Analytic and numerical output response example in Matlab

In the following example, we explore the output response derived both analytically and numerically in Matlab.

Example 7.4

Consider a state-space model with the following standard matrices.

A = [...
  -1, 3, 5, 7;...
  0, -2, 0, 6;...
  -2, 1, -3, 0;...
  0, 1, 3, -4;...
];
n = length(A); % order
B = [...
  0; 1; 0; 2; ...
];
C = eye(n);
D = zeros([n,1]);

Solve for the unit step response output y given the following initial condition.

x0 = [2;0;2;0];

Analytic solution

We use the solution of : $$\begin{aligned} \bm{y}(t) &= C \Phi(t) \bm{x}(0) + C \int_0^t \Phi(t-\tau) B \bm{u}(\tau) d\tau + D \bm{u}(t). \end{aligned}$$

First we need Φ(t). The “primed” basis requires the eigendecomposition.

[M,L] = eig(A);

We can find Φ from the primed-basis version Φ′, which is easy to compute.

Phi_p = @(t) diag(diag(exp(L*t)));

Now the basis transformation.

M_inv = M^-1; % compute just once, not on every call
Phi = @(t) M*Phi_p(t)*M_inv;

Declare symbolic variables.

syms T tt

Apply .

y_sym = C*Phi(tt)*x0 + C*int(Phi(tt-T)*B*1,T,0,tt) + D*1;

Convert this to a numerically evaluable function.

y_num = matlabFunction(y_sym);

Plot it; the result is shown in .

figure
t_num = linspace(0,8,200);
plot(t_num,y_num(t_num),'linewidth',1)
xlabel('time (s)')
ylabel('analytic output response')
legend('y_1','y_2','y_3','y_4')

Numerical solution

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

Using `lsim`

First, use lsim to compute the response numerically.

u_s = ones(size(t_num)); % a one for every time
y_lsim = lsim(sys,u_s,t_num,x0); % simulate

Now plot it; the result is shown in .

figure
plot(t_num,y_lsim,'linewidth',1)
xlabel('time (s)')
ylabel('numerical output response')
legend('y_1','y_2','y_3','y_4')
hgsave(h,'figures/temp');

Now take the difference between the two solutions and plot the error. As shows, the differences are minimal.

figure
plot(t_num,y_lsim-y_num(t_num).','linewidth',1)
xlabel('time (s)')
ylabel('error in output response')
legend('y_1','y_2','y_3','y_4')

Using the `step` and `initial` commands with superposition

Just for fun, here’s how we could use step and initial (instead of lsim) with superposition to numerically solve.

y_step = step(sys,t_num); % forced response
y_initial = initial(sys,x0,t_num); % free response
y_total = y_initial + y_step; % (superposition)

Online Resources for Section 7.6

No online resources.