Control Notes 2010

The University of Birmingham Control Engineering

Introduction to the Module of Control Systems

Learning objectives: (a) to understand the teaching arrangement of the module; (b) to understand

clearly what a control system and its related theory are used for; (c) to be able to draw block

diagrams of ordinary systems; (d) to understand the relationship between the main content blocks of

the module.

1.1 Delivery of the module

Aim of Control Engineering: to provide an introduction to the knowledge and skills of engineering

feedback systems.

Requirements to students:

• understand the concept of automatic control correctly;

• be able to model common dynamic systems;

• be able to analyse systems using various techniques;

• be able to design basic control systems.

Reference books:

1) Control Engineering, Second edition, by William Bolton, Prentice Hall, 1998, ISBN 0-582-

32773

2) Control Systems Engineering, Third Edition, by Norman S Nise, John Wiley & Sons, 2000,

ISBN 0-471-36601-3

3) Modern Control Systems, Ninth Edition, by Richard C Dorf and Robert H Bishop, Prentice-

Hall-Wesley, 2001, ISBN: 0-13-031411-0

Delivery: Lectures and tutorials: 18 h; Laboratory: 2 h, see separate lab timetable.

Assessment: 2 hour examination (90%); lab report (10%)

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1.2 Introduction to Control Engineering:

Other terms of control engineering: theory of automatic control, feedback systems, (linear) dynamic

systems.

Typical examples of dynamic systems:

Example 1: Drebble’s incubator, 1624

Working mechanism

Control Systems: A single-output control system consists of subsystems assembled for controlling

the output of the system process by causing a system variable to conform to a desired value.

General form of a control system:

Example 2: Watt’s fly-ball governor

Working mechanism:

Kyle Jiang 2010/20112


Drawing a system block diagram: (1) identify the control parameter; (2) the block diagram starts

from the expected control parameter as input, and ends with the system actual output.

Example 3: Antenna Azimuth Position Control System.

Working mechanism:

1. Set desired ;

Block diagram:

Control systems feature in feedback subsystems. A control system can be an analogue system,

digital system, or a mixed system of the both.

Example 4: Inverted pendulum – to keep the rod upwards by reducing .

How is an inverted pendulum controlled?

Steps to construct an automatic system:

1. Build the mathematical model of the system;

2. Analyse the system to see whether its performance is within

requirements;



3. If not, redesign the system based on the system model and the requirements.

Outline of the module contents:

a) Model physical systems into mathematical models, i.e. equations;

b) Analyze the mathematical models to anticipate the performance of the system. A variety

of techniques are introduced for this purpose.

c) Design a control system using analogue or digital methods.

Use of Matlab in this module: necessary for one

laboratory, but not compulsory in tutorials.

Reading materials: Ref. 1, page 1 – 17.

Exercises: Solve problems 1.1 to 1.3, and read to

understand problems 1.4 and 1.5 and solutions.

Problems:

1.1 In the past, control systems used a human operator as a part of a closed-loop control system. Sketch the block diagram of the valve control system shown in Fig. P1.1

1.2 In a chemical process control system, it is valuable to control the chemical composition of the product. To do so, a measurement of the composition can be obtained by using an infrared stream analyzer, as shown in Fig. P1.2. The Valve on the additive stream can be controlled. Complete the control feedback loop, and sketch a block diagram describing the operation of the control loop.

1.3 A light-tracking control system is shown in Fig. P1.3. The output shaft, driven by the motor through a worm reduction gear, has a bracket attached on which are mounted two photocells. Complete the closed-loop system so that the system follows the light source.

1.4 An automatic turning gear for windmills was invented by Meikle in about 1750. The fantail gear shown in Fig. P1.4 automatically turns the windmill into the wind. The fantail windmill at right angle to the mainsail is used to turn the turret. The gear ratio is of the order of 3000 to 1. Discuss the operation of the windmill, and establish the feedback operation that maintains the main sails into the wind.


Fig. P 1.2 Chemical composition control

Fig. P1.3 A photocell is mounted in each tube. The light reaching each cell is the same in both only when the light source is exactly in the middle as shown.

Fig. P 1.1 Fluid –flow control

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1.5 An aircraft’s attitude varies in roll, pitch and yaw as defined in Fig. 5. Draw a functional block diagram for a closed loop system that stabilizes the roll as follows: the system measures the actual roll angle with a gyro and compares the actual roll angle with the desired roll angle. The ailerons respond to the roll-angle error by undertaking an angular deflection. The aircraft responds to the angular deflection, producing a roll angular rate. Identify the input and output transducers, the controller and the plant. Further more identify the nature of each signal.

2010/2011 Kyle Jiang

Figure P1.4. An automatic turning gear for windmills.

Figure P1.5. Aircraft control

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2. Mathematical Models of Systems

Learning objectives: (a) to be able to model mechanical, electrical, and mixed dynamic systems

into differential equations; (b) to be able to model dynamic systems in transfer functions; (c) to be

able to model dynamic systems in block diagrams and manipulate block diagrams.

2.1 Laplace Transform

Why do we use Laplace transform?

Laplace transform is used to transfer a differential equation into an algebraic equation, so the equation can be solved conveniently.

Example 2.1: Given the following differential equation, solve for y(t) if all initial conditions are zero.

d2 ydt2

+12dydt

+32 y=32 u( t )

Solution:

Definitions: Laplace transform:

F (s )=L {f ( t )}=∫0−

∞f ( t )e−st dt

Inverse Laplace transform:

f ( t )= 12πj∫σ− jω

σ + jωF ( s )e+st ds

Note: the Laplace transform is correct only when all initial conditions are zero.

Properties of Laplace transforms: see Appendix A.

Example 2.2: Find the Laplace transform for the following expressions:

Differentiation:

f ( t )=dm y ( t )

dtm

; L{f(t)}= , where all the initial conditions are 0.

Integration:

f ( t )=∫ y ( t )dt; L{f(t)}=



Impulse function transform: (t)=

{1 when t=00 otherwise

;

L{(t)}=

f (t) =e-at, L{f(t)}=

2.2 Mathematical Models of Dynamic Systems

2.2.1 Mechanical System Transfer Functions

Transfer Function: The transfer function of a linear system is defined as the ratio of the Laplace

transform of the output variable to the Laplace transform of the input variable, with all initial

conditions assumed to be zero.

For example, for an equation in time domain, y(t) = g [ r(t)], its transfer function is

Y (s )R (s )

=outputinput

=G(s ) .

System modelling steps: (a) build up each system differential equation by applying Newton’s law:

ma=∑ F i ; (b) change the differential equation into a transform function; (c) rearrange the

transform equation to get the transfer function..

