Example 46.2
Let’s consider a 3rd order control system which is described by the following differential equation
We may choose the following state variables
Then we may describe this system in matrix form
and
If the state variable feedback matrix is
and
then the closed-loop system is
The state feedback matrix is
and the characteristic equation is
(46.1)
If we seek a rapid response with a low overshoot, we choose a desired characteristic equation such as (see Equation 5.18 and Table 5.2)
We choose
= 0.8 for minimal overshoot and
to meet the settling time requirement.
If we want a settling time (with a 2% criterion) equal to 1 second, then
If we choose = 6, the desired characteristic equation is
(46.2)
Comparing Equations (46.1) and (46.2) yields the three equations
Therefore, we require that
= 9.4,
= 79.1, and
=
170.8. The step response has no
overshoot and a settling time of 1 second, as desired.
Example 46.3
Let’s consider a control system with full-state
feedback control (output variable
).
This system is using a DC motor which is controlled by an exciting circuit.
The transfer function of this DC motor is as follows
where
=
the complex transfer coefficient of the DC motor.
We suppose that
=1
and
=5.
As you may see from Fig.46.2 the control system has 3 feedback loops: on position, motor speed, and excitation current.
Suppose, that the coefficient of feedback loop on position is equal to -1 (look at Fig.46.3).
Fig.46.2 A control system with full-state feedback control (output variable ).
Fig.46.3 A simulation circuit for Example 46.3
If , then
(46.3).
At
the
system will be unstable.
But if we use all three feedback loops, then it’s possible to make this system stable and provide for desirable step response characteristic.
Fig.46.4 The block diagram of the control system with full state feedback for Example 46.3
(46.4)
and
(46.5)
where
and
.
Since a designer may define the values of
and
independently
of one another, then it’s possible to choose the desirable zeroes
of function
.
We may to choose the
zeroes of function
in
order to reduce them with real poles of
.
(46.6)
To do so it’s necessary to have
=
1/5,
=
6, and
= 1.
In this case we have
(46.7)
where .
In this case the closed-loop transfer function will be as follows
.
(46.7)
In this case we may choose
to
guarantee the system’s stability, but step response characteristic
of the closed-loop system will be defined by the poles s = -1 and s =
-5.
But we note that the zeroes of it’s necessary to choose in order to provide for the desirable location of the roots of the closed-loop system characteristic equation in the left half-plane of s-plane.
2 Ackermann’s formula
Fig.46.5 Compensator design method using the pole placement method
Classical compensator design methods are using a control system transfer function.
But there are special methods connected with pole placement using state-space models of control systems.
For any LTI control system we may write the following equation
(46.8)
In the general case the input signal
is
the function of state variables:
(46.9)
This equation is called a control law.
The use of the pole placement methods will result in the following control law:
(46.10)
where K –
is the vector of constant coefficients with
dimension
at any given points of closed-loop system (the order of a controlled
object).
We may write the control law in the following form:
(46.11)
So we may see that the control system input signal represents linear combination of all state variables.
Combining (46.10) and (46.8) yields
(46.12)
where
=
the coefficient matrix of the closed-loop loop system.
In this case we may rewrite the characteristic equation the closed-loop loop system
(46.13)
Suppose, that according to design requirements the
roots of the system characteristic equation must have the following
values
.
The we may write the desirable system characteristic equation in the following form
(46.14)
According to the procedure of compensator design it’s necessary to define such a matrix K in order to equal (46.13) and (46.14), that is
(46.15)
This equation contains n
unknown quantities (
).
Equating the coefficients at equal powers of s variable we obtain n equations regarding n unknown quantities. Thus, solving these equations we obtain the elements of feedback matrix K.
Ackermann had proposed the formula to calculate the elements of feedback matrix K according to Eq.( 46.15).
We consider a simulation circuit in the controllability canonical form (Fig.46.6).
Each block of this circuit has the transfer
function
.
Thus, we have the following transfer function
(46.16)
