Voltage Stabilizer

Prepared by Abeer ALNuaimi Balqees ALDaghar Afra Ebrahim Lila Abdullah

United Arab Emirates University College of Engineering Electrical Engineering Department

Project AdvisorDr. Abdulla Ismail

200203257 200324560200310882

ContentsContents Introduction

• Summary about our project.• Review GP1 task.• Gantt chart for GP2

optimal control:• LFC with PI & optimal control• AVR with PI & optimal control

ContentsContents Combination of LFC and AVR LFC with Fuzzy logic control LFC with Robust control Comparison for three controllers

• PI, Fuzzy & Robust Conclusion

AGC OverviewAGC Overview• The system:

– Power Generation system.

• The problems:– Frequency and voltage variations

• The consequences:– Machine damage.– Blackouts, or outages.

GP1 OverviewGP1 Overview• The Project:

– Automatic Generation Control system• The Advantages:

– Limits the variations.– Avoide machine damages– Avoide blackouts– Enhance the system reliability and security.

GP1 OverviewGP1 Overview

GP1 OverviewGP1 Overview

Gp1

Gantt chart GP2 PlanGantt chart GP2 Plan

Gantt chart GP2 PlanGantt chart GP2 Plan

Optimal Linear Control Systems

optimal control is a set of differential equations describing the paths of the control variables that concerned with operating a

dynamic system to minimize the cost functional with weighting factors supplied by a engineer.

Example for optimal control

Optimal Linear Control Systems • Application of optimal control:

– Mechanics of motion.– Economics.– Medicals.– Populations.

The targets for using the optimal linear control system:

1. Stable closed-loop system.2. Reduce steady state errors.3. Reach standard performance

measures:– Peak Time, Tp.– Percent of overshoot.– Percent of under shoot.– Settling time, Ts.– Rise time, Tr.

• Minimization cost equation:

0,0)()()()(21

2121)(

min

RRdttuRtutxRtxJif

io

TT

tu

State variable Input

The LFC with the I and OPC

l

f1

0.875

scontroller

1

0.3s+1Turbine

PL

1

0.08s+1Governor

1.428

0.37s+1Generator

0.4

Gain

• Model1: With the integral control.

The LFC with the I and OPC• MATLAB: Defining the Matrices.

– A=[-12.5 0 -12.5 -5;3.33 -3.33 0 0;0 3.86 -2.70 0;0 0 0.87 0];– B=[12.5;0;0;0]; F=[0;0;-1.93;0];– C=[0 0 1 0];– D=[0];


0 5 10 15 20 25 30-0.045

-0.04

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0Model1 output respronse without integral controller

Time (s)

df/d

t (H

z)

US: 4%

SSR: t=25s

The LFC with the I and OPC• Model1: With the integral control and optimal control.

f2

f1

-0.4s-0.875

scontrol ler1

-0.875

s

control ler

1

0.3s+1Turbine1

1

0.3s+1

Turbine

PL1

PL

1

0.08s+1Governor1

1

0.08s+1

Governor

1.428

0.37s+1Generator1

1.428

0.37s+1

Generator

-K-

Gain4

-K-

Gain3

-K-

Gain2

-K-

Gain1

-K-Gain

du/dt

Derivative

-K-

-1.143

The LFC with the I and OPC• MATLAB: Defining the Matrices.

– Q=[10 0 0 0; 0 10 0 0 ;0 0 10 0 ;0 0 0 10];% – R=1;– [K,P,ev]=lqr(A,F,Q,R)– Ao=A-(F*K)– sys1=ss(Ao,F,C,D);– yo=lsim(Ao,F,C,0,u,t);


0 5 10 15 20 25 30-0.045

-0.04

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0Model1 output respronse without integral controller(Blue) and Optimal control(green)

Time (s)

df/d

t (H

z) US: 1.5%

SSR: t=10s

The LFC with the I and OPC• The Integral and the Optimal control Advantages:

– Undershoot Reduction from 4% to 1.5%– The steady state response deducted faster– The integral control helps in enhancing the steady state response

from t=25s to t=10s.– The Optimal control helps in enhancing the Transient response.