Example 2.3: Find the transfer function

X ( s )F (s ) for the mass, spring and damper system, Fig. 2.1.

Figure 2.1 Mass, spring, and Damper system, and force analysis diagram

Solution:

Example 2.4: Find the transfer function

Θ(s )M (s ) for the rotational mass, spring and damper system,

Fig. 2.2.

Figure 2.2 A rotational mass spring and damper system.

Solution:



Relationships of Fundamental mechanical dynamic components:

Transfer Functions for Systems with Gears

• A gear system:

• Gear transmission relationships:

x1 = x2; (distance travelled)

v1 = v2; (velocity)

a1 = a2; (acceleration)

r11= r22 (based on distance travelled), or

where Ni is the number of teeth of Gear i, and the diameter D1= mN1

Considering the equal force at the contact point of the gears:

Question: If the output of the system is 1 and the input is T1, find the transfer function Q(s)/M(s) of the shown system.

Working steps:

(a) Analyse the torques on J: T2, Tk, TD.

(b) Build equation.

(c) Introduce T1 and 1.

(d) Take Laplace transform and manipulate.

Solution:


θ2

θ1

=r1

r2

=N1

N2


2.2.2 Electrical System Transfer Functions

Kirchhoff’s laws to be used to build equations:

1) Law 1. The total current flowing towards a junction is equal to the total current flowing from that

junction, i.e. iin = iout

2) Law 2. In a closed circuit or loop, the algebraic sum of the voltages across each part of the circuit

is equal to the applied voltage, i.e. vappl = vcomp

Example 2.5. Demonstration of using Kirchhoff’s laws on the circuit given in Fig. 2.3.

From Law 1, we can get i0 = i1 + i2;

From Law 2, we can get v (t) = vR1 + vL

Relationships of Fundamental electrical dynamic components:

Figure 2.4 RLC network

Example 2.6: Find the transfer function relating the capacitor voltage, Vc(s), to the input voltage

V(s), in Figure 2.4.

Solution:

Example 2.5: Find the transfer function I2(s)/V(s), in a two loop circuit shown in Figure 2.3.


Figure. 2.3 Two loop electric circuit.

Figure. 2.3 Two loop electric circuit.

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Solution:

2.2.3 Transfer function of a DC motor

Figure 2.7 A DC motor (a) wiring diagram, and (b) sketch

Find the transfer functions of a field controlled DC motor and an armature controlled DC motor, sVf(s) and sVa(s).

Solution:

Mechanical function:

tm=J

d2θdt 2

+bdθdt (2.1)

Electromechanical function: Torque tm = K1 ia(t) = K1Kf if (t) ia(t) (2.2)

where the flux = Kf if (t) ,

combining (2.1) and (2.2) yields

Jd2θdt2

+bdθdt = K1Kf if (t) ia(t) (2.3)

In a field controlled motor, the armature current ia(t) is constant, while the field current if (t)

varies for control the speed of the motor. Then eq. (2.3) can be written as

Jd2θdt2

+bdθdt = [K1Kf ia(t)] if (t) = Km if (t) (2.4)

where Km is defined as the motor constant. In reference to Fig. 2.7, we can get the electric equation

v f=if R f +Ldi f

dt (2.5)


if(t) Field

ia

Angle

ia

10


Taking Laplace Transform on (2.4) and (2.5) yields

(Js2 + bs) (s) = Km If (s) (2.6)and Vf(s) = Rf If (s)+Ls If (s) = [Rf + Ls] If (s)

or I f=

V f (s )Rf +Ls (2.7)

Substituting (2.7) to (2.6) and rearranging it, we get

(Js2 + bs) (s) = Km

V f ( s )R f +Ls ,

or

Gf ( s )=Θ (s )V f (s )

=Km

s( Js+b )( Ls+R f )=

Km/bR f

s (τ f s+1 )(τ L s+1 )where field constant f = J/b, and inductor constant L = L/Rf . L>f , and f can be ignored.

In an armature controlled motor, the field current if(t) is constant, while the armature current ia(t) varies for control the speed of the motor. Then eq. (2.3) can be written as

Jd2θdt2

+bdθdt = [K1Kf if(t)] ia (t) = Kn ia (t) (2.8)

where Kn is a constant related to the magnetic material. It is noted in an armature controlled motor that the back electromotive force (b.m.f) vb exists, and its direction is against the supplied voltage,

vb = Kb

dθdt

.

Similar to (2.5), we can get the electric equation

va−vb=ia Ra+Ldia

dt

or va=ia Ra+ L

dia

dt+Kb

dθdt (2.9)

Taking Laplace Transform on (2.8) and (2.9) yields

(Js2 + bs) (s) = Kn Ia (s) (2.10)

and Va(s) = Ra Ia (s)+Ls Ia (s) + Kbs (s)

orI a=

V a( s )−Kb sΘ( s )Ra+La s (2.11)

Substituting (2.11) to (2.10) and rearranging it, we can get

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G( s )=Θ( s )V a( s )

=Km

s [ Ra( Js+b )+Kb K m]=

[ Km / (Ra b+ Kb Km )]s (τ1 s+1)

where the time constant 1= RaJ/(Rab + KbKm).

2.3 Block Diagram Models

Block diagram example: A DC motor block diagram.

Block diagram transformation:

Parallel form



Example 2.7 Reduce the system shown in Figure 2.8 to a single transfer function.

Solution:

Reading materials: Ref. 1, pages 36 – 51, 106-118, 145 – 155; Ref.2. pages 250 – 261.

Problems:

2.1 Derive a transfer function relating the input, force F, with the output, displacement x of the mass, for each of the systems described by Fig. P2.1.

2.2 Derive the transfer functions between the output circuits shown in Fig. P2.2.


Figure 2.8 Block diagram for Example 2.7

(a) (b)Fig. P2.1

(a) (b) (c)Fig. P2.2

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2.3 An electric circuit is shown in Figure P2.3. Obtain a set of simultaneous integrodifferential equations representing the network.

2.4 Consider the mass-spring system depicted in Fig. P2.4. Determine a differential equation to describe the movement of the mass, mm. Obtain the system response to an initial displacement x(0) = 1. Assume motion only in the vertical plane.

2.5 Find the transfer function for Y(s)/R(s) for the idle speed control system for a fuel injected engine as shown in Figure P2.5.

2.6 Find the equivalent transfer function, T(s)=C(s)/R(s), for the system shown in Figure P2.6.


Figure P2.3

Figure P2.5

Figure P2.6

Figure P2.4 Suspended mass-spring system.

r(t)Speed control

y(t)engine speed

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3. The Performance of Feedback Control Systems

Learning objectives: (a) to understand the characteristics of a feedback system; (b) to describe quantitatively the transient response of second-order systems; (c) to understand the effects of poles and zeros on a system; (d) to be able to calculate the steady-state error of feedback systems.

3.1 Introduction

Open- and closed-loop control systems

Figure 3.1 An open system Figure 3.2 A closed-loop system

Test input signals

Table 3.1 Test input signals and functions


G(s)

G(s)

H(s)

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3.2 First-order systems

A first-order system transfer function can be generally shown in Fig. 3.3. if the input is a unit step R(s)=1/s, the laplace transform of the response is C(s), where

C (s )=R (s )G(s )= as( s+a)

.

After the inverse Laplace transform, the step response is given by

c(t) = cf (t) + cn(t) = 1 e-at (3.1)

where the forced response cf(t) = 1, and the system

natural response cn(t) = eat. , time constant. = 1/a;

Tr, rise time. It is defined as the time for the waveform to go from 0.1 to 0.9 of its original value.

Tr =

2. 31a

−0. 11a

=2 .2a

.

Ts, settling time. A settling time is defined as the time for the response to reach and stay within a percentage value. A 2% settling time can be found by letting c(t) = 0.98 in Eq.(3.1) and solving for time, t.

Then

T s=4a

.

3.3 The performance of second-order systems

Given a closed-loop system shown in Fig. 3.5, the system output is

Y ( s )= K

s2+ ps+KR(s )

Y (s )R (s )

= Ks2+ ps+ K

=p( s )q (s )


Figure 3.3 A first-order

Figure 3.4 First-order system response to a unit step.

Figure 3.5 A second order system

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Characteristic equation: The denominator polynomial q(s) of a Laplace transform function, when set equal to zero, is called the characteristic equation, i.e. q(s) = 0.

Poles: The roots of the characteristic equation q(s) = 0 are called poles.

Zeros: The roots of the numerator polynomial p(s) = 0 are called zeros.

System performance in relation with the location of the poles.


Figure 3.6 Second-order systems, poles, and step response.

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3.4 The parameters of second-order systems

A second-order system function

Y ( s )= K

s2+ ps+KR(s )

can be written in general form:

Y ( s )=ωn

2

s2+2 ζωn s+ωn2

R( s )

(3.2)

n , natural frequency: n =√ K ;

, damping ratio:

p2 ωn ;

d , damped frequency: d = n √1−ζ 2

When a unit step input u(t) is given, the system transient output is

y ( t )=1− 1

√1−ζ2e−ζωn t

sin(ωn√1−ζ 2t +cos−1 ζ )

where 0<<1.

System performance in relation with the damping ratio.


Figure 3.8 Transient response of a second order system for a step input.

Figure 3.7 Time functions of the second-order system, associated with pole locations in the s-plane, in response to an impulse input.

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a) Overdamped responses:

Poles: two real at –, –

b) Underdamped responses:

Poles: two complex at –d±jd .c) Undamped responses:

Poles: two imaginary at ±jd .

d) Critical damped responses:

Poles: two real at –A well balanced under damped response is the solution to most control problems.

Parameters associated with underdamped second-order systems:

n , Natural frequency, n =

√ K;

, damping ratio, p

2 ωn

Tp, peak time,

T p=π

ωn√1−ζ2

Ts, settling time,

T s=4

ζωn

(2% criterion)

P.O., Percent overshoot,


Figure 3.10 Step response of a control system and measuring parameters

Figure 3.9 Step responses for second-order system damping cases.

19

P .O .=100e

−ζπ

√1−ζ 2

ς=−ln( PO /100 )

√π2+ln2 (PO /100 )


Tr, rise time,

T r=2. 16 ζ +0 .60

ωn

The graphical relationship between , n, d and where cos .

3.5 Effects of a third pole and zero on the second-order system response

Effect of third pole:

C (s )= As

+B( s+ζωn )+Cωd

(s+ζωn )2+ωd

2+ D

s+α r

c ( t )=Au( t )+e−ζωn t

(B cos ωd t+C sin ωd t )+De−α r t

Dominant poles: The poles of the transfer function that cause the dominant transient response of the system.

Three cases are discussed, in which the third pole is located in different position.


Figure 3.11 Relationship between , n, d and

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In case I, the real pole’s transient response is still significant at Tp and Ts generated by the second

order pair, thus the transient response cannot be ignored.

In case II, the real pole’s transient response decays to an insignificant value at Tp and Ts of the

dominated poles. So the response can be ignored.

The third pole is negligible when r > 10|n|, or r >5|n|.

In case III, the pole has not contributed much in its transient response, and is not considered.

Effect of a zero:

(1) The closer the zero is to the dominant poles, the greater the effect on the transient response.

(2) For a system with the transfer function T (s )=

(ωn2 /a )(s+a )

s2+2 ζωn s+ωn2

R (s ), the percent overshoot for

a step input as a function of an, when 1 , is given in Fig. 15.

(3) If a zero at –z is very close to a pole at –p3, then both the zero and the pole can be cancelled.


Figure 3.12. Component responses of a three-pole system: a. pole plot; b. component responses of three cases.

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3.6 Steady-state error of unity feedback systems

System error: E(s) = R(s) – Y(s)


Figure 3.13. Effect of adding a zero to a two-pole system with poles at (-1j2.828)

Figure 3.15. Closed loop control system

Figure 3.14. Percent over shoot as a function of

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When H(s) = 1, the system error is equal to the signal Ea(s).

The steady-state error is then

limt →∞

e ( t )=ess=lims→0

sR ( s )1+G(s )

(3.3)

Static error constants

For a step input, the steady-state error is

e (∞)=estep(∞)= 11+lim

s→0G(s )

Position constant

K p=lims→0

G( s )

For a ramp input, the steady-state error is

e (∞)=eramp(∞)= 1lims→0

sG( s )

Velocity constant

K v=lims→0

sG( s )

For a parabolic input, the steady-state error is

e (∞)=e parabola(∞)= 1

lims→ 0

s2 G( s )

Acceleration constant

Ka=lims→0

s2 G (s )


Table 3.2 Relationship between input, system type, static error constants and steady-state errors.

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Reading materials: Ref. 1: Pages 125 – 143; Ref. 2 Pages 179 – 208, 367 – 387;

Ref 3. Pages 223-244.

Problems:

3.1 A U.S. experimental magnetic levitation train, Trans rapid-06, is shown in Fig P3.1(a). the use of magnetic levitation and electromagnetic propulsion to provide contactless vehicle movement makes the Transrapid-06 technology rapidly different from the existing Metroliners. The underside of the Transrapid-06 carriage (where the wheels trucks would be on conventional car wraps around a guideway. Magnets on the bottom of the guideway attract electromagnets on the “wraparound”, pulling it upwards the guideway. This suspends the vehicles about one centimeter above the guideway.

The levitation control is represented by Fig. P3.1(b). (a) Select K so that the system provides an optimum response with coefficients s2+1.4ns+n

2 . (b) Determine the expected percent overshoot to a step input.

Figure P3.1 Levitated train control

3.2 A feedback system with negative unity feedback has a plant G( s )=

2(s+8)s (s+4 ) . (a) Determine

the closed-loop transfer function T(s)=Y(s)/R(s). (b) Find the time response y(t) for step input r(t)=A for t>0. (c) Using the final-value theorem, determine the steady-state value of y(t).

3.3 A second-order control system has the closed-loop transfer function T(s) = Y(s)/R(s). The system specifications for a step input follow:

a) Percent shoot 5% .b) Settling time < 4 second.c) Peak time Tp<1 second.


Y(s) gas spacing

I(s), coil current

24


Show the permissible area for the poles of T(s) in order to achieve the desired response. Use a 2% settling time.

3.4 A system is shown in Fig P3.4(a). The response to a unite step, when K=1, is shown in Figure P3.4(b). Determine the value K so that the steady-state error is equal to zero. (Answer: K = 1.25)

Figure P 3.4

3.5 An important problem for television system is the jumping or wobbling of the pictures due to the movement of the camera. This effect occurs when the camera is mounted on a moving truck or airplane. The Dynalens system has been designed to reduce the effect of rapid scanning motion; see Fig. P3.5(a). a maximum scanning motion of 250/s is expected. Let Kg= Kt=1 and assume that g is neglectable. (a) Determine the error of the system E(s). (b) Determine the necessary loop gain, Ka Km Kt when a 10/s steady-state error is allowable. (c) The motor time constant is 0.40 s. Determine the necessary loop gain so that the settling time to within 2% of the final value of vb is less than or equal to 0.03 s.

Figure P3.5 Camera wobble control

3.6 The open-loop transfer function of a unity negative feedback system is G( s )= K

s (s+2) . A system response to a step input is specified as follows: peak time Tp=1.1 s, percent overshoot = 5%. (a) Determine whether both specifications can be met simultaneously. (b) if the specifications cannot be met simultaneously, determine a compromise value for K so that the peak time and percent shoot specifications are relaxed the same percentage.

3.7 A closed-loop transfer function is T (s )=

Y ( s )R( s )

=96( s+3 )(s+8 )(s2+8 s+36 ) .

i. Determine the steady-state error for a unit step input R(s) = 1/s. ii. Assume that the complex poles dominate, and determine the overshoot and settling time

to within 2% of the final value. iii. Plot the actual system response, and compare it with the estimates of part b).

3.8 A closed-loop system is shown in Fig. P3.8. Plot the response to a unit step input for the system for z=0, 0.05, 0.1, and 0.5. Record the percent overshoot, rise time and settling time


Torque motor

vc Camera speed

vb

Bellows speed

25


(2% criterion) as z varies. Describe the effect of varying z. Compare the location of the zero, 1/z, with the location of the closed-loop roots.

Figure P3.8 System with a variable zero

3.9 A block diagram model of an armature-current-controlled dc motor is shown in Fig. P3.9. a) Determine the steady-state error to a ramp input, r(t)=t, t0, in terms of K, Kb, and Km. b) Let Km = 10 and Kb = 0.05, and select K so that steady-state error is equal to 1.c) Plot the response to a unit step input and a unit ramp input for 20 seconds. Are the

response acceptable?

Figure P3.9 dc motor control



4. The stability of linear feedback systems

Learning objectives: (a) to be able to determine a system stability presented in a transfer function using Routh-Hurwitz stability criterion; (b) to be able to determine a state variable system stability.

4.1 The concept of stability

Bounded-input, bounded-output (BIBO) definition of stability:

1) A system is stable if every bounded input yields a bounded output.

2) A system is unstable if any bounded input yields an unbounded output.

A necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function are located in the left half-plane (LHP).

4.2 The Routh-Hurwitz stability criterion

The denominator of a system transfer function can be written as a characteristic function

(s) = q(s) =ansn + an-1sn-1 + … + a1s + a0 =0

The Routh-Hurwits criterion:

For an nth-order system in which the denominator is given in a polynomial form:

q(s)=ansn + an-1sn-1+ an-2sn-2+ … + a1s+ a0 ,

i. A necessary and sufficient condition for the system to be is that all of the elements in the first

column of the Routh array be positive.

ii. The number of roots of q(s) with positive real parts is equal to the number of changes in sign

of the first column elements of the Routh array.

Routh’s Array:

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sn an an-2 an-4 …

sn-1 an-1 an-3 an-5 …sn-2 bn-1 bn-3 bn-5 …sn-3 cn-1 cn-3 cn-5 …. . . .. . . .. . . .s0 hn


Wherebn−1=

(an−1 )( an−1 )−an (an−3 )an−1

= −1an−1

|an an−2

an−1 an−3

|

bn−3=− 1an−1

|an an−4

an−1 an−5

|

and

cn−1=−1bn−1

|an−1 an−3

bn−1 bn−3

|

Example 4.1 Find the stability of the system given in Fig. 4.1, using the Routh Array.

Figure 4.1

Special case 1. If only the first element in one of the rows is zero, then it may be replaced by a small positive number that is allowed to approach zero, 0, after completing the array.

Example 4.2: q(s)=s5 + 2s4 +2s3 +4s2 +11s + 10 .

Special case 2. An entire row of the Routh array is zero. This happens when factors such as (s+)(s) or (s+j)(sj) occur.

Solution: if the ith row is zero, we form the following auxiliary equation from the previous non-zero row: U(s) = 1si+1+2si1+3si3 + … ,

where {i} are the coefficients of the (i+1)th row in the array. We then replace the ith row by the coefficients of the derivative of the auxiliary polynomial U(s), and complete the array. The roots of the auxiliary polynomial U(s) are also roots of the characteristic equation, and these must be tested separately.

Example 4.3 q(s)=s5 + 7s4 +6s3 +42s2 +8s + 56



Special case 3. Repeated roots of the characteristic equation on the j-axis.

If the j-axis roots of the characteristic equation are simple, the system is neither stable nor unstable; it is instead called marginally stable, since it has an undamped sinusoidal mode. If the j-axis roots are repeated, the system response will be unstable, with a form t[sin(t+)]he Routh-Hurwitz criterion will not reveal this form of instability.

Reading materials: Ref. 1. Pages 167 –180; Ref 2. Pages 290 – 304; Ref 3. Pages 324 – 349.

