Title: ANTENNA AZIMUTH POSITION CONTROL SYSTEM MODELLING, ANALYSISAND CONTROLLER DESIGN

Course: CONTROL SYSTEMS IV (SECOND SEMESTER 2012)DEPARTMENT: (ELECTRICAL ENGINEERING)

INTRODUCTION

This will include everything in page 14‐15 (Include the diagrams)

CHALLENGE 1

Evaluation of the transfer function of sub systems‐

v i (s )θi (s )

=10π

=3.18

V p(s )V e(s)

=K

Ea(s)V p(s )

= 150s+150

Jm=J a+J L (N 1

N 2)2

=0.25

Dm=Da+DL( N1

N2)2

=0.13

θm(s )Ea(s)

=

K t

Ra Jm

s [s+ 1Jm

(Dm+K t Kb

Ra

)]= 0.8

s(s+1.32)

θo(s )Ea(s)

=0.2θm(s)Ea(s )

= 0.16s(s+1.32)

Parameter Configuration2Kpot 3.18K -K1 150a 150Km 0.8am 1.32K g 0.2

CHALLENGE 2

State space representation of each dynamic sub systems‐

G (s )=Ea(s)V p(s)

= 150(s+150)

( s+150 )Ea(s)=150V p(s )

d ea

dt+150ea=150V p(t)

d ea

dt=−150ea+150V p(t)

y=ea

ea (t )=ia (t )Ra+K b

dθm

dt

T m (t )=K t ia ( t )=Jm

d2θm

d t 2+Dm

dθm

dt

ea ( t )=( Ra Jm

K t) d

2θm

d t 2+(

Dm Ra

K t

+Kb)dθm

dt

x1=θm

x2=dθm

dt

ea (t )=( Ra Jm

K t) d x2

dt+(

DmRa

K t

+Kb)x2

d x2dt

=−1Jm

(Dm+K t K b

Ra)x2+( K t

Ra Jm)ea(t )

d x1dt

=0+x2

d x2dt

=0− 1Jm

(Dm+K t Kb

Ra)x2+( K t

Ra Jm)ea(t)

y=0.2x1

x=[0 1

0−1Jm

(Dm+K t Kb

Ra)] x+[ 0

K t

Ra Jm]ea(t)

y= [0.2 0 ] x

x=[0 10 −1.32] x+[ 00.8 ]ea( t)

y= [0.2 0 ] x

CHALLENGE 3

a. Prediction of the open-loop angular velocity response of the power amplifier, motor, and load to a step voltage at the input to the power amplifier.

b. Finding the damping ratio and natural frequency of the open-loop system.

c. Derivation of the open-loop angular velocity response of the power amplifier, motor, and load to a step-voltage input using transfer functions.

d. The open-loop state and output equations.

e. Using MATLAB to obtain a plot of the open-loop angular velocity response to a step-voltage input.

a.V p(s) Ea(s) θo(s) wo(s)

V p(s) wo(s)

a.wo (t )=A+Be−150 t+C e−1.32t

b.

G (s )= 24

s2+151.32 s+198

wn=√b=14.07

ζ= a2wn

= 151.322×14.07

=5.38(overdamped )

c.

24(s+150 )(s+1.32)

= As+ Bs+150

+ cs+1.32

24(s+150 )(s+1.32)

=0.121s

+ 1.08e−3

s+150+−0.122s+1.32

wo (s )=0.121s

+ 1.08e−3

s+150− 0.122

s+1.32

wo (t )=0.121+1.08×10−3 e−150t−0.122e−1.32 t

d.

