ODE response vs CCF response of mass-damper-spring system

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J AI
J AI 2023-2-1
评论： Sam Chak 2023-2-2
Ran in:
Could someone please explain why is that the (zero-input) response obtained using ode45 solver is different than the CCF (controllable canonical form) response as shown below? However, if the C matrix is changed to [1 0], the responses match.
clear all
syms x(t)
x0 = [0;2];
m = 3;
b = 0.6;
k = 1.5;
Fa = 0;
eqn = m*diff(x,2) + b*diff(x,1) + k*x == Fa;
V = odeToVectorField(eqn);
M = matlabFunction(V,'vars',{'t','Y'});
ySol = ode45(M,[0 50],x0);
time = 0:0.1:50;
yValues_pos = deval(ySol,time,1);
s = tf('s');
G = 1/(m*s^2+b*s+k);
% CCF
Accf = [0, 1;
-k/m, -b/m];
Bccf = [0;
1];
Cccf = [1/m, 0];
Dccf = 0;
for i = 1:length(time)
yccf(i) = Cccf*expm(Accf*time(i))*x0;
end
figure()
hold on
plot(time,yccf,'b')
plot(time,yValues_pos,'r--')
legend('ccf','ODE45')
clear all
syms x(t)
x0 = [0;2];
m = 3;
b = 0.6;
k = 1.5;
Fa = 0;
eqn = m*diff(x,2) + b*diff(x,1) + k*x == Fa;
V = odeToVectorField(eqn);
M = matlabFunction(V,'vars',{'t','Y'});
ySol = ode45(M,[0 50],x0);
time = 0:0.1:50;
yValues_pos = deval(ySol,time,1);
s = tf('s');
G = 1/(m*s^2+b*s+k);
% CCF
Accf = [0, 1;
-k/m, -b/m];
Bccf = [0;
1];
Cccf = [1, 0];
Dccf = 0;
for i = 1:length(time)
yccf(i) = Cccf*expm(Accf*time(i))*x0;
end
figure()
hold on
plot(time,yccf,'b')
plot(time,yValues_pos,'r--')
legend('ccf','ODE45')

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回答（1 个）

Sam Chak
Sam Chak 2023-2-2
编辑：Sam Chak 2023-2-2
Hi @J AI
Because the output matrix really means for this output equation
where is the output vector and is the state vector (come from your ODE).
Thus, if and , then it implies that the measured output is
.
In fact, the division by m only occurs at the dynamical level.
.
Solving this ODE returns the solution for the and , the states of the system.
So, if you want to compare the ODE result with the result using the state-space model, naturally you want to make for only or for both and .
  2 个评论
J AI
J AI 2023-2-2
My question - to be more precise - is that why the CCF has 1/m instead of 1 in the C matrix? Why does CCF scale the output by 1/m?
Sam Chak
Sam Chak 2023-2-2
Ran in:
As I explained previously, the division by m only occurs at the dynamical level; now as shown symbolically below. It is unnecessary to scale the output by , or .
syms x(t) m b k
x0 = [0;2];
% m = 3;
% b = 0.6;
% k = 1.5;
Fa = 0;
eqn = m*diff(x,2) + b*diff(x,1) + k*x == Fa;
V = odeToVectorField(eqn)
V = 

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