Page 63 - Servo Motors and Industrial Control Theory -

3.3 Eigenvalues, Eigenvectors, and Characteristic Equation 55

d X:= (X tT+ − X )
t
dt T
d
X :T= ⋅ X X+
tT+
dt t
In the above equation, T is the time interval and t is the time. The process can be
started from t = 0 and the calculation can be repeated to obtain the solution over
the required time interval. In the above equation, X can be considered as the state
variable vector. Substituting for dt d X from Eq. (3.6) into above equation giving the
solution for state variables,

X tT+ = : T AX +( − Bu +) X t (3.18)

In Eq. (3.18), A is the system matrix and B is the input matrix. The time interval
T must be sufficiently small to avoid numerical instability. Then, sufficiently large
time span must be selected to give the response. The time interval and the time span
can be selected with respect to eigenvalues. The input variable may be a single vari-
able or could be a vector U if more than one input variable exist.
The eigenvalues of matrix A can be calculated making use of various software
such as MathCad. In this method, the eigenvalues can be calculated as
In the first example for K = 2,

0 1 0
A := 0 0 1
−4 −2 −3
w := eigenvals A( )
− . 0 102 +1 .192 i
w =−0 .102 +1 .192 i
− − 2 796.

w is the eigenvalues of A for k = 2. It can be seen that the eigenvalues are the same as
roots of characteristic equation. The eigenvectors corresponding to each eigenvalue
can obtained as follows:
v : eigenvec(A, 0.102 1.192i)= − +
− 0.228 0.414i−
v = 0.516 0.23i−
0.221 0.639i+
As it was discussed earlier, for each eigenvalue there is a corresponding eigenvec-
tor. This process can be repeated for all eigenvalues as follows: