CONTOL SYTEM DESIGN ASSIGNMENT – ALICJA WASIK

Contents

Question 1. Design of the phase lead & lag compensation using root locus approach.............................................................1

Question 2. Design of phase lag compensation using frequency domain approach.................................................................5

QUESTION 1. DESIGN OF THE PHASE LEAD & LAG COMPENSATION USING ROOT LOCUS APPROACH

STEP I : DETERMINATION OF THE REQUIREMENTS OF THE SYSTEMG (s )= Ks (s+6 ) (s+10 )

From the design specifications of the system the damping ratio and undamped natural frequency may be obtained:

P .O .=15%=100e−πζ

√1−ζ 2 ζ= −ln0.15

√π2+( ln 0.15)2=0.5169

T s<1≅4ωn ζ

ωn≥4T sζ≈7.738

Therefore the choice is ζ =0.52 and ωn=7.74 .

The desired root positions are the following:

r1,2=−ωn ζ ± j ωn√1−ζ 2

r1,2=−4± j6.6

It can be clearly seen from the root locus that the desired roots cannot be obtained using the simple gain adjustment.

ST E P I I : INTRODUCTION OF THE LEAD COMPENSATOR TO MEET THE REQUIREMENTS :

GLEAD(s)=s+zs+ p

Since the desired roots have to appear on the root locus of the compensated system, the angle condition has to be satisfied:

θ=≮� GLEAD ( s)G (s )=−1800

Placing the zero just to the left of the first two real poles 0 and 6 gives z = 6.2.

≮� r z−≮� r0−≮� r 6−≮� r10−≮� r p=−180∘

≮� r z=≮� (−4+ j 6.6+6.2 )=tan−1 6.62.2

=71.60

≮� r0=≮� (−4+ j6.6 )=1800−tan−1 6.64

=121.20

≮� r6=≮� (−4+ j6.6+6 )=tan−1 6.62

=73.10

≮� r10=≮� (−4+ j6.6+10 )=tan−1 6.66

=47.70

≮� r p=≮� (−4+ j6.6+p )=tan−1 6.6p−4

≮� r z−≮� r0−≮� r 6−≮� r10−≮� r p=71.60−121.2∘−73.1∘−47.7∘−≮� r p=−1800

≮� r p=9.6∘

p= 6.6

tan 9.6o+4=39

GLEAD(s)=s+6.2s+39

The gain K0 can be obtained from the magnitude condition to assure that the desired roots are placed appropriately on the root locus branches of the compensated system:

|GLEAD (s )G ( s)|=1

| K (r1+4 )(r1)(r1+6)(r 1+10)(r1+16.4)|=1

K=|(−4+ j 6.6)(2+ j6.6)(6+ j 6.6)(35+ j6.6)(2.2+ j 6.6) |=2430However K=2272 has to be chosen in order not to change the dominant poles of the closed-loop system.

GLEAD (s )G ( s)= 2272(s+6.2)s(s+6)(s+10)(s+39)

Closed - loop transfer function:

2271.8521(s+6.2)(s+40.8)(s+6.301)(s2+7.901 s+54.79)

=2271.8521(s+6.2)

(s+40.8 ) ( s+6.301 ) (−3.9505+6.2597 i )(−3.9505−6.2597 i)

STEP III : INTRODUCTION OF THE LAG COMPENSATOR TO ASSURE THAT THE SYSTEM MEETS

THE VELOCITY FACTOR CONDITION K v=40:

This steady-state specification cannot be obtained by adding simple gain compensator since it would shift the position of the roots on the root locus branches and the satisfying transient response of the desired roots would be lost.

GLAG (s )= s+z'

s+ p'=α

1+ sz '

1+ sp'

α= z'

p'

K vo=lims→0sGLEAD (s )G ( s )=lim

s→0s

2272(s+6.2)s (s+6)(s+10)(s+39)

=6

K v=lims→0sGLAG (s )GLEAD (s )G ( s )=lim

s→0sGLAG (s )∗K vo=αK vo

α=K vK vo

=406

=6.67

Choose the values of zero and pole for a lag compensator close to the origin of s-plane to ascertain that the negative angle of GLAG(s) will not affect considerably the root locus achieved by adding the lead compensator.

z '<ωn ζ

10= 410p '= z

'

α=0.06

GLAG (s )= s+0.4s+0.06

10-4

10-2

100

102

104

-270

-225

-180

-135

-90

P.M.: 50.1 degFreq: 5.23 rad/sec

Frequency (rad/sec)

Pha

se (

deg)

-150

-100

-50

0

50

100

G.M.: 17.9 dBFreq: 19.1 rad/secStable loop

Open-Loop Bode Editor for Open Loop 1 (OL1)

Mag

nitu

de (

dB)

-10 -8 -6 -4 -2 0-10

-8

-6

-4

-2

0

2

4

6

8

100.090.20.30.420.540.68

0.84

0.95

0.090.20.30.420.540.68

0.84

0.95

2

4

6

8

10

2

4

6

8

10

Root Locus Editor for Open Loop 1 (OL1)

Real Axis

Imag

Axi

s

X: -3.771Y: 6.115Z: 3

Step Response

Time (sec)

Am

plitu

de

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4System: clI/O: Input to Out(1)Peak amplitude: 1.21Overshoot (%): 20.6At time (sec): 0.529

System: clI/O: Input to Out(1)Time (sec): 0.891Amplitude: 1.04

From: Input To: Out(1)

Unfortunately this lag compensator decreased the desired phase margin influencing significantly the design. For further improvement and achievement of better performance the positions of zero and pole of the lag compensator can moved

even closer to the imaginary axis.