System ModelsSystem Models (AGC)(AGC)

Valve Control mechanism

Load frequency control (LFC)

Frequency sensor

Voltage sensor

Automatic Voltage Regulator (AVR)

Excitation system

Gen. field

Turbine

GP

Steam

tieP

VP

GG QP ,

G

Shaft

Automatic Voltage RegulationAutomatic Voltage Regulation

For efficient and reliable operation of Power Systems, the control of voltage should satisfy the following objective:

• Voltages at the terminals of all equipment in the system are within acceptable limits. Maintaining voltages within the required limits is complicated due to the fact that:

1) The power system supplies power to vast number of loads and fed from many generating units.

2) System voltage is closely related to the system reactive power which is a reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power.

3) The proper selection and coordination of equipment for controlling the system voltage and the reactive power are among the major challenges in power system operation and control.


• Reactive Power (QV) is one of the two main elements in the power system must be controlled.

• Any voltage error in the system is sensed, measured, and transformed into reactive-power command signal.

• The objective of the AVR is to keep the system terminal

voltage at the desired value by means of feedback control


Block diagram of AGC model

AVR Model

Block diagram of a simple automatic voltage regulator (AVR)

KKEE=200=200 T TEE =0.05 =0.05KKGG=1=1 T TGG =0.2 =0.2KKRR=0.05=0.05 T TRR =0.05 =0.05KA=0.15 TA=10KA=0.15 TA=10

• Voltage error is improved by controlling the rotor field-current generator EMF.

• The steady state voltage error can be eliminated using an integral controller.

• The AVR has a substantial effect on transient stability when varying the field voltage to maintain the terminal voltage constant.

Block diagram of a simple automatic voltage regulator (AVR)

31

AVR Model

Case 1: AVR without PI (Proportional and Integral ) controller.

Case 2: AVR with PI controller.

Case 3: AVR with optimal control.

Case 1: AVR without PI (Proportional and Integral ) controller.

Block diagram of AVR model without PI controller

The output voltage response without controller

Steady State errorOvershoot error

Time (s)

∆V

Overshoot

Steady state error

The output voltage response when Ka of the amplifier was changed to 0.1

The output voltage response when Ka of the amplifier is 0.15

Case 2: AVR with PI controller.

Block diagram of AVR model with Ki and Kp gains

The output voltage response when Ki=0.2 and Kp= 1.5

Overshoot

Time (s)

V

The output voltage response with PI controller

Case 3: AVR with optimal control.

Block diagram of AVR model with feedback gains

4)( xsU 3x

)(01.001.01.03 43 tuxxx

2x

32 4000202 xxx

3x

2x1x

211 55 xxx

4x

144 2020 xxx

1x

Step1: Find the state variables and output equations:

State differential Equation:

Output Equation:

A=[-5 5 0 0; 0 -20 4000 0; 0 0 -0.1 -0.01;20 0 0 -20] B=[0;0;0.01;0]C=[1 0 0 0]D=[0]

)()()( tuBtxAtx mnnn

)()()( tuDtxCty mpnp

Step2: Find A,B, C, D matrices

Step 3: MATLAB command to find the feedback gains

MATLAB command

A=[-5 5 0 0; 0 -20 4000 0; 0 0 -0.1 -0.01;20 0 0 -20] Q=[5 0 0 0; 0 5 0 0; 0 0 5 0; 0 0 0 5]B=[0;0;0.01;0]R=5[F,P,ev]=lqr(A,B,Q,R)

Result of running the program:

F = -0.0230 0.0582 206.0278 -0.0899 P = 1.0e+005 * 0.0000 0.0000 -0.0001 0.0000 0.0000 0.0000 0.0003 0.0000 -0.0001 0.0003 1.0301 -0.0004 0.0000 0.0000 -0.0004 0.0000 ev = -20.6225 + 3.4261i -20.6225 - 3.4261i -2.9577 + 3.1290i -2.9577 - 3.1290i

values of feedback gains k1,k2,k3,k4

MATLAB Function

The output voltage response with optimal and integral control

Time (s)

V

AVR

LFC

AGC system

x7x6x5

x9

x8

x1 x2 x3

x4

df/dt

Vref (s)

V(t) + df/dt

V(t)

1

0.3s+1Turbine

1

0.05s+1Sensor

PL

1.5

Kp

0.2

sKI

1K5

0 K4

0.1K3

1 K2

0.8K1

0.875

sIntegral control 1

0.08s+1Gov ernor

1

0.2s+1Generator1

1.428

0.37s+1Generator

200

0.05s+1Exciter

0.1

10s+1Amplifier

0.4

1/R

x3

Load Frequency Control

Auto Voltage Regulator

AVR and LFC Combination

• Forming A, B, C, D and F Matrices:– State Differential Equation.State Differential Equation.– Output Equation.Output Equation.– MATLAB.MATLAB.