Problems:

P4.1 Utilizing the Routh-Hurwitz criterion, determine the stability of the systems with the following characteristic equations:

(a) s2+5s+2=0(b) s3+4s2+6s+6=0(c) s3+2s24s+20=0(d) s4 +s3+2s2+10s+8=0(e) s4 +s3+3s2+2s+K=0(f) s5 +s4+2s3+s+5=0(g) s5 +s4+2s3+s2+s+K=0

Determine the number of roots, if any, in the right-half plane. Also when it is adjustable, determine the range of K that results in a stable system.

P4.2 Determine the systems with the following characteristic equations are stable or unstable:(a) s3 s2+6s+100=0(b) s4 –6 s3s2 –17s –6 =0(c) s2+6s+3=0

2010/2011 Kyle Jiang 29


5. Frequency Response Method

Learning objectives: (a) to understand the concept of frequency response; (b) to be able to plot

frequency response; (c) to be able to use frequency response to analyse system stability.

Frequency response method is a useful tool for control system analysis and design. It is often used

together with other tools for achieving satisfactory results.

5.1 Frequency Response Concept

In analysis of a system steady-state frequency response,

(a) the input is a sinusoidal signal;

(b) the output of the linear system is also sinusoidal;

(c) the output differs from input in amplitude and phase angle.

Example 5.1 Frequency response of a mass-spring-damper system.

Figure 5.1 A mass-spring-damper system and its frequency response

Magnitude frequency response: M (ω )=

M o (ω )M i(ω )

Phase frequency response:

Frequency response: M()



In frequency response, the system transfer function is obtained through the Fourier transform. The

Fourier transform pair, Eq.5.1, are similar to the Laplace transform pair, referring pp 5. In practice,

Laplace transform is used in frequency response method, but the variable s is replaced by j, i.e.

T(j) = T (s)|sj .

F ( jω)=F {f ( t )}=∫−∞

∞f ( t )e− jωt dt

(5.1)

f ( t )= 12 π∫−∞

∞F ( jω)e jω t dω

5.2 Bode Diagram

The transfer function of a system can be described in the frequency domain by the relation

G(j) = G(s)|s=j=R() +jX() (5.2)

where R() = Re [G(j)], and X() = Im[G(j)]

Alternatively the transfer function can be represented by a magnitude |G(j)| and a phase () as

G(j) = | G(j)|() (5.3)

where φ (ω )=tan−1 X (ω )

R(ω ) , and | G() |2 = [R()]2 + [X()]2

5.2.1 Concept of Bode diagram

The log-magnitude and phase frequency response curves as functions of log are called Bode

diagrams or plots, in which the logarithm of the magnitude is expressed as

Logarithmic gain = 20 log10|G(j)|

where the units are decibels (dB). The angle is given by Eq.(5.3). The changes of the logarithmic

gain and angle versus are plotted in separate coordinates.

Example 5.2: Plot both the separate magnitude and phase diagrams for system G(s) = 1/(s+2)

Solution:

2010/2011 Kyle Jiang 31


5.2.2 Bode Diagram by Asymptotic Approximations

Magnitude response: Consider the general transfer function

G( s )=K ( s+z1 )(s+ z2)⋯(s+zk )

sm(s+p1 )(s+ p2)⋯( s+ pn ) .

Converting the magnitude response into dB yields

20 log|G(j)| = 20 log K +20 log |(s+z1)| +20 log |(s+z1)| + … 20 log |sm| 20 log |(s+p1)|…|sj

Thus, if we knew the response of each term, the algebraic sum of the sum would yield the total

response in dB. Further, if we could make an approximation of each term that would consist only of

straight lines, graphic addition of terms would be greatly simplified.

Phase response: The phase response is the sum of the phase frequency response curves of the zero

terms minus the sum of the phase frequency response curves of the pole terms. Again the general

phase response curve of a given system can be produced by summing the phase response of each

individual term.

MAGNITUDE AND PHASE RESPONSES OF TYPICAL FOUR KINDS OF FACTORS


Figure 5.1 Bode diagrams of example 5.1

32


Bode diagrams for constant gain G(s) = Kb

The logarithmic magnitude 20 log Kb = constant in dB

The phase angle () = 0

Bode diagrams for poles or zeros at the origin

A pole at the origin:

The logarithmic magnitude

The phase angle

A zero at the origin:


The phase angle

Bode diagrams for poles or zeros on the real axis

A pole on the real axis:


The asymptotic curve:

The exact curve:

The phase angle


The exact curve:


Figure 5.2 Bode diagrams of transfer functions with poles or zeros at the origin.

33


A zero on the real axis:


The phase angle


Figure 5.4 Bode diagrams of transfer a function with a zero on the real axis.

Figure 5.3 Bode diagrams of transfer a function with a pole on the real axis.

34


Bode diagrams for complex conjugate poles or zeros

A pair of complex conjugate poles:G( jω )= 1

1+ j 2 ζω/ωn+( jω /ωn )2

Let u = n, G(u )= 1

1+ j 2ζu−u2

|G(u )|2= 1

(1−u2 )2+(2 ζ u)2(5.4)

∠G(u )=−tan−1( 2 ζ u

1−u2 )(5.5)



The phase angle


Unlike the first order frequency response approximation, the difference between the asymptotic

approximation and the actual frequency response can be great for < 0.707 . So corrections are

needed.

2010/2011 Kyle Jiang 35


Resonant frequency ωr=ωn√1−2 ζ2, <0.707 (5.6)

The maximum value of the magnitude |G()| is

M pω=|G(ωr)|=

1

2ζ √1−ζ 2, <0.707 (5.7)

Example 5.3 Draw the log-magnitude and phase characteristic diagrams of

G( s )= s+3

(s+2)( s2+2 s+25) .

Solution:

(1) Convert G(s) to show the normalized components that have unity low-frequency gain.

(2) Make the magnitude diagram slope table, and draw the asymptote for each factor, then

compose the final diagram.


Figure 5.4 Bode diagram for a system with two complex conjugate poles.

36


(3) Make the phase diagram slope table, and draw the asymptote for each factor, then compose

the final diagram.


Figure 5.6 Magnitude characteristic for G(s) =(s + 3)/[(s + 2) (s2 + 2s + 25)]a. components; b. composite

37


5.3 Performance Specifications in the Frequency Domain

As the performances of a second order system can be specified in terms of settling time, and

overshoot, it is also specified in the frequency domain. The closed-loop transfer function of a

second order system is

T (s )=ωn

2

s2+2 ζωn s+ωn2

.

Maximum Magnitude Mp: At the resonance frequency r, a

maximum value of the frequency response Mp is obtained.

The Bandwidth: it is the frequency, B, at which the frequency response has declined 3 dB from its

low-frequency value.

