# How to find the damping ratio of this transfer function

Discussion in 'Homework Help' started by lukus08, May 24, 2009.

1. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
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40K / s(s+3)(s+5)

How do i find the damping ratio and also the gain...K?

thanks

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
How about you take a stab at computing the two parameters and then posting your effort here?

hgmjr

3. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
0
i need help, which is why i am posting.

I can only get up to:

40k / s^3 + 3s^2 + + 15s + 40k

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Were you supplied the circuit diagram to which the transfer function applies? Or were you just given the transfer function?

hgmjr

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
Im not sure of your knowledge level, so I'm going to dump quite a bit of information out without really answering your question directly.

This is a third order system. Before you try to solve this, ask yourself if you have a good level of understanding of a second order system. Do you fully understand second order systems and the classifications; under-damped, over-damped and critically-damped? Do you know how to determine natural frequency and damping factor for a second order system?

If you understand a second order system, then I'm sure you understand a first order system.

Dealing with more complicated systems is a bit of an art. If you have poles and zeros, you can think about pole/zero cancelation, even if they are not perfectly matched in frequency. In other words a third order system with a pole and zero near each other, may act pretty much like a second order system without that pole/zero pair.

The understanding of higher order systems, (like your example) with no zeros present, is made easier if it can be viewed as a combination of first and/or second order systems. A third order system will have 3 poles. If these poles are separated by a large frequency, then write the transfer function as the multiplication of three separate first order systems. If two poles are near each other, with the other far away, then write the transfer function as the multiplication of a first order system with a second order system. If all poles are near each other, then it's much harder to understand the system behavior. But if you have two complex conjugate poles, then you can view this as an underdamped response and define a natural frequency and damping ratio for those two poles as if they are a second order system. I believe you are dealing with this last case.

In general, the system gain can be found when you normalize the first and second order system in the usual ways. A single poles system will be normalized with unity gain at zero frequency. The second order system is normalized to have unity gain at the natural frequency. Any gain factor left over can be called the gain factor K. However, gain is frequency dependent, so you could define the gain in other ways, such as; DC gain, high frequency gain, maximum gain etc. It's kind of arbitrary and based on whatever definition you find useful.

It's possible to get into a debate over the exact definition of damping ratio for a third order system like this. Is it simply the damping ratio of the second order portion of the system? Or, is it possible to define a more encompassing damping factor (which includes the effect of all poles) based on the way the full system response decays over time? I think both are possible and useful to consider. However, I don't know which you (or your teacher) are considering.

Last edited: May 24, 2009
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6. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
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I do understand second order system but as this is a third it is different.

the open loop transfer function is attached.

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7. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, it is different, but it's good that you do understand 2'nd order systems. Otherwise, it would be too difficult to understand this system or any other higher order systems. Is the information I provided in any way useful? Do you see the value in representing your 3'rd order system as the concatenation (i.e. multiplication of transfer functions) of a first order system with a second order system? Do you understand how to determine that your system contains a second order underdamped systems?

Referring to the feedback that you show in the figure, - is it positive or negative feedback? The diagram shows a summing node with no indication of positive or negative signs. This usually means all inputs are added, but your total transfer function is only correct if the feed back is negative.

EDIT: Sorry, hgmjr is right, the first transfer function is not correct with either positive or negative feedback assumed.

Last edited: May 24, 2009
8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If you take a look at the material at this link you should be able to spot the flaw in your interpretation of the expression for the transfer function in your original post.

hgmjr

POSTSCRIPT: Steveb's point about the need for the inclusion of the correct polarity signs on the arrows is also important. You do need to include the polarity marks on your diagram for it to be complete.

Last edited: May 24, 2009
9. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
0
yes steveb it is useful. thanks

which orginal post are you refering to? that was the question i was given.

As for the negative or positive. the postive sign is on at the top and negative on the lower.

thanks for your keep responses guys

10. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, makes sense now. Your first post is the open loop transfer function and then later you give the closed loop transfer function with negative feedback.

The link by hgmjr appears to be what you need. He was smart enough to see what the original question is really asking. I missed it. Basically, the gain factor K they want is the maximum value of K that results in a stable system.

11. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Above is the original post that started this thread. The formula it contains is the one to which I was referring.

hgmjr

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I can see now that I should have stated my assessment more clearly. Indeed Steveb is correct in that the transfer function you stated in the first post in this thread is the open-loop gain. It was not until we were supplied with the diagram that this point became clear.

hgmjr

13. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The typical expression that applies in this case is:

$\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)H(s)}$

where G(s) = the forward gain expression
and H(s) = the feedback gain expression

hgmjr

14. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
0
I still am having trouble in working out the expression

15. ### steveb Senior Member

Jul 3, 2008
2,433
469
It would help to know exactly what the issue is. Otherwise we will just be guessing and not helping you in the area that is giving you trouble.

For example, do you understand the method provided in hgmjr's link. That is called the root-locus method. You basically trace the locus (curve) of the roots (system poles) as the parameter K is varied from 0 to infinity. The value of K that causes the poles to move into the right half plane of the complex frequency plane is the maximum gain for stability.

