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Simplifying a transfer function to find overshoot

by soyks
Tags: overshoot, transfer function
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soyks
#1
Jun11-12, 12:08 AM
P: 5
Hi,

I have question where I am asked to find the % overshoot of a system based only on its transfer function.

T(s) = 15(s+2.1)
(s+2)(s2+4s+29)(s2+2s+50)

I'm assuming I need to use second order assumptions and then find the damping ratio (zeta) to find the %OS, but this equation looks a little complicated for that.

I noticed that the numerator (s+2.1) could be cancelled with the (s+2) term in the denominator because they're similar. Am I allowed to do that?

Then I'm still left with two quadratic equations on the bottom.. I still can't see how you could approximate the %OS.
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NascentOxygen
#2
Jun11-12, 10:39 AM
HW Helper
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P: 5,141
Hi soyks!

The only approach I can think of is to use your knowledge of the Laplace Method to determine the time response Vout(t) from Vout(s), and then carefully analyze that equation for Vout(t) to determine its overshoot.

I can't see any shortcuts; maybe someone else can.
soyks
#3
Jun11-12, 07:34 PM
P: 5
Thanks for your reply NascentOxygen. I'm not too sure what you mean though, could you possibly begin explaining the method you'd use, then I'll see if I can get it from there.

In the mean time, anyone else have any ideas?

NascentOxygen
#4
Jun11-12, 10:49 PM
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P: 5,141
Simplifying a transfer function to find overshoot

You determine Vo(s) using T(s) and the Laplace Transform of a unit step input. Then consult a table that mathematicians have provided (or otherwise) to deduce the sinusoidal and exponential components (or whatever) that make up that particular Vo(t). You could plot Vo(t), but calculus allows you to calculate when dV/dt =0, and the first occurrence will probably be of the overshoot.

If you have studied the Laplace Method, examples of this will be in any textbook, illustrating how to determine the step response of a system.
soyks
#5
Jun12-12, 07:59 PM
P: 5
I should probably mention this is a question from a past exam. So I assume it should be solvable within a few minutes
viscousflow
#6
Jun13-12, 05:51 AM
P: 273
Based on your knowledge of control systems there are some really quick things you can determine!


[itex]T(s) = \frac{15\left(s+2.1\right)}{\left(s+2\right)\left(s^2+4s+29\right)\left (s^2+2s+50\right)}[/itex]

Firstly

[itex]T(s) = \frac{\left(s+2.1 \right)}{\left(s+2\right)}[/itex]
are really close so they can be cancelled right away.

And are left with
[itex]T(s) = \frac{15}{ \left(s^2+4s+29 \right) \left(s^2+2s+50 \right)}[/itex]

Secondly, one basic rule is: the zeros closest to the imaginary line have the strongest influence on the dynamics of the system. Thus, you want to find the closest roots and throw away the others. That step is easy enough to do it yourself right :)
viscousflow
#7
Jun13-12, 09:33 PM
P: 273
Quote Quote by viscousflow View Post
Secondly, one basic rule is: the zeros closest to the imaginary line have the strongest influence on the dynamics of the system. Thus, you want to find the closest roots and throw away the others. That step is easy enough to do it yourself right :)
Err this should be poles, or roots of the denominator.
soyks
#8
Jun13-12, 10:23 PM
P: 5
Yeah I figured that was the case :P

Ok thanks heaps for that reply! You really explained it well.

I just found the roots of the quadratics... the first one is -25j and the second is -17j. As you said, the closer one to the imaginary axis has the larger effect, but I was taught that it had to be at least 5 times smaller than the other poles to dominate.. I'd say this is pushing it. I checked with MATLAB, finding the overshoot of the system with both sets of quadratics and then the first one only, and the overshoot dropped from 78% to 28% - not too accurate.

Maybe they don't expect any better? Since this is an exam question.. I can't see of any way of doing it better but if you do, let me know.
soyks
#9
Jun13-12, 10:28 PM
P: 5
Sorry I just re-read the question - it says 'comment on the validity of second order assumptions'. I suppose this would be one where its pretty uh.. ****. :P
Ratch
#10
Jun17-12, 02:33 PM
P: 315
soyks,

I don't know what % overshoot means. That transfer function is an exponentially decaying sine wave which has a high point of 0.05, and is mostly dead after 6 secs. You can get its inverse LaPlace by a partial fraction expansion, after which you can look up each term in a table. Then do a quick plot and all will be revealed.

Ratch


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