control system basics

8/13/2019 Control System basics

1/69

1

Lecture Notes

Control Systems

Xiaoping Liu

Department of Electrical Engineering

Lakehead University


2/69

2


3/69

Contents

1 Mathematical Models: Differential Equations 1

1.1 Electrical Systems: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Translational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.2 Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Thermal Systems: Temperature Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Laplace Transform 7

2.1 Partial Fraction Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Using Laplace Transforms to Solve DE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Mathematical Models: Transfer Function Representation 11

3.1 Transfer Function for the 1st Order System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2 Transfer function for the 2nd order system: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4 Mathematical Models: Block Diagrams 15

5 Dynamic Responses 21

5.1 First Order Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.1.1 Impulse Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.1.2 Step Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.1.3 Ramp Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.1.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.2 Second-Order Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.2.1 Step Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.2.2 Performance Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

6 Steady-State Errors 29

6.1 Type of System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.2 Steady-State Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6.2.1 Steady-State Errors for Closed-Loop Systems with Unity Feedback . . . . . . . . . . . . . . . . . . . . 32

6.2.2 Steady-State Errors for General Closed-Loop Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.2.3 Steady-State Errors for Systems with Disturbances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3


4/69

0 CONTENTS

7 The Routh Stability Criterion 39

7.1 Poles and Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.2 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.3 The Routh Stability Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8 Root Locus Analysis 43

8.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

8.2 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

9 PID Controllers 49

10 Frequency Response 55

10.1 S teady-State Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

10.2 Bode Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

11 Control of Discrete Processes and PLC 61

11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6111.2 PLC Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

11.3 Internal Relays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

11.4 Timers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

11.5 Counters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64


5/69

Chapter 1

Mathematical Models: Differential

Equations

1.1 Electrical Systems:

Basic Building Blocks

Resistor v= Ri i= vR

Capacitor v= 1C

idt i=Cdv

dt

Inductor v= L

di

dt i=

1

L

vdt

Example 1Consider an RC circuit shown to the left.Input: v; Output: vC.It follows from Kirchhoffs voltage law that

v= iR + vC

Replacingi with CdvCdt

gives

v= RCdvC

dt + vC

Moving all the terms related to the output to the left-hand side and the rest to the right-hand side, we get

RCdvCdt

+ vC=v

which is a first-order system.

General form for the first order system is given by

a1dy

dt + a0y= b0u

with input u and output y .

1


6/69

2 CHAPTER 1. MATHEMATICAL MODELS: DIFFERENTIAL EQUATIONS

Example 2Consider an RL circuit shown to the left.Input: v; Output: iL.It follows from Kirchhoffs voltage law that

v= iLR+ vL

ReplacingvL with LdiLdt

gives

v= RiL+ LdiLdt


LdiLdt

+ RiL= v

which is also a first-order system.

Example 3Consider an RLC circuit shown to the left.Input: v; Output: vC.

It follows from Kirchhoffs laws that

i = iL+ iC

v = iR+ vC

Substituting the first equation to the second equation and replacing iL with 1L

vCdt and iC with C

dvCdt

gives

v= R

1

L

vCdt+ C

dvCdt

+ vC

Differentiating the equation with respect to t produces

dv

dt =

R

LvC+ RC

d2vC

dt2 +

dvC

dt


RCd2vCdt2

+dvC

dt +

R

Lvc=

dv

dt

which is a second-order system.

General form for a second-order system is given by

a2d2y

dt2 + a1

dy

dt + a0y = b1

du

dt + b0u

1.2 Mechanical Systems

The basic building blocks for mechanical systems are shown below.


7/69

1.2. MECHANICAL SYSTEMS 3

1.2.1 Translational Motion

Translational Damper

Damper represents the type of forces experienced when we endeavor to push an object through a fluid or move an objectagainst frictional forces. The faster the object is pushed, the greater the opposing forces.

In the ideal case, the damping or resistive force Ffis proportional to the velocity v of the piston or object, that is ,

Ff =cv = c dxdt

c is a damping coefficient constant. The larger the value ofc the greater the damping force at a particular velocity.

Translational Spring:

Suppose a spring is fixed at one end. When you try to pull or to push the spring at the other end, you are experiencingsome kinds of resistive forces exerted by the stretched or compressed spring. This force will be in the opposite direction andequal in size to the force used to stretch or compress the spring, which is proportional to the extension or compression, thatis, displacement

Fs= kx

where k is a spring constant called stiffness. The bigger the value of k , the greater the forces have to be to stretch orcompress the spring so the greater the stiffness.

Translational Mass

The mass building block exhibits the property that the bigger the mass the greater the force required to give it a specificacceleration. Newtons second law gives

F =ma =md

dt

dx

dt

= m

d2x

dt2

where the constant of proportionality between the force and the acceleration is the constant called the mass m. ais the rateof change of velocity, i.e., a = dv

dt, and velocity v is the rate of change of displacement x, i.e., v = dx

dt.

1.2.2 Rotational Motion

Rotational Damper:

A disc is rotated in a fluid and the resistive torque fis proportional to the angular velocity (radians/second)

f =c = cd

dt

is the rate of change of angular displacement (rotated angle) .

Torsional Spring:

The tortue produced by a rotation spring is directly proportional the angle rotated .

s= k

Moment of Inertia:

Moment of inertia exhibits the property that the greater the moment of inertia Jthe greater the torque needed to produce

an angular acceleration (radians/second2

)

T=J = Jdw

dt =J

d

dt

d

dt

= J

d2

dt2

Example 4Consider a mechanical system shown to the left.Input: F; Output: x.since the net force is Fnet= Fkxcv= Fkxcdxdt . It followfrom Newtons second law that

Fnet= ma =md2x

dt2


8/69


Replacing Fnet withF kx cdxdt yields

F kx c dxdt

= md2x

dt2


md2x

dt2 + c

dx

dt + kx = F

Example 5Consider a mechanical system shown to the left.Input: ; Output: .Since the net torque is net = k c = k cddt . follows from Newtons second law that

net= J =Jd2

dt2

Replacing net with k cddt yields

k c ddt

= Jd2

dt2


Jd2

dt2 + c

d

dt + k =

Example 6Armature Controlled DC MotorsIn the armature controlled dc motor, vf is kept constant, so

if= VfRf

. The input to the system is va and the output is . Th

load torque is considered as a disturbance. The equivalent circufor the armature controlled dc motor is shown in the figure to thleft. The equations described the armature controlled dc motare given below.

va = Raia+ La dia

dt + ea (1.

ea = Keif= K3 (1.

d = Kifia = K4ia (1.

f = c (1.

d L f = Jddt

(1.

Substituting (1.3) and (1.4) into (1.5) gives

K4ia L c= Jddt

Solving this equation for ia yields

ia = J

K4

d

dt +

c

K4+

1

K4L (1.6)

Differentiatingia produces

diadt

= J

K4

d2

dt2 +

c

K4

d

dt +

1

K4

dLdt

(1.7)

Substituting (1.6), (1.7) and (1.2) into (1.1), it follows that

va = Ra

J

K4

d

dt +

c

K4+

1

K4L

+ La

J

K4

d2

dt2 +

c

K4

d

dt +

1

K4

dLdt

+ K3


9/69


10/69


Example 7 Consider a thermal system shown to the left. The ambienttemperature isTaand the temperature inside the box is T. The heat producedby the heater flows into the box at a rate qi. The thermal capacitance of theair inside the box is Cand the thermal resistance of the box wall is R. Findthe model for the system with input qi and output T.Solution: The heat flow rate out of the box through the box wall is

qo =

T

Ta

R

The net heat flow rate into the box is qi qo. The heat required to raise theinside temperature fromT to T+ dT is C dT, which implies that the heat raterequired is CdT

dt. As a result,

CdT

dt = qi q0 = qi T Ta

R

Rearranging the equation gives

CdT

dt +

T

R = qi+

TaR

whereTa can be considered a disturbance to the system.

Example 8Consider a thermometer at temperatureTwhich has just been insertedinto a liquid at temperature TL.If the thermal resistance to heat flow from the liquidto the thermometer is R, then the heat flow rate from the liquid to the thermometeris

q=TL T

R

which is used up to raise the thermometer temperature from T to T+ dT. Then wehave

q=TL T

R =C

dT

dt

that is,TL T

R =C

dT

dt

Rearranging the equation gives

CdT

dt +

T

R=

TLR

which is a first-order system with input TL and output T .


11/69

Chapter 2

Laplace Transform

The Laplace transform is a method of transforming differential equations into more easily handled algebraic equations.. Letf(t) be a function defined on [0, T]. Then, its Laplace transform is defined by

L[f(t)] =F(s) =

0

f(t)estdt

The inverse Laplace transform is defined by

f(t) = L1[F(s)] = 12j

b+jbj

F(s)estds

The lower case letters are used to denote the functions in time domain. The upper case letters are used to denote thefunctions in s domain.

