Section 9-

9-1

9-7 Design with Lead-Lag Controller• Transfer function of a simple lead-lag (or lag-lead)

controller:

• The phase-lead portion is used mainly to achieve a shorter rise time and higher bandwidth, and the phase-lag portion is brought in to provide major damping of the system.

• Either phase-lead or phase-lag control can be designed first.

7, p. 574

)1,1(1

1

1

1)( 21

2

22

1

11

aasT

sTa

sT

sTasGC

lead lag

Section 9-

9-2

Example 9-7-1: Sun-Seeker SystemExample 9-5-3: two-stage phase-lead controller design

Example 9-6-1: two-stage phase-lag controller design

• Phase-lead control:From Example 9-5-3 a1 = 70 and T1 = 0.00004

• Phase-lag control:

7, p. 575

Section 9-

9-3

Example 9-7-1 (cont.)

7, p. 575

Section 9-

9-4

9-8 Pole-Zero-Cancellation Design:Notch Filter

• The complex-conjugate poles, that are very close to the imaginary axis of the s-plane, usually cause the closed-loop system to be slightly damped or unstable.

Use a controller to cancel the undesired poles

• Inexact cancellation:

8, p. 576

Section 9-

9-5

Inexact Pole-Zero Cancellation

• K1 is proportional to 11, which is a very smaller number. Similarly, K2 is also very small.

• Although the poles cannot be canceled precisely, the resulting transient-response terms will have insignificant amplitude, so unless the controller earmarked for cancellation are too far off target, the effect can be neglected for all practical purpose.

8, p. 577

Section 9-

9-6

8, p. 577

Section 9-

9-7

8, p. 578

Section 9-

9-8

Second-Order Active Filter

8, p. 579

Section 9-

9-9

Frequency-Domain Interpretation

“notch” at the resonant frequency n.

• Notch controller do not affect the high- and low-frequency properties of the system

•

8, p. 580

n

Section 9-

9-10

Example 9-8-1

8, p. 581

Section 9-

9-11

Example 9-8-1 (cont.)• Loop transfer function:

• Resonant frequency 1095 rad/sec

• The closed-loop system is unstable.

8, p. 582

Section 9-

9-12

Pole-Zero-Cancellation Design with Notch Controller• Performance specifications:

– The steady-state speed of the load due to a unit-step input should have an error of not more than 1%

– Maximum overshoot of output speed 5%

– Rise time 0.5 sec

– Settling time 0.5 sec

• Notch controller:

to cancel the undesired poles 47.66 j1094

• The compensated system:

Example 9-8-1: Pole-Zero Cancellation

8, p. 582

Section 9-

9-13

Example 9-8-1 (cont.)G(s): type-0 system:

• Step-error constant:

• Steady-state error:

• ess 1% KP 99

• Let n = 1200 rad/sec and p = 15,000– Maximum overshoot = 3.7%

– Rise time tr = 0.1879 sec

– Settling time ts = 0.256 sec

9910198.1

2

8

n

8, p. 583

Section 9-

9-14

Example 9-8-1: Two Stage Design• Choose n = 1000 rad/sec and p = 10

the forward-path transfer function of the system with the notch controller:

maximum overshoot = 71.6%

• Introduce a phase-lag controller or a PI controller toEq. (9-167) to meet the design specification given.

8, p. 583

Section 9-

9-15

Example 9-8-1: Phase-Lag Controller

Second-Stage Phase-Lag Controller Design• Phase-lag controller:

8, p. 584

Section 9-

9-16

Example 9-8-1: PI Controller

Second-Stage PI Controller Design

• PI controller:

• Phase-lag controller (9-169)

• KP = 0.005 and KI/KP = 20 KI = 0.1maximum overshoot = 1%rise time tr = 0.1380 secsettling time ts = 0.1818 sec

8, p. 584

Section 9-

9-17

Example 9-8-1: Pole-Zero Cancellation

Sensitivity due to Imperfect Pole-Zero Cancellation• Transfer function:

maximum overshoot = 0.4%rise time tr = 0.17 secsettling time ts = 0.2323 sec

8, p. 585

Notch Controller

Section 9-

9-18

Example 9-8-1: Unit-Step Responses

8, p. 585

Section 9-

9-19

Example 9-8-1: Freq.-Domain Design

8, p. 586

attenuation = 44.86 dB

g

Section 9-

9-20

Example 9-8-1: Notch Controller

• PM = 13.7°

• Mr = 3.92

0435.0z

612.7p

8, p. 586

Section 9-

9-21

Example 9-8-1 (cont.)

8, p. 587

Section 9-

9-22

Example 9-8-1: Notch-PI ControllerDesired PM = 80°• New gain-crossover frequency:

(9.32)

(9.25)

8, p. 586

sec43 radg

Section 9-

9-23

Example 9-8-1 (cont.)

8, p. 588

Section 9-

9-24

9-9 Forward andFeedforward Controllers

9, p. 588

Forward compensation:

Section 9-

9-25

Example 9-9-1Second-order sun-seeker with phase-lag control (Ex. 9-6-1):

Time-response attributes:

maximum overshoot = 2.5%, tr = 0.1637 sec, ts = 0.2020 sec

• Improve the rise time and the settling timewhile not appreciably increasing the overshoot add a PD controller Gcf(s) to the system (forward)

add a zero to the closed-loop transfer functionwhile not affecting the characteristic equation

maximum overshoot = 4.3%, tr = 0.1069, ts = 0.1313

9, p. 589

Section 9-

9-26

Example 9-9-1 (cont.)

9, p. 590

Forward controller

Feedforward controller

Section 9-

9-27

9-10 Design of Robust Control Systems• Control-system application:

1. the system must satisfy the damping and accuracy specifications.2. the control must yield performance that is robust (insensitive) to external disturbance and parameter variations

• d(t) = 0

• r(t) = 0

10, p. 590

Section 9-

9-28

Sensitivity

• Disturbance suppression and robustness with respect to variations of K can be designed with the same control scheme.

10, p. 591

Section 9-

9-29

Example 9-10-1Second-order sun-seeker with phase-lag control (Ex. 9-6-1)

• Phase-lag controller low-pass filterthe sensitivity of the closed-loop transfer function M(s) with respect to K is poor

10, p. 591

a = 0.1T = 100

Section 9-

9-30

10, p. 592

Section 9-

9-31

10, p. 593

Section 9-

9-32

Example 9-10-1 (cont.)• Design strategy: place two zeros of the robust controller

near the desired close-loop poles

• According to the phase-lag-compensated system,s = 12.455 j9.624

• Transfer function of the controller:

• Transfer function of the system with the robust controller:

10, p. 594

Section 9-

9-33

10, p. 595

Section 9-

9-34

Example 9-10-1 (cont.)

10, p. 596

Section 9-

9-35

10, p. 596

Section 9-

9-36

Example 9-10-1 (cont.)

10, p. 597

Section 9-

9-37

10, p. 597

Section 9-

9-38

Example 9-10-2Third-order sun-seeker with phase-lag control (Ex. 9-6-2)

Phase-lag controller: a = 0.1 and T = 20 (Table 9-19) roots of characteristic equation: s = 187.73 j164.93

• Place the two zeros of the robust controller at

180 j166.13

• Forward controller:

10, p. 597

Section 9-

9-39

Example 9-10-2 (cont.)

10, p. 598

Section 9-

9-40

Example 9-10-2 (cont.)

10, p. 599

Section 9-

9-41

Example 9-10-3Design a robust system that is insensitive to the variation of

the load inertia.

Performance specifications: 0.01 J 0.02

Ramp error constant Kv 200

Maximum overshoot 5%

Rise time tr 0.05 sec

Settling time ts 0.05 sec

10, p. 599

s = 50 j86.6s = 50 j50

Section 9-

9-42

Example 9-10-3 (cont.)• Place the two zeros of the robust controller at

55 j45 • K = 1000 and J = 0.01:

K = 1000 and J = 0.02:

• Forward controller:

10, p. 600

Section 9-

9-43

Example 9-10-3 (cont.)

10, p. 600

Section 9-

9-44

9-11 Minor-Loop Feedback Control

Rate-Feedback or Tachometer-Feedback Control

• Transfer function:

• Characteristic equation:

• The effect of the tachometer feedback is the increasing of the damping of the system.

11, p. 601

Section 9-

9-45

Steady-State Analysis• Forward-path transfer function:

type 1 system

• For a unit-ramp function input:tachometer feedback ess = (2+Ktn)/n

PD control ess = 2/n

• For a type 1 system, tachometer feedback decrease the ramp-error constant Kv but does not affect the step-error constant KP.

