PART-A

1. What are constant M and N circles? [Nov/Dec-2015, Nov/dec-2013]Ans:

Closed loop magnitude plots and closed loop phase plots are called M

circles and N circles respectively.

2. Determine the frequency domain specification of a second order system

when closed loop transfer function is given by C(s)/R(s)¿64

s2+10 s+64

[April/May-2010]Ans:

3. Give the specifications used in frequency domain analysis. [Nov/Dec-2015]Ans:

i) Resonant peak. ii) Resonant frequency.

iii)Bandwidth iv)cut-off rate

iv) Phase margin vi) Gain margin

4. Derive the transfer function of a lead compensator network. [April/May-2010]Ans:

5. Draw the polar plot of an integral term transfer function. [May/June-2013]Ans:

6. Write the MATLAB statement to draw the Bode plot of the given system. [May/June-2013]Ans:

The program is,clcnum= [ ] …………..Enter numeratorden = [ ] ………….Enter denominatorsys = tf(num, den) ……Systembode(sys) ……Obtain Bode plot

7. List the advantages of Nichol’s Chart? [Nov/dec-2010]Ans:

The complete closed loop frequency response can be obtained. The value of resonant peak Mr of closed loop system with given G(jw) can

be obtained. The frequency ωr, corresponding to the Mr for the closed loop system can

be obtained. To design the value of K for the given Mr.

8. Define Gain and phase margin. [Nov/dec-2013, Nov/dec-2014]Ans:Gain margin:

The gain margin, kg is defined as the reciprocal of the

magnitude of the open loop transfer function at phase cross over frequency.

Phase margin:

The phase margin, γ is the amount of phase lag at the gain cross over

frequency required to bring system to the verge of instability.

)(1

PCg jG

K

9. What is meant by ‘Corner frequency’ in frequency response analysis? [Nov/dec-2012, April/May-2011, May/June-2014]Ans:

The frequency at which the meeting points of two asymptotic lines in a

magnitude plot is called corner frequency.

10.What is Nichols chart? [Nov/dec-2012]Ans:

The constant magnitude loci (M circles) and constant phase angle loci (N circles) are transferred to the gain phase plot to obtain the resultant chart called Nichol’s chart.

11.Draw the circuit of lead compensator and draw its pole zero diagram. [April/May-2011]Ans:

Pole zero diagram:

12.Write the MATLAB command for plotting Bode diagram Y(s)/U(s)

¿ 4 s+6s3+3 s2+8 s+6

[Nov/dec-2011]

Ans:clcnum= [ 4 6] …………..Enter numeratorden = [1 3 8 6 ] .………….Enter denominatorsys = tf(num, den) ….……Systembode(sys) ……Obtain Bode plot

PART-B

1. Sketch the bode plot for the following transfer function and determine the system gain K for the gain cross over frequency to be 5 rad/sec. [Nov/Dec-2010]

G (s )= Ks2

(1+0 .2 s ) (1+0 .02 s ) .

Ans:Substitute s=jω

G ( jω )= K ( jω )2

(1+0 .2 jω ) (1+0 .02 jω )

Magnitude plot:

The corner frequencies are ωc1=

10.2

=5 rad /sec ωc1=1

0. 02=50 rad /sec

ωl=0. 5 rad /sec ωh=100 rad /sec

TermCorner frequencyin rad/sec

Slope indB/dec

Change in Slopein dB/dec

( jω)2 - 40

1(1+0 .2 jω)

5 -20 20

1(1+0 . 02 jω)

50 -20 0

Atω=ωl , A=20 log (ω )2=20 log (0 .5 )2=−12 db

Atω=ωl , A=20 log (ω )2=20 log (5 )2=−28 db

Atω=ωc2 , A=[slope from ωc1 to ωc2×log

ωc2

ωc1 ]+ A at ωc1=20×log 505

+28=48 db

At ω=ωh , A=[slope from ωc2 to ωh×log

ωh

ωc2 ]+ A at ωc2=0×log 10050

+48=48 db

Phase plot: φ=180−tan−10 .2 ω−tan−1 0.02ω

ω 0.5 1 5 10 50 100

φ 174 168 130 106 50 30

From the graph,

20 log K=−28

log K=−2820

log K=−1. 4K=10−1.4=0. 0398

2. Explain in detail the design procedure of lead compensator using Bode plot. [May/June-2013]Ans:Design procedure of lead compensator:Step 1: Choose the value of K for uncompensated system to meet the steady

state error requirement

Step 2: Sketch the bode plot uncompensated system

Step 3: Determine the phase margin of the uncompensated system.