As an example, let us calculate the Laplace transform for a unit step function

1(t) =

1 t 00 t < 0

Its Laplace transform is

L[1(t)] =0

estdt=

1

sest

0

=1

s

Generally speaking, its not easy to calculate the Laplace transform and the inverse Laplace transform for a given function.Fortunately, for those functions which are commonly used in this course, we dont need to calculate them ourself. There isLaplace Transform Table available, where you can find what you want. But it is useful to remember some commonly usedLaplace transforms, for instance:

L[(t)] = 1, L[t] = 1s2

, L[eat] = 1s + 1

, L[sin t] = s2 + 2

, L[cos t] = ss2 + 2

Basic Rules:

L[f

1(t) + f

2(t)] =F

1(s) + F

2(s)

L1[F

1(s) + F

2(s)] =f

1(t) + f

2(t)

L[f1(t) f2(t)] =F1(s) F2(s) L1[F1(s) F2(s)] =f1(t) f2(t)L[af(t)] =aF(s) L1[aF(s)] =af(t)L[f(t T)] = eTsF(s) L1[eTsF(s)] =f(t T)L[ d

dtf(t)] =sF(s) f(0)

L[ d2dt2

f(t)] =s2F(s) sf(0) df(0)dt

L[ dndtn

f(t)] =snF(s) sn1f(0) dn1f(0)dtn1

L[t0

f(t)dt] = 1s

F(s)Limitt0

f(t) =Limits

sF(s) The initial value theorem

Limitt

f(t) =Limits0

sF(s) The final value theorem

7


12/69


13/69

2.2. USING LAPLACE TRANSFORMS TO SOLVE DE 9

2.2 Using Laplace Transforms to Solve DE

1. Determine initial conditions

2. Take Laplace transform to change the differential equations to algebraic equations

3. Solve the algebraic equations for unknown variables

4. Carry out partial fraction expansion to put the resulting equations into the sum of fractions which can be found in

the Laplace transform table.5. Take inverse Laplace transform to change s-domain functions into t-domain functions

Example 3Solve the differential equation

RCdvCdt

+vC=v

with zero initial condition and step input v =V 1(t), which is a model forthe RC circuit to the left.Solution: Taking Laplace transform on both sides gives

RCsVC(s) + VC(s) =V

s

Solving forVC(s) yields

VC(s) = V

s(RCs + 1)

Converting it into the form that you can find in Laplace transform table, it follows that

VC(s) =V1

RC

s

s + 1RC

Taking the inverse transform produces

VC(t) =V

1 e tRC

Example 4Solve the differential equation

CdTdt

+ 1R

T= 1R

TL, T(0) = 0

which is the mathematical model for the thermal system shown in the figure to theleft.Solution: TakingLtransform gives

CsT(s) + 1

RT(s) =

1

RTL(s)

Solving forT(s) produces

T(s) =1R

Cs + 1R

TL(s) = 1

RCs + 1TL(s)

Case 1 Unit-impulse responseTL(t) =(t) = TL(s) = 1. The temperature in s-domain is given byT(s) =

1

s+ 1=

1

s + 1

with =RC. TakingL1 transform yieldsT(t) =

1

e

t

Case 2 Unit-step responseTL(t) = 1(t) = TL(s) = 1s . The temperature in s-domain is given by

T(s) = 1

s+ 1

1

s =

1

s(s + 1

)


14/69

10 CHAPTER 2. LAPLACE TRANSFORM

TakingL1 transform givesT(t) = 1 e t

Case 3Unit ramp response TL(t) =t = TL(s) = 1s2 . The temperature in s-domain is given by

T(s) = 1

s+ 1

1

s2 =

1

s(s + 1

)

TakingL1 transform producesT(t) = t (1 e

t )

1


15/69

Chapter 3

Mathematical Models: Transfer Function

Representation

Suppose we have a system where the inputs u is related to the output y by the differential equation:

a2d2ydt2

+ a1dydt

+ a0y = b1dudt

+ b0u

wherea2, a1, a0, b1,and b0 are constants. If all the initial conditions are zero, then the Laplace transform of this equation is

a2s2Y(s) + a1sY(s) + a0Y(s) =b1sU(s) + b0U(s)

Solving forY(s) gives

Y(s) = b1sU(s) + b0U(s)

a2s2 + a1s +a0=

b1s+ b0a2s2 + a1s + a0

U(s)

Hence,

Y(s)

U(s) = b

1s+ b

0a2s2 + a1s + a0

This is the transfer function of the system. The transfer functionG(s) of a linear system is defined as the ratio of the Laplacetransform of the output variable Y(s) to the Laplace transform of the input variableU(s),with all initial conditions assumedto be zero.

G(s) =Y(s)

U(s)

For the system described by the differential equation above

G(s) =Y(s)

U(s)=

b1s + b0a2s2 + a1s+ a0

3.1 Transfer Function for the 1st Order System

General Form

Differential Equation a1dydt

+ a0y= b0u zero initial conditionsLaplace Tansform a1sY(s) + a0Y(s) =b0U(s)

Transfer Function G(s) = Y(s)U(s) =

b0a1s+a0

Standard Form

Differential Equation dydt

+ y= Ku zero initial conditionsLaplace Tansform sY(s) + Y(s) =KU(s)


Ks+1

11


16/69

12 CHAPTER 3. MATHEMATICAL MODELS: TRANSFER FUNCTION REPRESENTATION

3.2 Transfer function for the 2nd order system:

General Form

Differential Equation a2d2ydt2

+ a1dydt

+ a0y= b0u zero initial conditionsLaplace Tansform a2s2Y(s) +a1sY(s) + a0Y(s) =b0U(s)


b0a2s2+a1s+a0

Standard Form

Differential Equation d2y

dt2 + 2n

dydt

+ 2ny= K2nu zero initial conditions

Laplace Tansform s2

Y(s) + 2nsY(s) + 2

nY(s) =K 2

nU(s)Transfer Function G(s) =

Y(s)U(s) =

K2ns2+2ns+2n

3.3 Examples

Example 1Find the transfer function for the differential equation

RCdvCdt

+ vC=v

which is a model for the RC circuit to the left.Solution: Taking Laplace transform on both sides, with zero initial conditions,gives

RCsVC(s) + VC(s) = V(s)

Solving forVC(s) yields

VC(s) = V(s)

RCs + 1

Then the transfer function is given by

G(s) =VC(s)

V(s) =

1

RCs + 1


L

R

diLdt

+ iL= v

which is a model for the RL circuit to the left.Solution: Taking Laplace transform on both sides, with zero initial conditions,gives

L

RsIL(s) + IL(s) =V(s)

Solving forIL(s) produces

IL(s) = 1LR

s+ 1V(s)

which implies that

G(s) = IL(s)

V(s) =

1LR

s+ 1


RCd2vCdt2

+dvC

dt +

R

LvC=

dv

dt

which is a model for the RLC circuit to the left.Solution: Taking Laplace transform on both sides, with zero initialconditions, gives

RCs2VC(s) + sVC(s) +R

LVC(s) = sV(s)


17/69

3.3. EXAMPLES 13

Solving forVC(s) produces

VC(s) = s

RCs2 + s+ RL

V(s)

which implies that

G(s) =IL(s)

V(s) =

s

RCs2 + s + RL

Example 4 Find the transfer function for the differential equatio

LaJ

K4

d2

dt2 +

RaJ+ Lac

K4

d

dt +

Rac + K3K4K4

= va RaK4

L LK

which is a model for the armature controlled dc motor shown tthe left.Solution: Taking Laplace transform on both sides, with zeinitial conditions, gives

LaJ

K4

s2(s) +RaJ+ Lac

K4

s(s) +Rac + K3K4

K4

(s)

= Va(s) RaK4

L(s) s LaK4

L(s)

Solving this equation for (s) produces

(s) = K4

LaJs2 + (RaJ+ Lac)s + Rac+ K3K4Va(s) Las + Ra

LaJs2 + (RaJ+ Lac)s + Rac+ K3K4L(s)

= K4

(Las + Ra)(Js+ c) + K3K4Va(s) Las + Ra

(Las + Ra)(Js+ c) + K3K4L(s)


CdT

dt +

1

RT=

1

RTL

which is the mathematical model for the thermal system shown in the figure to theleft.Solution: TakingLtransform gives

CsT(s) + 1

RT(s) =

1

RTL(s)

Solving forT(s) produces

T(s) =1R

Cs + 1R

TL(s) = 1

RCs + 1TL(s)

Then, the transfer function is given by

G(s) = T(s)TL(s)= 1RCs + 1


18/69

14 CHAPTER 3. MATHEMATICAL MODELS: TRANSFER FUNCTION REPRESENTATION


19/69

Chapter 4

Mathematical Models: Block Diagrams

Consider a transfer functionG(s) = Y(s)U(s) withU(s) andY(s) being its input and output. It is obvious thatY(s) =G(s)U(s),

which can be represented by a block diagram as shown in the figure below.

