How do I calculate the gain at a point on the root locus

I am trying to generate the gain "K" at a point (anypoint) on the root locus using the charastic equation. 1 + KGsHs = 0. So K = -(D(s)/N(s))

Example from Control Systems Engineering (Norman Nice)

Steps, 1) generate num and den as normal

2) Give s a reference value eg 1.6 + 1.6i

3) multiply the coeficient by the required value of s

This where I have the problem:

The den is a row vector, that is the coefficients are listed in decreasing powers of s. Having multiplied the coefficients by the point value, that must now be praise to the level of the elements power. eg the first element of pg is 1.6+1.6i which is now the new coefficient of s^2

So K = s^2+s^1+s^0

I have attempted varios ways to achieve that but have failed. My last effort was the for loop as shown below.

If any one can show where I have gone wrong I would appreciate it

Thanks and regards.

clear all;
close all;
clc;
syms s
s = tf('s');
H = 1;

num = 1+0j

num = 1

den = [1 3 0]                       %den =(s^2 + 3s)

den = 1×3

1 3 0

G = tf(num,den)

G =
 
      1
  ---------
  s^2 + 3 s
 
Continuous-time transfer function.

pG1 = [0]

pG1 = 0

P1 = -1.6+1.6i                       % the point at which I want the gain value

P1 = -1.6000 + 1.6000i

                                    %therefore "s" = P1

%denPoly = poly2sym(den, s)
K = 0

K = 0

pg = P1.*den                        %  I can multiply the den elements by 

pg = 
  -1.6000 + 1.6000i  -4.8000 + 4.8000i   0.0000 + 0.0000i

                                    % the point value P1
                                    %  The output looks correct, as a row
                                    %  vector I know the final value should
                                    %  be 3


for n = length(den) : 1
    n
    m = (length(den) - n)         % I have tried to calculate j to
                                  % represent the power from placement
    pG1(n) = pg(n).^m
end

Answer 1

[Paul Buckman]

I hope that I am not messing people arround but:

I got around the problem above this way: new array psden to hold the powers of den from sC to 0

clear all; close all; clc;
syms s
s = tf('s');
H = 1;

num = 1+0j

num = 1

den = [1 3 0]                       %den =(s^2 + 3s)

den = 1×3

1 3 0

G = tf(num,den)

G =
 
      1
  ---------
  s^2 + 3 s
 
Continuous-time transfer function.

[sR, sC] = size(den)

sR = 1

sC = 3

psden = sC-1:-1:0              % 

psden = 1×3

2 1 0

pG1 = zeros(sR,sC)

pG1 = 1×3

0 0 0

P1 = -1.6+1.6i                       % the point at which I want the gain value

P1 = -1.6000 + 1.6000i

                                    %therefore "s" = P1

%denPoly = poly2sym(den, s)
K = 0

K = 0

pg = P1.*den                        %  I can multiply the den elements by 

pg = 
  -1.6000 + 1.6000i  -4.8000 + 4.8000i   0.0000 + 0.0000i

                                    % the point value P1
                                    %  The output looks correct, as a row
                                    %  vector

pG1 = pg.^psden

pG1 = 
   0.0000 - 5.1200i  -4.8000 + 4.8000i   1.0000 + 0.0000i

Now pG1(3) does not look correct as any number^0 = 0 can anyone explain this please or am I in error again.