Step 4: Determine the amount of phase angle to be contributed by the lead

network by using formula φm=γ d−γ+ε

φm=Maximum phase lead angle of the compensatorε =Additional phase lag to compensate = 5°

γd =Desired phase margin

γn =Phase margin of the uncompensated system

If φm is more than 60° then realize the compensator as cascade of two lead

compensators with each compensator contributing half of the required angle.

Step 5: Determine the transfer function of the lead compensator

Calculate α using the equation α=

1−sin φm

1+sin φm

From the bode plot, determine the frequency at which the magnitude of G (jω) is

−20 log 1√α db. This frequency is ωm.

Calculate T from the relation,ωm=

1T √α

∴T= 1ωm √α

Transfer function of the lead compensator=GC( s )=s+ 1

T

s+ 1αT

=α( 1+sT1+sαT )

Step 6: Determine the open loop transfer function of the compensated system

open loop transfer function of the compensated system

=Go( s )=1α

GC( s )G( s )=1α

α (1+sT1+sαT )G( s )=1+sT

1+sαTG( s )

Step 7: Verify the designFinally the Bode plot of the compensated system is drawn and verify

whether it satisfies the given specifications. Otherwise repeat the procedure from

step 4 by taking ε as 5° more than previous design.

3. The open loop transfer function of a unity feedback system is given by

G (s )= 1s2 (1+s ) (1+2 s ) .Sketch the polar plot and determine the gain margin

and phase margin. [ Nov/Dec-2015]Ans:Substitute s=jω

G ( jω )= 1jω2 (1+ jω ) (1+2 jω)

The corner frequencies are ωc1=

12=0. 5 rad /sec ωc2=

11=1 rad /sec

G ( jω )=1jω2 (1+ jω ) (1+2 jω )

=1ω2∠180 °√1+ω2∠ tan−1ω √1+4 ω2∠ tan−12 ω

G ( jω )=1ω2√1+ω2 √1+4 ω2

∠−180 °− tan−1ω−tan−1 2ω

|G ( jω )|=1

ω2 √(1+ω2) (1+4ω2)=

1

ω2√ (1+ω2+4 ω2+4 ω4 )=

1

ω2 √(1+5ω2+4 ω4)∠G ( jω )=−180 °− tan−1ω−tan−12 ω

ω 0.45 0.5 0.55 0.6 0.65 0.7 0.75 1.0

G(jω) 3.3 2.5 1.9 1.5 1.2 1 0.8 0.3

φ -246 -256 -256 -261 -265 -269 -273 -288

Gain margin = Kg=0Phase margin= 90°

4. Sketch the Bode plot for the following transfer function and determine the phase margin and gain margin. [Nov/Dec-2015]

G (s )=20s (1+3 s ) (1+4 s)

Ans:The sinusoidal T.F of G(s) is obtained by replacing s by jw in the given T.F

G(jw) = 20 / [jw (1+j3w) (1+j4w)]

Corner frequencies: wc1= 1/4 = 0.25 rad /sec ;

wc2 = 1/3 = 0.33 rad /sec

Choose a lower corner frequency and a higher Corner frequency

wl= 0.025 rad/sec ;

wh = 3.3 rad / sec

Calculation of Gain (A) (MAGNITUDE PLOT)

A @ wl ; A= 20 log [ 20 / 0.025 ] = 58 .06 dB

A @ wc1 ; A = [Slope from wl to wc1 x log (wc1 / wl ] + Gain (A)@wl

= - 20 log [ 0.25 / 0.025 ] + 58.06

= 38.06 dB

A @ wc2 ; A = [Slope from wc1 to wc2 x log (wc2 / wc1 ] + Gain (A)@ wc1

= - 40 log [ 0.33 / 0.25 ] + 38

= 33 dB

A @ wh ; A = [Slope from wc2 to wh x log (wh / wc2 ] + Gain (A) @ wc2

= - 60 log [ 3.3 / 0.33 ] + 33

=-27 dB

Calculation of Phase angle for different values of frequencies [PHASE PLOT]

Ø = -90- tan -13w - tan -1 4w

When

Calculations of Gain cross over frequency The frequency at which the dB magnitude is Zero wgc = 1.1 rad / sec

Calculations of Phase cross over frequencyThe frequency at which the Phase of the system is - 180 ̊ wpc = 0.3 rad / sec

Gain Margin The gain margin in dB is given by the negative of dB magnitude of G(jw) at phase cross

over frequency

GM = - { 20 log [G( jwpc )] = - { 32 } = -32 dB

Phase Margin Ґ = 180 ̊+ Øgc= 180 ̊+ (- 2400 ̊) = -60 ̊

Conclusion For this system GM and PM are negative in values. Therefore the system is unstable in

nature.

5. Discuss in detail about lead and lag network. [Nov/Dec-2013]Ans:

Circuit of the lag compensator:

Transfer function of lag compensator:

Pole zero plot:

Design procedure for a lag compensator:Step 1: Choose the value of K for uncompensated system to meet the steady state

error requirement

Step 2: Sketch the bode plot uncompensated system

Step 3: Determine the phase margin of the uncompensated system. If the phase

margin does not satisfy the requirement, then the lag compensation is required

Step 4: Choose a suitable value for the phase margin of the uncompensated

system

γn=γd+ε

ε =Additional phase lag to compensate = 5°

γd =Desired phase margin

γn =Phase margin of the uncompensated system

Step 5: Determine the new gain cross over frequency ωgcn

γn=180 °+φgcnφgcn=γ n−180 °

Step 6: Determine the parameter βAgcn=20 log β

log β=Agcn

20

β=10Agcn

20

Step 7: Determine the transfer function of the lag compensator

Zero of the lag compensator=Zc=1T

=ωgcn

10

T=10ωgcn

Pole of the lag compensator Pc=1βT

Transfer function of the lag compensator=GC (s )=s+1

T

s+1βT

=β (1+sT1+sβT )

Step 8: Determine the open loop transfer function of the compensated system

open loop transfer function of the compensated system

=Go( s )=1β

GC( s )G( s )=1β

β (1+sT1+sβT )G(s )=1+sT

1+sβTG(s )

Step 9: Determine the actual phase margin of the compensated system

γ0=180 °+φgco

If the actual phase margin satisfies the given specifications then the design

is accepted. Otherwise repeat the procedure from step 4 to 9 by taking ε as 5°

more than previous design.

Bode plot of a typical lag compensator:

Frequency response of lead compensator:

Design procedure of lead compensator:Step 1: Choose the value of K for uncompensated system to meet the steady

state error requirement

Step 2: Sketch the bode plot uncompensated system

Step 3: Determine the phase margin of the uncompensated system.

Step 4: Determine the amount of phase angle to be contributed by the lead

network by using formula φm=γ d−γ+ε

φm=Maximum phase lead angle of the compensatorε =Additional phase lag to compensate = 5°

γd =Desired phase margin

γn =Phase margin of the uncompensated system

If φm is more than 60° then realize the compensator as cascade of two lead

compensators with each compensator contributing half of the required angle.

Step 5: Determine the transfer function of the lead compensator

Calculate α using the equation α=

1−sin φm

1+sin φm

From the bode plot, determine the frequency at which the magnitude of G (jω) is

−20 log 1√α db. This frequency is ωm.

Calculate T from the relation,ωm=

1T √α

∴T= 1ωm √α

Transfer function of the lead compensator=GC( s )=s+ 1

T

s+ 1αT

=α( 1+sT1+sαT )

Step 6: Determine the open loop transfer function of the compensated system

open loop transfer function of the compensated system

=Go( s )=1α

GC( s )G( s )=1α

α (1+sT1+sαT )G( s )=1+sT

1+sαTG( s )

Step 7: Verify the designFinally the Bode plot of the compensated system is drawn and verify

whether it satisfies the given specifications. Otherwise repeat the procedure from

step 4 by taking ε as 5° more than previous design.