The bandwidth of a two-pole system can be found as


Figure 5.6 Magnitude characteristic of the second-order system

Figure 5.6 Phase characteristic. a. components; b. composite

38


ωBW=ωn√(1−2 ς2 )+√4 ς4−4 ς2+2 (5.8)

The relationship between BW and settling time:

ωBW= 4T s ς

√(1−2 ς2)+√4 ς4−4 ς2+2(5.9)

The relationship between BW and peak time:

ωBW= π

T p√1−ς2√(1−2 ς2 )+√4 ς4−4 ς2+2

(5.10)

Desirable frequency domain specifications are:

(1) Relatively small resonant magnitudes: Mp< 1.5, for example.

(2) Relatively large bandwidths so that the system time constant n is sufficiently small.

5.4 Other Frequency Response Diagrams

Polar plots

In the polar plot, the coordinates of the polar plot are the real and imaginary parts of G(j) in Eq.

(5.2). The polar plot represents the transfer function graphically with w changing from - to +.

Example 5.4: The transfer function of an RC filter is G( s )=

V 2( s )V 1(s )

= 1RCs+1 . The polar plot for

the RC filter is shown in Fig. 5.8.

Log- magnitude-phase diagrams

Log-magnitude-phase diagrams are usually transferred from Bode diagrams.


Figure 5.7 An RC filter

Figure 5.8 Polar plot for the RC filter

39


Example 5.3 Given a transfer function

GH (s )= 5s(0 . 5 s+1)( s/6+1 ) , its log magnitude and phase

diagram is shown in Fig. 5.9.

5.5 Stability in the Frequency Domain

5.5.1 Nyquist plot:

The Nyquist plot is the plot of an open loop function GH(jw) when frequency w varies

from – ∞ to + ∞ .

Features of the Nyquist plot

– The plot is formed by two parts. One part is created when w is from 0 ® + ∞ , while

the other is created when w is from - ∞ ®0. One is the mirror of the other.

– The two parts meet at the frequencies = + ∞ and = - ∞.

– In a Nyquist plot, varies from – ∞ to + ∞, which implies that j in function GH(j)

varies along the imaginary axis.

Nyquist Plot Examples:

• GH(s) = 100/[(s+2)(s+5)]

• A MATLAB example:


Figure 5.9 Log-magnitude-phase diagram for GH(j).

40


GH(s) = 100(s-2)/[(s+1)(s+5)]

5.5.2 The Nyquist stability criteria

Original statement: Let Z = number of zeros of 1+GH in the RHP,

P = number of poles of GH in the RHP,

N = number of counter clockwise encirclements of (-1,0),

then Z = P - N. If Z > 0, the system is unstable.

Case 1: A feedback system is stable if and only if the Nyquist plot in the GH(jw) plane does

not encircle the (-1,0) point when the number of poles of GH(s) in the right half plane is zero, ie

P=0.

Case 2: A feedback system is stable if and only if, for the the Nyquist plot GH(jw), the number

of counter clockwise encirclements of the (-1,0) point is equal to the poles of GH in the RHP,

ie P = N.

Rationale of the Nyquist stability criteria

– The Nyquist criteria uses GH(s) to judge the stability of the whole feedback system

T(s):

– The idea of the Nyquist stability criteria is that when you draw a loop to cover the

complete right half s-plane, the Nyquist plot will encircle the poles of the complete

system T(s).

– Although Z is the number of zeros of GH in the RHP, it is effectively the number of

poles of the complete system T(s) in the RHP

Find the system stability of the following open-loop functions:

2010/2011 Kyle Jiang 41

T (s )=G( s )

1+G( s )H ( s )


Example 1: GH(s) = 100/[(s+2)(s+5)]

Solution:

Example 2: GH(s) = 100(s-2)/[(s+1)(s+5)]

Solution:

Example 3: GH(s) = K/[s(s-1)]

Solution:

5.5.2 Relative Stability and Nyquist Criterion



In a Nyquist diagram, the stability condition is

|KG(j)| < 1 at G(j) = -180o.

If the plot passes through the (-1,0) point, the system is called marginally stable. The margin for a

system plot to pass through the (-1, 0) point is used as a measure of the system stability.

The gain margin is the change in open-loop gain, expressed in decibels (dB), required at

1800 of phase shift to make the closed-loop system unstable.

The phase margin is the change in open-loop phase shift required at unity gain to make the

closed loop system unstable.

Damping ration from

Phase Margin

mp =

tan−1 2ξ

√−2 ζ2+√1+4 ζ 4,

or = 0.01mp (5.12)

Reading materials: Ref.1. Pages 252 – 285; Ref. 2. 586-636; Ref. 3. 406 – 439.

Problems


Figure 5.10 Nyquist diagram showing gain and phase margin.

Fgiure 5.11 Bode diagram for GH(j)=1/[j(j+1)(0.2j+1)] showing gain and phase margins.

43


5.1 Sketch the Bode diagram representation of the frequency response for the following transfer

functions.

(a) GH (s )= 1

(1+0. 5 s )(1+2 s )

(b)GH (s )=

(1+0.5 s )s2

(c)GH (s )= s−10

s2+6 s+10

(d)GH (s )=

30 (s+8 )s( s+2 )(s+4 )

5.2 A control system for controlling the pressure in a closed chamber is shown in Figure P5.2. the

transfer function for the measuring element is H ( s )=100

s2+15 s+100 , and the transfer function

for the valve is G1( s )= 1

(0 . 1 s+1 )(s /15+1) . The control function is Gc(s) = s +1. Obtain the

frequency response characteristics for the system loop transfer function Gc(s)G1(s)H(s)[1/s].

Figure P 5.2 (a) Pressure controller. (b) Block diagram model.

5.3 A feedback control system is shown in Fig.

P5.3. the specification for the closed-loop

system requires that the overshoot to a step

input be less than 10%. (a) Determine the

corresponding specification in the frequency domain Mpw for the closed-loop transfer function

Y(j) = Y(j)/R(j). (b) Determine the resonant frequency, wr. (c) Determine the bandwidth of the

closed-loop system.


Fig. P7.3

44


5.4 The control of one of the joints of a space robot can be represented by the loop transfer function

GH (s )=781 (s+10)s2+22 s+484 . (a) Sketch the Bode diagram of GH(j). (b) Determine the maximum

value of 20 log |GH|, the frequency at which it occurs, and the phase at that frequency.

5.5 The frequency response of a process G(j) is shown in Fig. P5.5. Determine G(s).

Figure P5.5

5.6 The Bode diagram of a closed-loop film transport system, T(s), is shown in Fig. P5.6. Assume

that the system transfer function T has two dominant complex conjugate poles. (a) Determine

the best second-order model for the system. (b) Determine the system bandwidth. (c) Predict the

percent overshoot and settling time (2% criterion) for a step input.


