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VOL. 6, NO. 11, NOVEMBER 2011 ISSN 1819-6608

ARPN Journal of Engineering and Applied Sciences

2006-2011 Asian Research Publishing Network (ARPN). All rights reserved.

www.arpnjournals.com

136

CONTROL SYSTEM DESIGN OF GUIDED MISSILE

Muhammad A. R. Yass1, Mohammed Khair Aldeen Abbas

2and Ismail Ibrahim Shabib

1

1Electro-Mechanical Engineering Department, University of Technology, Iraq

2Mechanical Engineering Department, Al-Nahrain University, IraqE -Mail: mkaskar79@yahoo.com

ABSTRACTIn the present work, analytical performance of dynamic motion and transfer function were calculated in roll mode

of flying body. The analysis of the control circuit was made by using four methods, Routh Criterion to determine whether

the system is stable or not, Root-Locus method used to determine the limitation of the stability for different value of rate

gyro (0.1 to 0.3) and different value of gain (1 to 10), Frequency Response method used to find best transfer function

which have shortest time setting and less amount of overshoot and last method used the compromising method which was

done first with aileron deflection to determine the limited value of gain, secondly with capability actuator swelling rate.Final transfer function selected was with rate of gyro equal to 0.3 and gain equal to 8 for best steady state behavior. The

method above is a perfect solution for flying body control system in all modes and gives an excellent result.

Keywords:guided missile, flying body control system, Routh criterion, Root-Locus method, frequency response method.

INTRODUCTION

It can determine whether our design of a controlsystem meets the specification if the desired time response

of the controlled variable determined. by deriving the

deferential equation for the system solving them anaccurate solution of the system's performance can be

obtained, but this approach is not feasible for other than

simple system if the response doesn't meet the

specifications 'it is not easy to determine from this solutionjust what physical parameters in the system should be

changed to improve the response.The ability of prediction the system's

performance by an analysis that does not require the actual

solution of the differential equation. Also, we would liketo indicate this analysis readily the manner or method by

which this system must be adjusted or compensated to

produce the desired performance characteristics.The first thing beat we want to know about a

given is whether or not it is stable. This can be determined

by examining the root obtained from the involved indetermine the root can be tedious, a simpler approach is

desirable. By applying Rouths criterion to the

characteristic equation it is possible in short or unstable.Yes it doesn't satisfy us because it doesn't indicate the

degree of stability of the system, i.e., the amount of

overshoot must be maintained, within prescribed limit andtransients must die out in a sufficiently short time. The

graphical met holds to be described in this test not onlyindicate whether a system is stable or not but, for a stable

system also a low degree of stability.

These are two basic methods available to us, wecan choose to analyze and interpret the steady state

suicidal response of the transfer function of the system

obtains an idea of the system's response. This method isbased upon the interpretation of a SyQuest plot. Although

this frequency-response approach doesn't yield an exact

quantitative prediction of the system's performance, i.e.,

the poles of the control ratio c(s)/r(s) can not be determine,enough information can be obtained to indicate whether

the system need to be the system should be compensated.

This deals with the record method that roots locus

method, which incorporates the more desirable features ofboth, the classical method and the frequency - response

method. The root locus is a plot of the root of the

characteristic equation of the closed loop system as thefunction of the gain. This graphical approach yield a clear

indication of the effect of gain adjustment with solving

small effort compared with other method. The underlying

principle is that the poles of C(s)/R(s) (transient response)are released to the source and poles of the open loop

transfer function G(s) H(s) and also to the gain. Animportant advantage of the root locus method is that the

root of the characteristic equation of the system can be

obtained directly, this result in a complete and accuratesolution of the transient and steady state response of the

controlled variable. Another important feature is that an

approximate solution can be obtained with a reduction ofthe work required. As with any other design technique, a

person who has obtained a coefficient experience with this

method is able to apply it and to synthesize acompensating network, if one is required, with relative

ease.

MATHEMATICAL ANALYSIS

In the present work, the analytical performance of

dynamic motion and transfer function were calculated inroll mode of flying body. The analysis of the control

circuit was made by using four methods, Routh Criterionto determine whether the system stable or not, Root-Locus

method used to determine the limitation of the stability for

different value of rate gyro (0.1 to 0.3) and different valueof gain (1 to 10), Frequency Response method used to find

best transfer function which have shortest time setting and

less amount of overshoot and last method used thecompromising method which was done first with aileron

deflection to determine the limited value of gain, secondly

with capability actuator swelling rate. Final transfer

function selected was with rate of gyro equal to 0.3 andgain equal to 8 for best steady state behavior. The method

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137

above is a perfect solution for flying body control systemin all modes and give an excellent result.

