Index in position 1 is invalid. Array indices must be positive integers or logical values. Error in pid_optimized1 (line 17) solution(K,:) = [K a m ts];

Hi all

I am trying to run this code but I am getting an error 'Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in pid_optimized1 (line 17)

solution(K,:) = [K a m ts];'

Can someone help me to run this code.

The code is given by

clear all

clc

%Required to optimize a plant with a transfer function 4/s^3+6s^2+8s+4 with

%a PID controller whose transfer function is given by K(s+a)^2/s.

t = 0:0.01:8;

K = 0;

for K =3:0.2:5

for a =0.1:0.1:3

num =[4*K 8*K*a 4*K*a^2];

den = [1 6 8+4*K 4+8*K*a 4*K*a^2];

y = step(num,den,t);

s = 801;while y(s)>0.98&&y(s)<1.02;s = s-1;end

ts = (s-1)*0.01;%ts = settling time;

m = max(y);

if m<1.15 && m>1.10; if ts<3.00

K = K+1;

solution(K,:) = [K a m ts]; %this sorts the solution of K a m ts in a column array

end

end

end

end

Answer 1

[Alex Mcaulley]

You are using K as a logical indexer, but K take non positive integers values, throwing, then, an error. I gues that the solution is:

t = 0:0.01:8;
k = 0;
for K =3:0.2:5
    for a = 0.1:0.1:3
        num = [4*K 8*K*a 4*K*a^2];
        den = [1 6 8+4*K 4+8*K*a 4*K*a^2];
          y = step(num,den,t);
          s = 801;while y(s)>0.98&&y(s)<1.02;s = s-1;end
        ts = (s-1)*0.01;%ts = settling time;
        m = max(y);
        if m < 1.15 && m > 1.10
            if ts<3.00
               k = k + 1;
               solution(k,:) = [K a m ts]; %this sorts the solution of K a m ts in a column array
            end
        end
    end
end

[polycarp onyebuchi]

The error still persist with this code you gave me. thanks

[Bob Thompson]

Did you make the adjustment he suggested? I copied your code in, changed it to be like Alex has posted and it ran fine for me.

[polycarp onyebuchi]

thanks. it is working perfectly.hoping for more help from you guys

Answer 2

[Bob Thompson]

You are recieving the error because you are trying to use K as an index value, but K is not always a positive integer.

for K =3:0.2:5
solution(K,:) = [K a m ts];

Values of K include 3, 3.2, 3.4, 3.6, ...

Matlab cannot define an element as being in position 3.2, or any other value which isn't a positive integer.

As a side note, you should really not change the value of yoru loop index within the loop itself. It is generally considered bad practice, and is very prone to errors.

[polycarp onyebuchi]

please, what is then the solution. thanks

Answer 3

[polycarp onyebuchi]

The error still persist. someone help please