Scene 1

Scene 2

Figure 9.7 Ideal integral compensated  system response and the uncompensated system response of Example 9.1

Note the followings : C ss  (t)  of  the Comp Sys = 1   C ss  (t)  of  the Uncomp Sys = 0.892 After 3 sec. integrator in the comp slowly compensates for the  e(t)  until zero error is finally reached. It takes 18 s for the compensated system to reach to within ±2% of  C(∞)   of 1, while the uncomp sys takes about 6 s to settle to within ±2% of its  C(∞)  of 0.892. The compensated system reaches the uncompensated system's final value in about the same time. The remaining time is used to improve the steady-state error over that of the uncompensated system.

Scene 3

the objective is to design a response that has a desirable percent overshoot and a shorter settling time than the uncompensated system.

Improving Transient Response via Cascade Compensation

It avoids pure differentiation, passive network

The transient response of a system can be selected by choosing an appropriate closed-loop pole location on the s-plane.

If this point is on the root locus, then a simple gain adjustment is all that is required in order to meet the transient response specification.

If the closed-loop pole location is not on the root locus, then the root locus must be reshaped so that the compensated (new) root locus goes through the selected closed-loop pole location.

Scene 4

-2.437 +j5.583
K 35.34
Third
pole
—6
—4
-3
-5
X = Closed-loop pole
x
Open-loop pole
-1.869 +j4.282
K 20.76
Third
pole
—6
—4
-3
-5
X = Closed-loop pole
x
Open-loop pole
— 0.4
-1
04
-2
-1
-2
(c)
(d)
jo
j6
j 4 s-plane
113.60
0
jo
j6
j 4 s-plane
113.60
0

One way to speed up the original system that generally works is to add a single zero to the forward path.

04
-0.939 +j2.151
K 23.72
Third
pole
—6
—4
-1
-5
-3
-1
-3
jo
j6
0
jo
j7
j6
X = Closed-loop pole
x
Open-loop pole
= 0.4
-3 +j6.874
K 51.25
—6
—4
-5
X = Closed-loop pole
x
— Open-loop pole
-2
(a)
-2
(b)
s-plane
113.60
s-plane
113.60
0

Scene 5

In summary, although compensation methods  c  and  d  yield slower responses than method  b,  the addition of ideal derivative compensation shortened the response time in each case while keeping the percent overshoot the same.

Scene 6

Figure 9.16   Uncompensated system and ideal derivative compensation solutions from Table 9.2

An added benefit is the improvement in the steady-state error, even though lag compensation was not used.

the compensated responses are faster and exhibit less error than the uncompensated response.

Scene 7

Figure 9.17 Feedback control system for Example 9.3

Example: design an ideal derivative compensator to yield a 16% overshoot, with a threefold reduction in settling time.

Scene 8

Figure 9.18  Root locus for uncompensated system shown in Figure 9.17

At  K  = 43.35 third pole at —7.59, which is over six times as far from the  Jw -axis

Since 16% overshoot is equivalent to  ζ  = 0.504

search along that damping ratio line for an odd multiple of 180

Scene 9

Scene 10

Figure 9.19   Compensated dominant pole superimposed over the uncompensated root locus for Example 9.3

find the location of the compensated system's dominant poles.

the compensated system's settling time will be 1.107

Next, design the location of the compensator zero.

Scene 11

Figure 9.20   Evaluating the location of the compensating zero for Example 9.3

Scene 12

Figure 9.21   Root locus for the compensated system of Example 9.3

Scene 13

Figure 9.22 Uncompensated and compensated system step responses of Example 9.3

The second-order approximation for the compensated system, however, may be invalid. The percent overshoot differs by 3% between the uncompensated and compensated systems. there is approximately a threefold improvement in speed as evaluated from the settling time.

Scene 14

Scene 15

Implementation of the ideal derivative, or PD controller

Scene 16

Figure 9.24   Geometry of lead compensation

The advantages  of a passive lead network over an active PD controller are that no additional power supplies are required  noise due to differentiation is reduced.

When passive networks are used, a single zero cannot be produced ;  rather, a compensator zero and a pole result.

if the pole is farther from the imaginary axis than the zero, the angular contribution of the compensator is still positive and thus approximates an equivalent single zero

Scene 17

In other words, the angular contribution of the compensator pole subtracts from the angular contribution of the zero but does not preclude the use of the compensator to improve transient response, since the net angular contribution is positive, just as for a single PD controller zero.

Scene 18

Figure 9.26 Lead compensator design, showing evaluation of uncompensated and compensated dominant poles for Example 9.4

The real part of the desired pole location is

Arbitrarily assume a compensator zero at -5

Design lead compensators for the system of Figure 9.17  that will reduce the settling time by a factor of 2 while maintaining  30% overshoot.

Scene 19

Figure 9.27 s  -plane picture used to calculate the location of the compensator pole for Example 9.4

The angular contribution required from the compensator pole

Scene 20

Figure 9.28 Compensated system root locus

search for the third and fourth closed-loop poles

for the gain equal to that of the compensated dominant pole, 1423

20 times the real part of the dominant pole

Scene 21

Figure 9.29 Uncompensated system and lead compensation responses for Example 9.4

How do the possible lead compensators differ? The differences are in the values of static error constants, The gain required to reach the design point on the compensated root locus,  The difficulty in justifying a second-order approximation when the design is complete, and the ensuing transient response.

Scene 22

TABLE 9.4
Plant and
Comparison of lead compensation designs for Example 9.4
compensator
Dominant
e(CX))
Other poles
Zero
Comments
-1.007 ±J2.627
63.21
0.358
2.813
30 (28)
3.972 (4)
1.196 (1.3)
2.634
0.380
-7.986
None
Second-order
approx. 0K
Compensation a
-l- 4)(s + 6)(s + 42.96)
-2.014 ± j5.252
1423
0.358
5.625
30 (30.7)
1.986 (2)
0.598 (0.6)
6.9
0.145
-43.8.-5.134
—5
Second-order
approx. 0K
Compensation b
-2.014±j5.252
698.1
0.358
5.625
30 (28.2)
1.986 (2)
0.598 (0.6)
5.791
0.173
-22.06
None
Second-order
approx. 0K
Compensation c
-2.014 ± j5.252
345.6
0.358
5.625
30 (14.5)
1.986 (1.7)
0.598 (0.7)
3.21
0.312
- 13.3, —1642
-2
No pole-zero
cancellation
Simulation results are shown in parentheses.

Chapter 9