150(s+150)

0.16s (s+1.32)

s24

(s+150 )(s+1.32)

wo(s)V p(s )

= 24s2+151.32 s+198

wo+151.32 ˙wo+198wo=24v p

x1=wo

x2=wo

x1=x2

x2=−198 x1−151.32x2+24 v p

y=x1

x=[ 0 1−198 −151.32] x+[ 024 ]v p(t)

y= [1 0 ] x

e.

num=24

den=poly([-150 -1.32])

G=tf(num,den)

Step(G)

CHALLENGE 4

a. Finding the closed-loop transfer function using block diagram reduction.

b. Representation of each subsystem with a signal-flow graph and finding the state-space representation of the closed-loop system from the signal-flow graph.

c. Using the signal-flow graph found in (b) along with Mason's rule to find the closed-loop transfer function.

d. Replace the power amplifier with a transfer function of unity and evaluate the closed-loop percent overshoot, settling time, and peak time for K = 5.

e. For the system used for(d), derive the expression for the closed loop step response.

f. For the simplified model in (d), find the value of preamplifier gain, K, to yield 15% overshoot.

a.

b.

x=[x1x2ea]

10π

150(s+150)

K 0.16s (s+1.32)

10π

76.39Ks (s+1.32 )(s+150)

0.16s (s+1.32)

10Kπ

150(s+150)

76.39K

s3+151.32 s2+198 s+76.39K

x1=0+x2

x2=0−1.32x2+0.8ea

ea=−95.49K x1−150ea+477.46K θi

y=θo=0.2 x1

x=[ 0 1 00 −1.32 0.8

−95.49K 0 −150] x+[ 00

477.46K ]θi

y= [0.2 0 0 ] x

c.

T 1=( 10π ) (K ) (150 )( 1s ) (0.8 )( 1s )( 1s ) (0.2 )=76.39Ks3

GL1 ( s )=−150s

GL2(s)=−1.32

s

GL3 (s )=(−10π )(K ) (150 )( 1s ) (0.8 )( 1s )( 1s ) (0.2 )=−76.39Ks3

GL1 ( s )GL2(s)=198

s2

∆=1− [GL1 (s )+GL2 ( s )+GL3 ( s) ]+[GL1 (s )GL2 (s )]

¿1+ 150s

+ 1.32s

+ 76.39Ks3

+ 198s2

∆1=1

T ( s )=C (s)R (s)

=T 1∆1∆

= 76.39Ks3+151.32 s2+198 s+76.39K

d.

G (s )= 2.546s (s+1.32)

T ( s )= 2.546

s2+1.32 s+2.546 s

wn=√b=√2.546=1.596

ζ= a2wn

=0.41

T ( s )= 4−0.41×1.596

=−6.113 s=6.113 s

T p=π

wn√1−ζ 2=2.158 s

%OS=e−( ζπ

√1−ζ 2)

×100=24.37%

e.

C ( s )= 2.546

s (s2+1.32 s+2.546)

C ( s )=1s− 125.5

25.5+33s2+1.32 s+2.546

c (t )=1−e−0.66(cos [1.454 t ]+0.454 sin [1.454 t ] )

f.

G (s )= 0.00255ks (s+1.32)

T ( s )= 0.00255K

s2+1.32 s+0.00255K

ζ=−ln(%OS

100)

√π2+ln2(%OS100

¿)=0.517¿

wn=√0.00255K∧2 ζ wn=1.32

ζ= 1.322√0.00255K

=0.517 therefore K=660.8

For 15% overshoot

CHALLENGE 5

Finding the range of preamplifier gain required to keep the closed-loop system stable.

T ( s )= 76.39K

s3+151.32 s2+198 s+76.39K

Routh table

s3 1 198

s2 151.32 76.39K

s1 29961.36-76.39K 0

s0 76.39K 0

Therefore29961.36-76.39K=0K=392.22For stability 0< K < 392.22

CHALLENGE 6

a. Find the steady-state errors in terms of gain, K, for step, ramp, and parabolic inputs.

b. Find the value of gain, K, to yield a 20% error in the steady state.

a. e (∞ )=lims→0

sE ( s )=¿lims→0

sR(s)

1+G(s)¿

G (s )= 76.39Ks (s+1.32 )(s+150)

for step input R (s )=1s

e (∞ )=0

for ramp input R ( s)= 1s2

e (∞ )= 19876.39K

for ramp input R ( s)= 1s3

e (∞ )=∞

b. e (∞ )=0.2= 1K v

=(1.32 )(150)76.39K

=198

76.39K=2.59K

K=12.96

For stability 0< K < 392.22

CHALLENGE7

a. Find the preamplifier gain, K, required for an 8-second settling time.

b. Repeat, using MATLAB.

a.