Arrows are used to represent the directions of signal flow.

Comparator can be represented by a summing point.

Where a signal is taken off from some point in a signal path, the take-off point is used.

Closed-loop system

Forward path is used for those elements through whichsignal passes when moving in the direction from input

output for a system as a whole, as shown in the figure tthe left. The transfer function of the forward path is callethe feedforward transfer function.Feedback path is used for those elements through whichsignal passes when being fed back from the output towarthe input, as shown in the figure to the left. The transffunction of the feedback path is called the feedback transffunction.

Series Connection: The transfer function of a system composed of several subsystems connected in series is the productof transfer functions of individual subsystems, as shown in the figure below.

Parallel Connection: The transfer function of the system composed of two systems connected in parallel, as shown inthe figure below, is the sum of transfer functions of individual subsystems.

15


20/69

16 CHAPTER 4. MATHEMATICAL MODELS: BLOCK DIAGRAMS

Feedback Connection: The open-loop transfer function is defined as the ratio of the feedback signal Uf(s) to theactuating error signalU(s) Uf(s), which is given by

Go(s) = Uf(s)

U(s) Uf(s) =G(s)H(s)

The closed-loop transfer function is defined as the ratio of the output signal Y(s) to the input signal X(s), which is derivedas follows:

Y(s) = G(s)X(s)

= G(s)[U(s) Uf(s)]= G(s)[U(s) H(s)Y(s)]= G(s)U(s) G(s)H(s)Y(s)

which implies that

[1 + G(s)H(s)]Y(s) =G(s)U(s)

Therefore, the transfer function is

Gc(s) = G(s)

1 + G(s)H(s)

Example 1: Find transfer functions for the systems shown in thfigure to the left.Solution: The transfer function for the first system is

G(s) = G1(s) + G2(s) = 1

s + 1+ 5 =

1

s+ 1+

5(s+ 1)

s + 1 =

5s +

s +

The transfer function for the second system is

Gc(s) = G(s)

1 G(s)H(s) =2

s+1

1 2s+1 5s

=2

s+1s+110s

s+1

= 2

9s + 1Example 2 Armature controlled dc motor


21/69

17

The equivalent circuit for the armature controlled dc motor shown in the figure to the left. The system can be divided infour parts, armature winding,Ea versus,d versusia, and loaThe transfer functions for all parts are derived below.Armature Winding: It follows from Kirchhoffs voltage la

thatva ea= Raia+ La dia

dt

Taking Laplace transform on both sides gives

Va(s) Ea(s) =RaIa(s) + LasIa(s)

that is,

Ia(s)

Va(s) Ea(s) = 1

Las+ Ra

Ea versus : It is obvious thatea= K3

Taking Laplace transform on both sides givesEa(s) = K3(s)

that is,Ea(s)

(s) =K3

d versus ia: It is easily seen thatd = K4ia

Taking Laplace transform on both sides givesd(s) = K4Ia(s)

that is,

d(s)Ia(s)

=K4

Load: It follows from Newtons second law that

d L f =Jddt

Replacingf byc , we get

d L c= Jddt

which is equivalent to

d L = Jddt

+ c

Taking Laplace transform on both sides gives


22/69


d(s) L(s) = J s(s) + c(s)that is,

Ia(s)

d(s) L(s) = 1

Js + c

These blocks are shown in the figure in previous page.

Now we can put all four block diagrams together to create a block diagram for the armature controlled dc motor, whichis shown in the figure below.

In the following, we will determine the transfer function Ga(s) from Va(s) to (s) and the transfer function G(s) fromL(s) to (s). Ga(s) can be determined by letting L(s) = 0. The process for determining Ga(s) is shown in the figurebelow, from which we get

Ga(s) = K4

(Las + Ra)(Js+ c) + K3K4


23/69

19

G(s) can be determined by letting Va(s) = 0. The process for determining Ga(s) is shown in the figure below, fromwhich we get

G(s) = Las+ Ra

(Las+ Ra)(Js+ c) + K3K4

The output from the armature controlled dc motor is given by

(s) = Ga(s)Va(s) + G(s)[L(s)] = K4(Las+ Ra)(Js+ c) +K3K4

Va(s) Las+ Ra(Las+ Ra)(Js+ c) + K3K4

L(s)


24/69



25/69

Chapter 5

Dynamic Responses

There are four kinds of responses, namely impulse response, step response, ramp response, and frequency response, whichare commonly used in control systems. The impulse response is the output of the system corresponding to an impulse input.The step response is the output of the system corresponding to a step input. The ramp response is the output of the systemcorresponding to a ramp input. The frequency response is the output of the system corresponding to a sinusoidal input.

5.1 First Order Systems

A first-order system can be described by a differential equation or a transfer function as shown below.

= a1a0

, K= b0a0

a1dydt

+ a0y= b0u dydt + y = KuL1 L L1 LG(s) = b0

a1s+a0 G(s) = K

s+1

a1 = , a0 = 1, b0 = K

5.1.1 Impulse Response

Inputu = A(t) or U(s) =AOutput in s-domain is given by

Y(s) =G(s)U(s) = KA

s+ 1

Output in t-domain is given by (taking inverse Laplace transform)

y(t) =KA

e

t

5.1.2 Step Response

Inputu = A 1(t) or U(s) = AsOutput in s-domain is given by

Y(s) =G(s)U(s) = KA

s( s+ 1)=K A

1

s(s + 1

)


y(t) = KA(1 e t ) = KA KAe t

where the first term is called the forced response or steady-state response, which is caused by the input, whereas the secondterm is referred to as transient response or dynamic response, which represents the transient process experienced by thesystem after an input is applied.

21


26/69

22 CHAPTER 5. DYNAMIC RESPONSES

Kis called a steady-state gain, which is the ratio between the steady-state output and input. is referred to as a timeconstant, which shows how fast the system responds to the step input. The output will reach to 60% of its steady-state valueat onemoment, 95% at 3 times .

5.1.3 Ramp Response

Inputu = At or U(s) =

A

s2

Output in s-domain is given by

Y(s) =G(s)U(s) = AK

s2( s+ 1)=K A

1

s2(s + 1

)


y(t) =K A

t

1 e t

5.1.4 Example

Example: A thermocouple has the dynamics of

10dTodt

+ To= 30 106Ti

with output To in volt and input Ti inC. Answer the following questions:

1. Transfer function.

2. Time constant and steady-state gain.

3. The final steady-state value when there is a step input of 100C.

4. What will be the output of the thermocouple 5s after it was subject to a temperature impulse of 100C by suddenlyand very briefly coming into contact with a hot subject of 100C.

5. When the thermocouple is subject to a steadily rising temperature input of 5C/s, what will be the thermocoupleoutput after 12s.

Solution:

1. Taking Laplace transform on both sides of the differential equation gives

10sTo(s) + To(s) = 30 106Ti(s)

Solving this equation for To(s) produces

To(s) =30 106

10s + 1 Ti(s)

which implies that the transfer function is

G(s) =30 106

10s + 1

2. Since G(s) is in the standard form, = 10sand K= 30 106V /C.

3. The input is Ti(t) = 100 1(t) and its Laplace transform is Ti(s) = 100s . It follows from the final value theorem that

To() = lims0

sTo(s) = lims0

sG(s)Ti(s) = lims0

s30 106

10s + 1

100

s = lim

s03000 106

10s + 1 = 3000 106V


27/69

5.2. SECOND-ORDER SYSTEMS 23

4. The input is Ti(t) = 100(t) and its Laplace transform is Ti(s) = 100. Then, the output in s-domain is

To(s) =G(s)Ti(s) =30 106

10s + 1 100 =3000 10

6

10s + 1 = 3000 106

110

s + 110

Taking inverse Laplace transform gives

To(t) =

3000

106

10 e t10

ThereforeTo(t) at t=5s is given by

To(5) = 3000 106

10 e

5

10 = 1.8 104V

5. The input is Ti(t) = 5tand its Laplace transform is Ti(s) = 5s2

. Then, the output in s-domain is

To(s) =G(s)Ti(s) =30 106

10s + 1

5

s2 =

150 106s2(10s + 1)

= 150 106110

s2(s + 110 )

Taking inverse Laplace transform gives

To(t) = 150 106

t 10

1 e t10

ThereforeTo(t) at t=12s is given by

To(12) = 150 106

12 10

1 e1210

= 7.5 104V

5.2 Second-Order Systems

A second-order system can be described by a differential equation or a transfer function as shown below.