• Tuning K1,K2,K3,K4 and K5 Between 0 and 1:– Trial and Error:

• K1 has no affect on either one of the two systems.• K2 has an affect on the LFC response.• K3 has an affect on both the LFC and the AVR system

stability.• K4 and K5 both have an affect on the AVR overshoot.


• Tuning K1,K2,K3,K4 and K5 Between 0 and 1:– K at which the responses of both AVR and LFC are behaving normally:

• K1= 1• K2= 0.8• K3= 0.1• K4= 0• K5= 1


0 5 10 15 20 25 30-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005Model1(reordered) output respronse with the integral controller

Time (s)

df/d

t (H

z)

LFC response

AVR response

AGC response

0 100 200 300 400 500 600 700 800 900 10000

20

40

60

80

100

120

x7x6x5x9

x8

x1 x2 x3

x4

df/dt

Vref (s)

V(t) + df/dt

V(t)

1

0.3s+1Turbine

1

0.05s+1Sensor

PL

0

Kp

0.09

sKI

-K-

K9

-K-

K8

-K-

K7

-K-

K6

1K5

0 K4

0.1K3

1 K2

-K-

K10

0.8K1

0.875

sIntegral control 1

0.08s+1Governor

1

0.2s+1Generator1

1.428

0.37s+1Generator

200

0.05s+1Exciter

0.1

10s+1Amplifier

0.4

1/R

x3

Load Frequency Control

Auto Voltage Regulator


0 5 10 15 20 25 30-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0


Time (s)

df/d

t (H

z)

LFC response

AVR response

AGC response

0 50 100 150 200 250 300 350 400 450 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Model1 output respronse with integral and optimal controllers

Time (s)

Vol

ts (v

)

0 1 2 3 4 5 6 7 8 9 10-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

Model1yv and ydf output respronse in an AVR & LFC combination

Time (s)

Vol

t/Df/d

t (vs

)

0 5 10 15 20 25 30-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0


Time (s)

df/d

t (H

z)

LFC response

AVR response

AGC response

0 50 100 150 200 250 300 350 400 450 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Model1 output respronse with integral and optimal controllers

Time (s)

Vol

ts (v

)

• The Combination of both AVR and LFC systems might cause slight changes in their responses.

• Fortunately the undershoots and overshoots never exceeded 20%.– AVR stand alone system

• Overshoot = 2.2%• With the optimal control the overshoot almost eliminated.

– AVR within AGC system• Overshoot= 0.2%

– LFC stand alone system • Undershoot= 2.8%

– LFC within AGC system • Undershoot= 2.5%


Fuzzy logic controlFuzzy logic control

It is the process of formulating the mapping from a given input to an

output using fuzzy logic. The mapping of FL done based on human

operator’s behavior.

Fuzzy logic controlFuzzy logic control

FuzzFuzzy y

logiclogicNatural language

s

Model nonlinear function

Cheaper

Faster

Flexible

Easy to understand

Fuzzy Logic Control Fuzzy Logic Control ApplicationApplication

• Automatic control • Data classification• Decision analysis• Expert systems• Computer vision

• Cameras• washing machines• microwave ovens• Industrial process control• Medical instrumentation

Fuzzy logic control processFuzzy logic control process

• Fuzzify Inputs• Apply Fuzzy Operator• Apply Implication Method• Aggregate All Outputs• Defuzzify

Modeling one area LFC with Modeling one area LFC with Fuzzy logic controlFuzzy logic control

FIS EditorFIS Editor

Membership FunctionMembership Function

FIS variablesFIS variables

de\e LN MN SN VS SP MP LP

LP VS SP MP LP LP LP LP

MP SN VS SP MP MP LP LP

SP MN SN VS SP SP MP LP

VS MN MN SN VS SP MP MP

SN LN MN SN SN VS SP MP

MN LN LN MN MN SN VS SP

LN LN LN LN LN MN SN VS

FIS variablesFIS variables

Variables Linguistic Term Range of linguistic term

Near +0.05 LP (large positive) [+0.04 +0.05 +0.05]