G (s )= 76.39Ks (s+1.32 )(s+150)

T s=8 s

T s=4

−ζ wn

=8 s

−ζ wn=0.5

Closed loop pole= -0.5+ j6.4 the gain value yields 76.39K=6210.507Therefore K=81.3

b.

Programnumg= 1; deng=poly([0 -150 -1.32]); G=tf(numg,deng) rlocus(G) axis([-2,0,-10,10]); grid on [K1,p]=rlocfind(G) K=K1/76.39

Computer ResponseTransfer function: 1-----------------------s^3 + 151.3 s^2 + 198 s Select a point in the graphics window

selected_point =

-0.5134 + 6.0735i

K1 =

5.5870e+003

p =

1.0e+002 *

-1.5025 -0.0054 + 0.0607i -0.0054 - 0.0607i

K =

73.1383

CHALLENGE 8

a. For your solution to the challenge in Chapter 8, evaluate the percent overshoot and the value of the appropriate static error constant.

b. Design a cascade compensator to reduce the percent overshoot by a factor of 4 and the settling time by a factor of 2. Also, improve the appropriate static error constant by a factor of 2.

c. Repeat Part b using MATLAB.

a. Closed loop poles=-0.5+ j6.4

G (s )= 76.39Ks (s+1.32 )(s+150)

ζ=cos¿¿

%OS=e−( ζπ

√1−ζ 2)

×100=78.21%

K v=76.39×81.31.32×150

=31.37

b.

new%OS=78.21%4

=19.55%

ζ=−ln (%OS

100)

√π2+ln2(%OS100

¿)=0.461¿

Lead compensator design

new settling time isT sn=T s

2=4

an=−ζ wn=4T sn

=1

wn=1ζ=2.169 rad / s

bn= j wn√1−ζ 2= j 1.925

Dominant second order pole is -1+ j 1.925Assuming a zero at -1.5 yields -56.69 which is the angular contribution required from compensator pole

1.925pc−1

=tan (56.69 )

pc=2.265

Glead (s )= 76.39K (s+1.5)s (s+1.32 ) ( s+150 )(s+2.265)

Lag compensator designK=9.53 for the lead compensated

K v=76.39×9.53×(1.5)(1.32 ) (150 )(2.265)

=2.43

the amount of improvement required is2×31.372.43

=25.82

Choosing pc=−0.0001 therefore zc=25.82×0.0001=0.002582

Glag (s)=(s+0.002582)(s+0.0001)

GLLC=76.39K (s+1.5 )(s+0.002582)

s ( s+1.32 ) (s+2.265 ) (s+150 )(s+0.0001)

c.

Program and computer response>> num=76.39den=poly([0 -1.32 -150])G=tf(num,den)rlocus(G)zun=0.078sgrid(zun,0)numld=[76.39 114.585]den=poly([0 -1.32 -150])Pn=-1+1.925iangle_Pn=angle(polyval(numld,Pn)/polyval(den,Pn))*(180/pi)O=-angle_Pn-180den1=poly([0 -1.32 -2.265 -150])Gld=tf(numld,den1)rlocus(Gld)z=0.461sgrid(z,0)numlg=[1 0.002581]denlg=[1 0.0001]Glg=tf(numlg,denlg)Gllc=series(Gld,Glg)rlocus(Gllc)

z=0.461sgrid(z,0)K=8.61T=feedback(K*Gllc,1)step(T)