2n = a1a2

, 2n= a0a2

, K2n = b0a2

a2d2ydt2

+ a1dydt

+a0y= b0u d2y

dt2 + 2n

dydt

+ 2ny= K2nu

L1 L L1 LG(s) = b0

a2s2+a1s+a0 G(s) = K2n

s2+2ns+2na2= 1, a1 = 2n, a0 =

2n, b0 = K

2n

The equation obtained by letting the denominator of a transfer function equal to zero is called the characteristic equation.For the second order system, the characteristic equation is as follows:

s2 + 2ns + 2n = 0

Roots of this characteristic equation are

s1,2 =2n

(2n)2 42n2

(5.1)

=2n

422n 42n2

(5.2)

=n n

2 1 (5.3)

Case 1: >1. The system is called overdamped. In this case, the characteristic equation has two different real roots.Case 2: = 1. the system is called critically damped. In this case, the characteristic equation has two equal real roots.Case 3:


28/69


29/69

5.2. SECOND-ORDER SYSTEMS 25

Now let a = s1= n. Then,y(t) =K

1 ent ntent

Case 3:


30/69


The percent overshoot is

%Overshoot = e

12 100% =e 0.4410.442 100% = 21%

The 5% settling time is

ts = 3

n=

3

0.44 2 = 3.4s

Example 2An armature controlled dc motor speed system is described by the following differential equation:

d2

dt2 + 5

d

dt+ 100= 100va

What is the 5% settling time for the system?

Solution: Comparing the model above with the standard form

d2

dt2 + 2n

d

dt + 2n= K

2nva

produces

2n= 100 n = 102n = 5 = 5

2n= 0.25

K2n = 100 K=100

2n= 1

Thus, the 5% settling time is

ts= 3

n=

3

0.25 10= 1.2s

Example 3A second-order system has an overshoot of 10% and a rise time of 0.4s when subject to a step input. Findthe damping ratio, damped frequency and natural frequency.

Solution: It follows from the definition of the percent overshoot that

e

12 100% = 10%

that is

e

12 = 0.1 1 2 = ln 0

.1 = 2.3

Multiplying both sides by

1 2 gives

= 2.3

1 2 22 = 2.32(1 2)

Solving foryields

= 2.32 + 2.32

= 0.6

Since the rise time is given by tr = 2d

, it is obtained that

d=

2tr=

2 0.4= 3.9rad/s

It follows from d= n

1 2 thatn =

d1 2 =

3.91 0.62 = 4.9rad/s

Example 4A second-order system has a damping ratio of 0.4, a natural frequency of 10rad/s and a steady-state gain of10. Find the transfer function, percent overshoot, rise time, and unit step response.


31/69


32/69


Example 6A robot arm has a transfer function of

G(s) = 9

(s+ 3)2

Find the unit ramp response.

Solution: The input is u(t) = t and its Laplace transform is U(s) = 9s2

. Hence, the output in s-domain is

Y(s) =G(s)U(s) = 9

s2(s+ 3)2

Carrying out partial fraction expansion, we have

9

s2(s+ 3)2 =

A

s +

B

s2+

C

s+ 3+

D

(s + 3)2

Multiplying both sides by s2(s + 3)2 gives

9 =As(s + 3)2 + B(s+ 3)2 + Cs2(s + 3) + Ds2

Let s = 3. Then, 9 = 9D, that is D = 1.Let s = 0. Then, 9 = 9B, that is, B = 1.Let s = 1 and s = 1, respectively. Then, we get

9 = 16A + 16B+ 4C+ D

9 =4A + 4B+ 2C+ D

Replacing B and D by 1 produces

8 = 16A + 4C 2 = 4A + C4 =4A + 2C 2 = 2A + C

Subtracting the second equation from the first equation gives 4 = 6A, that isA = 23 . It follows from the second equationthat C= 2 + 2A= 2 43 = 23 . As a result, we have

Y(s) = 2

3

s +

1

s2+

23

s+ 3+

1

(s + 3)2

By taking inverse Laplace transform, the output in t-domain is given by

y(t) = 23

1(t) + t+2

3e3t + te3t


33/69

Chapter 6

Steady-State Errors

6.1 Type of System

Consider a closed-loop system with unity feedback described in the figure below.

Note that the open-loop transfer function of the closed-loop system is Go(s), which is the same as the feedforward transferfunction. Generally speaking, annth order system can be put into the form of

Go(s) = K(sm + bm1sm1 + + b1s+ b0)

sq(snq + anq1snq1 + + a1s + a0)The integer qis called the type or class of the system. The closed-loop transfer function is given by

Gc(s) = Go(s)

1 + Go(s)

ExampleIdentify the types of the closed-loop systems with unity feedback, which have the following open-loop transferfunctions:

5s+2 Type 02(s+1)

s2+2s+1 Type 06

s(s+3) Type 12

s2+4s = 2s(s+4) Type 1

2(s+3)s2(s2+2s+1) Type 2

29


34/69

30 CHAPTER 6. STEADY-STATE ERRORS

The types of general closed-loop systems, as shown in the figure below, can be identified by the following procedure.

1. Convert the system to one with unity feedback as shown in the figure above.

2. Calculate the open-loop transfer function for the equivalent closed-loop system with unity feedback.

3. Convert the open-loop transfer function into the form as in the definition of the type of a system

ExampleSee the figures on the next page.

6.2 Steady-State Errors

The following constants are useful for the calculation of steady-state errors.The proportional error constant kp is defined as

kp = lims0

Go(s)

The velocity error constant kv is defined as

kv = lims0 sGo(s)

The acceleration error constant ka is defined as

ka = lims0

s2Go(s)

It is not difficult to verify the table below.

Type 0 Type 1 Type 2

kp Kb0a0

kv 0 K

b0a0

ka 0 0 K

b0a0


35/69

6.2. STEADY-STATE ERRORS 31


36/69


Steady-state error is defined as

ess(t) = limt

e(t) = lims

0sE(s)

wheree(t) =u(t) y(t) and E(s) =U(s) Y(s) with input u(t) or U(s) and output y(t) or Y(s).

6.2.1 Steady-State Errors for Closed-Loop Systems with Unity Feedback

In the closed-loop system with unity feedback as shown in the figure below, the error in s-domain is given by

E(s) =U(s) Y(s) =U(s) Go(s)E(s)

which implies that

E(s) = 1

1 + Go(s)U(s)


37/69


38/69


ExampleSee the figure below.

Go(s) = 6(s + 3)

s3 + 5s2 12kp = lim

s0Go(s) = lim

s06(s + 3)

s3 + 5s2 12= 18

12= 3

2

kv = lims0

sGo(s) = lims0

6s(s + 3)

s3 + 5s2 12= 0

12= 0

ka = lims0

s2Go(s) = lims0

6s2(s+ 3)

s3 + 5s2 12= 0

12= 0

ess =

11+kp

= 11 3

2

= 2 U(s) = 1s

1kv

= U(s) = 1s2

1ka = U(s) = 1s3


39/69


It is important to note that steady-state errors can also be calculated directly, as shown below.

Gc(s) = Y(s)

U(s) =

6s(s+2)

1 + 6s(s+2)

1s+3

= 6(s + 3)

s(s + 2)(s+ 3) + 6=

6s + 18

s3 + 5s2 + 6s + 6

E(s) = U(s) Y(s) =U(s) Gc(s)U(s) = [1 Gc(s)]U(s) = 1 6s + 18s3 + 5s2 + 6s + 6U(s)=

s3 + 5s2 + 6s + 6 6s 18s3 + 5s2 + 6s + 6

U(s) = s3 + 5s2 12s3 + 5s2 + 6s + 6

U(s)

ess = lims0

sE(s)

=

lims0 s s3+5s212

s(s3+5s2+6s+6) U(s) = 1

s

lims0 s s3+5s212

s2(s3+5s2+6s+6) U(s) = 1s2

lims0 s s3+5s212

s3(s3+5s2+6s+6) U(s) = 1

s3

=

126 = 2 U(s) = 1s120 = U(s) = 1s2120 =

U(s) = 1

s3

6.2.3 Steady-State Errors for Systems with Disturbances

The direct method is used to calculate steady-state errors for systems with disturbances as shown in the figure below.

To this end, the output in s-domain is computed first.

Y(s) = G2(s)[Ud(s) + X1(s)] =G2(s)[Ud(s) + G1(s)X2(s)]

= G2(s){Ud(s) + G1(s)[Ui(s) X3(s)]} =G2(s){Ud(s) + G1(s)[Ui(s) H(s)Y(s)]}= G2(s)Ud(s) + G2(s)G1(s)Ui(s) G2(s)G1(s)H(s)Y(s)

from which one gets

Y(s) = G2(s)

1 + G2(s)G1(s)H(s)Ud(s) +

G2(s)G1(s)

1 + G2(s)G1(s)H(s)Ui(s)

The error in s-domain is given by

E(s) = Ui(s) Y(s) =Ui(s)

G2(s)

1 + G2(s)G1(s)H(s)Ud(s) +

G2(s)G1(s)

1 + G2(s)G1(s)H(s)Ui(s)

= 1 + G2(s)G1(s)H(s) G2(s)G1(s)

1 + G2(s)G1(s)H(s) Ui(s) G2(s)

1 + G2(s)G1(s)H(s)Ud(s)

The steady-state error is ess= lims0 sE(s).


