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Hi I was given a closed system as:

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

Where

$$H(s) = \frac{s}{s+1}$$ $$G(s) = \frac{k(s+4)}{(s+2)(s^2+s+6)}$$

This is where I'm unsure of. When I calculated the centroid It was 0, which MATLAB doesn't seem to agree with. What I did to get the poles and zeros was just put G(s) and H(s) into the characteristic equation.

$$1+ G(s)H(s) = 0$$

I feel like this is where I'm messing up. To get the poles and zeros should I plug G(s) and H(s) in to T(s) and simplify from there?

EDIT: Ok so I think my problem was entering the transfer function in MATLAB. Correct me if I'm wrong but I can obtain the poles and zeros of the system using the characteristic equation.

Poles resulting In: -1, -2, -0.5 + 2.4j, -0.5 - 2.4j

Zeros resulting In: 0, -4

centroid = (-1 + -2 + -0.5 + 2.4j + -0.5 - 2.4j + 4) /2 = 0

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  • $\begingroup$ btw what do you want to know exactly? Do you need to draw the characteristic loci? Or do you only need the transfer function which describes the poles? Or you need the poles as a function of $k$? $\endgroup$ – WG- Mar 25 '16 at 10:32
  • $\begingroup$ Need to draw the root loci has k goes to infinity $\endgroup$ – user108698 Mar 25 '16 at 15:39
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The root locus gives the closed loop poles trajectories as a function of a system parameter, often the feedback gain $k$.

Now your open-loop transfer function is given by: $OL(s) = G(s)H(s)$. The closed loop transfer function you tell me is given by: $CL(s) = \frac{G(s)}{1 + OL(s)}$. However, I will assume $CL(s) = \frac{OL(s)}{1 + OL(s)}$. The result boils down to the same.

enter image description here

$CL(s) = \frac{G(s)}{1 + OL(s)}$

enter image description here

$CL(s) = \frac{OL(s)}{1 + OL(s)}$

Now the stability of the closed loop is governed by $1 + OL(s)$. Assume that $OL(s) = \frac{a(s)}{b(s)}$. Then $1 + OL(s) = 1 + \frac{a(s)}{b(s)} = \frac{a(s) + b(s)}{b(s)}$. As a result, the poles of the open-loop are given by $b(s) = 0$. But the zeros of $1 + OL(s)$, which are given by $a(s) + b(s) = 0$, contains the closed-loop poles.

As a result your characteristic loci is given by: $a(s) + b(s) = 0$

For your equations $1 + OL(s) = \frac{s^4 + 4\, s^3 + \left(k + 11\right)\, s^2 + \left(4\, k + 20\right)\, s + 12}{s^4 + 4\, s^3 + 11\, s^2 + 20\, s + 12}$. Your closed loop poles are thus goverend by $s^4 + 4\, s^3 + \left(k + 11\right)\, s^2 + \left(4\, k + 20\right)\, s + 12$. The result is fairly complex so therefor I just simply plot them in Matlab...

%% Symbolic computation
clear all;
close all;
clc;

s = sym('s');
k = sym('k');

H = s/(s + 1);
G = k*(s+4)/((s + 2)*(s^2 + s + 6));

% Open-loop
OL = G*H;

% Closed loop denominator
L = (1 + G*H);

% Closed loop
CL = G/L;

% Fetch numerator and denominator of L
% num contains the closed loop poles
% den contains the open loop poles
[num,den] = numden(L);

% Try to solve...
poles = solve(num == 0,s);

%% Transfer function
clear all;
close all;
clc;

s = tf('s');

H = s/(s + 1);
G = (s+4)/((s + 2)*(s^2 + s + 6));

% Open-loop
OL = G*H;

% Closed loop denominator
L = (1 + G*H);

% Closed loop
CL = G/L;

% Compute rlocus
rlocus(OL);

In case you need to draw the root locus by yourself, you can use the procedure described here: https://en.wikibooks.org/wiki/Control_Systems/Root_Locus#The_Root-Locus_Procedure.

enter image description here

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  • $\begingroup$ thank you for the very detailed answer but this question is from an old final. We aren't aloud to use MATLAB. It is all hand written. $\endgroup$ – user108698 Mar 25 '16 at 9:46
  • $\begingroup$ Ah okay. But still you have the poles of the closed loop, which are the roots of $s^4+4s^3+(k+11)s^2+(4k+20)s+12$. I will edit my answer a bit such that you have a nicer form of your poles. $\endgroup$ – WG- Mar 25 '16 at 10:19
  • $\begingroup$ I don't think they are. Wouldn't the poles and zeros be from the open loop transfer function G(s)H(s) $\endgroup$ – user108698 Mar 25 '16 at 15:38
  • $\begingroup$ Maybe I'm just not understanding what your saying. But looking at the graph I can see 2 zeros at -4 and 0. And the poles at -1,-2,-0.5 + 2.4j, -0.5 - 2.5j. Which is what i first calculated. $\endgroup$ – user108698 Mar 25 '16 at 16:58

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