num =

76.3900

den =

1.0000 151.3200 198.0000 0

Transfer function: 76.39-----------------------s^3 + 151.3 s^2 + 198 s

zun =

0.0780

numld =

76.3900 114.5850

den =

1.0000 151.3200 198.0000 0

Pn =

-1.0000 + 1.9250i

angle_Pn =

-123.3133

O =

-56.6867

den1 =

1.0000 153.5850 540.7398 448.4700 0

Transfer function: 76.39 s + 114.6-------------------------------------s^4 + 153.6 s^3 + 540.7 s^2 + 448.5 s

z =

0.4610

numlg =

1.0000 0.0026

denlg =

1.0000 0.0001

Transfer function:s + 0.002581------------ s + 0.0001 Transfer function: 76.39 s^2 + 114.8 s + 0.2957---------------------------------------------------s^5 + 153.6 s^4 + 540.8 s^3 + 448.5 s^2 + 0.04485 s

z =

0.4610

K =

8.6100

Transfer function: 657.7 s^2 + 988.3 s + 2.546--------------------------------------------------------s^5 + 153.6 s^4 + 540.8 s^3 + 1106 s^2 + 988.3 s + 2.546

CHALLENGE 9

a. Find the range of gain for stability. ___

b. Find the percent overshoot for a step input if the gain, K, equals 3.

C. Repeat Parts a. and b. using MATLAB

K1=v i (s )θi (s )

=10π

=3.18

V p(s )V e(s)

=K

G1 (s )=Ea(s)V p(s )

= 100s+100

Jm=J a+J L (N 1

N 2)2

=0.25

Dm=Da+DL( N1

N2)2

=0.13

Gm (s )=θm(s)Ea(s)

=

K t

Ra Jm

s [s+ 1Jm

(Dm+K t K b

Ra

)]= 0.8

s (s+1.32)

K2=50250

=0.2

G(s)=K1KG1Gm K2=50.88K

s ( s+1.32 )(s+100)

Parameter Configuration3Kpot 3.18K -K1 100a 100Km 0.8am 1.32 and 100K g 0.2

a. Phase response is -180 at w=11.8 rad/sThe magnitude plot is -48.9dB

K=10( 48.920

)=278.61

Therefore the system is stable if 0 ‹ K ‹ 278.61

b. If K=3The magnitude curve is moved up by 20log3 = 9.54 dBThe adjusted magnitude curve goes through zero dB at w=1The phase angle (∅) is -128 yielding phase margin 53.96

∅=tan−1 2 ζ

√−2 ζ 2+√1+4 ζ 4

ζ=0.527

%OS=e−( ζπ

√1−ζ 2)

×100=14.25%

c.Program and computer response

num=50.88den=poly([0 -1.32 -100])G=tf(num,den)bode(G)grid onnum1=152.64den1=poly([0 -1.32 -100])G1=tf(num1,den1)bode(G1)grid on[Gm,Pm,Wc,Wcp]=margin(G1)

num =

50.8800

den =

1.0000 101.3200 132.0000 0

Transfer function: 50.88-----------------------s^3 + 101.3 s^2 + 132 s

num1 =

152.6400

den1 =

1.0000 101.3200 132.0000 0

Transfer function: 152.6-----------------------s^3 + 101.3 s^2 + 132 s

Gm =

87.6195

Pm =

53.9644

Wc =

11.4891

Wcp =

0.9414

>> for z=0:.01:1 Pme=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); if Pm-Pme<=0; break end end z percent=exp(-z*pi/sqrt(1-z^2))*100

z =

0.5300

percent =

14.0366

CHALLENGE 10

Gain Design

a. Find the value of K to yield 25% overshoot for a step input.

b. Repeat Part a using MATLAB.

a.

G (s )= 50.88Ks (s+1.32 )(s+100)

Let K=1

%OS=25%

Phase margin= 43.63˚Therefore the phase angle=-180+43.63=-136.537 at w=1.34 rad/s where the gain is -13.9 dB

K=10( 13.920

)=4.95(13.9dB)

b.