40/69


ExampleIn the armature controlled dc motor as shown in the figure below, Va(s) = 1s

and L(s) = 5s

. Find ess|L=0,ess|Va=0, ess.

Solution: First let G1(s) = 500.1s+1 , G2(s) =

1s+1 , and H(s) = 0.2. Then we have

1 + G2(s)G1(s)H(s) = 1 + 1

s + 1

50

0.1s + 1 0.2

= (0.1s + 1)(s+ 1) + 10

(0.1s + 1)(s + 1) =

0.1s2 + 1.1s + 11

(0.1s + 1)(s + 1)

1 + G2(s)G1(s)H(s) G2(s)G1(s) = 0.1s2 + 1.1s + 11

(0.1s + 1)(s+ 1) 1

s+ 1

50

0.1s + 1

= 0.1s2 + 1.1s 39

(0.1s + 1)(s+ 1)

1 + G2(s)G1(s)H(s) G2(s)G1(s)1 + G2(s)G1(s)H(s)

=

0.1s2+1.1s39(0.1s+1)(s+1)

0.1s2+1.1s+11(0.1s+1)(s+1)

= 0.1s2 + 1.1s 390.1s2 + 1.1s + 11

G2(s)

1 + G2(s)G1(s)H(s) =

1s+1

0.1s2+1.1s+11(0.1s+1)(s+1)

= 0.1s + 1

0.1s2 + 1.1s + 11

E(s) = 1 + G2(s)G1(s)H(s) G2(s)G1(s)

1 + G2(s)G1(s)H(s) Va(s) G2(s)

1 + G2(s)G1(s)H(s)[L(s)]

= 1 + G2(s)G1(s)H(s) G2(s)G1(s)

1 + G2(s)G1(s)H(s) Va(s) +

G2(s)

1 + G2(s)G1(s)H(s)L(s)

E(s) |L=0 = 1 + G2(s)G1(s)H(s) G2(s)G1(s)

1 + G2(s)G1(s)H(s) Va(s)

= 0.1s2 + 1.1s 39s(0.1s2 + 1.1s + 11)

ess|L=0 = lims0 sE(s) |L=0= lims00.1s2 + 1.1s

39

0.1s2 + 1.1s + 11= 39

11

E(s) |Va=0 = G2(s)

1 + G2(s)G1(s)H(s)L(s)

= 5(0.1s + 1)

s(0.1s2 + 1.1s + 11)

ess|Va=0 = lims0

sE(s) |Va=0= lims0

5(0.1s + 1)

0.1s2 + 1.1s + 11=

5

11

E(s) = 1 + G2(s)G1(s)H(s) G2(s)G1(s)

1 + G2(s)G1(s)H(s) Va(s) +

G2(s)

1 + G2(s)G1(s)H(s)L(s)

= E(s) |L=0 +E(s) |Va=0


41/69


Therefore, we have

ess= lims0

sE(s) = lims0

sE(s) |L=0 + lims0

sE(s) |Va=0=ess|L=0 +ess|Va=0= 39

11+

5

11=

3411


42/69



43/69

Chapter 7

The Routh Stability Criterion

7.1 Poles and Zeros

In general, the transfer function G(s) of a system can be represented by

G(s) = K(sm + bm1sm1 + + b1s + b0)

sn + bn1sn1 + + a1s + a0whereKis the gain of the system.

Now assume that z1, z2, , zm are roots of the numerator, and p1, p2, , pn are roots of the denominator. Then, z1,z2, ,zm are called zeros of the system, and p1, p2, , pn are referred to as poles of the system. In addition, the transferfunctionG(s) can be written as

G(s) =K(s z1)(s z2) (s zm)

(s p1)(s p2) (s pn)We have the following observations:

1. The zeros are the values of s for which the transfer function becomes zero and the poles are the values of s for which

the transfer function is infinite.2. The transfer function can be specified by the values of zeros, poles, and gain.

The poles and zeros of a transfer function are complex numbers, so can be represented by points on the complex plane,which results in a pole-zero plot. In a pole-zero plot, a pole is marked with a cross and a zero is marked with a circle.

7.2 Stability

A system is called to be stable if the output dies away as t ; unstable if the output tends to infinity ast ; criticallystable if the output does not die away or increase to infinity but tends to some finite but non-zero values when subject to animpulse input.

ExampleCheck the stability of the systems described by

G1(s) = 1

s + 2 G2(s) =

1

s 2Solution: Let input be a unit impulse function, that is, U(s) = 1. Then the output in s-domain is

Y1(s) =G1(s)U(s) = 1

s+ 2, Y2(s) =G2(s)U(s) =

1

s 2Taking the inverse Laplace transform gives

y1(t) = e2t 0, t

y2(t) = e2t , t

39


44/69

40 CHAPTER 7. THE ROUTH STABILITY CRITERION

So, G1(s) is stable andG2(s) unstable.

Stability Test:

1. A system is stable if all the poles are in the open left-hand side of the s-plane, that is, all the poles have negative realparts.

2. A system is unstable if one or more poles are in the open right-hand side of the s-plane, that is, one or more poles have

positive real parts.3. A system is critically stable if one or more poles are on the imaginary axis and the rest are in the open left-hand side

of the s-plane, that is one or more poles have zero parts and the rest have negative real parts.

Example 1Consider the following systems

(a) s 1s2 4s + 4 (b)

2(s + 1)

(s+ 1)(s + 2)(s 3) (c) (s+ 3)(s 1)

s(s+ 2)(s + 3)(s+ 4) (d)

s+ 4

s2 + s + 3 (e)

1

s2 + s + 1

Find poles and zeros for the systems, plot them in s-plane, and check the stability.

Solution

(a) s1s24s+4 =

s1(s2)2 . p1 = 2, p2 = 2, z1 = 1. It is unstable.

(b) p1= 1, p2 = 2, p3 = 3, z1 = 1. It is unstable.(c) p1= 0, p2 = 2, p3 = 3, p4 = 4, z1 = 3, z2 = 1. It is stable.(d) s+4

s2+s+3 = s+4

[s( 12+j1

2

11)][s( 12j

1

2

11)]

. p1 = 12 +j 12

11, p2 = 12j 12

11, z1 = 4. It is stable.

(e) 1s2+s+1 =

1

[s( 12+j 123)][s( 12j 12

3)]

. p1 = 12 + j 12

3, p2= 12j 12

3. It is stable.

Example 2Find transfer functions and check stabilities for the systems

(a) poles: -1, -2; zeros: no; gain: 2;

(b) poles: 1, -2; zeros: 0; gain: 10;

(c) poles: -2+j1, -2-j1; zeros: 1; gain: -5;

(d) poles: 1+j2, 1-j2; zeros: -1; gain: 7.

Solution:(a) 2(s+1)(s+2) is stable.

(b) 10s(s1)(s+2) is unstable.

(c) 5(s1)

[s(2+j)][s(2j)] = 5(s1)(s+2)2+1 =

5(s1)s2+4s+5 is stable.

(d) 7(s+1)[s(1+j2)][s(1j2)] = 7(s+1)(s1)2+4 =

7(s+1)s22s+5 is unstable.

7.3 The Routh Stability Criterion

We have discussed that the stability of a system can be determined by checking the positions of poles in the s-plane, which

means that we need to find roots of the denominator of a transfer function. In general, it is not a easy thing to find roots ofthe polynomialans

n + an1sn1 + an2s

n2 + + a1s + a0with n >3.

The Routh criterion can be used to determine the stability of a high order system without finding roots of its denominator.The procedure is given below.

1. Write the characteristic equation in the form of

ansn + an1s

n1 + an2sn2 + + a1s + a0

with an > 0


45/69

7.3. THE ROUTH STABILITY CRITERION 41

2. If any of the coefficients are zero or negative, then there is a root or roots that are imaginary or have positive realparts, that is, the system is either critically stable or unstable.

3. If all coefficients are positive, arrange the coefficients of the polynomial according to following pattern.

sn an an2 an4 sn1 an1 an3 an5 sn2 b1 b2 b3

sn3 c1 c2 c3 ...

......

...s2 x1 x2 x3s1 y1 y2s0 z1

where

b1 = an2

anan1

an3

b2 = an4

anan1

an5

c1 = an3

an1b1

b2

c2 = an5

an1b1

b3

Routh stability criterion states that the number of roots of the characteristic equation with positive real parts is equalto the number of changes in sign of the coefficients of the first column of the Routh array. If all elements in the firstcolumn are positive, then the system is stable. If there are negative elements in the first column, then the system isunstable.

Example 1Check the stability ofG(s) = 2s+1s4+2s3+3s2+4s+1 .

Solution:

1. All coefficients of the characteristic equation are positive.

2. Routh array is given below.s4 1 3 1s3 2 4s2 1

= 3 12 4

1

= 1 12 0

s1 2

= 4 21 1

s0 1

= 1 12 0

All the elements in the first column are positive, so the system is stable.