11, p. 602

Section 9-

9-46

Example 9-11-1Second-order sun-seeker system:

11, p. 603

Section 9-

9-47

Example 9-11-1• Characteristic equation:

• Kt = 0.02:maximum overshoot = 0tr = 0.04485 sects = 0.06061 sectmax = 0.4 sec

11, p. 604

Section 9-

9-48

9-12 A Hydraulic Control System

12, p. 605

two-stage electro-hydraulic valve

double-acting single rod linear actuator

Section 9-

9-49

Modeling Linear Actuator

• The applied force:f: the force efficiency of the

actuator

• Volumetric efficiency:

for an ideal case

12, p. 606

Section 9-

9-50

Four-Way Electro-Hydraulic ValveTwo-stage control valve:

• The first stage is an electrically actuated hydraulic valve, which controls the displacement of the spool of the second stage of the valve.

• The second stage is a four-way spool valve, which controls the fluid flow and pressure into and out of ports A and B of the actuators.

At the nominal operating conditions:

12, p. 606

Section 9-

9-51

Orifice Equation• The classic orifice equation for the fluid flow:

– The discharge coefficient:

12, p. 607

Section 9-

9-52

Liberalized Flow Equations forFour-Way Valve

• Orifice equation:

• Taylor series:

Take x0 = 0 and

Kq: flow gain

Kc: pressure-flow coefficient

12, p. 608

Section 9-

9-53

Rectangular valve-port geometry

For the rectangular geometry:For the open-center valve:

12, p. 609

Section 9-

9-54

Equations for Four-Way Valve

• Volumetric flow rates into and out of the actuator:

for critical centered valves

mxx

12, p. 610

Section 9-

9-55

Input Voltage & Main Spool Displacement

• Flow equation of pilot spool:(the control valve is critically centered)

• Assuming an incompressible fluid:

• Displacement of pilot spool:

12, p. 611

Section 9-

9-56

Transfer Function of Two-Stage Valve

12, p. 611

Section 9-

9-57

Modeling the Hydraulic System

12, p. 612

Section 9-

9-58

Mathematical Equations• Force balance equation for an ideal linear actuator:

• Expressing pressure level PA and pressure level PB:

• Expressing the pressure difference two sides of linear

actuator:

• General equation:

12, p. 612

Section 9-

9-59

Applications: Translational Motion• The voltage fed back from the actuator displacement z:

• The input voltage to the two-stage valve Verror:

• Desired input zdesired and desired voltage Vdesired:

General equation:

Transfer function of two-stage valve:

12, p. 614

Section 9-

9-60

Transfer Func. of Translational System

12, p. 615

rfaq

mc

cf KKKK

AT

sTK

,

)1(

1

Section 9-

9-61

Applications: Rotational System

12, p. 616

Section 9-

9-62

Transfer Function of Rotational System• The translational displacement of the rod in terms of the

angular displacement:

• Main valve displacement:

• System transfer function:

12, p. 615

Section 9-

9-63

Applications: Variable Load

12, p. 617

Section 9-

9-64

9-13 Control DesignP Control:

13, p. 617

Section 9-

9-65

P Control: transfer function• Simplified hydraulic system transfer function:

(neglecting the pole at 142350)

• Apply a P controller:

closed-loop transfer function:

13, p. 618

Section 9-

9-66

P Control: root loci & responses

13, p. 619

Section 9-

9-67

PD Control

• Closed transfer function:

PD control

13, p. 621

Section 9-

9-68

PD Control: design• Steady-state error for a unit-ramp input:

ess 0.00061 KP 5

• Damping ratio for KP = 5:

= 1 KD = 0.0066

• Setting time:

ts 0.005 KD 0.0044

• Stability requirement: KP 0 and KD 0.00307

13, p. 622

Section 9-

9-69

PD Control: root loci & root contour• Characteristic equation (KD = 0):

13, p. 622

Section 9-

9-70

PD Control: responses

13, p. 623

Section 9-

9-71

PI Control

13, p. 626

PI control

Section 9-

9-72

PI Control: design• Steady-state error for a ramp input:

ess 0.2 KI 0.015

• Characteristic equation:

stable

• KI/KP = 5

13, p. 626

Section 9-

9-73

PI Control: root loci & responses

13, p. 627

Section 9-

9-74

PI Control: attributes

PI

13, p. 627

Section 9-

9-75

PID Control

• Transfer function:

13, p. 628

Section 9-

9-76

PID Control: design

• PD controller:

Table 9-28 KD1 = 0.0066, KP1 = 5

• PI controller:

KI2/KP2 = 5

13, p. 629

Section 9-

9-77

PID Control: attributes

13, p. 629

Section 9-

9-78

PID Control: responses

13, p. 630