Figure P5.6

45


6. The Design of Feedback Control Systems

Learning objectives: (a) to understand the working mechanism of popularly used controllers and

compensators; (b) to be able to use frequency response method to design these

controllers/compensators.

6.1 Introduction

6.1.1 Configurations

It is quite often that we get a set of parameters that do not satisfy all the expected values.

Unless we compromise in some of the expected values, we have to alter the system to improve the

performance. Te alternation or adjustment of a control system in order to provide a suitable

performance is called compensation.

Several types of compensation are shown in Fig. 6.1. The compensator placed in the

forward path is called a cascade or series compensator. Similarly, the other compensators are called

feedback, output, and input compensation, as shown in Fig. 6.1 (b), (c), and (d).

6.1.2 Controllers and Compensators

Controller and compensator are two terms used alternately to indicate a similar type of

device. There is no fundamental difference between the two. For some devices, people are used to

call them as controllers, and for others, compensators.

The compensating device may be electric, mechanical, hydraulic, pneumatic, or some other

type of device or network, and is often called a compensator. A compensator is an additional

component or circuit that is inserted into a control system to compensate for a deficient

performance.

6.1.3 Compensators versus System Improvements


Figure 6.1 Types of compensation. (a) Cascade compensation. (b) Feedback compensation. (c) Output, or load, compensation. (d) Input compensation.

46


Cascade compensators are widely used, and can affect the system performance in different ways. A

summary of the most commonly used controllers and compensators, and their functions are listed

below.

6.2 Phase-Lead and Phase-Lag Compensators


Table 6.1 Controllers, compensators and their functions and characteristics.

47


Both the phase-lead and phase-lag compensators can be described using the transfer function

Eq. (6.1):

Gc( s )=K ( s+z )α( s+ p ) (6.1)

where K is the gain of the compensator and is the ratio of the

pole and zero, α= p

z

Phase-lead compensator:

When |z| < |p|, Eq.(6.1) is called a phase-lead compensator.

Phase-lag compensator:

When |z| > |p|, Eq.(6.1) is called a phase-lag compensator.

In the both compensators,

Phase angle m :sin φm=α−1

α+1 (6.2)

Hardware implementation

Both lead and lag compensators can be implemented by either integrating a piece of hardware into

the original system, or using a microprocessor/computer. Figure 6.4 (a) and (b) are two electrical

circuits for lead and lag compensators.

6.3 Phase-lag Design Using the Bode Diagram


Figure 6.3 Pole-zero diagram of the phase-lag compensator

V1(s)V1(s) V2(s)

V2(s)

Figure 6.4 (a) Phase-lead network; (b) Phase-lag network

Figure 6.2 Pole-zero diagram of the phase-lead compensator

48


Visualizing Lag Compensation

The function of the lag compensator as seen in Bode diagram is to (1) improve the static error

constant by increasing the overall system gain without any resulting instability, and (2) increase the

phase margin of the system to yield the desired transient response. These concepts are illustrated in

Figure. 6.5.

Figure 6.5 Visualizing lag compensation

The uncompensated system is unsuitable, since the gain at 1800 is greater than 0 dB. The lag

compensator, while not changing the low-frequency gain, does reduce the high-frequency gain.

Thus the low-frequency gain of the system can be made high to yield a large Kv without creating

instability. This stabilizing effect of the lag compensator comes about because the gain at 1800 of

phase is reduced below 0 dB. Through judicious design, the magnitude curve can be reshaped, as

shown in Figure 6.5, to go through 0 dB at the desired phase margin. Thus, both Kv and the desired

transient response can be obtained.

Designing Procedure

The lag compensator can be readily accomplished on the Bode diagram. The advantage of

the Bode diagram is that we can simply add the frequency response of the compensator to the Bode

diagram of the uncompensated system in order to obtain a satisfactory system frequency response.

The design procedure for a phase-lag compensator on the Bode diagram is as follows:

(1) Determine the K in the given system plant G(s), using the expected steady state error ess.

2010/2011 Kyle Jiang 49


(2) Obtain the Bode diagram of the uncompensated system with the gain adjusted for the desired

error constant.

(3) Determine the phase margin Pm of the uncompensated system and, if it is insufficient, proceed

with the follow steps.

(4) Determine the frequency where the phase margin requirement would be satisfied if the

magnitude curve crossed the 0 dB line at this frequency, ’c (allow for 5o phase lag from the

phase lag compensator when determine the new crossover frequency).

(5) Place the zero of the compensator one decade below the new crossover frequency ’c and thus

ensure only 5o of additional phase lag at ’c .

(6) Measure the necessary attenuation Mp at ’c to ensure that the magnitude curve crosses at this

frequency.

(7) Calculate by noting that the attenuation Mp introduced by the phase lag compensator is

Mp=20log at ’c.

(8) The zero is z = ’c /10, and the pole is p = z/.

(9) Modify the system gain by multiplying 1/, to count for the contribution of the compensator to

ess, and the design is completed.

Example 6.1 The system transfer function is GH (s )= K

s( s+2 ) .

It is required that the damping ratio = 0.45, and a system velocity constant Kv = 20 .

Solution:

(1) Determine the constant error Kv. K v=lim

s→0sGH ( jω)=lim

s→ 0s

Ks( s+2 )

=K2

=20

So K = 40. The uncompensated Bode diagram is given in Fig. 6.6.

(2) The system Bode diagram is given as follows.

(3) Find the phase margin. From the diagram, it can be seen that the phase margin PM=20o.

(4) Determine the new crossover frequency 'c , allowing 5o phase lag.

('c)= 130o , and 'c= 1.7

(5) The attenuation Mp necessary to cause 'c to be the new crossover frequency is Mp=20log at

’c. Calculate , 20dB = 20 log , or = 10.

(6) z = 'c/10 = 0.17, and p = z/ = 0.017

(7) Multiply 1/ to Gc(s)

The compensated system is Gc( s )GH ( s )=

4 ( s+0 .17)s (s+2)( s+0 .017 )



6.4 Phase-Lead Design Using the Bode Diagram

Visualizing Lead Compensation


’c

Figure 6.6 Design of a phase-lag compensator on the Bode diagram for Example 6.1

Figure 6.7 Time response to a step input for the uncompensated system(solid line), and compensated system (dashed line) of example 6.1

51


The lead compensator increases the bandwidth by

increasing the gain crossover frequency. At the

same time, the phase diagram is raised at higher

frequencies. The result is a larger phase margin and

a higher phase-margin frequency. In the time

domain, lower percent overshoots (larger phase

margins) with smaller peaktimes (higher phase-

margin frequencies) are the results. The concepts

are shown in Figure 6.8.