Near -0.05 LN (large negative) [-0.05 -0.05 -0.04]

So far from +0.05 MP (medium positive) [+0.025 +0.035 +0.045]

So far from -0.05 MN (medium negative) [-0.045 -0.035 -0.025]

Very far from +0.05 SP (small positive) [+0.005 +0.015 +0.03]

Very far from -0.05 SN (small negative) [-0.03 -0.015 -0.005]

∆f=0 VS (very small) [-0.007 0 +0.007]

Rule EditorRule Editor

If-And-Then rules

The response of LFC one The response of LFC one area Fuzzy controlarea Fuzzy control

Modeling two area LFC with Modeling two area LFC with Fuzzy logic controlFuzzy logic control

Area Control ErrorArea Control Error

• Area control error is the difference between the actual

power flow out of area, and scheduled power flow. ACE

also includes a frequency component.

fff

PPP

PPP

sch

tieschtietie

gschgg

The response of LFC two The response of LFC two areas Fuzzy controlareas Fuzzy control

∆ F1

∆ F2

The response of Tie line for The response of Tie line for LFC two areas Fuzzy controlLFC two areas Fuzzy control

LFC Model with Robust controllerLFC Model with Robust controller

What is Robust control ?

Why we need Robust control in our model (AGC)?

Applications of Robust control

Robust controller design

Robust controllerRobust controller

The dynamic behavior of electric power systems is heavily affected by disturbances and changes in the operating points.

An industrial plant such as power systems always contains parametric uncertainties.

In many control applications, it is expected that the behavior of the designed system will be insensitive (robust) to external disturbance and parameter variations

Applications of Robust control

Robust control of Temperature

Disk drive read system

Mobile ,Remote-Controlled video camera

Spacecraft

Control of a(Digital audio tape) DAT player

Elevator

Microscope control

Robust controller designRobust controller design

∆Pd(t) : load disturbance (P.u. MW)Tg : governor time constant (s)Kg : governor gainTt : turbine time constant (s)Kt : Turbine gainTp : Generator time constant (s)K p : Generator gainR : speed regulation due to governor action (HZ p.u. MW-1)KI: Integral control gain

LFC Block diagram of power system

Our robust load-frequency controller design procedure is as follows:

Step 1:Step 1: Find the range of the system parametersFind the range of the system parameters• State equation:

• Output equation:

• Where :

)()()( tuDtxCty mpnp

)](),(),(),([)( 4321 txtxtxtxtx

)()()()( tPFtButAxtx d

14 xKx I

1x

143 )(31 xRT

ktu

Tk

xTgkgx

Tx

g

g

g

g

g

3221 x

Tkx

Tx

t

t

t

d

P

P

P

P

P

PTkx

Tkx

Tx 211

1

A = -1/Tp Kp/Tp 0 0 0 -1/Tt 1/Tt 0 -1/RTg 0 -1/Tg -1/Tg K 0 0 0

• The range of the system parameters is:

0,/1,0,0 GTB

0,0,0,/ PP TKF

],[/1

],[/1

],[/1

],[/

],[/1

55

44

33

22

11

aaRT

aaT

aaT

aaTK

aaT

G

G

T

PP

P

]639.10,081.3[/1]857.17,615.9[/1

]762.4,564.2[/1]12,4[/

]1.0,033.0[/1

G

G

T

PP

P

RTTT

TKT

Step 2:Step 2: Choose the nominal parameters for the system andChoose the nominal parameters for the system and decide the bound of the uncertaintiesdecide the bound of the uncertainties

• The nominal parameters are from the original model of LFC:

And ,

)()()()()( tuBBtxAAtx

AAA BBB

A = -2.7030 3.8595 0 0 0 -3.3330 3.3330 0 -31.2500 0 -12.5000 -12.5000 0.8800 0 0 0

B=[0;0;12.5;0]

F=[3.8595;0;0;0]

• Now, let decide the bound of the uncertainties:

• Hence, the parametric uncertainties are:

AA

BB FF

7.0&5.0,3.0

A = -0.8109 1.1579 0 0 0 -0.9990 0.9999 0 -9.3750 0 -3.7500 -3.7500 0.2640 0 0 0

B=[0;0;6.25;0]

F=[2.70165;0;0;0]

• After this change in the system the new matrices are as follow:

AAA BBB

A = -3.5139 5.0174 0 0 0 -4.3320 4.3320 0 -40.6250 0 -16.2500 -16.2500 1.1440 0 0 0

B=[0;0;18.75;0]

F=[6.56115;0;0;0]

Step 3:Step 3: Choose the design constants εChoose the design constants ε and the design constant and the design constant matrices matrices Q Q and and RR

And because the algebraic Riccati equation is nonlinear equation we use MATLAB program to solve it.