6. The open loop transfer function of a unity feedback system is given by

G (s )= 1s (1+s ) (1+2 s ) .Sketch the polar plot and determine the gain and phase

margin. [Nov/Dec-2014]Ans:Substitute s=jω

G ( jω )= 1jω (1+ jω ) (1+2 jω )

The corner frequencies are ωc1=

12=0. 5 rad /sec ωc2=

11=1 rad /sec

G ( jω )=1jω (1+ jω ) (1+2 jω)

=1ω∠90 °√1+ω2∠ tan−1ω √1+4ω2∠ tan−1 2ω

G ( jω )=1ω√1+ω2 √1+4 ω2

∠−90 °− tan−1ω−tan−1 2ω

|G ( jω )|=1

ω√ (1+ω2) (1+4 ω2 )=

1

ω √(1+ω2+4 ω2+4 ω4)=

1

ω√ (1+5ω2+4 ω4 )∠G ( jω )=−90°−tan−1ω− tan−1 2ω

ω 0.35 0.4 0.45 0.5 0.6 0.7 1.0

G(jω) 2.2 1.8 1.5 1.2 0.9 0.7 0.3

φ -144 -150 -156 -162 -171 -180 -198

Gain margin = Kg=1.4286Phase margin= 12°

7. Write short notes on constant M and N circles. [Nov/Dec-2014]Ans:

M CIRCLES:

N CIRCLES:

8. Design a lead compensator for a unity feedback system with open loop transfer function, G(s) = K/s(s+1)(s+5) to satisfy the following specifications. (i) Velocity error constant, Kv ≥ 50 (ii) phase margin is ≥ 20◦

Ans:

Step-1 : Determine the calculation of gain K Given that Kv≥50, Let Kv = 50

By definition of velocity error constant, Kv = Lt

s →0s K

s( S+1 )(S+5)= K

5

∴K=5×K v=5×50=250

Step-2: Draw the bode plot for the uncompensated system

G( s )=250

s (s+1)(s+5 )=50

s(1+s )(1+0.2 s )

let s= jω, ∴G( jω)=50

jω(1+ jω)(1+0 .2 jω )

Magnitude Plot: The corner frequency are

ωc1 = 1 rad/sec ωc2 = 1/0.2 = 5 rad/sec

TermCorner frequency rad/sec

SlopedB/dec

Change in slopedB/dec

50jω

- -20 -

11+ jω ωc1 = 1 -20 -40

11+0 .2 jω ωc2 = 5 -20 -60

ωl = 0.5 rad/sec and ωh = 10 rad/sec

At ω = ωl ,

A=20 log|50jω

|=20 log 500 . 5

=40db

At ω = ωc1 ,

A=20 log|50jω

|=20 log 501

=34 db

At ω = ωc2, A=[ slope fromωc1 to ωc 2×log

ωc2

ωc1 ]+ A at(ω=ωc1 )

= −40×log 5

1+34=6db

At ω = ωh, A=[ slope fromωc2 to ωch×log

ωh

ωc2 ]+ A at (ω=ωc 2)

= −60×log 10

5+6=−12db

Phase Plot :

φ =∠G( jω)=−90 °−tan−1 ω−tan−1 0 .2 ω

Step 3: Determine the phase margin

Let φgc = Phase margin of G(jω ) at gain crossover frequency.

and γ = Phase margin of uncompensated system.

From the bode plot of uncompensated system we get,φgc = -224.

Now, γ = 180º+φgc=180 °−220 °=−44 °

The phase margin of the system is negative and so the system is unstable.Hence

lead compensator is required to make the system stable and to have a phase

margin of 20 °

Step 4:

The desired phase margin, γd=¿20°

Let the additional phase lead required, Є = 5 °

Maximum lead angle , φm=γ d−γ+ Є = 20 ° -(-44 ° )+5 °=69 °

ω in rad/sec 0.1 0.5 1.0 5 10

φ in degree -96 -122 -146 -214 -238

Since the lead angle required is greater than 60 ° , we have to realize the

lead compensator as cascade of two lead compensators with each compensator

providing half of the required phase lead angle. ∴φm=69 °

2=34 .5 °

Step-5: Determine the transfer function of lead compensator

α=

(1−sin φm)(1+sin φm)

= 1−sin 34 .5 °1+sin 34 . 5°

=0 .28

The dB magnitude corresponding to

ωm=−20 log 1√α

=−20 log 1√0. 28

=−5 . 5db

From the bode plot of uncompensated system the frequency, ωm

corresponding to a db gain of -5.5db is found to be 7.8 rad/sec.

