Файл: Microcontroller based applied digital control (D. Ibrahim, 2006).pdf

220 DISCRETE CONTROLLER DESIGN

Figure 9.9 Controller response

Given the pole positions of a system, (9.3) gives the required transfer function of the controller as

T (z) is the required transfer function, which is normally in the form of a polynomial. The denominator of T (z) is constructed from the positions of the required roots. The numerator polynomial can then be selected to satisfy certain criteria in the system. An example is given below.

Example 9.3

The open-loop transfer function of a system together with a zero-order hold is given by

HG(z) = 0.03(z + 0.75) . z2 − 1.5z + 0.5

Design a digital controller so that the closed-loop system will have ζ = 0.6 and wd = 3 rad/s. The steady-state error to a step input should be zero. Also, the steady-state error to a ramp input should be 0.2. Assume that T = 0.2 s.

Solution

The roots of a second-order system are given by

√

z1,2 = e−ζ ωn T ± j ωn T 1−ζ 2 = e−ζ wn T (cos ωn T 1 − ζ 2 ± j sin ωn T 1 − ζ 2).

We now have to determine the parameters of the numerator polynomial. To ensure realizability, b0 = 0 and the numerator must only have the b1 and b2terms. Equation (9.6) then becomes

The other parameters can be determined from the steady-state requirements. The steady-state error is given by

E (z) = R(z)[1 − T (z)].

For a unit step input, the steady-state error can be determined from the final value theorem, i.e.

If Kv is the system velocity constant, for a steady-state error to a ramp input we can write

222 DISCRETE CONTROLLER DESIGN

From (9.9) and (9.11) we obtain,

b1 = 0.596 and b2 = −0.243.

Equation (9.10) then becomes

Equation (9.12) is the required transfer function. We can substitute in Equation (9.3) to find the transfer function of the controller:

The step response of the system with the controller is shown in Figure 9.10.

9.1.4 Pole-Placement Control – Graphical

In the previous subsection we saw how the response of a closed-loop system can be shaped by placing its poles at required points in the z-plane. In this subsection we will be looking at some examples of pole placement using the root-locus graphical approach.

When it is required to place the poles of a system at required points in the z-plane we can either modify the gain of the system or use a dynamic compensator (such as a phase lead or a phase lag). Given a first-order system, we can modify only the d.c. gain to achieve the required time constant. For a second-order system we can generally modify the d.c. gain to achieve a constant damping ratio greater than or less than a required value, and, depending on the system, we may also be able to design for a required natural frequency by simply varying the d.c. gain. For more complex requirements, such as placing the system poles at specific points in the z-plane, we will need to use dynamic compensators, and a simple gain adjustment alone

Figure 9.14 Block diagram for Example 9.5

drawn from the point where the damping factor is 0.7. At the required point the roots are z1,2 = 0.7191 ± j 0.2114. The value of K at this point is calculated to be K = 0.0618.

In this example, the required specification was obtained by simply modifying the d.c. gain of the system. A more complex example is given below where it is required to place the poles at specific points in the z-plane.

Example 9.5

The block diagram of a digital control system is given in Figure 9.14. It is required to design a controller for this system such that the system poles are at the points z1,2 = 0.3 ± j 0.3.

Solution

In this example, we will draw the root locus of the system and then use a dynamic compensator to modify the shape of the locus so that it passes through the required points in the z-plane.

The root locus of the system without the compensator is shown in Figure 9.15. The point where we want the roots to be is marked with a × and clearly the locus will not pass through this point by simply modifying the d.c. gain K of the system.

The angle of G(z) at the required point is

G(z) = − (0.3 + j 0.3 − 0.2) − (0.3 + j 0.3 − 0.9)

Since the sum of the angles at a point in the root locus must be a multiple of −180◦, the compensator must introduce an angle of −180 − (−45) = −135◦. The required angle can be obtained using a compensator with a transfer function

There are many combinations of p and n which will give the required angle. For example, if we choose n = 0.5, then,