Page 61 - Servo Motors and Industrial Control Theory -

3.3 Eigenvalues, Eigenvectors, and Characteristic Equation 53

3.3 Eigenvalues, Eigenvectors, and Characteristic
Equation

The state variable equations, which are in the form of n first order differential equa-
tion, can also be solved like ordinary differential equations. Because state equations
are written in matrix form some extra terms must be taken into account. Like for
ordinary differential equation two solutions of transient and steady state can be
assumed. To find the transient solution, the input vector are set to zero and state
equations become
d X := AX (3.11)
dt
Without loss of generality, a system with three state variables is assumed. A solution
in the following form is assumed:

x 1
X := x ⋅ e st (3.12)
2
x 3
In Eq. (3.12), it is clear that the values of x , x , x give the amplitude of each state
2
3
1
variable and should be distinguished from the state variables. The second term in
Eq. (3.12) determines weather the system is unstable, stable, overdamped, or un-
derdamped. Substituting from Eq. (3.12) in Eq. (3.11) and after some algebraic
manipulation gives
x x
1 1
s x e⋅ ⋅ st = : A x e⋅ st (3.13)
2 2
x x
3 3
In Eq. (3.13), A is a matrix of n × n, in this case 3 by 3. Simplifying the above equa-
tion gives
x 1
(A sI x− ) ⋅ 2 = : 0 (3.14)
x 3
I is unit matrix with diagonal terms 1 and off diagonal terms zero. It is clear that
Eq. (3.14) is an eigenvalue problem. The eigenvalues are the roots of characteristic
equation. This will be shown for the above example. The matrix ( A − sI) is known as
dynamic matrix. The eigenvector associated with each eigenvalue has some mean-
ing in vibrations as it gives the mode shape but in control has only meaning when
the output has to be calculated. With numerical integration method there is no need
to go into details of other methods.
Equation (3.14) has a nontrivial solution if the determinant of the dynamic ma-
trix is zero. The first example in the above sections will be considered, this gives