The uncompensated system has a small phase

margin (B) and a low phase margin frequency (A).

using a phase lead compensator, the phase angle

plot is raised for higher frequencies. At the same

time, the gain crossover frequency is the magnitude

plot is increased from A rad/s to C rad/s. These effects yield a larger phase margin (D), a higher

phase-margin frequency (C), and a larger bandwidth.

Design steps:

(1) Evaluate the uncompensated system phase margin when the error constants are satisfied.

(2) Allowing for a small amount of safety, determine the necessary additional phase lead, m.

(3) Evaluate a from Eq.(6.2).

(4) Evaluate 10log and determine the frequency where the uncompensated magnitude curve is

equal to –10log dB. Because the compensator provides a gain of 10log at m, this

frequency is the new 0-dB crossover frequency and m simultaneously.

(5) Calculate the pole p=m√α, and z = p/.

(6) Draw the compensated frequency response, check the resulting phase margin, and repeat the

steps necessary. Finally, for an accepted design, raise the gain of the amplifier in order to

account for the attenuation (1/..

Example 6.2: A single loop feedback system has an open-loop uncompensated transfer function

GH (s )= K

s2+2 s .


Figure 6.8 Visualizing lead compensator

52


Design a lead compensator to meet the system specifications:

A steady state error for a ramp input equal to 5%; the phase margin of the system be at least 45o.

Solution:

(1) Evaluate the error constant:

Kv = 1/ess = 1/0.05 = 20.

Evaluate the uncompensated system phase margin:

GH ( jω)= Kjω( jω+2)

=K v

jω(0 . 5 jω+1)

The phase margin of the uncompensated system can be calculated as:

PM = |-180o - GH(jw)| = |-180o –[-90o- tan-1(0.5)]|

At the crossover frequency, = c =6.2 rad/s. we have PM = 18o.

The phase margin can also be found from the Bode diagram, see Fig. 6.9.

(2) Determine the phase lead

As specified, the initial phase lead is 45o –18o = 27o

With consideration o 10% increment, the phase lead is = 27o +3o =30o

(3) Evaluate .

α−1α+1

=sin 30o=0 . 3, and therefore = 3.

(4) Determine the frequency where the uncompensated magnitude curve is equal to –10log

Because the compensation network provides a gain of 10log at m, this frequency is the new 0-dB

crossover frequency and m simultaneously.

The magnitude at m is 10log10log3 = d The compensated crossover frequency is then

evaluated where the magnitude of GH(j) is –4.8 dB, the thus m=c=8.4.

(5) Calculate the pole and zero.

z = 4.8, and p = z = 14.4 . The compensation network is

Gc( s )=13

1+s /4 .81+s /14 . 4

(6) Draw the compensated frequency response to check the results. And raise the gain of the

amplifier in order to account for the attenuation (1/).

The total dc loop gain must be raised by a factor of 3 in order to account for the factor 1/Then the

compensated loop transfer function is

Gc( s )GH ( s )=20( s /4 .8+1)s (0 .5 s+1 )(s /14 .4+1 ) .

2010/2011 Kyle Jiang 53


Reading materials: Ref. 2. Pages 684 – 710; Ref.3. 553-567, 580 – 584.

Problems

6.1 A negative feedback control system has a transfer function G( s )= K

(s+3 ) . We select a

compensator, Gc( s )= s+a

s , in order to achieve zero steady-state error for a step input. Select a

and K so that the overshoot to a step input approximately 5% and the settling time (2%

criterion) is approximately 1 second. (Answer: K=5, a = 6.4)

6.2 A unity feedback system has G( s )=1350

s (s+2)( s+30 ) .

A lead network is selected so that Gc( s )=

(1+0 .25 s )(1+0 .025 s ) . Determine the peak magnitude and

the bandwidth of the closed-loop frequency response using a plot of the closed-loop

frequency response.

6.3 The control of an automobile ignition system has unity negative feedback an da loop transfer

function Gc(s)G(s), where G( s )= K

s (s+10) , and Gc(s) = K1 +K2/s. A designer selects K2/K1=

0.5, and asks you to determine KK1 so that the dominant roots have a of 1/√2 and the settling

time (2% criterion) is less than 2 seconds.


Compensated phase angle

Compensated magnitude

Figure 6.9 Bode diagram for the lead compensator for Example 6.2

54


6.4 The attitude control system for the lunar vehicle is shown in Fig. P6.4. The vehicle damping is

negligible, and the attitude is controlled by gas jets. The torque, as a first approximation, will be

considered to be proportional to the signal V(s) so that T(s) = K2V(s). The loop gain may be

selected by the designer to provide a suitable damping. A damping ratio of = 0.6 with settling

time (2% criterion) of less than 2.5 seconds is required. Using a lead network compensation,

select the necessary compensator Gc(s) by using frequency response techniques.

Figure P6.4 Attitude control system for a lunar excursion module.

6.5 A block diagram of a numerical path controlled machine turret lathe is shown in Fig. P6.5. The

gar ratio is n = 0.1, J = 10-3, and b = 10-2. It is necessary to attain an accuracy of 5X10-4 in., and

therefore a steady-state position accuracy of 2.5% is specified for a ramp input. Design a

cascade compensator to be inserted before the silicon-controlled rectifiers in order to provide a

response to a step command with an overshoot of less than 5%. A suitable damping ratio for the

system is 0.7. The gain of the silicon –controlled rectifiers is KR = 5. Design a suitable lag

compensator by using the Bode diagram method.

Figure P6.5 Path-controlled turret lathe.

6.6 A system with a cascade compensator and a unity feedback has G( s )= K

(s+3 )2. It is desired

that the steady-state error to a step input be approximately 4% and that the phase margin of the

system be approximately 45o. Design a lag network to meet these specifications.

2010/2011 Kyle Jiang 55


Appendix A: Complex Number

Definition: j2= 1 , j = √−1

A complex number is the sum of a real number and an imaginary number, so that

c = a + jb

where a and b are real numbers, and jb is the imaginary number.

Expressions: Given c = a + jb,

Re{c}= a,

Im{c}= b.

Exponential form: c = r ewhere r = √a2+b2

θ=tan−1( ba )

Polar form: c = r

Mathematical operations:

To add or subtract two complex numbers, we add (or subtract) their real parts and their imaginary

parts. Therefore if c = a + jb and d = f + jg, then

The multiplication of two complex numbers is obtained as follows (note j2 = 1):

Alternatively we use the polar form to obtain

Division of one complex number by another complex number is easily obtained usingthe polar form as follows:


Figure A1. An expression of a complex number in a plane.

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Appendix B: Properties of Laplace transform:

2010/2011 Kyle Jiang 57


Appendix C: Table of Laplcae Transforms:


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Control Notes 2010