Does this make sense to you? Calculations are one thing, but you need to understand the basic principle before you can do the calculations.

The calculations you need to do here are a little tricky actually, and usually people use Matlab to make it easy. The root equation for your system poles is a cubic function which results in 3 roots (poles). You are probably familiar with solving the roots of a quadratic equation, right? $x_{1,2}={{b^2\pm \sqrt{b^2-4ac}}\over{2a}}$
That's pretty easy. However, without a good computer tool, developing expressions by hand requires solving for roots of the cubic equation in the denominator of your equation. Any good math handbook will have the equations for this, but it's not as straightforward as the quadratic case. Once you have mathematical expressions for the roots as a function of K, you can write the transfer function as a product of a first order system with a second order system. You can then pick off the damping factor from the second order portion of the system (if that is what you really need).

By the way, since this is a homework problem, could you post the exact question as it was given to you. Also, what class material have you covered that you are expected to apply to the problem? Are you allowed to use Matlab, or do you have to derive by hand? These things help set the proper context to answer the question.

16. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
0
ok, yes im farmilair with quadratic equation.

I have to use both matlab and hand calculations for it.
I have to firstly find the gain K, given steadystate error 0.15. I guess with the K value i can easily plot the root locus but i need to find K which is what i am asking.

From the root locus i need to find the undamped namtural freq, damping ratio and settling time.

I then need to find the K to get a damping ratio of 0.5.

the problem arises when we have not really been taught matlab so it is exceptionally hard for me. I know a little but not enough to derive what i need.

You have the diagram which is part of the question.

i forgot to add can use matlab to get the root locus.

So i need to find K to insert it into matlab. Then i cant get the root locus and work out the rest. So my question is how do i get the gain value K from 40K / s(s+3)(s+5).

and i would like to know if my first step is correct.
40K / s(s+3)(s+5)+40K

thank you

Last edited: May 25, 2009
17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The expression you have given for the closed-loop transfer function is correct. I would however recommend that you proceed and multiply the s-terms in the denominator.

$\frac{C(s)}{R(s)} = \frac{40K}{s(s+3)(s+5)+40K}$

You should get

$\frac{C(s)}{R(s)} = \frac{40K}{s^3+As^2+Bs+40K}$

Where A and B are the coefficients of the $s^2$ and the s term respectively.

hgmjr

18. ### steveb Senior Member

Jul 3, 2008
2,433
469
It's good that you can use Matlab. It makes things go faster. Note that it's good to learn how to do all the calculations by hand, but once you know the method, there is nothing wrong with using a tool to speed up the process.

You have many steps, so let's go one at a time.

In the above statement you said something that is not quite correct. You don't need the value of K to plot the root locus because the root locus plot is made from the curves (loci) of all poles (roots), as K varies over a range of values (usually 0 to infinity). The location of the poles move as K is varied and the location of the poles tells you how the system behaves. The root locus plot gives you a feel for how to tune the system to the proper value of K. For example, you may want to know what values of K make the system unstable. Instability occurs when at least one pole moves into the right half plane (i.e. real part of the complex value is greater than one). As another example, you may want the system to have critical damping which is also obvious in the plot if two curves meet at one point on the real axis and split off into two complex conjugate poles.

Anyway, the above discussion is not too critical yet, but it's good to understand it for later.

OK, for this first step, you want the value of K that results in a steady state error of 0.15. This requires additional information. In particular you need to know the steady state input value. I'm going to assume that the steady state input value is one. A trick to figure out the steady state gain is to remember that steady state means time equals infinity which corresponds to frequency of zero. The gain at zero frequency is one, hence the steady state error is zero for any input and for any gain K, unless the system is unstable.

This means that something is wrong with the question as you have stated it. Or, perhaps I'm misunderstanding what you meant.

Another way to see why the steady state error must be zero if the system is stable is to look at the open loop transfer function. It includes an integrator (1/s, or a pole at zero). Basically, the control loop is a proportional-integral feedback system which drives the error to zero.

So we need to clear this confusion up before we can go any further. It really would be a good idea if you state the problem exactly as it was given to you. The diagram is fine, but can you state the question in the exact words?

Last edited: May 25, 2009
19. ### lukus08 Thread Starter Active Member

Mar 14, 2009
34
0
thanks. that is the question above.

what we are given is steady state input error to unit velocity = 0.15
damping ratio=0.5
and settling time within 2%<4 sec.

20. ### steveb Senior Member

Jul 3, 2008
2,433
469
I'm having trouble understanding this specification. I don't know what unit velocity means.

Is this an input ramp of unit slope? This would make sense from one point of view, since a PI control would have an non-zero error for a ramp input. However, then the term steady state would not make sense. Usually steady state is a term applied to sinusoids or any signal that reaches a constant DC level. In both cases the assumption is that any transients have decayed at time equal to infinity.

Can you or anyone else shed some light on this terminology?