(s+a)/(s+b)] = n/d.

Expand n + d into s3 + cs2 + ds + e

and equate coefficients of like powers of s with

(s2 + 2ζωns + ωn2)(s + αωn)
which you also have to expand as above.

By equating coefficients of like powers of s you get 3 equations with 3 unknowns (a, b and α).
Solve for a and b.

(k was not given numerically so you can assume any value which will agree with one of the four answer choices).

3. Sep 24, 2013

Why do you have used (s + αωn) i mean how do you have "αωn" as a root of the characteristic equation ?

4. Sep 24, 2013

### rude man

Good question.

α is a new, dimensionless variable for a 3rd-order system. If you compare the inverse-Laplace transform for the 3rd-order chas. equation with the same xfr function for the 2nd order system, i.e. without the extra (s + αωn) term, you would get a similar time response to a delta function input except for a modified coefficient in front of the sine term, plus a second, non-sinusoidal, term. The second term decays as exp(-αωnt) whereas the sinusoidal part decays as exp(-ζωnt), same as for the 2nd-order system. The argument of the sine is the same for both 2nd and 3rd order systems
= (ωn√(1 - ζ2)t + ψ).

And the phase angle ψ(3rd order) = ψ(2nd order) - a term including α.

So the bottom-line answer is that the two systems behave somewhat similarly if for the 3rd order system you retain the 2nd order expression multiplied by (s + αωn).

The complete time response expression is a mess to write out & I'm not going to do it here, with or without the extra (s + αωn) in the chas. equation. I suggest you get hold of a very extensive Laplace transform table which includes the time responses to both cases.

5. Sep 24, 2013

I am still not very much clear about the part the root of 3 rd order system including natural frequency.

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6. Sep 24, 2013

### rude man

Alpha is not arbitrarily chosen. You get 3 equations to solve for a, b and alpha.

Can you get hold of a really good Laplace transform table?

BTW your photos are 90 degrees twisted and very hard to read.

7. Sep 24, 2013

sorry for the photos.

8. Sep 24, 2013

I got 3 equations but what value of "k" should i take ?

9. Sep 24, 2013

### rude man

Like I said, a constant that will make one of your answer choices correct. I think it was dumb of them not to give you a value for k. It was obviously an incompletely stated problem.

I haven't done the work and don't want to, so here you're a bit on your own.

EDIT: wait, did you solve for a(k) and b(k), and if so, what did you get?

Last edited: Sep 24, 2013
10. Sep 25, 2013

I have posted the Image.

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11. Sep 25, 2013

### rude man

OK, so then a = 5 - 125/k = (5k - 125)/k and
b(k) = k/5

So if you start with b(k), choice (a) would seem closest: b(k) = 9.9, k = 49.5, then a(k) ~ 2.5.

As I said, I think it was dumb of them not to have given you k numerically since it certainly determines a and b.

12. Oct 13, 2013

Thanks a lot and sorry for such a late reply, since i wasn't able to check my thread for a long time.But how did you come to conclusion of "k" being 49.5 ?

13. Oct 13, 2013

### rude man

b(k) = k/5 so
k = 5b(k)
But b = 9.9
Therefore k = 5*9.9 = 49.5

14. Oct 15, 2013