# Homework Help: Root loci - transfer function

1. May 23, 2010

### greg997

Hi, I ve got two problems.
1) The problem i have is how to find an open loop transfer function that gives poles at -3, -6, -7+_ j5 and zero at -5.
So i think that the numerator is (s+5), and denominator has (s+3),(s+6). and now I dont know how to carry on. Is it ( s+7+j5) and (s+7-j5)?. Am I right so far? It would be now a closed loop transfer function, so how to go back to open loop one?

2.)I need to fine open loop transfer function, gain that gives critical stability, what is the frequency at that critical staility. I can only see that there are 3 poles and no zeros. How to proceede.?

Thanks

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2. May 26, 2010

### super sky

That's right so far. With your first question, they might want you to rearrange your complex poles. To do that, you'd just expand (s+7+j5)*(s+7-j5). Otherwise, you'd be correct in using those as part of your denominator to have it as (s+3)*(s+6)*(s+7+j5)*(s+7-j5)

How do you know that that's the closed loop transfer function? If you were reading the poles and zeros from a root locus, then those poles and zeros are the open loop ones.

When you've got a root locus, the visible X's and O's are your open loop poles and zeros. The lines are what happens to the closed loop poles as the open loop gain changes.

For your second question, do you know where the closed loop poles need to be for the system to be stable? Hint - one side of the s-plane gives a stable response, and if there are any on the other side it becomes unstable.
Once you've worked out that, you can look at your root locus and see where the pole is when it reaches that axis. Now to calculate the gain, you find the distance on the s-plane to this chosen pole each open loop pole or zero (where the distance is sqrt(IM{s}^2+RE{s}^2) ...). Then K = (distances of the poles all multiplied together)/(distances of the zeros all multiplied together).

This website has a really good example:
http://www.facstaff.bucknell.edu/mastascu/eControlHTML/RootLocus/RLocus1A.html#Problems

3. May 26, 2010

### greg997

:) thnks, the system is marginnaly stable when pole(s) is at 0. So the real part must be zero. I will try to solve that. And thanks for the link.