State Space Simulation in Matlab

Since matlab has first-class support for matrices and vectors, it is quite simple to implement a state-space model in Matlab using no support functions whatsoever, e.g.,

% Define the state-space system parameters:
A = [0 1; -1 0]; % State transition matrix
B = [0; 1];  C = [0 1];  D = 0; % Input, output, feed-around

% Set up the input signal, initial conditions, etc.
x0 = [0;0];             % Initial state (N=2)
Ns = 10; 	        % Number of sample times to simulate
u = [1, zeros(1,Ns-1)]; % Input signal (an impulse at time 0)
y = zeros(Ns,1);        % Preallocate output signal for n=0:Ns-1

% Perform the system simulation:
x = x0;                % Set initial state
for n=1:Ns-1           % Iterate through time
  y(n) = C*x + D*u(n); % Output for time n-1
  x = A*x + B*u(n);    % State transitions to time n
end
y' % print the output y (transposed)
% ans =
%  0   1   0  -1   0   1   0  -1   0   0

The restriction to indexes beginning at 1 is unwieldy here, because we want to include time $n = 0$ in the input and output. It can be readily checked that the above examples has the transfer function

$$H(z) = \frac{z^{-1}}{1 + z^{-2}},$$

so that the following matlab checks the above output using the built-in filter function:

NUM = [0 1];
DEN = [1 0 1];
y = filter(NUM,DEN,u)
% y = 
%   0   1   0  -1   0   1   0  -1   0   1

To eliminate the unit-sample delay, i.e., to simulate $H(z)=1/(1+z^{-2})$ in state-space form, it is necessary to use the $D$ (feed-around) coefficient:

[A,B,C,D] = tf2ss([1 0 0], [1 0 1])
% A =
%    0   1
%   -1  -0
% 
% B =
%   0
%   1
% 
% C =
%   -1   0
% 
% D = 1

x = x0;                % Reset to initial state
for n=1:Ns-1
  y(n) = C*x + D*u(n);
  x = A*x + B*u(n);
end
y'
% ans =
%  1   0  -1   0   1   0  -1   0   1   0

Note the use of trailing zeros in the first argument of tf2ss (the transfer-function numerator-polynomial coefficients) to make it the same length as the second argument (denominator coefficients). This is necessary in tf2ss because the same function is used for both the continous- and discrete-time cases. Without the trailing zeros, the numerator will be extended by zeros on the left, i.e., ``right-justified'' relative to the denominator.