AERODYNAMIC ANALYSIS

Roll attitude control system

The non dimensional Lateral equation [4]

( (1)

A multi-loop feed-back control system applied to

flying-body, a main feedback of roll angle an attitude gyro

and an inner loop of roll rate feed-back [3], see Figure-1.

Figure-1.Roll attitude control system.

(2)

Re-write eq. (1)

(3)

and

Then

(4)

L = and L =

So

=

(5)

The inner loop transfer function

(6)

=

While the overall transfer function

(7)

If the actuator dynamic included the presentation

can be seen in Figure-2 the inner loop transfer function

= (8)

While the overall transfer function was [8]

Figure-2.Roll attitude control system with actuator.

(9)

(10)

Open loop transfer function (K-Root Locus)From open loop transfer function which was

equal to [4]

(s)= (11)

So the char- equation became equal to

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138

( ) s + ( + ) s+ (1+GR ) s = 0 (12)

Dividing equation (12) by ) we will get

S+ S+ S = 0 (13)

Solving equation above for diferent[ k] and [GR]

and all the complex root found was studied to determune

the effect of ( k).

Colse loop teanfer function (GR Root Locus)

The close loop t.s which was equal to

(s)=

Divide equation (15) by

(s)=

Also it can be written as

D

(s)=))()((

31 pspspss

D

s +++

By partial fraction

2

3

12

21

11

211

gsipgs

i

ipgS

i

S +

+

+

+

+

++

The Inverse Laplace Transformation

D(t) =

)sincos(2113123

12 tptpee tgtg

+ (16)

Solving equation above for diferent[ k] and [GR]

and all the complex root found was studied to determune

the effect of [GR].

Dynamic response of aileron deflection

from equation

ST

K

a

+=

1

and it can be also written as

)()1())(( sSTKsa += 16a

By taking the laplace inverse of equation (16a)

))(( Ksa = )()( tTt oo + .16b

Divided equation (16b) by 0

0

)(0)())((

tTtktS

D

a

oo

+=

From equation

tgtg

D

egegt 12 123 2)( ++=

( )cossin(2)sincos 1312111312

1 tptpePgtptp tg +++

)sincos()cossin(2 13122

113121111 tptpePtptpegP

tgtg +++++

)

)

++

+++

+++

++

)sincos(2

)1cossin(2)1cossin(2

)sin3cos2(2(cos3sin(

P2-)sincos(23

1312

2

1

3121131211

112122

23112

1131212

1

11

1

112

tptpep

tptpegptptpepg

tptpegegtptp

etptpegeg

tg

tgtg

tgtg

tgtgtg

26By varying time it can be solve for Aileron deflection.

Actuator slewing rate

It is the time required for actuator to deflect the aileron.

D

Sa

(t)=

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139

By differentiate equation (26).

[

++

++

++

++++

++++

=

)cossin(2)sin

cos(2)cossin(2)sincos(

2)cossin(2)sincos(2

)sincos(2)1cossin(2

)1cossin(2)sincos(2

1

13121

2

113

121

2

11312

2

111312

22

1113121

2

11312

3

3

231312

2

131211

312111312

2

1

2

23

1

11

11

111

11

tptpegptp

tpegptptpegptptp

epgtptpepgtptpeg

egTtptpeptptpegp

tptpepgtptpegg

k

aS

tg

tgtg

tgtgtg

tgtgtg

tgtg

D

So by solving the above equation at different GR, K and at

t = 0, become max actuator slewing rate which occurs at

t = 0, we will get different actuator slewing rate at

different combinations of GR and K.

RESULTS AND DISCUSSIONS

DISCUSSIONS

The purpose of this design to obtain largest gain

attainable required and the smallest peak overshoot andshortest time setting. From Table-1 and Figures 1 and 2,

GR = 0.1, maximum K is equal to 4 and for more than that

the system is not stable.

It can be observe from Table-1 that by increasing

K at constant value of GR, the real part of complex rootwill decrease and the imaginary part will increase, this

mean that the overshoot will increase too and setting time

will also increase, so it must have a choose between thelargest gain attainable, shortest peak overshoot and

shortest time setting. Also by increasing at constant K the

real part of complex root will increase and imaginary partwill increase too and that increase the peak overshoot and

better stability.

See Figures (2-10)

It can be observed from Table-2 and ailerondeflection curve (Figures 1 to 10) that by increasing K at

constant GR, aileron deflection will also increase but it

decrease by deceasing increasing GR at constant K.

