3MATLAB Examples
The following examples shows the essential commands needed in this project and some of the simulation tools. Additional
help can be found using the MATLAB help tool.
Example 1: a=[0 1;2 1];
b=[0;1];
c=[1 0];
d=[0];
k = -place(a,b,[-1 -2]) (solves the eigenvalues assignment problem)
ak=(a+b*k);
t=0:0.01:6;
(defines a vector t such that 0 ≤ t ≤ 6 in 0.01 sec intervals.
sys=ss(ak,[],eye(2),[]);
(builds the state space realization with A = ak, B = 0, C = I, and D = 0. the matrix C = I so that the output is equal to
each of the states. This is for the purpose of simulation.)
z=initial(sys,[2;-4],t);
(Simulates x˙ = ak x(0) with x(0) = [2;−4]). The result is the vector z that has 2 columns and as many rows as the vector t
x1=[1 0]*z’;
x2=[0 1]*z’;
(Extracts x1 and x2 as the first and second column of z, respectively.
subplot(2,1,1);plot(t,x1),grid
title(’Response to Initial Condition’)
ylabel(’State x1’)
subplot(2,1,2);plot(t,x2),grid
title(’Response to Initial Condition’)
ylabel(’State x2’)
(The function subplot(x,y,z) splits the plot into x rows and y columns. z indicated the number of each plot. Thus, for example,
subplot(3,2,z) subdivides the page into 3 rows and 2 columns. The last number will index each plot from 1 tp 6.
Example 2: a=[0 1;2 1];
b=[0;1];
c=[1 0];
d=[0];
k = -place(a,b,[-1 -2]) (solves the eigenvalues assignment problem)
ak=(a+b*k);
t=0:0.01:6;
[x, y, t]=step (ak, b, eye(2), d, 1, t)
(Computes the closed loop step response. x is the state (two columns in this example), y is the output (single column in this
example)).
x1 = [1 0] ∗ x′;
x2 = [0 1] ∗ x′;

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