Hi everybody. I wrote a code but something is going wrong, I need at that point.
I need to stop the code at the gain of 0.5 and I need to take the values that I need as P and D. But The code does not do what I need. It always work around i = 65 and j = 65. What is the error ?
clear all; clc;
Kp = -10:0.31:10;
Kd = -10:0.31:10;
% n = [Kd Kp];
% d = [(1+Kd) (Kp+1)];
%
% [r,p,k] = residue(n,d);
A = zeros(length(Kp), length(Kd));
for i = 1:length(Kp)
for j = 1:length(Kd)
n = [Kd(j) Kp(i)];
d = [(1+Kd(j)) (Kp(i)+1)];
[r,p,k] = residue(n,d);
S = sum(r);
TF = tf(n,d);
[z, gain] = zero(TF);
if gain == 0.5
break;
end
% if S<0.6 && S>0.4
% A(i,j) = 1;
% else
% A(i,j) = 0;
% end
end
end

Antworten (1)

James Tursa
James Tursa am 20 Mär. 2018

0 Stimmen

It is often better to use tolerances for floating point comparisons, unless you wrote the underlying code and you know that the result must be exactly 0.5. So this test
if gain == 0.5
should probably look something like this instead
if abs(gain - 0.5) < tol % for some appropriate value of tol

5 Kommentare

Stephen23
Stephen23 am 21 Mär. 2018
Bearbeitet: Stephen23 am 21 Mär. 2018
Berkcan Oz's "Answer" moved here:
What do you mean ? What should I change ?
Stephen23
Stephen23 am 21 Mär. 2018
Bearbeitet: Stephen23 am 21 Mär. 2018
@Berkcan Oz: your code contains exactly one instance of the operator ==. Change it as James Tursa recommended.
Berkcan Oz
Berkcan Oz am 21 Mär. 2018
What is tol ? How to define ?
Stephen23
Stephen23 am 21 Mär. 2018
You could start with tol=1e-5;
Berkcan Oz
Berkcan Oz am 21 Mär. 2018
Bearbeitet: Walter Roberson am 21 Mär. 2018
Now, I did this but this give me a system that starts at gain almost 0.5 and goes 1. I need to make it from 0 to 0.5. Not from 0.5 to 1. What should I do ?
clear all; clc;
Kp = -10:0.31:10;
Kd = -10:0.31:10;
tol = 1e-5;
A = zeros(length(Kp), length(Kd));
for i = 1:length(Kp)
for j = 1:length(Kd)
n = [Kd(j) Kp(i)];
d = [(1+Kd(j)) (Kp(i)+1)];
[r,p,k] = residue(n,d);
S = sum(r);
TF = tf(n,d);
[z, gain] = zero(TF);
if gain <0.6 && gain>0.4
break;
end
end
end
sys = tf(n,d)
step(sys);
grid;

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Gefragt:

am 20 Mär. 2018

Bearbeitet:

am 21 Mär. 2018

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