So it cant go beyond GR = 0.3 and K = 8 because it is

limited by aileron deflection and case study design datawhich was equal to 5 rad.

After compromising all the values of K and GR

searching for our design requirement which is givenabove, it is found that GR = 0.3, K = 8 is the best selectionwhich must be checked with actuator slewing time (time

of actuator to deflect). From Table-3 it can be notice that

the maximum value of actuator slewing rate at GR = 0.3and K = 10 was equal to 43.668 rad /sec and the design

value of actuator slewing rate is equal to 90 rad/sec, that

mean that the selection value was within the limit. Also itcan be observed that as K increases the actuator slewing

time will also increase and for increasing GR at constant K

the actuator slewing time remains the same.

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140

CONCLUSIONSIt can conclude that the response of a control

system is examined by root locus method and it depend on

real root of the characteristic equation of close loop

transfer function and this root depend on K and GR so by

increasing K the root shift to right of root locus whichdecrease the stability but by increasing GR the root shift to

the left which increases the stability also increasing K and

GR depend on max aileron deflection.

REFERENCES

[1] Roskam J. 1971. Methods for Estimating Stability andControl Derivatives of Conventional Subsonic

Airplanes. Roskam Aviation and Engineering

Corporation, Lawrence, Kansas.

[2] Hoak D.E. et al. 1978. USAF Stability and Control

DATCOM, Flight Control Division. Air Force Flight

Dynamics Laboratory, WPAFB, Ohio.

[3] Torenbeek E. 1982. Synthesis of Subsonic AirplaneDesign, Delft Univ. Press, Delft, the Netherlands.

[4] Hanke C.R. 1971. The Simulation of a Large Jet

Transport Aircraft. Vol. I: Mathematical Model.

NASA CR-1756, March.

[5] Abbott I.H. and von Doenhoff A.E. 1959. Theory of

Wing Sections, Dover, New York.

[6] Nelson R. C. 1989. Flight Stability and Automatic

Control. McGraw-Hill, New York.

[7] Mac Millin P.E., Golovidov O.B., Mason W.H.,

Grossman B. and Haftka R.T. 1996. Trim Control and

Performance Effects in Variable-Complexity High-Speed Civil Transport Design, MAD 96-07-01, July.

[8] Heffley R.K. and Jewell W.F. Aircraft Handling

Qualities Data, NASA CR.

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141

Table-1.Roll angle response at different values of K and GR.

stabilityMax peak

over shootSteady timeImaginary rootReal rootCharacteristics equationsKGR

stable1.29990.9999- -6.99412.053j-36.01110.1

stable1.41210.9925-4.37217.61j-41.25420.1

unstable1.60980.9591-2.60221.34j-44.79430.1

unstable1.86700.1842-0.21024.056j-47.57940.1

unstable1.84780.1495-0.04126.292j-49.91750.1

unstable1.78540.3195-0.97728.214j-51.95560.1

unstable1.75010.3195-1.88729.913j-53.77470.1

unstable1.47790.4961-2.71231.444j-55.42580.1

unstable1.49840.5106-3.47132.843j-56.94390.1

unstable1.61900.5752-4.17534.136j-58.351100.1

unstable1.29990.1360-21.90225.188j-6.197901.3610.2

unstable1.29990.6129-16.69327.668j-16.61220.2

stable1.29990.9915-11.49125.154j-27.07130.2

sable1.29990.9670-8.35427.557j-33.29040.2

stable1.29990.8732-6.21529.662j-37.56850.2

stable1.482020.7698-4.55731.496j-40.88460.2

unstable1.56900.5990-3.18733.124j-43.62470.2

unstable1.65500.4922-2.02934.593j-45.98180.2

unstable1.72120.3995-0.96935.936j-48.06190.2

unstable1.79560.2237-0.03337.177j-49.932100.2

unstable0.67010.1349-23.19137.014j-3.61710.3

unstable0.913540.8902-21.03836.051j-7.92220.3

unstable0.98370.9401-18.51035.391j-12.97930.3

unstable1.000480.9730-15.75535.282j-18.48940.3

Stable1.011730.9553-13.11935.778j-23.76150.3

stable1.022650.9804-10.83636.666j-28.32860.3

stable1.06090.8704-8.91837.715j-32.16470.3

stable1.10450.8148-7.29438.803j-35.41280.3

stable1.16190.7619-7.29438.803j-38.41290.3

stable1.461030.6737-5.89239.882j-38.215100.3

unstable1.29990.1075-23.69345.613j-2.61810.4

unstable1.29990.1274-22.26345.016j-5.47220.4

unstable1.29990.1046-20.70544.513j-8.52030.4

unstable1.29990.6192-19.02844.149j-11.94440.4

unstable1.29990.8044-17.26943.972j-15.46150.4

unstable1.29990.9151-15.48744.007j-19.02560.4

stable1.29990.9983-13.57044.251j-22.50270.4

stable1.29990.9732-12.10944.662j-25.78280.4stable1.29990.9568-10.59145.201j-28.81790.4

