# Need Help to design compensator for 4th order system

Discussion in 'Homework Help' started by Rb9797, Sep 29, 2014.

1. ### Rb9797 Thread Starter New Member

Sep 29, 2014
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The transfer function of system is G=1/(s+1)(s+2)(s^2+0.2s+4.01)

%overshoot <8%
settling time< 8 seconds

I need to design a compensator for this system to meet this criteria, the question is "How do we calculate the dominant close loop pole that will satisfy this criteria?" or "How do i find damping factor (zeta) and natural frequency(Wn) for given over shoot and settling time?"

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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What methods have you been taught for compensator design?
First point to note is that for zero steady state error you will need (at the very least) an open loop pole at the origin of the complex plane.
Second point - since this is homework you are expected to show some work you have done thus far.

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3. ### Rb9797 Thread Starter New Member

Sep 29, 2014
4
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yes, you are absolutely right for the real pole at origin I have tried to solve this with trial and error method in matlab. In the given transfer function the complex poles are very close to the imaginary axis so, i have placed a complex pair of zeroes very close to them to reduce their effects on the system.

i have studied proportional, PI, PD and PID controller till now...

I just need help to find the close loop poles through which our system must pass to meet this performance criteria....

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You may recall for a simple second order pole pair that the overshoot and settling time can be deduced from the damping factor and the natural frequency.
Given the characteristic polynomial ..

$s^2 + 2 \zeta \omega_n s +\omega_n^2$

The overshoot is

$M_p=exp(\frac{-\pi \zeta}{\sqrt{1- \zeta^2}})$

For approx 5% settling time

$t_s=\frac{4.6}{\zeta \omega_n}$

So you can find a damping factor value simply from the overshoot alone. Given the damping factor you can obtain an initial estimate for the natural frequency.

I'll leave you to recall where the corresponding (dominant) pole pair would lie on the complex plane.

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5. ### Rb9797 Thread Starter New Member

Sep 29, 2014
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the formulas are for 2nd order system , will they apply for the 4th order system too?!!!!

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Depends how isolated the resulting dominant closed loop 2nd order poles are from the remainder of the other closed loop poles.
You mentioned cancelling the two complex conjugate open loop poles closest to the imaginary axis. This isn't a bad idea - in theory anyway. As you stated, this hopefully removes the influence of troublesome open loop poles. It then remains to be seen if this works in practice. This is where the Root Locus technique comes in handy.
As a hint, I think if you use a PID compensator to do the pole cancellation and to add a pole at the origin, you may well have the solution at hand. It may be a matter of finally tweaking the overall loop gain. Remember you don't have to have 8% overshoot - you simply need to meet the <8 seconds settling time criterion. The latter means you have be within a +/- 5% band of the final value from <8 seconds and onwards in time.
If a more dynamic response (better rise time) is required then some judiciously placed phase lead could be added to the already implemented PID compensation via a lead function of the form

$G_{lead}=\frac{s+a}{s+b}$

Giving a total compensator of the form

$G_{comp}=K_{loop}G_{lead}\times $$K_p+K_ds+\frac{K_i}{s}$$$

Where a < b with a & b chosen accordingly and the overall loop gain Kloop is further "tweaked". Of course one could consolidate the terms into a single equivalent compensator transfer function where the Phase lead & PID components are transparent from an observer's viewpoint.

Once you come up with a workable design it would be worth considering whether an optimal solution can be obtained either by direct analysis or by a semi-heuristic / trial and error process.

Last edited: Sep 29, 2014
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7. ### MrAl Distinguished Member

Jun 17, 2014
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From post #5:
"The formulas are for 2nd order system , will they apply for the 4th order system too?!!!! "

Hello there,

Just to add a little...

From what i understand there are a few different methods to convert a higher order system to a lower order system with a similar but not exact response. There were publications as far back as 1966 that i happen to know of.

Using one of these methods, the following 3rd order system:
6/(s^3+6*s^2+11*s+6)

converts to the second order system:
1.6/(s^2+2.584*s+1.60)

and the responses of the two systems are similar (note right away the two gains are exactly equal).

The conversion requires quite a few little steps, and although each step is not super difficult (requiring a fairly simple derivative) each result must be used carefully to get the final equation. Im not sure how interested anyone is in seeing this procedure.

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8. ### Rb9797 Thread Starter New Member

Sep 29, 2014
4
0
Thanks everyone actually , I discussed this with my professor and he said that there is no need to calculate values of damping factor and natural frequency for higher order system, all we need to do is cancel out the unwanted poles with zeroes and add new desired poles to the system those are far away from imaginary axis and a integrating pole at the origin for zero steady state error.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I'm not entirely convinced by your report of the professor's comments. Notwithstanding the agreed addition of the pole at the origin and the proposed pole-zero cancellation, there remains a specification (post #1) which must presumably be achieved. Nothing in your professor's plan suggests how the full specification will be met other than a rather vague notion of a complex pole pair placement "far away from the imaginary axis".
Perhaps you have misunderstood something in your conversation with him/her.
If I have some time I will attempt to review the problem and offer a (hopefully) more consistent method of solution.

10. ### MrAl Distinguished Member

Jun 17, 2014
2,556
515
Hi,

I am not convinced either but of course the good professor may be trying to reduce this exercise to a more manageable one for the students depending on their previous level of education.

But there's another point to consider, and that's the real life situation where pole cancellation is a very tricky issue where it is not always going to be an acceptable solution. That's because component values are not exact, and they also change value over time. The pole that is completely and utterly canceled today may or may not be completely canceled tomorrow because the components may have varied. Depending how much they vary could mean the system goes unstable the next day after tomorrow. It's even possible that the company's parts department can not even order a part of the specified value, and even if it is ordered when the tech puts it in the circuit it may already be too far off to cancel the pole effectively due to normal parts tolerances. This kind of thing has to be investigated very carefully.

Again however, if the good professor told you to do it one way in particular then you will have to do it that way because he'll be correcting your paper (ha ha) and he wont want to see anything but what he already taught you

In any case, i wish you the best of luck with it.

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11. ### MrAl Distinguished Member

Jun 17, 2014
2,556
515
Hello again,

I couldnt resist trying out a few different methods for the transfer function reduction for this function.
Here are the graphical results for three different methods. The RED plot is the original function and the others are for different ways of converting to a 2nd order system.

From this we see that it's not hard to meet the 8 second criterion, and the simple pole cancellation is the slowest.
It is also possible to vary the partial pole cancellation to allow a little overshoot and thus a faster rise time due to the complex pole.

• ###### TransferFunctionReduction-1.gif
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Last edited: Oct 4, 2014
12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi Mr Al,
My understanding of the OP's original question and their repsonse in post #3 was that the transfer function given is the open loop plant function.The design process involves the specification of a compensator that would give the required transient and steady state response parameters when the combined compensator and plant were configured in (unity negative feedback) closed loop. That's why an open loop pole at the origin would give zero steady state error in closed loop mode.
I'm not sure your most recent post takes account of these matters.
Your earlier comments about the unwanted complex pole pair compensation by complex zero pair cancellation unfortunately went unacknowledged by the OP. Whilst the idea seems sound in theory it is potentially problematic in real physical systems. I'm not sure professors make this important distinction clearly to their students.
Regards,
t_n_k

Last edited: Oct 4, 2014