∴ωm =7.8 rad/sec. Now, T =

1ωm√α =

17 .8√0 .28 = 0.24

Transfer function of the lead compensator Gc(s) =

( s= 1T

)2

( s+ 1αT

)2=α2 (1+st )2

(1+sαT )2

= (0.28)2

(1+0. 24 )2

(1+0 . 28∗0 . 24 s )2=0 .0784 (1+0.24 s )2

(1+0 .067 s )2

Step 6: Open loop transfer function of compensated system Open loop transfer function of compensated system is G0(s)

=

0 .0784 (1+0 .24 s )2

(1+0 .067 s )2 ×50s(1+ s )(1+0. 2 s ) =

4 (1+ j 0 .24 ω )2

s (1+s )(1+0 . 2 s )(1+0 .067 s )2

Step 7: Draw the bode plot of compensated system to verify the design.

Put s =jω in G0(s), ∴G0 (s )=

4(1+ j0 . 24 )2

jω(1+ jω)(1+ j0 . 2ω )(1+ j0 . 067 ω )2

Magnitude plot: The corner frequency are ω c1=1 rad/sec,

ω c2 =

10 .24 =4.2 rad/sec

ω c3 =

10 .2

=5 rad/sec

ω c4 =

10 .067

=15 rad/sec

Term Corner frequencyrad/sec

Slopedb / dec

Change in slopedb / dec

4jω

- -20

11+ jω

ωc1=1 -20 -20-20=-40

(1+j0.24ω ) ωc2=1

0. 24=4 . 2 40 -40+40=0

1(1+ j 0 .2 ω )

ωc3=1

0 .2=5 -20 0-20=-20

1(1+ j 0 . 067ω )2

ωc4=1

0 . 067=15 -40 -20-40=-60

ωl = 0.5 rad/sec and ωh= 30 rad/sec.

At ω = ω l A0 = 20log | 4

jω|= 20log

40 .5 =18db.

At ω=ωc1 , A0 =20log | 4

jω| =20log

41 = 12db

Atω=ωc2 , A0= [slope from ωcl to ωc2 ×log

ωc 2

ωc 1 ] + A0 at (ω=ωcl)

= -40× log

4 .21 +12= -13db.

Atω=ωc2 , A0= [slope from ωcl to ωc2 ×log

ωc 2

ωc 1 ] + A0 at (ω=ωcl)

= -40* log

4 .21 +12= -13db.

Atω=ωc1 , A0= [slope from ωc1 to ωc2 *log

ωc 2

ωc 1 ] + A0 at (ω=ωcl)

= -40×log

4 .21 +12= -13db.

At ω=ωc 3 , A0=[slope from ωc2 to ωc3 ×log

ωc 3

ωc2 ] + A0 at (ω=ωc2 )

= -40× log

4 .21 +12= -13db.

At ω=ωc 4 , A0=[slope from ωc3 to ωc4 ×log

ωc 4

ωc3 ] + A0 at (ω=ωc3 )

= -20× log

155 -13=-23db. .

At ω=ωch , A0= [slope from ωc4 to ωch×log

ωh

ωc 4 ] + A0 at (ω=ωc4 )

= -60×log

3015 -23= -41 db.

Phase plot:

φ0=∠G0( jω) =2 tan−1 0. 24ω−tan−1ω−90−tan−10.2ω−2 tan−1 0 .067 ω

ω rad/sec 0.1 0.5 1 2 5 10 15

∠G0 ( jω) deg -94 -112 -126 -139 -150 -170 -188

Let, φgc 0= Phase of G0(jω ) at new gain crossover frequency(ωgcn )

And γ0 = Phase margin of compensated system.

From the bode plot of compensated system we get, φgco = -140 °

Conclusion: The phase margin of the compensated system is satisfactory. Hence the

design acceptable.