First iteration: z’=0.3, p’=z 'α

=0.045

Therefore

GLAG (s )= s+0.3s+0.045

Open loop transfer function is the following:

GLAG (s )GLEAD (s )G ( s )=2272 (s+6.2 )(s+0.3)

s ( s+6 ) (s+10 ) ( s+39 )(s+0.045)

Closed loop transfer function:

T ( s )=GLAG (s )GLEAD (s )G ( s)1+GLAG ( s )GLEAD ( s)G (s )

=2272(s+6.2)(s+0.3)

(s+40.79)(s+6.307)(s+0.3134)(−3.8185+6.1505 i)(−3.8185−6.1505 i)

Step Response

Time (sec)

Am

plitu

de

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

1

1.2

1.4System: xI/O: Input to Out(1)Peak amplitude: 1.19Overshoot (%): 19At time (sec): 0.536

System: xI/O: Input to Out(1)Settling Time (sec): 0.84

System: xI/O: Input to Out(1)Rise Time (sec): 0.225

System: xI/O: Input to Out(1)Final Value: 1

From: Input To: Out(1)

Overshoot of the step response is 19%.Settling time is 0.84s.

Steady state error of the unit ramp is 100-99.98=0.02, so less than 1K v

= 140

=0.025 and this specification is satisfied.

The compensated system does not meet all the design specifications; however the responses are satisfactory enough for those compensators to be chosen and implemented.

QUESTION 2. DESIGN OF PHASE LAG COMPENSATION USING FREQUENCY DOMAIN APPROACH

ST E P I : DE T E R M I N A T I O N O F R E Q U I R E M E N T S O F T H E S Y S T E M

K AG (s )=K A

s (s+6 ) (s+10 )

By converting the time-domain specification of percentage overshoot to frequency domain obtain the value of the required

damping ratio and hence the phase margin:

P .O .=100e−πζ

√1−ζ 2=15%

ζ=0.5169

PM≈100 ζ=52°

It can be clearly seen on the bode diagram that the uncompensated system G(s) does not meet this requirement.

Therefore the gain KA has to be adjusted to shift the magnitude plot up so that it intersects 0dB where the desired PM occurs.

ST E P I I : IN T R O D U C T I O N O F T H E G A I N K A :First find the frequency that gives φ=−180 °+52°+5 °=−128 ° which will become the new gain crossover

frequencyωC.

≮� G ( j ωC )=−≮� ¿ j ωC−≮� ( j+6)−≮� ( j ωC+10)=−1280

90 °+ tan−1ωC6

+ tan−1ωC10

=1280

ωC=−162 tan 380

±√162+4∗60¿¿¿¿

ωC=2.6 rad / s

Total gain of the gain compensated system KAG(s) should be 0 dB, which implies:

0 20 40 60 80 1000

20

40

60

80

100X: 100Y: 99.98

|K AG ( jωC )|=| K A( j ωC)( jωC+6)( j ωC+10)|=1

K A=ωC√ ( j ωC )2+62√ ( jωC )2+102=175.7

K AG (s )= 175.7s (s+6 ) (s+10 )

With dominant roots at: r1 K ,2 K=−1.35± j 2.22

STEP III: DESIGN OF THE PHASE LAG COMPENSATOR

Design of the phase lag compensator is needed to assure that the system meets the velocity factor condition Kv=40, maintaining required transient response characteristic of 15% percentage overshoot:

GLAG2 (s )=s+zLs+ pL

=β1+ sz L

1+ spL

β=zLpL

K v=lims→0s K AGLAG2 (s )G (s )=lim

s→0ss+ zLs+ pL

175.7s ( s+6 ) ( s+10 )

=175.760β

β=60K v175.7

=13.66

Placing the zero at a decade below ωC gives: zL=ωC10

=0.26 rad /s pL=zLβ

=0.019

GLAG2 (s )= s+0.26s+0.019

Open loop transfer function is:

K AGLAG2 (s )G ( s)=175.7(s+0.26)s ( s+6 )¿¿

Closed loop transfer function is:

T 2 ( s)=K AGLAG 2 ( s)G (s )1+K AGLAG2 (s )G ( s)

=175.7(s+0.26)

(s+12.25)(s+0.2837)(s+0.322)(s2+3.485 s+13.14)

With dominant roots at:

r1 KLAG, 2KLAG=−1.74 ± j3.18

The compensated system does not meet the overshoot specifications. The overshoot of a step response rises to 26% which is high above requested 15%.

Adding a lag compensator introduces a zero in the closed-loop transfer function T2(s) which moves the dominant roots thus affecting the transient response of the system. Its negative effect can be decreased by addition of about 5 – 12 degrees to the phase margin requirement and iteration of the design steps.

The other solution is using a pre-filter which cancels this negative effect of zero.

ST E P IV: INTRODUCTION OF THE LEAD COMPENSATOR TO MEET THE REQUIREMENTS :

F ( s )=s+ pFs+ pF

= 2.8s+2.8

Y (s)R (s)

=F (s )T2 (s )=

2.8s+2.8

∗175.7(s+0.26)

(s+12.25)(s+0.2837)(s+0.322)(s2+3.485 s+13.14)

The compensated system meets all the design specifications.