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Hi all hope someone can help me here.

My problem involves the transfer function

 G(s)=(s+32.8)/((s+2.8)(s+5)(s+27.8) )

As far as im concerned the centroid is calculated using

σ_a= (Ʃpoles - Ʃzeros) / (# of poles - # of zeros)

From my calculations

σ_a=((-2.8+-5+-27.8)-(-32.8))/(3-1)=-1.4

However when i input the following code into MATLAB the solution from the graph shown the centroid will be -1.55

s=tf('s');
G=(s+32.8)/((s+2.8)*(s+5)*(s+27.8));
rlocus(G)

plot produced by matlab

Im not 100% whats the issue but if im doing something wrong could someone please correct me and help me understand.

Thank you for taking your time out to read this.

dorudrakky
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1 Answers1

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I used Octave and I got that the asymptotes goes at -1.4, just as you calculated. You should double check the Matlab result.

asymptote of the root locus plot

I think you might be misunderstanding the meaning of the centroid, it is the point where the asymptote lines meet at the real line, it is not the breakaway point (where the poles leave the real line). Check slide 3 of these notes

jDAQ
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  • Thanks for the clarification thats good to know. MATLAB doesnt make it clear where the centroid is and im not sure theres a command which allows me to see it. But if my hand written calc is correct thats all i need to know. Again i really appreciate the help. – dorudrakky May 29 '20 at 23:03
  • If this answers your question, please consider upvoting and\or selecting it as the answer to the question. – jDAQ May 29 '20 at 23:05
  • Also yeah i understand the difference just thought the line on MATLAB should have been much closer to -1.4 than it was shown so i wasnt sure if there was an issue with my answer. – dorudrakky May 29 '20 at 23:08
  • Ok, I see what you are doing, but looking at the pole position will not get you a very precise value of the centroid (because the pole will only meed the asymptote at infinity), it is easier to just zoom in on the intersection of the asymptote and real line and look at the value. – jDAQ May 29 '20 at 23:12