Result: The transfer function of lead compensator is

Gc(s) =

0 .0784 (1+0 .24 s )2

(1+0 . 067 )2 =

(s+4 . 17 )2

( s+14 . 92)2

The open loop transfer function of lead compensated system is

G0(s) =

4 (1+0. 24 s )2

s (1+s )(1+0 . 2 s )(1+0 .067 S )2

9. Consider the unity feedback system whose open loop transfer function

is G( s )= K

s (s+3)( s+6 ) . Design a lag-lead compensator to meet the following specifications. (i) Velocity error constant, Kv = 80 (ii) phase margin, γ ≥ 35º.

Ans: Step 1: Determine the calculation of gain K

For unity feedback system, Kv = Lt

s →0sG (s )

Given that, Kv = 80

∴ Lt

s→0sG(s )

= Lt

s →0s K

s( s+3 )(s+6 )=80

K3×6

=80 (or) K = 80¿ 3¿ 6 = 1440

∴G( s )=1440

s (s+3 )( s+6)=80

s (1+0 .33 s )(1+0 .167 s )

Step2: Draw the bode plot of uncompensated system.

In G(s), put s= jω ∴G( jω)=80

jω(1+0 . 33 jω)(1+0 . 167 jω)

Magnitude plot: The corner frequencies are ωc1 and ωc2.

ωc1 =

10 .33

=3 rad /sec

ωc2 =

10 .167

=6 rad /sec

TermCorner frequency rad/sec

Slopedb/dec

Change in slopedb/dec

80jω - -20 -

11+0 .33 jω

ωc1=3 -20 -40

11+0 .167 jω

ωc2=6 -20 -60

Let ωl = 0.5 rad/sec and ωh = 20 rad/sec

At ω = ωl ,

A=20 log 80ω

=20 log800 .5

=44 db

At ω = ωc1,

A=20 log 80ω

=20 log803

=28 .5db≈28db

At ω = ωc2, A=[ slope fromωc1 to ωc 2×log

ωc2

ωc1 ]+ A at(ω=ωc1 )

= −40×log 6

3+28=16 db

At ω = ωh, A=[ slope fromωc 2 to ωch×log

ωh

ωc2 ]+ A at (ω=ωc2 )

= −60×log 20

6+16=−15db

Phase Plot:

Φ=∠G( jω)=−90ο−tan−1 0 .33ω−tan−1 0 .167ω

ω rad/sec 0.5 1.0 3.0 6 10 20

∠G( jω )deg -104 -118 -160 -198 -222 -244

Step 3: Find phase margin of uncompensated system.

Let φgc = Phase margin of G(jω) at gain crossover frequency

γ = Phase margin of uncompensated system

From the bode plot of uncompensated system we get, φgc = -226º

γ=180+φgc

¿ =180°−226=−¿ 46°¿

Step 4: Choose a new phase margin The desired phase margin, γd = 35º

The phase margin of compensated system,γn=γd+∈

Let initial choice of ∈=5ο

∴ γn=γd+∈

=35ο+5ο=40ο

Step 5: Determine new gain crossover frequency

Let ωgcn = New gain crossover frequency and

Фgcn =Phase of G(jω) at ωgcn

Now, γn = 180º+Фgcn,

Фgcn = γn -180º = 40º - 180º = -140º

From the bode plot we found that the frequency corresponding to a phase of -

140º is 1.8 rad/sec.

Let ωgcl = Gain crossover frequency of lag compensator.

Choose ωgcl such that, ωgcl> ωgcn

Let ωgcl = 4 rad/sec.

Step 6: Calculate β of lag compensator. From the bode plot we found that the db magnitude at ωgcl is 23 db.

∴|G ( jω)| in db at (ω=ωgcl ) =Agcl = 23db.

Also, Agcl = 20logβ ; ∴ β=10 Agcl¿20

= 1023 /20

= 14.

Step 7: Determination the transfer function of lag section.

The zero of the lag compensator is placed at a frequency one-tenth of ωgcl .