Example 2Check the stability ofG(s) = 2s+1s4+s3+s2+4s+1 .

Solution:

1. All coefficients of the characteristic equation are positive.

2. Routh array is given below.s4 1 1 1s3 1 4s2 3 = 1 11 4 1 = 1 11 0s1 13

3

= 4 13 1

s0 1

= 1 313

3

0

The first column has a negative element, so the system is unstable.


46/69

42 CHAPTER 7. THE ROUTH STABILITY CRITERION

Example 3The systemG(s) = 2s+1s54s4+3s3+2s2+5s+2 is unstable because there is a negative coefficient in the denominator.

Example 4Find k so that the system with G1(s) = 10s+1 andG1(s) =

1s(s+4) , as shown in the figure below is stable.

Solution:

1. The closed-loop transfer function is computed below.

G(s) = G(s)

1 + G(s)H(s)=

10s+1

1s(s+4)

1 + 10s+1

1s(s+4)

k

= 10

(s+ 1)s(s + 4) + 10k =

10

s(s2 + 5s + 4) + 10k

= 10

s3

+ 5s2

+ 4s + 10k

2. It follows from Routh-Hurwitz criterion that 10k >0, that is k >0.

3. Routh array is given below.s3 1 4s2 5 10ks1 4 2k = 4 15 10ks0 10k

= 10k 542k 0

It follows from Routh-Hurwitz criterion that 4 2k >0, that is, k 0, that is k >0. So 0 < k


47/69

Chapter 8

Root Locus Analysis

8.1 Definition

Consider a closed-loop system described by the following block diagram.

The open-loop transfer function for the system is Go(s) = ks+1 with pole s = 1. The closed-loop transfer function is

G(s) = Go(s)

1 + Go(s)=

ks+1

1 + ks+1

= k

s+ 1 + k

with pole(1 + k).It is evident that the closed-loop pole changes as k changes. For instance, the closed-loop pole is -1, which is the same as

the open-loop pole, when k = 0, -2 for k = 1, -3 fork = 2, and so on. A curve of the closed-loop pole versusk for k >0 canbe plotted in s-plane. Such a curve is called the root locus, which is shown below.

Now consider a second-order system with the open-loop transfer function Go(s) = k

s2+4s+1 with open-loop poles 2 +

3

and2 3. Its closed-loop transfer function is computed below.

G(s) = Go(s)

1 + Go(s)=

ks2+4s+1

1 + ks2+4s+1

= k

s2 + 4s + 1 + k

43


48/69

44 CHAPTER 8. ROOT LOCUS ANALYSIS

The closed-loop poles, that is, roots of the characteristic equation s2 + 4s + 1 + k= 0, are

p=4

16 4(1+ k)

2 = 2

4 (1 + k) = 2

3 k

As k = 0, the closed-loop poles are2 + 3 and2 3, which are the open-loop poles. The system is overdamped.As 0< k 3, the closed-loop poles are2 +jk 3 and2 jk 3. The system is underdamped.The root locus plot for this system looks like the graph shown above.

From these two example, we can make the following observations.

1. The root locus is a locus of the closed-loop poles in s-plane as the gain increases from 0.

2. The root locus starts at the open-loop poles.

8.2 Construction

Now consider a closed-loop system with unity feedback. Assume that the open-loop transfer function is given by

Go(s) =k(s z1)(s z2) (s zm)

(s p1)(s p2) (s pn)with open-loop gaink, open-loop zerosz1, z2, , zm, and open-loop poles p1, p2, , pn. Its closed-loop transfer function isgiven by

G(s) = Go(s)

1 +Go(s)

The closed-loop characteristic equation is

1 + Go(s) = 1 +k(s z1)(s z2) (s zm)

(s p1)(s p2) (s pn) = 0

that is,k(s z1)(s z2) (s zm)(s p1)(s p2) (s pn) = 1 = 1(odd multiple of)

which implies that the phase condition

[(s z1) + (s z2) + + (s zm)] [(s p1) + (s p2) + + (s pn)] = odd multiple of

and magnitude condition| (s z1) || (s z2) | | (s zm) || (s p1) || (s p2) | | (s pn) | = 1

must be satisfied.

Construction of Root Loci

1. The number of root loci is equal to the degree n of the open-loop characteristic equation or the order of the closed-looptransfer function.

2. The root loci start at the open-loop poles and end at the open-loop zeros. If there are more poles than zeros, then mloci end at the open-loop zeros and the remaining (n-m) loci end at infinity.

3. The root loci are symmetrical with respect to the real axis.

4. Portions of the real axis are sections of root loci if the number of poles and zeros lying on the axis to the right of theportion is odd.


49/69

8.2. CONSTRUCTION 45

5. Those loci terminating at infinity tend towards asymptotes at angles to the positive real axis of

n m , 3

n m , 5

n m , ,[2(n m) 1]

n m6. The asymptotes intersect the real axis at a point, called the center of gravity or centroid of the asymptotes, given by

(p1+p2+ +pn) (z1+ z2+ + zn)n m

7. The intersection of root loci with the imaginary axis can be found by calculating those values ofk which result in theexistence of imaginary roots, that is, solutions to the closed-loop characteristic equation with s =j.

8. The term breakaway point is used for the case that two or more loci meet at a point and then subsequently breakaway from that point along separate paths. The breakaway point can be found by solving the closed-loop characteristicequation for k and solving

dk

ds = 0

for s.

9. The angle of departure of a locus at k = 0 from a complex pole and the angle of arrival of a locus at k = at acomplex zero can be determined by using the phase condition.

Example 1Draw root loci for the unity feedback system with open-loop transfer function

Go(s) = k(s+ 1)

(s+ 2 +j3)(s + 2 j3)

Solution:

1. Find the closed-loop characteristic equation:

G(s) = Go(s)

1 + Go(s)=

k(s+1)(s+2+j3)(s+2j3)

1 + k(s+1)

(s+2+j3)(s+2j3)=

k(s + 1)

(s+ 2 +j3)(s + 2 j3) + k(s + 1)

which means the closed-loop characteristic equation is

(s + 2 +j3)(s+ 2 j3) + k(s + 1) = 0

2. Since Go(s) is of second order, there are two loci.

3. Open-loop poles are2 +j3 and2 j3 and zeros are -1.4. The portion of the root loci on the real axis is from -1 tobecause to its right there is only one zero.5. The angle of the asymptote is

nm = 21 = = 180

, which implies that the asymptote is the real axis.

6. Since the asymptote is the real axis, the intersection with the real axis has no significance.

7. The intersection of loci with the imaginary axis can be determined by solving the characteristic equation withs = jfor k and , that is

(j + 2 +j3)(j + 2 j3) + k(j + 1) = 0(j + 2)2 (j3)2 +jk+ k = 0

(j)2 + 4j + 4 + 9 + jk+ k =

2 +j(4 + k)+ 13 + k = 0which means that

(4 + k) = 0

2 + 13 + k = 0


50/69


Solving these equations for k and gives k = 4 and = 13 + k = 13 4 = 3. In this course, the root lociare restricted to the case with positive k . So a negativek means there are no intersections between the root loci andthe imaginary axis, that is, the root loci does not cross the imaginary axis.

8. The arrival point is determined by solving

(s + 2 +j3)(s+ 2 j3) + k(s + 1) = 0

for k , which is

k= (s + 2 +j3)(s + 2 j3)s + 1

= s2 + 4s + 13

s + 1

and letting dkds

= 0, that is,

0 =dk

ds = (2s + 4)(s + 1) (s

2 + 4s + 13)

(s+ 1)2

This is equivalent to

(2s + 4)(s + 1)

(s2 + 4s + 13) = 2s2 + 6s + 4

s2

4s

13 = s2 + 2s

9 = 0

Solutions to this equation are

s=2 4 + 36

2 = 1

10 = 4.1623, 2.1623

Onlys = 4.1623 is on the root loci.

9. There is one pair of complex poles and no complex zeros. The angle of departure of the locus at k = 0 from the complexpoles = 2 +j3 is determined by using the phase condition

(s z1) [(s p1) + (s p2)] = odd multiple of 180


51/69

8.2. CONSTRUCTION 47

with s being a point on the root locus very close to s = 2 +j3.

(s z1) = 180 tan1

3

1

= 108.4

(s p2) = 90

Thus,

108.4 90 (s p1) = 180Solving this equation gives (s p1) =180 108.4+ 90 = 198.4, 161.6. These two angles are actually thesame because 198.4+ 161.6= 360.

Example 2 Given a closed-loop system with unity feedback, which has an open-loop transfer function with poles -1,-2, -3, no zero and gain k. Draw root loci for this system and find the ranges ofk for the closed-loop system to be stable,critically stable and unstable.

