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I am looking at a control system which has an unstable pole in the process. The transfer functions of process and controller are the following:

G = 10/((s+10)*(s-1));
K1 = 4*(s+1)/s;

where G is the process and K1 is the controller. I have found with the Routh Criterion that the closed loop is stable for $K>1.125$ and so I have chose the value 4 since it gives good performances. Now, when I plot the root locus of the closed loop:

T1 = K1*G/(1+K1*G)
figure;
rlocus(T1)

I get the following:

enter image description here

What I don't undestand is from where the zero that cancels the unstable pole comes from, since the closed loop transfer function is:

$\frac{(40(s + 1))}{(40 - 10s + 40s + 9s^2 + s^3)}$

can somebody help me?

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  • $\begingroup$ The question you aren't asking is "will my system actually work in the real world". The answer is no, because that unstable pole will still be there, it'll get excited by something, and the plant output will grow as $e^t$ until something hits a stop, breaks, or burns up. $\endgroup$ – TimWescott Dec 20 '19 at 21:49
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rlocus() takes the open loop transfer function as an argument, not the closed loop. i.e. G not T. Look at the documentation for rlocus here https://www.mathworks.com/help/control/ref/rlocus.html

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