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  #1  
Old 05-24-2009, 01:20 PM
lukus08 lukus08 is offline
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Default How to find the damping ratio of this transfer function

40K / s(s+3)(s+5)

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

thanks
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  #2  
Old 05-24-2009, 02:06 PM
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hgmjr hgmjr is offline
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How about you take a stab at computing the two parameters and then posting your effort here?

hgmjr
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  #3  
Old 05-24-2009, 02:46 PM
lukus08 lukus08 is offline
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i need help, which is why i am posting.

I can only get up to:

40k / s^3 + 3s^2 + + 15s + 40k
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Old 05-24-2009, 02:56 PM
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Were you supplied the circuit diagram to which the transfer function applies? Or were you just given the transfer function?

hgmjr
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Old 05-24-2009, 03:46 PM
steveb steveb is offline
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Quote:
Originally Posted by lukus08 View Post
i need help, which is why i am posting.

I can only get up to:

40k / s^3 + 3s^2 + + 15s + 40k
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 by steveb; 05-24-2009 at 03:54 PM.
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  #6  
Old 05-24-2009, 04:25 PM
lukus08 lukus08 is offline
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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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File Type: jpg TF.jpg (6.3 KB, 59 views)
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Old 05-24-2009, 04:50 PM
steveb steveb is offline
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Quote:
Originally Posted by lukus08 View Post
I do understand second order system but as this is a third it is different.
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?

Quote:
Originally Posted by lukus08 View Post
the open loop transfer function is attached.
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 by steveb; 05-24-2009 at 05:30 PM.
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  #8  
Old 05-24-2009, 04:52 PM
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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 by hgmjr; 05-24-2009 at 05:33 PM. Reason: Added postcript comment.
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  #9  
Old 05-24-2009, 05:29 PM
lukus08 lukus08 is offline
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yes steveb it is useful. thanks

Quote:
Originally Posted by hgmjr View Post
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
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
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  #10  
Old 05-24-2009, 05:36 PM
steveb steveb is offline
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Quote:
Originally Posted by lukus08 View Post
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
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.
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