Modern Control Systems (MCS)

Modern Control Systems (MCS)

Dr. Imtiaz HussainAssistant Professor

email: [email protected] :http://imtiazhussainkalwar.weebly.com/

Lecture-12-13-14-15Lead Compensation

Lecture Outline

• Introduction to Lead Compensation

• Electronic Lead Compensator

• Electrical Lead Compensator

• Mechanical Lead Compensator

Lead Compensation

• Lead Compensation essentially yields an appreciable improvement in transient response and a small change in steady state accuracy.

• There are many ways to realize lead compensators and lag compensators, such as electronic networks using operational amplifiers, electrical RC networks, and mechanical spring-dashpot systems.

Lead Compensation

• Generally Lead compensators are represented by following transfer function

• or

, ()

, ()

Lead Compensation

, ()

-10 -8 -6 -4 -2 0-1

-0.5

0

0.5

1Pole-Zero Map

Real Axis

Imag

inar

y A

xis

-20

-15

-10

-5

0

Mag

nitu

de (d

B)

10-2

10-1

100

101

102

103

0

30

60Ph

ase

(deg

)

Bode Diagram

Frequency (rad/sec)

Electronic Lead Compensator• Following figure shows an electronic lead compensator using

operational amplifiers.

𝐸𝑜(𝑠)𝐸𝑖 (𝑠)

=𝑅2𝑅4

𝑅1𝑅3

𝑅1𝐶1𝑠+1𝑅2𝐶2𝑠+1

Electronic Lead Compensator

• This can be represented as

• Where,

• Then,

𝐸𝑜(𝑠)𝐸𝑖 (𝑠)

=𝑅2𝑅4

𝑅1𝑅3

𝑅1𝐶1𝑠+1𝑅2𝐶2𝑠+1

𝐸𝑜(𝑠)𝐸𝑖 (𝑠)

=𝑅4𝐶1

𝑅3𝐶2

𝑠+ 1𝑅1𝐶1

𝑠+ 1𝑅2𝐶2

𝑇=𝑅1𝐶1 𝑎𝑇=𝑅2𝐶2 𝐾 𝑐=𝑅4𝐶1

𝑅3𝐶2

, ()

𝑅1𝐶1>𝑅2𝐶2

Electronic Lead Compensator

• Pole-zero Configuration of Lead Compensator

𝑅1𝐶1>𝑅2𝐶2

Lead Compensation Techniques Based on the Root-Locus Approach.

• The root-locus approach to design is very powerful when the specifications are given in terms of time-domain quantities, such as

– damping ratio– undamped natural frequency– desired dominant closed-loop poles– maximum overshoot– rise time– settling time.

Lead Compensation Techniques Based on the Root-Locus Approach.

• The procedures for designing a lead compensator by the root-locus method may be stated as follows:

– Step-1: Analyze the given system via root locus.

Step-2

• From the performance specifications, determine the desired location for the dominant closed-loop poles.

Step-3• From the root-locus plot of the uncompensated system

(original system), ascertain whether or not the gain adjustment alone can yield the desired closed loop poles.

• If not, calculate the angle deficiency.

• This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles.

Step-4• Assume the Lead Compensator to be:

• Where α and T are determined from the angle deficiency.

• Kc is determined from the requirement of the open-loop gain.

Step-5• If static error constants are not specified, determine the

location of the pole and zero of the lead compensator so that the lead compensator will contribute the necessary angle.

• If no other requirements are imposed on the system, try to make the value of α as large as possible.

• A larger value of α generally results in a larger value of Kv, which is desirable.

• Larger value of α will produce a larger value of Kv and in most cases, the larger the Kv is, the better the system performance.

Step-6

• Determine the value of Kc of the lead compensator from the magnitude condition.

Final Design check

• Once a compensator has been designed, check to see whether all performance specifications have been met.

• If the compensated system does not meet the performance specifications, then repeat the design procedure by adjusting the compensator pole and zero until all such specifications are met.

