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
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
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.
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°
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 )