Solving Control System Problems: Natural Frequency and K Linkage Explained

AI Thread Summary
The discussion focuses on understanding the relationship between natural frequency and gain (K) in a third-order control system. Participants clarify that the characteristic equation can be expanded to derive coefficients for the system's transfer function, leading to equations that relate variables a, b, and α. The variable α is introduced as a dimensionless factor that modifies the system's response, allowing for a comparison between second and third-order systems. There is confusion regarding the value of K, which is not provided in the problem statement, but it is suggested that any constant value can be used to match answer choices. Overall, the conversation emphasizes the importance of Laplace transforms and the need for clear problem statements in control systems analysis.
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Homework Statement


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.


Homework Equations





The Attempt at a Solution


1 st image in the attachments is the attempt, and second image is the question.

 

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

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

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

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.
 
sorry for the photos.
 
I got 3 equations but what value of "k" should i take ?
 
lazyaditya said:
I got 3 equations but what value of "k" should i take ?

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?
 
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I have posted the Image.
 

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  • #11
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
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
b(k) = k/5 so
k = 5b(k)
But b = 9.9
Therefore k = 5*9.9 = 49.5
 
  • #14
can i ask another question over here only regarding electromagnetic fields or should i post another thread ?
 
  • #15
lazyaditya said:
can i ask another question over here only regarding electromagnetic fields or should i post another thread ?

I would start a new thread since control systems and e-m are very different disciplines.
 
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