Why won't my rlocus function work

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John Smith
John Smith on 3 Nov 2016
Answered: Star Strider on 3 Nov 2016
close all; clc; clear
% Q1
A=[0 1 0 0; -4 -56 4 56; 0 0 0 1; 31.25 437.5 -62.5 -437.5];
B=[0 0; 1 0; 0 0; 0 31.25];
C=[1 0 -1 0];
D=[0 0];
[n1,d1]=ss2tf(A,B,C,D,1) % nominator and denominator of P
[n2,d2]=ss2tf(A,B,C,D,2) % nominator and denominator of F
P=tf(n1,d1) % transfer function of P=Y/U
F=tf(n2,d2) % transfer function of F=Y/W
% Q2
D=F/P % transfer function of D
Dmin=minreal(D) % simplified D
% Q3
P_poles=pole(P) % poles of plant P
%Q4
pole_far=P_poles(1)
s=tf('s');
P_s=minreal(((pole_far-s)/(pole_far))*P)
%Q5
rlocus(P_s)
%Q6
z1=-1+(-4)^0.5;
z2=-1-(-4)^0.5;
p1=-20;
p2=-50;
N=(P_s*(s-z1)*(s-z2))/((s-p1)*(s-p2));
figure
rlocus(N)
axis([-4 0.5 -6 6])
%Q7
[k,poles]=rlocfind(N);
%Q8
C=k*(s-z1)*(s-z2)/((s-p1)*(s-p2));
H=Dmin*feedback(P,C);
step(0.2*H)
if true
% code
end
Also H isnt producing anything, where have I gone wrong?

Answers (1)

Star Strider
Star Strider on 3 Nov 2016
The numerator of ‘N’ is complex, although the imaginary parts are vanishingly small (on the order of approximation error), so you can safely delete them. Add the ‘N.numerator’ assignment to your code just after the initial ‘N’ assignment, and it works (at least as far as I’ve tested it):
N=(P_s*(s-z1)*(s-z2))/((s-p1)*(s-p2));
N.numerator = real(N.numerator{:})); % <— ADD THIS ASSIGNMENT

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