)1(....01121

1

1 EqQUPTBRBPPAAP TT

Where , Q > 0 and R > 0

ε & ε1 > 0 , very small value

• Algebraic Riccati equation:

T & U are the rate change of the generation

Step 4:Step 4: Use the algorithm given eq. (1) to solve Riccati Use the algorithm given eq. (1) to solve Riccati equationequation

and obtain the solution and obtain the solution PP• By using the command from the MATLAB we can found P as

follow:MATLAB Command MATLAB Command

A=[-3.5139 5.0174 0 0;0 -4.332 4.332 0; -40.625 0 -16.25 -16.25; 1.144 0 0 0]Q=[5 0 0 0; 0 5 0 0; 0 0 5 0; 0 0 0 5]B=[0;0;18.75;0]R=0.01[F,P,ev]=lqr(A,B,Q,R)

Result of the command Result of the command F = 9.6166 17.6707 21.6925 21.5108P = 1.1042 0.6104 0.0051 1.7927 0.6104 0.9237 0.0094 1.1636 0.0051 0.0094 0.0116 0.0115 1.7927 1.1636 0.0115 7.7951ev = 1.0e+002 * -4.1956 -0.0522 + 0.0168i -0.0522 - 0.0168i -0.0083

• By using MATLAB the output frequency response was drawn without considering the feedback gains:

MATLAB CommandMATLAB Commandclct=[0:0.1:20];u=-0.1*ones(length(t),1);x0=[0 0 0 0];A=[-3.5139 5.0174 0 0;0 -4.332 4.332 0; -40.625 0 -16.25 -16.25; 1.144 0 0 0];eig(A)B=[0;0;18.75;0];C=[1 0 0 0];D=[0];sys=ss(A,B,C,D);[y,x]=lsim(sys,u,t,x0); plot(t,y)Title('The output frequency respronse');xlabel('Time');ylabel('f');grid

0 2 4 6 8 10 12 14 16 18 20-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01The output frequency respronse

Time(s)

Del

ta f

(HZ)

The output frequency response with uncertainties parameters and without feedback gains

Step 5:Step 5: Construct the feedback gainConstruct the feedback gain

• Also, by using the same command from MATLAB we found the optimal gains:

F = 9.6166 17.6707 21.6925 21.5108

LFC Block diagram of power system for the proposed robust controller

The output frequency response for the proposed robust controller

Comparison between robust and integral controller

With proposed robust controller

With Integral controller



Figure 1: With nominal parameters 1/Tp= 2.7030, Kp/Tp= 3.8595, 1/TT = 3.333, 1/TG =12.5, 1/RTG= 31.25, KI= 0.88

Figure 2: With 1/Tp=1.05, Kp/Tp=1.494, 1/TT=1.3, 1/TG=1.79,1/RTG=0.7143,KI=1.144



Figure 3 : With 1/Tp = 0.8, Kp/Tp= 1.1424, 1/TT=1.031, 1/TG=1.52,1/RTG=0.61,KI=0.88



Figure 4: With 1/Tp = 0.033, Kp/Tp= 4, 1/TT=2.564, 1/TG=9.615,1/RTG=3.081,KI=0.88

Comparison Comparison The use of the PID algorithm for control does not guaranteeoptimal control of the system or system stability, that’s whyin our designs of LFC and AVR we used the optimal linearcontrol systems.

The Fuzzy-logic controller can be seen as a heuristic andmodular way of defining nonlinear system but the fuzzylogic controller failed in considering the uncertainties.

Comparison Comparison The proposed robust controller is simple, effective and canensure that the overall system is asymptotically stable forall admissible uncertainties.

ConclusionConclusionOur goal in the end is to design a control system that serves

the power network in the UAE for better performance and better power services in terms of consumption and supplement.

Enhance our skills and understanding of Engineering project design and management.

Achieve the best as an outcome of a successful group work.

Documents

Voltage Stabilizer