Page 62 - Servo Motors and Industrial Control Theory -

54 3 State Variable Feedback Control Theory

0 1 0 s0 0 −s 1 0
0 0 1 − 0s 0 : = 0 −s 1 (3.15)
− 2K − 2 − 30 0 s − 2K − 2 ( −−
3 s)
The determinant of the matrix becomes
−s(( − −−s( 3 s) − ))2 − K :2 = 0

Expanding gives,
s + s 3 + s 2 + 2 K =: 0 (3.16)
3
2
It can be seen that the determinant of the dynamic matrix is the same as that of char-
acteristic equation. There are computer programs such as MathCAD which gives
the eigenvalues of the dynamic matrix and there is no need to calculate the charac-
teristic equation. The MathCAD may be used to obtain the characteristic equation
in symbolic form and polyroots (v) statement may be used to calculate the roots for
various parameters.
When the stability is assured, the steady state value of the output variables can be
obtained by setting all derivatives to zero and solving n linear equation as follows:
+
0 := AX BU
−
1
X :=−A BU (3.17)
y := CX
An example will clarify the above equations. The first example in previous section
will be considered. The matrix equation of (3.17) becomes

=
K: 2
=
u: 1
−
x 1 0 1 0 ( 1) 0
x :=− 0 0 1 · 0 ⋅ u
2
x 3 − 2K − 2 − 3 2K

x 1
⋅
=
y : (1 0 0) x 2
x 3
y = (1)

It can be seen that for k = 2 and input u = 1 the output is also 1, which means that
the steady state error is zero. For input other than step, it is better to solve the dif-
ferential equation. One method is solving the differential equation numerically. In
this case by definition,