Why my graphic always on straight line?

clc

clear

%% Nilai Data

Ip = 0.004026

Mp = 0.419

L = 0.108

Ir = 0.000189

Mr = 0.204

Im = 0.00548

Mm = 0.112

g = 10

%Equation

A1 = ((Mp+Mr+Mm)*g*L)/((Im+Ir+Ip)-Ir+(Mp+Mr+Mm)+L^2);

B1 = 1/((Im+Ir+Ip)-Ir+(Mp+Mr+Mm)+L^2);

B2 = ((1/Ir)+1/((Im+Ir+Ip)-Ir+(Mp+Mr+Mm)+L^2));

%Matriks

A = [0 0 1 0; 0 0 0 1; 0 A1 0 0; 0 A1 0 0];

B = [0;0;B1;B2];

C = [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1];

D = [0];

%% controllability

sys = ss(A,B,C,D);

Co = ctrb(sys)

Con = rank(Co)

%% stability

%tambahan

eig(A)

%% Polynomial Characteristic ITAE Method

Wn = 1.5

ITAE = [1 (2.1*Wn) (3.4*Wn^2) (2.7*Wn^3) (Wn^4)]

Roots1 = roots(ITAE)

K = place(A,B,Roots1)

Acl = A-B*K

E2 = eig(Acl)

Pcl = place(A,B,E2)

%% Call Simulink

[Out] = sim('simulasi.slx')

figure(1)

plot(Out.tout, Out.XITAE1);

title('Grafik response');

xlabel('Time');

ylabel('radian');

grid on;

[Walter Roberson]

We have no information about what your model does.

[Sulaymon Eshkabilov]

Post your simulink model simulasi.slx. Nobody can guess what you have in your Simulink model.

Answer 1

[Sam Chak]

Hi @Nugroho Wisnumurti

Although you haven't provided the Simulink file 'simulasi.slx', I believe your result indicates that the system is in equilibrium (0, 0, 0, 0). What happened is that you simulated the model in Simulink with zero initial conditions and injected zero input. Naturally, the states will remain at the equilibrium value throughout the simulation unless you either inject a non-zero external input or use non-zero initial conditions. Three cases are demonstrated below, and yours most likely belongs to Case 2.

%% Nilai Data
Ip = 0.004026;
Mp = 0.419;
L  = 0.108;
Ir = 0.000189;
Mr = 0.204;
Im = 0.00548;
Mm = 0.112;
g  = 10;

% Equation
A1 = ((Mp+Mr+Mm)*g*L)/((Im+Ir+Ip)-Ir+(Mp+Mr+Mm)+L^2);
B1 = 1/((Im+Ir+Ip)-Ir+(Mp+Mr+Mm)+L^2);
B2 = ((1/Ir)+1/((Im+Ir+Ip)-Ir+(Mp+Mr+Mm)+L^2));

% Matriks
A = [0  0  1  0; 
     0  0  0  1; 
     0 A1  0  0; 
     0 A1  0  0];
B = [ 0;
      0;
     B1;
     B2];
C = [1 0 0 0; 
     0 1 0 0; 
     0 0 1 0; 
     0 0 0 1];  % Sam: this is identity matrix, eye(4)
D = [0];

%% controllability
sys = ss(A, B, C, D);
Co  = ctrb(sys);
Con = rank(Co);

%% stability
%tambahan
eig(A)

ans = 4×1

0 0 1.0246 -1.0246

%% Polynomial Characteristic ITAE Method
Wn   = 1.5;
ITAE = [1 (2.1*Wn) (3.4*Wn^2) (2.7*Wn^3) (Wn^4)];   % Sam: most likely from Dorf & Bishop
Roots1 = roots(ITAE);
K    = place(A,B,Roots1)

K = 1×4

0.0009 0.0016 0.0016 0.0006

Acl  = A-B*K;
E2   = eig(Acl);
Pcl  = place(A,B,E2);

%% Compensated system via pole-placement method
csys = ss(A-B*K, B, C, D);
N    = 1/dcgain(csys);
csys = ss(A-B*K, B*N(1), C, D)

csys =
 
  A = 
               x1          x2          x3          x4
   x1           0           0           1           0
   x2           0           0           0           1
   x3   -0.001205       1.048    -0.00217  -0.0007866
   x4      -4.824      -7.649      -8.683      -3.148
 
  B = 
             u1
   x1         0
   x2         0
   x3  0.001205
   x4     4.824
 
  C = 
       x1  x2  x3  x4
   y1   1   0   0   0
   y2   0   1   0   0
   y3   0   0   1   0
   y4   0   0   0   1
 
  D = 
       u1
   y1   0
   y2   0
   y3   0
   y4   0
 
Continuous-time state-space model.

%% Case 1: Time response when input u is Unit-Step & initial condition is zero
step(csys), grid on, title('Case 1: Time Response with step input & zero x0')

%% Case 2: Time response when both input u & initial condition are zero
t  = linspace(0, 12, 1201);
u  = zeros(1, numel(t));
lsim(csys, u, t), grid on, title('Case 2: Time Response with zero input & zero x0')

%% Case 3: Time response when input u is zero & initial condition is non-zero
t  = linspace(0, 12, 1201);
u  = zeros(1, numel(t));
x0 = [1 0 0 0];             % initial condition
lsim(csys, u, t, x0), grid on, title('Case 3: Time Response with zero input & non-zero x0')

%% Call Simulink
% [Out] = sim('simulasi.slx')
% figure(1)
% plot(Out.tout, Out.XITAE1);
% title('Grafik response');
% xlabel('Time');
% ylabel('radian');
% grid on;