∴ Zero of lag compensator, Zcl =

−1T1 =

ωgcl

10

Now,T1 =

10ωgcl =

104 =2.5

Pole of lag compensator ,Pcl =

1βT1 =

114∗2 .5 =

135

Transfer function of lag compensator G1(s) =β

(1+sT 1 )(1+sβT 1)

=14

(1+2 .5 s )(1+35 s )

Step 8: Determine the transfer function of lead section.

Let α =

1β ;

∴α= 114

=0 .07

The db gain (magnitude) corresponding toωm=-20log

1√α

= - 20log

1√0.07

=−11.5db≈12 db.

From the bode plot uncompensated system the frequency ωm corresponding to a

db pair of -12 db is found to be 17 rad/sec.

∴ωm=17 rad/sec. ∴T 2=

1ωm √α =

117√0 .07 = 0.22.

Transfer function of lead compensator G2(s) =α

(1+sT 2 )(1+sαT 2 ) =

(1+. 22 s )(1+. 0154 s )

Step 9: Determine the transfer function of lag-lead compensator. Transfer function of lag-lead compensator Gc(s) = G1(s)×G2(s)

=14

(1=2 . 5 s )(1+35 s )

×0 . 07 (1+0 .22 s )(1+0. 0154 s ) =

(1+2. 5 s)(1+. 22 s )¿ (1+35 s )(1+0. 0154 s) ¿¿

¿

Step 10: Determine open loop transfer function of compensated system.Open loop transfer function of compensated system is

G0(s) =

80(1+2 .5 s )(1+0 .22 s )s (1+35 s )(1+0 . 0154 s )(1+0 . 167 s )

Step 11: Bode plot of compensated system. Put S = jω in Go(s)

∴G0 ( jω ) =

80(1+ j2 .5ω )(1+ j0 .22 ω)s (1+ j 35ω)(1+ j0 . 0154ω )(1+ j 0 . 33ω)(1+ j0 .167 ω )

Magnitude plot:

ωci=1

35=0 . 03

rad / sec ; ωc2=

12.5

=0 .4 rad/sec ;

ωc3=1

0 .33=3

rad/sec;

ωc4=1

0 . 22=4 . 5

rad/sec; ωc5=

10 .167

=6 rad/sec ;

ωc6=1

0 .0154=

65 rad/sec.

Term

Corner frequency rad/sec

Slopedb/dec

Change in slope Db/dec

80jω

11+ j 35 ω

1+j2.5ω

11+ j 0. 33ω

1+j0.22ω

11+ j 0. 167 ω

11+ j 0. 0154 ω

-

ωcl=1

35=0 .03

ωc2=1

2.5=0 .4

ωc3=1

0 .33=3

ωc4=1

0 . 22=4 . 5

ωc5=1

0 .167=6

ωc6=1

0 .0154=65

-20

-20

+20

-20

+20

-20

-20

-

-40

-20

-40

-20

-40

60

Let ωl = 0.01 rad/sec and ωh = 80 rad/sec

At ω =ωl , A = 20 log 80

0.01=78db

At ω =ωc1 , A = 20 log 80

0. 03=68 db

At ω =ωc2 , A = −40×log 0. 4

0 .03+68=23db

At ω =ωc3 , A = −20×log 3

0 . 4+23=5db

At ω =ωc4 , A = −40× log 4 . 5

3+5=−2db

At ω =ωc5 , A = −20×log 6

4 . 5+(−2 )=−4db

At ω =ωc 6 , A = −40× log65

6+(−4 )=−45db

At ω =ωh , A = −60×log 80

65+(−45)=−50db

Phase plot:Ф=

∠Gο ( jω)=tan−12.5ω+ tan−10 .22ω−90ο−tan−135ω−tan−10 . 0154ω−tan−10 .33 ω−tan−10 .167ω

ω rad/sec

0.01 0.03 0.1 0.4 1 4 10 65 80

∠G( jω )deg

-108 -132 -152 -138 -126 -144 -168 -220 -228

From the bode plot of compensated system we get Фgco = -144º

γ◦ = 180 -144 =36º

Open loop transfer function of compensated system

G0(s) =

80(1+2 .5 s )(1+0 .22 s )s (1+35 s )(1+0 .0154 s )(1+0 . 33 s )(1+0 .167 s )