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Hello everyone, I have a Matlab program that I got and want to run it on my PC. It gives me errors even though I'm 100% sure of the program and the data entry.My problem is in the function calling somewhere please help. My final is due this monday :( .

I have attached the code below, thank you for all the support.

```
% Program name: zerodynamics_tracking_comparison_Ex6_6_5_main
% The purpose of this example is to compare the tracking performance of a
% fully and partially linearizable system resulting from different choices
% of the output equations.
clc, clear all, close all, format short g
global g K KK ro rro ncon
%% Normal-form system matrices in controllable canonical form
Ac=[0 1; 0 0]; % Controllable Ac in z-domain
Bc=[0; 1]; % Controllable Bc in z-domain
Cc=[1 0]; % Output matrix
D=0; % Throughput matrix
g=[0; 1]; % g(x) matrix in xdot=f (x) +g (x) u
%% Pole placement to design for normal form
zeta=0.7; wn=6; wd=wn*sqrt (1-zeta^2)
p=[-zeta*wn+j*wd -zeta*wn-j*wd] % Chosen complex poles
K=place(Ac, Bc, p) % Poles placement gain matrix K
Acl=Ac-Bc*K; % CL system matrix in z-domain
% Step response of y(t)=x1 (t) in the z-coordinates with r(t)=ro (rad)
ro=8;
t=0: 0.01:2;
step (Acl,Bc*K(1)*ro,Cc,D,1,t) ;grid % Get y(t)=x1 (t) step response statistics
title ('Step response of fully linearizable system: \rho=2=n: y(t)=x_1(t)', 'fontsize',12)
xlabel ('t', 'fontsize',12), ylabel ('y(t)=x_1(t) ', 'fontsize',12)
%% Case 1: Step responses of fully linearizable system
ro=8; % Step input r(t)=ro
x0=[0; 0.2]; % Initial condition of x
t=0:0.01:2;
ncon=1; % Case 1
[t,x]=ode23 ('zerodynamics_tracking_comparison_Ex_6_6_5_fn', t, x0);
figure,plot(t,x(:,1),t,x(:,2), '--', 'linewidth',1.5),grid
xlabel('t (sec) ', 'Fontsize',12)
title('Step response of fully linearizable system:\rho=2:x_1(t)==>ro=8,x_2(t)','fontsize',12)
legend('x_1(t)', 'x_2(t) ', 'location', 'best')
%% Case 2: Step responses of normal form – note that this system is not robust
rro=2;
KK=4.2; % Control gain = closed-loop pole
t=0:0.01:6;
x0=[0; 0.2]; % Initial condition of x
ncon=2; % Case 2
[t,x]=ode23('zerodynamics_tracking_comparison_Ex_6_6_5_fn',t,x0);
figure,plot(t,x(:,1),t,x(:,2), '—','linewidth',1.5)grid
xlabel('t(sec)', 'Fontsize',12)
title('Step responses of normal form: \rho=1: x_1(t), x_2(t) ==>r(t)=2',
'fontsize',12)
legend('x_1(t) '',x_2(t) ', 'location','best')
n=2*[4.2]; d = [1 4.2]; % Get the settling time of Case 2
figure,step(n,d),grid
% Function name: zerodynamics_tracking_comparison_Ex6_6_5_fn
% function xdot=zerodynamics_tracking_comparison_Ex_6_6_5_fn
% The purpose of this example is to compare the tracking performance of a
% fully and partially linearizable system resulting from different choices
% of the output equations.
function xdot=zerodynamics_tracking_comparison_Ex_6_6_5_fn(t,x)
global g K KK ro rro ncon
f=[-x(1)+x(2)^3;-x(2)];
alpha=x(1)-4*x(2)^3;
Dx=3*x(2)^2; % Decoupling matrix
%% Case 1: Step input for fully linearizable system
if ncon==1 % Case 1
z=[x(1);-x(1)+x(2)^3]; % Transformation z=T(x)
v=-K(1)*z(1)+K(2)*z(2)+K(1)*ro; % z-domain linear tracking control law
u=inv(Dx)*(-alpha+v); % Linearizing feedback tracking control law
else % Case 2
%% Case 2: Step input for normal form
alpha=-x(2);
Dx=1;
v=-KK*(x(2)-rro);
u=inv(Dx)*(-alpha+v); % Linearizing feedback tracking control law
end
xdot=f+g*u;
```