stable1.29990.9994-9.20245.827j-31.595100.4

unstable1.29990.1144-23.97252.251j-2.055S3+50s2+3455.92s+6901.3610.5

unstable1.29990.1191-22.8852.331j-4.231S3+50s2+3455.92s+13802.7220.5

unstable1.29990.1121-21.73651.958j-6.526S3+50s2+3455.92s+20704.0830.5

unstable1.29990.40454-20.53151.640j-8.936S3+50s2+3455.92s+27605.4440.5

unstable1.29990.6363-19.27651.401j-11.446S3+50s2+3455.92s+34506.850.5

unstable1.29990.73411-17.98451.262j-14.030S3+50s2+3455.92s+41408.1660.5

unstable1.29990.74412-16.67253.302j-16.655S3+50s2+3455.92s+48309.5270.5

unstable1.29990.61467-15.36054.084j-19.280S3+50s2+3455.92s+55210.8880.5

stable1.29990.9728-14.06054.872j-21.869S3+50s2+3455.92s+62112.2490.5

stable1.29990.5494-12.81255.556j-24.375S3+50s2+3455.92s+69013.61100.5

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142

Table-2.Roll angle response at different values of K and GR.

Steady state time(sec)Time rise(sec)Overshoot(rad)Gain(K)GR

0.550.2050.1510.10.600.1020.4220.1

10.10.630.1

More than 10.10.7540.1

0.330.10.2540.2

0.580.0750.4360.2

0.970.070.6580.2

More than 10.060.8100.2

0.25Over dampingOver damping40.3

0.40.090.1660.3

0.460.750.3280.30.80.650.45100.3

Table-3.Result of maximum aileron deflection required for different values of K and CR.

GR Kmax ) max(rad)

1 0.72 0.062

2 1.38 0.120

3 2.04 0.1780.1

4 3.19 0.278

4 2.42 0.2111

6 3.56 0.3118 4.65 0.397

0.2

10 5.18 0.452

4 2.17 0.189

6 3.29 0.287

8 4.29 0.3740.3

10 6.11 0.533

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143

Table-4.Result of maximum actuator slewing rate required for different K and CR.

GR Kmax (1/Sec) )max(rad/sec)

1 49.6884.336

2 99.910 8.719

3 149.870 13.078

4 199.760 17.4380.1

4 199.760 17.437

6 299.649 26.149

8 399.593 34.8710.2

10 499.501 43.589

4 199.816 17.437

6 299.79 26.162

8 399.796 34.8640.3

10 500.398 43.668

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0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

1.40

1.80

0.0000

0.4000

0.8000

1.2000

1.6000

2.0000

RollAngleResponse

GR=0.1

K=1

K=2

K=3

K=4

Figure-1.Rolling response verses time response for GR = 0.1 and different gain value.

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

1.40

1.80

0.0000

0.4000

0.8000

1.2000

1.6000

2.0000

RollAngleResponse

K at constant GR

K=5

K=7

K=9


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145

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

1.40

1.80

0.0000

0.4000

0.8000

1.2000

1.6000

2.0000

RollAngleResponse

K at constant GR

K=6

K=8

K=10

Figure-3.Rolling response verses time response for GR=0.1 and Different Gain value.

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time(sec)

0.20

0.60

1.00

0.0000

0.4000

0.8000

1.2000

RollAngleResponse

K=1------GR=0.2

K=2------GR=0.2

K=3------GR=0.2


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146

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time(sec)

0.20

0.60

1.00

1.40

0.0000

0.4000

0.8000

1.2000

1.6000

RollAngleResponse

K=4------GR=0.2

K=5------GR=0.2

K=6------GR=0.2


0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time(sec)

