Please how can I plot the graph of the function it keep giving me Error in QCS2 (line 13), ode45 (line 115) odeargumen​ts(FcnHand​lesUsed, solver_name, ode, tspan, y0, options, varargin);

%%System Parameters
M = [400 0;0 40];
C= [12000 -12000;-12000 12125];
K = [80000 -80000;-80000 280000];
B = [0 0;200000 125];
[t,y] = ode45(@(t,y) MSsys(t,y,M,K,C,B),[0 20],[0 0 0 0])

x1 = y(:,1); x2 = y(:,2);
plot(t,x1,t,x2)
xlabel('Time (seconds)')
ylabel('x_1(t)and x_2(t) (m/sec)')
title('Speed Bump')

%%ODEs
function dydt= MSsys(t,y,M,K,C,B)
A = [zeros(2) eye(2);-inv(M)*K -inv(M)*C];
F = [0 0;200000 125]*0.03*sin(50*t);
dydt = A*y+[zeros(2);inv(M)*B]*F;
end

[Walter Roberson]

your state is length 2 so you need to return a 2x1 vector which you obviously do not do.

II suspect that your A*y fails.

Note that zeroes(2) is 2 x 2 so your A looks to be 4x4.

Your function only ever uses inv(M) and not M itself so pass in inv(M) so that you do not keep computing the inverse.

[Simisolaoluwa Adekoya]

Please, codes do I write to fix it.Thanks

[Simisolaoluwa Adekoya]

this is the part that fails

[t,y] = ode45(@(t,y) MSsys(t,y,M,K,C,B),[0 20],[0 0 0 0])

Please how can I correct it

Answer 1

[Walter Roberson]

Your code cannot be corrected. Your function needs to calculate exactly two derivatives but you are calculating numerous more (16 total I think). Your equations are wrong for the situation and you need to find the right ones.

I get the impression you might be working with a state-space system?

[Walter Roberson]

The result of [zeros(2);inv(M)*B] is 4 x 2

Your F is 2 x 2.

The result of * between a 4 x 2 and a 2 x 2 is 4 x 2, so [zeros(2);inv(M)*B]*F is 4 x 2.

Your A is 4 x 4. Your y is 4 x 1.

The result of * between a 4 x 4 and a 4 x 1 is 4 x 1, so A*y is 4 x 1.

You are adding a 4 x 1 + 4 x 2. In MATLAB before R2016b that would be an error. In R2016b and later, it is treated as if you had bsxfun, like

dydt = bsxfun(@plus, A*y, [zeros(2);inv(M)*B]*F);

so the x 1 of the A*y is replicated to 4 x 2, giving you a 4 x 2 + 4 x 2, which has a 4 x 2 result.

Therefore your dydt is a 4 x 2 matrix.

Now, you could get around your immediate error message by putting

dydt = dydt(:);

at the end of your function, so that the function returned 8 x 1: you would no longer get the error about "must return a column vector". Instead you would get an error:

@(T,Y)MSSYS(T,Y,M,K,C,B) returns a vector of length 8, but the length of initial conditions vector is 4. The vector returned by @(T,Y)MSSYS(T,Y,M,K,C,B) and the initial conditions vector must

have the same number of elements.

You need to rework the equations, or at the very least you need to rework the logic of

[zeros(2);inv(M)*B]*F

so that you end up with 4 x 1 as the result instead of 4 x 2.