Program and computer response

OS=0.25z=-log(OS)/sqrt(pi^2+(log(OS))^2)Pmc=atan(2*z/sqrt(-2*z^2+sqrt(1+4*z^4)))*(180/pi)phase_angle=-180+Pmcnum=50.88den=poly([0 -1.32 -100])G=tf(num,den)bode(G)grid onK=10^(13.9/20)K=4.95T=feedback(K*G,1)step(T)

OS =

0.2500

z =

0.4037

Pmc =

43.4630

phase_angle =

-136.5370

num =

50.8800

den =

1.0000 101.3200 132.0000 0

Transfer function:

50.88-----------------------s^3 + 101.3 s^2 + 132 s

K =

4.9545

K =

4.9500

Transfer function: 251.9-------------------------------s^3 + 101.3 s^2 + 132 s + 251.9

Cascade Compensation Design

a. Design a lag-lead compensator to yield a 15% overshoot and Kv =20. In order to speed up the system, the compensated system's phase margin frequency will be set to 4.6 times the phase-margin frequency of the uncompensated system.

b. Repeat Part a using MATLAB.

G (s )= 50.88Ks (s+1.32 )(s+100)

%OS=15%

ζ=−ln (%OS

100)

√π2+ln2(%OS100

¿)=0.5169¿

Therefore the phase margin=53.17 and the phase angle=-180+53.17=-126.83 therefore phase frequency w=0.945

K v=20

K v=50.88K

(1.32 )(100)

K=20×13250.88

=51.89

New phase frequency wmax=4.6 ×0.945=4.347 rad/s

Phase angle =-165 at new phase margin frequency

Phase margin=180-165=15

Therefore 53.17-15+5=43.17

Upper break =4.34710

=0.4347 rad /s

∅max=sin−1( 1−β1+β

)

0.684+0.684 β=1−β

β=0.188

y= 1β=5.32

Lower break frequency =1yT

=0.082

Glag=0.188(s+0.4347)(s+0.082)

T= 1wmax √β

=0.53

lower break frequency= 1T

=1.9

upper break frequency= 1βT

=10.04

Glag−lead−comp=50.88 (51.89 ) (s+0.4347 )(s+1.9)

s ( s+1.32 ) ( s+100 ) ( s+0.882 )(s+10.04)

b.Program and computer responseOS=0.15z=-log(OS)/sqrt(pi^2+(log(OS))^2)Pmc=atan(2*z/sqrt(-2*z^2+sqrt(1+4*z^4)))*(180/pi)phase_angle=-180+PmcK=51.89num=2640.1632den=poly([0 -1.32 -100])G=tf(num,den)bode(G)grid onnumc=poly([-0.4347 -1.8])denc=poly([0 -1.32 -100 -0.0748 -10.475])Glag_lead=tf(numc,denc)Kc=51.89*50.88T=feedback(Kc*Glag_lead,1)step(T)

OS =

0.1500

z =

0.5169

Pmc =

53.1718

phase_angle =

-126.8282

K =

51.8900

num =

2.6402e+003

den =

1.0000 101.3200 132.0000 0

Transfer function: 2640-----------------------s^3 + 101.3 s^2 + 132 s

numc =

1.0000 2.2347 0.7825

denc =

1.0e+003 *

0.0010 0.1119 1.2017 1.4720 0.1034 0

Transfer function: s^2 + 2.235 s + 0.7825-----------------------------------------------s^5 + 111.9 s^4 + 1202 s^3 + 1472 s^2 + 103.4 s

Kc =

2.6402e+003

Transfer function: 2640 s^2 + 5900 s + 2066-----------------------------------------------------s^5 + 111.9 s^4 + 1202 s^3 + 4112 s^2 + 6003 s + 2066

DISCUSSION

The objective of the report has been met, however some challenges that were encountered

while doing the report were that some of the formulas required to solve some of the

problems are not clearly easy to identify and the Matlab responses sometimes give

different answers to the examples given on the prescribed book just by a little. Another

challenge was to find mistakes done as a single mistake can make the entire system fail.