help, urgent: Error using DynamicSystem/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')

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency

Answer 1

[Star Strider]

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')

[Schmittna59039]

Thank you so much. After having tried and failed for the whole the code finally works. Greetings Nadja

[Star Strider]

My pleasure!

Answer 2

[Sebastian Castro]

First off, a tip: If you use "urgent" as part of your question, it makes people want to answer it less. Nobody gets special treatment here :)

Anyhow, I see two main issues with your code.

#1: I think the datevec portion where you define h1 and t is correct.

For example, the final values of t and h1 both end up being the vector 0 3 20 0 0 0, which creates the warnings you see. Time must be always increasing, but it goes from 20 back to 0!

My understanding is you want to generate a time vector that goes from 0 to 1.6 seconds. For the first 0.4 seconds, you will ramp up from 0 to 200*t = 80 and then hold the 80.

I would replace that whole section with:

t = 0:0.003:1.6;
h1 = 200*t;
h1(t>0.4) = 80;

#2: Your for-loop's end happens too late. I think you meant to add an extra end earlier in your code. Right now you're running the lsim command inside the for-loop, which is likely not what you want to do.

Of course, if you take my recommendation from #1, you won't need to use any for-loops here and you should just delete that end after the lsim command.

- Sebastian

[Schmittna59039]

Hello Sebastian, thank you for your tip and your help. I will consider it next time :). I tried your solution too and it also works :).

Greetings Nadja