MATLAB: Help, urgent: Error using DynamicSystem/lsim

lsim

Hello I have a problem. I have a power train system, as an input I have torque with a constant slope and afte 0.4 seconds i need a constant torque. I realized it with a loop and if conditions. It doesn't work and I don't understand how to change my Code that there won't be an Error anymore.
This is the ERROR:
Error using DynamicSystem/lsim (line 84)
In time response commands, the time vector must be real, finite, and
must contain monotonically increasing and evenly spaced time samples.
Error in Antriebsstrang_Schmitt (line 99)
   [y,t,x]=lsim(sys,h1,t);
THIS IS MY CODE:
% Antriebsstrang : Zustandsdarstellung
clear
clc
close all
% System-Parameter
    J4 = 0.0784;
    J2 = 0.000385;
    J1 = 0.000271;
      iG = 3.2;
      iA = 4.1;
      JM = 0.1322;     % Motor, ...
      cK = 1000;
      MM = 90;
      JA = J4+iA*iA*J2+iG*iA*iA*iA+J1    % Achswelle, ...
      cA = 7300;
      JF = 0.3426;     % RAd, ...
      JR = 1.0457;     % Reifen, ...
      cR = 31600;
% Systemmatrizen
      J = [ JM   0  0  0;...
             0   JA 0  0;...
             0   0  JF 0;...
             0   0  0  JR];
      C =  [cK          (-cK*iG*iA)            0       0;...
            (-iG*iA*cK) (-cK*iG*iG*iA*iA+cA)  -cA      0;...
            0            -cA                   cA+cR  -cR;...
            0            0                    -cR      cR];
% Lastverteilungsvektoren
      p1 = [ 1;...
             0;...
             0;...
             0  ];         %  zu MM
% Matritzen für System- und Ausgangsgleichung  &  Modellaufbau
      A = [  zeros(4,4)   eye(4) ;...
            -inv(J)*C    -inv(J)*zeros(4,4)   ]
      B = [  zeros(4,1) ;...
             inv(J)*p1  ]
      Ca = [  0      0      0     0    0     0      0    1  ] % phiR'
            % phiM' phiM  phiA' phiA  phiF' phiF  phiR' phiR
      Da = [  0  ]
%Loop and If-
      for x=0:0.003:1.6;
         if x< 0.4 
    t=datevec(x);
          h1= 200*t ;
         else 
          h1= 80;
          h1=datevec(h1);
      end
  hold on
      sys = ss(A, B, Ca, Da)
      figure (1)
      hold on
      bode(sys)
     %LSIM
     [y,t,x]=lsim(sys,h1,t);
      end
      figure(2)                   %Ausgabe nach Anregung
figure (2)
plot(t, y);
xlabel('Zeit [s]');
ylabel('Beschleunigung [m/s^2]');
title('Beschleunigung')

Best Answer

It is best to use a combination logical vector with conditions rather than a loop to create your input vector. My ‘h1’ assignment creates two logical vectors, one that is 1 (or true) for t<0.4, and zero elsewhere, multiplies it by 200*t (using element-wise multiplication) and adds it to another complementary vector calculated the same way to create the constant section. Note that logical true converts to numeric 1 in calculations, and false converts to numeric 0. (I plotted the ‘h1’ as a funciton of ‘t’ and it produces the result you describe as what you want.
The reason you are getting the error with your time vector is that datevec produces a 6-element vector for each time. This is not what you want for your simulation.
I added these lines to your code:
% Create ‘h1’ as logical vector with conditions on intervals — NEW CODE REPLACING ‘for’ LOOP AND ‘if’ BLOCK

x = 0:0.003:1.6;                            % Define ‘x’

t = x;                                      % Define ‘t’

h1 = (200*t).*(t < 0.4) + 80*(t >= 0.4);    % Create Input Vector

Your full code with my changes:
% System-Parameter
    J4 = 0.0784;
    J2 = 0.000385;
    J1 = 0.000271;
        iG = 3.2;
        iA = 4.1;
        JM = 0.1322;     % Motor, ...
        cK = 1000;
        MM = 90;
        JA = J4+iA*iA*J2+iG*iA*iA*iA+J1    % Achswelle, ...
        cA = 7300;
        JF = 0.3426;     % RAd, ...
        JR = 1.0457;     % Reifen, ...
        cR = 31600;
% Systemmatrizen
        J = [ JM   0  0  0;...
               0   JA 0  0;...
               0   0  JF 0;...
               0   0  0  JR];
        C =  [cK          (-cK*iG*iA)            0       0;...
              (-iG*iA*cK) (-cK*iG*iG*iA*iA+cA)  -cA      0;...
              0            -cA                   cA+cR  -cR;...
              0            0                    -cR      cR];
% Lastverteilungsvektoren
        p1 = [ 1;...
               0;...
               0;...
               0  ];         %  zu MM
% Matritzen für System- und Ausgangsgleichung  &  Modellaufbau
        A = [  zeros(4,4)   eye(4) ;...
              -inv(J)*C    -inv(J)*zeros(4,4)   ]
        B = [  zeros(4,1) ;...
               inv(J)*p1  ]
        Ca = [  0      0      0     0    0     0      0    1  ] % phiR'
              % phiM' phiM  phiA' phiA  phiF' phiF  phiR' phiR
        Da = [  0  ]
% Create ‘h1’ as logical vector with conditions on intervals — NEW CODE REPLACING ‘for’ LOOP AND ‘if’ BLOCK
x = 0:0.003:1.6;                            % Define ‘x’
t = x;                                      % Define ‘t’
h1 = (200*t).*(t < 0.4) + 80*(t >= 0.4);    % Create Input Vector
        sys = ss(A, B, Ca, Da)
        figure (1)
        hold on
        bode(sys)
       %LSIM
       [y,t,x]=lsim(sys,h1,t);
        figure(2)                   %Ausgabe nach Anregung
figure (2)
plot(t, y);
xlabel('Zeit [s]');
ylabel('Beschleunigung [m/s^2]');
title('Beschleunigung')
figure(3)
plot(t, h1)
grid
axis([xlim    0 100])
title('Input Signal')
xlabel('Time (s)')
ylabel('Amplitude')
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