Solution:

1. Find the open-loop transfer function

Go(s) = k

(s+ 1)(s+ 2)(s + 3)2. Find the closed-loop characteristic equation:

G(s) = Go(s)

1 + Go(s) =

k(s+1)(s+2)(s+3)

1 + k(s+1)(s+2)(s+3)=

k

(s + 1)(s+ 2)(s + 3) +k

which means the closed-loop characteristic equation is

(s+ 1)(s+ 2)(s + 3) + k= 0

3. Since Go(s) is of third order, there are three loci.

4. Open-loop poles are1,2,3 and zeros are none.5. The portion of the root loci on the real axis is from -1 to2 because to its right there is one pole and from3 to

because to its right there is three poles.

6. The angles of the asymptotes are nm =

30 =

3 and

3nm =

30 =.

7. The intersection with the real axis isp1+p2+p3

n m =1 2 3

3 0 = 2

8. The intersection of loci with the imaginary axis can be determined by solving the characteristic equation withs = jfor k and , that is

(j + 1)(j + 2)(j + 3) + k = 0(2 +j3+ 2)(j + 3) + k = 0

j3 32 +j2 32 +j9+ 6 + k = 06 + k 62 +j(11 3) = 0

which means that

6 + k 62 = 011 3 = 0

Solving these equations for k and gives = 11 and k = 62 6 = 66 6 = 60.


52/69


9. The breakaway point is determined by solving

(s+ 1)(s+ 2)(s + 3) + k= 0

for k , which isk= (s + 1)(s+ 2)(s + 3)

and letting dkds

= 0, that is,

0 =dk

ds = [(s +2)(s +3)+(s +1)(s +3)+(s +1)(s +2)] = [s2 + 5s + 6 + s2 + 4s + 3 + s2 + 3s +2] = [3s2 + 12s +11]

This is equivalent to3s2 + 12s + 11 = 0

Solutions to this equation are

s=12 144 132

6 = 1.42, 2.58

Onlys = 1.42 is on the root loci.

10. There is no complex poles and zeros, so there is no need to determine the departure or arrival angles.

11. Whenk = 60, the system is critically stable because the closed-loop poles are on the imaginary axis. When 0< k 60, the systemis unstable because two closed-loop poles are in the open right-hand side of s-plane.


53/69

Chapter 9

PID Controllers

Proportional Controller: In t-domain, y(t) =KPe(t); in s-domainY(s) =KpE(s)Integral Controller: In t-domain, y (t) = KI

e(t)dt; in s-domain Y(s) = KI

s E(s)

Derivative Controller: In t-domain, y(t) =KDde(t)dt

; in s-domain Y(s) =KDsE(s)

PID Controller: In t-domain, y (t) = KPe(t) + KI e(t)dt+ KD de(t)dt ;. In s-domainY(s) =KpE(s) + KIs E(s) + KDsE(s) = (Kp+

KIs + KDs)E(s).

The block diagrams for these controllers are shown below.

49


54/69

50 CHAPTER 9. PID CONTROLLERS

These controllers can be implemented by using op-amps, as shown below.

ExampleConsider a system with a transfer function ofGp(s) = 1s(s+1) . Answer the following questions.

1. Check the stability of the system.

2. Calculate the steady-state error with a unit step input and unit ramp input.

3. Design a proportional controller to stabilize the system and achieve the steady-state error with a ramp input of 10%and find 5% settling time.

4. Can you find an integral controller to stabilize the system?

5. Design a PI controller to stabilize the system and to reduce the steady-state error with a ramp input to 0.

6. Design a PID controller withKI= 0.5KP andKD = 0.5KPso that the overshoot is less than 25%.

Solution:

1. The system is critically stable.

2. The error in s-domain is

E(s) = U(s) Y(s) =U(s) 1s(s+ 1)

U(s) =

1 1

s(s+ 1)

U(s) =

s2 + s 1s(s + 1)

U(s)

The steady-state error is

ess= lims0

sE(s) =

lims0s

s2+s1s(s+1)

1s

= U(s) = 1s

lims0ss2+s1s(s+1)

1s2

= U(s) = 1s2


55/69

51

3. The closed-loop system with a proportional controller is shown below.

It is clearly seen that the open-loop transfer function is Go(s) = KPGp(s) = KPs(s+1) and the error in s-domain is

E(s) = 11+Go(s)U(s). So with U(s) = 1s2

andess= 0.1, we have

0.1 lims0

sE(s) = lims0

s

1 + Go(s)

1

s2= lim

s01

s + sGo(s)=

1

lims0sGo(s)=

1

lims0KPs+1

= 1

KP

which implies that KP 10. So, chooseKP= 10.The closed-loop transfer function is

Gc(s) = Go(s)

1 + Go(s) =

KPs(s+1)

1 + KPs(s+1)

= KP

s2 + s + KP

Comparing this with the standard form for a second-order system gives

2n = KP n =

KP =

102n = 1 = 12n = 1210

The 5% settling time is

ts= 3

n=

3

2

10 10 = 0.15s

4. The closed-loop system with an integral controller is shown below.

The closed-loop transfer function is

Gc(s) =

KIs

1s(s+1)

1 + KIs

1s(s+1)

= KI

s2(s + 1) + KI=

KIs3 + s2 + KI

Routh array is given as follows:s3 1 0s2 1 KIs1

K

I= 0 11KIs0 KI=KI 1KI 0

The system is stable only ifKI>0 andKI>0. It is impossible. So, there is no solution to the problem.

5. The closed-loop system with a PI controller is shown below.


56/69


The open-loop transfer function is

Go(s) =

KP+

KIs

1

s(s + 1)=

KPs + KIs2(s+ 1)

The error in s-domain is

E(s) = 1

1 + Go(s)U(s)

The steady-state error is

ess = lims0

sE(s)

=

lims0s

11+Go(s)

1s

= U(s) = 1s

lims0s 1

1+Go(s)1s2

= U(s) = 1s2

=

1

1+lims0Go(s)= 0 U(s) = 1

s1

lims0sGo(s) = 0 U(s) = 1

s2

So the PI controller with any KP andKIcan reduce errors to zero for both step and ramp input.


Gc(s) = Go(s)

1 + Go(s)=

KPs+KIs2(s+1)

1 + KPs+KIs2(s+1)

= KPs + KI

s3 + s2 + KPs+ KI

Routh array is given as follows:

s3 1 KPs2 1 KIs1 KP 11KI=KP KIs0 KI 1KPKI

KP

0 = KI

The system is stable only ifKP >, KP KI>0 and KI>0, that is KP > KI andKI>0.

6. The closed-loop system with a PID controller is shown below.


Gc(s) =

KP+

0.5KPs

+ 0.5KPs

1s(s+1)

1 + KP+ 0.5KP

s + 0.5KPs

1s(s+1)

= 0.5KPs2 + KPs+ 0.5KP

s2(s+ 1) + 0.5KPs2 + KPs+ 0.5KP

= 0.5KP(s + 1)2

s2(s+ 1) + 0.5KP(s + 1)2

= 0.5KP(s+ 1)

s2 + 0.5KPs+ 0.5KP

Comparing this with the standard form for a second-order system gives

2n= 0.5KP n =

0.5KP2n = 0.5KP =2n 2= n


57/69

53

It follows from the requirement on overshoot that

e

12 0.25

1 2 ln 0.25

1

2

ln 0.25 = ln

1

0.25

= ln 4

ln 4

1 222 (ln4)2 (1 2) = (ln4)2 (ln4)2 2

2 + (ln 4)2

2 (ln4)2

2 (ln4)2

2 + (ln 4)2

ln 42 + (ln 4)

2= 0.4037

Choose = 0.5. Thenn = 2= 1, so 0.5KP =2n = 1, that is, KP = 2.


58/69



59/69

Chapter 10

Frequency Response

10.1 Steady-State Response

Frequency response is the output of a system when subject to a sinusoidal input. The steady-state response of a system toa sinusoidal input u(t) =Usin(t) is also a sinusoidal function

yss(t) =Y sin(t + )

with Y =U| G(j)|and = [G(j )]. Here G(j ) is called frequency response function. This will be illustrated by thefollowing example.

Example 1Find the output of a system G(s) = 1s+2 with a sinusoidal input u(t) =a sin(t).

Solution: The Laplace transform of the input isU(s) = as2+2 and the output in s-domain is

Y(s) =G(s)U(s) = 1

s + 2

a

s2 + 2 =

a

(s + 2)(s2 + 2)

Performing the partial fraction expansion gives

Y(s) = A

s+ 2+

B s + C

s2 + 2 =

A(s2 + 2) + (Bs+ C)(s+ 2)

(s + 2)(s2 + 2)

As a result, we have

a= A(s2 + 2) + (Bs+ C)(s+ 2)

Let s = 2. Then, a = A(4 + 2), that is, A = a4+2 . Let s = 0. Then, a = A2 + 2C, that is,

C=a A2

2 =

a a34+22

= 4a

2(4 + 2)=

2a

4 + 2

Let s = 2. Then,a = A(4 + 2) + 4(2B+ C), that is, 0 = 4(2B+ C). As a result, we have

B= C2

= a4 + 2

Therefore, the output is

Y(s) =a

4+2

s + 2+

a4+2 s + 2a4+2s2 + 2

= a

4 + 21

s + 2 a

4 + 2s

s2 + 2+

2a

4 + 2

s2 + 2

Taking the inverse Laplace transform gives

y(t) = a

4 + 2e2t a

4 + 2cos(t) +

2a

4 + 2sin(t)

55


60/69

56 CHAPTER 10. FREQUENCY RESPONSE

Now let cos = 24+2

. Then, sin = 4+2

and tan = sincos = 2 , that is, = tan12 . The output can be

expressed as

y(t) = a

4 + 2e2t +

a4 + 2

[sin cos(t) + cos sin(t)] = a

4 + 2e2t +

a4 +2

sin(t+ )

When t goes to infinity, the first term will die away. Therefore, the steady-state response is given by

yss(t) = a4 + 2

sin(t+ )

Now let us look at G(j) = 1j+2 . Its magnitude and angle are given by

| G(j) | = 14 + 2

[G(j)] = tan1

2

= tan1

2

So, the steady-state output can be written as

yss(t) =a | G(j) | sin(t+ [G(j)])

As a matter of fact, the steady-state response of any linear system G(s) to a sinusoidal input u(t) =a sin(t) is given by

yss(t) =a | G(j) | sin(t+ [G(j)])

ExampleWhat will be the steady-state response of a second order system

d2y(t)

dt2 + 3

dy(t)

dt + 10y(t) = 5u(t)

when subject to the input u(t) = 2 sin(2t+ 70).

Solution: By taking Laplace transform on both sides of the differential equation, we get

s2

Y(s) + 3sY(s) + 10Y(s) = 5U(s)

Thus, the transfer function is

G(s) = 5

s2 + 3s + 10

The frequency response function is

G(j) = 5

2 +j3+ 10 = 5

10 2 +j3Its magnitude and phase angle at = 2 are given as follows:

| G(j) | = 5

[(10 22)2 + 9 22] =

5

6

2

[G(j)] = tan1 610 4

= tan1(1) = 45

Therefore, the solution to the problem is

yss(t) = 5

3

2sin(2t + 25)

10.2 Bode Plots

Bode plot consists of two graphs, one of magnitude plotted against the frequency and one of the phase angle plotted againstthe frequency. The magnitude and phase angle are plotted using logarithmic scales.


61/69

10.2. BODE PLOTS 57

It is common practice to express the magnitude in units of decibel (dB).

Magnitude in dB = 20 lg | G(j) |Note that ifG(j) = G1(j)G2(j) Gn(j), then

| G(j) |=| G1(j) |) | G2(j) | | Gn(j) |Taking logarithm to base 10 gives

20lg | G(j) |= 20lg | G1(j) | +20 lg | G2(j) | + + 20 lg | Gn(j) |In addition, the phase angle satisfies

[G(j)] = [G1(j)] + [G2(j)] + + [Gn(j)]Bode Plot for Constant Gain: G(s) = K, G(j) = K,| G(j)|= K, 20lg| G(j)|= 20lg K, and [G(j)] = 0.

The Bode plot is given below.

Bode Plot for

1

s : G(s) =

1

s , G(j) =

1

j ,| G(j) |= 1

, 20lg | G(j) |= 20lg , and

[G(j)] = 90.= 0.1rad/s 20lg | G(j) |= 20lg0.1 = 20dB= 1rad/s 20lg | G(j) |= 20lg1 = 0dB= 10rad/s 20lg | G(j) |= 20lg10 = 20dB



62/69


Bode Plot for s: G(s) =s, G(j) =j,| G(j) |=, 20lg | G(j) |= 20lg , and [G(j)] = 90.

= 0.1rad/s 20lg | G(j) |= 20 lg0.1 = 20dB= 1rad/s 20lg | G(j) |= 20 lg1 = 0dB= 10rad/s 20lg | G(j) |= 20lg 10 = 20dB


Bode Plot for 1s+1 : G(s) =

1s+1 , G(j) =

11+j ,| G(j)|= 11+22 , 20lg| G(j )|=20lg

1 + 22, and

[G(j)] = tan1 .

= 0.1 1

rad/s 20lg | G(j) |= 20lg 1 + 0.01 20lg1 = 0dB[G(j)] = tan1 0.1 0

= 1

rad/s 20lg | G(j) |= 20lg 1 + 1 = 20 lg 2 = 3dB 0dB[G(j)] =

tan1 1 =

45

= 10 1 rad/s 20lg | G(j) |= 20lg 1 + 10 20lg10 = 20dB[G(j)] = tan1 10 90

The point = 1

is called break-point. The Bode plot is given below.

Bode Plot for s+ 1: G(s) = s+ 1, G(j ) = 1 +j ,| G(j)|= 1 + 22, 20lg|G(j )|= 20lg 1 + 22, and


63/69

10.2. BODE PLOTS 59

[G(j)] = tan1 .

= 0.1 1

rad/s 20lg | G(j) |= 20lg 1 + 0.01 20lg1 = 0dB[G(j)] = tan1 0.1 0

= 1

rad/s 20lg | G(j) |= 20lg 1 + 1 = 20lg2 = 3dB 0dB[G(j)] = tan1 1 = 45

= 10 1

rad/s 20lg | G(j) |= 20lg 1 + 10 20lg 10 = 20dB

[G(j)] = tan1

10 90The point = 1

is called break-point. The Bode plot is given below.

ExampleDraw asymptotes of the Bode diagram for a system G(s) = 102s+1 .

Solution: G(s) = 10 12s+1 =G1(s)G2(s). The Bode plot is shown below.


64/69



65/69

Chapter 11

Control of Discrete Processes and PLC

11.1 Introduction

As a simple illustration of what is meant by discrete process control, consider a dc motor starting process. Because of alarge starting current, we need to insert some resistors into the armature circuit of a dc motor. In order to keep the startingcurrent under a certain level, we need to design a circuit to implement the following operations:

1. When the start button is pushed, the motors armature circuit is connected to its power supply and the machine startswith all resistorsR1, R2, andR3.

2. R1 is cut out from the armature circuit 10s later.

3. R2 is cut out 10s after R1 is cut out.

61


66/69

62 CHAPTER 11. CONTROL OF DISCRETE PROCESSES AND PLC

4. R3 is cut out 10s after R2 is cut out.

5. The motor will stop whenever the stop button is pushed.

This starting process can be implemented by a circuit composed of switches and relays.

The normally open button BO will be closed when pushed, whereas the normally closed button B Cwill be open whenpushed.

A relay contains a coil and some contacts, either normally open contacts or normally closed contacts, or both. Normallyopen contacts will be closed when the coil is energized and open when de-energized, whereas normally closed contacts willbe open when energized and closed when de-energized.

Time-delay relays are a kind of relays. Normally open contacts will be closed after a certain delay when the coil isenergized and open immediately when de-energized, whereas normally closed contacts will be open after a certain delaywhen energized and closed immediately when de-energized.

The starting process can also be implemented by using a Programmable Logic Controller (PLC).

A PLC is a microprocessor-based controller that uses a programmable memory to store instructions and implementfunctions such as logic, sequencing, timing, counting, and arithmetic.


67/69

11.2. PLC PROGRAMMING 63

11.2 PLC Programming

A PLC can be programmed by using a simple form of language, ladder diagram. Writing a program is equivalent to drawinga switching circuit.

A ladder diagram looks like a ladder. It consists of two vertical lines, representing the power rails, and several horizontallines, which are rungs of the ladder. In drawing a ladder diagram, the following conventions are adopted:

1. Vertical lines represent power rails between which circuits are connected.

2. Each rung (horizontal line) on the ladder defines one operation in the control process.

3. Each rung must start with an input or inputs and must end with at least one output.

4. A ladder diagram is read from top to bottom. The rung is read from left to right. When a PLC is in its run mode, itgoes through the entire ladder program to the end, and then promptly resumes at the start. This procedure of goingthrough all the rungs of the program is termed a cycle.

11.3 Internal Relays

In PLCs there are elements that are used to hold data and behave like relays, being able to be switched on or off or switchother devices on or off. These elements are termed as internal relays. Such internal relays do not exist as real-world switchingdevices, but are merely bits in the storage memory that behave the same way as an external relay.

Consider the dc motor starting circuit, a latch circuit is needed to start and stop the motor. Such a latch circuit can beimplemented by an internal relay, as shown below.


68/69


69/69

11.5. COUNTERS 65

value, that is, events are added until the number reaches the preset value. When the counters reach the preset value, theircontacts change state. As an example, let us analyze the ladder diagram shown below.

1. The counter is reset and ready for counting when there is a pulse to RST.

2. Start counting pulses to CTD.3. When the counter reaches its preset value, stop counting and its normally open contact changes state becomes closed.

The timing diagram for this ladder diagram is given below.

control system basics

Documents