0.20

0.60

1.00

1.40

1.80

0.0000

0.4000

0.8000

1.2000

1.6000

2.0000

RollAngleResponse

K=7------GR=0.2

K=8------GR=0.2

K=9------GR=0.2

K=10------GR=0.2


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147

0.20 0.60 1.00 1.400.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

0.0000

0.4000

0.8000

1.2000

RollAngleResponse

GR=0.3

K=1

K=2

K=3

K=4


0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

1.40

0.0000

0.4000

0.8000

1.2000

1.6000

RollAngleResponse

GR=0.3

K=5

K=6

K=7


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148

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

1.40

0.0000

0.4000

0.8000

1.2000

1.6000

RollAngleResponse

GR=0.3

K=8

K=9

K=10


0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

-0.20

0.20

0.60

1.00

-0.4000

0.0000

0.4000

0.8000

1.2000

RollAngleResponse

GR=0.4

K=1

K=2

K=3

K=4


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149

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

-0.20

0.20

0.60

1.00

-0.4000

0.0000

0.4000

0.8000

1.2000

RollAngleResponse

GR=0.4

K=5

K=6

K=7

Figure-11.Rolling response verses time response for GR=0.4 and different gain value.

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.20

0.60

1.00

1.40

0.0000

0.4000

0.8000

1.2000

1.6000

RollAngleResponse

GR=0.4

K=8

K=9

K=10


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150

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

-0.20

0.20

0.60

1.00

-0.4000

0.0000

0.4000

0.8000

1.2000

RollAngleResponse

GR=0.5

K=1

K=2

K=3

K=4

Figure-13.Rolling response verses time response for GR=0.5 and different gain value.

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

0.10

0.30

0.50

0.70

0.90

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

R

ollAngleResponse

GR=0.5

K=5

K=6

K=7


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151

0.20 0.60 1.00 1.40

0.00 0.40 0.80 1.20 1.60

Transient Time

-1.00

1.00

3.00

5.00

-2.0000

0.0000

2.0000

4.0000

6.0000

RollAngleResponse

GR=0.5

K=8

K=9

K=10


0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-6.0000

-2.0000

2.0000

6.0000

-8.0000

-4.0000

0.0000

4.0000

8.0000

Aileron

Deflection

(Rad)

GR=0.1

K=2

K=4

K=6

Figure-16.Aileron deflection verses time response for GR = 0.1 and different gain value.

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152

0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-6.0000

-2.0000

2.0000

6.0000

-8.0000

-4.0000

0.0000

4.0000

8.0000

Aileron

Deflection

(Rad)

GR=0.1

K=8

K=10


0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-6.0000

-2.0000

2.0000

6.0000

-8.0000

-4.0000

0.0000

4.0000

8.0000

AileronDeflection(Rad)

GR=0.2

K=2

K=4

K=6


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153

0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-3.3333

-2.6667

-1.3333

-0.6667

0.6667

1.3333

2.6667

3.3333

-4.0000

-2.0000

0.0000

2.0000

4.0000

Aileron

Deflection

(Rad)

GR = 0.3

K=2

K=4

K=6

Figure-19.Aileron deflection verses time response for GR = 0.2.

0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-6.0000

-2.0000

2.0000

6.0000

-8.0000

-4.0000

0.0000

4.0000

8.0000

Aileron

Deflection

(Rad)

GR=0.2

K=8

K=10


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154

0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-6.6667

-5.3333

-2.6667

-1.3333

1.3333

2.6667

5.3333

6.6667

9.3333

10.6667

-8.0000

-4.0000

0.0000

4.0000

8.0000

12.0000

Aileron

Deflection

(Rad)

GR = 0.3

K=8

K=10


0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-3.0000

-1.0000

1.0000

3.0000

-4.0000

-2.0000

0.0000

2.0000

4.0000

AileronDeflection(Rad)

GR=0.4

K=2

K=4K=6


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155

0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-6.0000

-2.0000

2.0000

6.0000

-8.0000

-4.0000

0.0000

4.0000

8.0000

Aileron

Deflection

(Rad)

GR=0.4

K=8

K=10


0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-1.0000

1.0000

3.0000

-2.0000

0.0000

2.0000

4.0000

A

ileronDeflection(Rad)

GR=0.5

K=2

K=4

K=6


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156

0.1000 0.3000 0.5000 0.7000 0.9000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000

Time Response (sec)

-1.5000

-0.5000

0.5000

1.5000

2.5000

-2.0000

-1.0000

0.0000

1.0000

2.0000

3.0000

Aileron

Deflection

(Rad)

GR=0.5

K=8

K=10


Appendix A

Figure-A1.Control system design simulation using Matlab.

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Figure-A2.Control system design result using Matlab, Gr = 0.3 K = 8.

jeas_1111_596

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