Final Design check

• If the selected dominant closed-loop poles are not really dominant, or if the selected dominant closed-loop poles do not yield the desired result, it will be necessary to modify the location of the pair of such selected dominant closed-loop poles.

Example-1• Consider the position control system shown in following

figure.

• It is desired to design an Electronic lead compensator Gc(s) so that the dominant closed poles have the damping ratio 0.5 and undamped natural frequency 3 rad/sec.

Step-1 (Example-1)• Draw the root Locus plot of the given system.

)1(10)()(

ss

sHsG

• The closed loop transfer function of the given system is:

• The closed loop poles are

1010

)()(

2

sssRsC

1225.35.0 js

Step-1 (Example-1)• Determine the characteristics of given system using root loci.

• The damping ratio of the closed-loop poles is 0.158.

• The undamped natural frequency of the closed-loop poles is 3.1623 rad/sec.

• Because the damping ratio is small, this system will have a large overshoot in the step response and is not desirable.

1010

)()(

2

sssRsC

Step-2 (Example-1)• From the performance specifications, determine the

desired location for the dominant closed-loop poles.

• Desired performance Specifications are: It is desired to have damping ratio 0.5 and undamped natural

frequency 3 rad/sec.

939

2)()(

222

2

sssssRsC

nn

n

5981.25.1 js

Step-2 (Example-1)• Alternatively desired location of closed loop poles can also

be determined graphically

Desired ωn= 3 rad/sec

Desired damping ratio= 0.5

1cos

60)5.0(cos 1

Desired Closed Loop

Pole

60

Step-3 (Exampl-1)• From the root-locus plot of the uncompensated system

ascertain whether or not the gain adjustment alone can yield the desired closed loop poles.

Desired Closed Loop

Pole

Step-3 (Exampl-1)• If not, calculate the angle deficiency. • To calculate the angle of deficiency apply Angle Condition at desired

closed loop pole.

-1

5981.25.1 js Desired Closed Loop Pole

-1

-2

-2

120o100.8o

8.100120180d

89.40d

Step-3 (Exampl-1)

• Alternatively angle of deficiency can be calculated as.

5981.25.1 js Where are desired closed loop poles

5981.25.1)1(10180

jsd ss

5981.25.15981.25.1)1(10180

jsjsd ss

8.100120180d

89.40d

Step-4 (Exampl-1)• This angle must be contributed by the lead

compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles.

• Note that the solution to such a problem is not unique. There are infinitely many solutions.

Step-5 (Exampl-1)• Solution-1

– If we choose the zero of the lead compensator at s = -1 so that it will cancel the plant pole at s =-1, then the compensator pole must be located at s =-3. 89.40

Solution-1

Step-5 (Example-1)• If static error constants are not specified, determine the

location of the pole and zero of the lead compensator so that the lead compensator will contribute the necessary angle.

89.40

Solution-1

Step-5 (Example-1)• The pole and zero of compensator are determined as

89.40

=

1𝑇=1 yields

→𝑇=1

1𝛼𝑇=3yields

→𝛼=0.333

• The Value of can be determined as

Solution-1

Step-6 (Example-1)

89.40

• The Value of Kc can be determined using magnitude condition.

𝐺𝑐 (𝑠 )=0.9 𝑠+1𝑠+3

|𝐾 𝑐(𝑠+1)𝑠+3

10𝑠 (𝑠+1)|𝑠=− 1.5+ 𝑗2.5981

=1

|𝐾 𝑐10

𝑠(𝑠+3)|𝑠=−1.5+ 𝑗2.5981=1

𝐾 𝑐=|𝑠(𝑠+3)10 |𝑠=−1.5+ 𝑗 2.5981

=0.9

Solution-1

Final Design Check• The open loop transfer function of the designed system

then becomes

• The closed loop transfer function of compensated system becomes.

𝐺𝑐 (𝑠 )𝐺(𝑠)= 9𝑠 (𝑠+3)

𝐶 (𝑠)𝑅(𝑠)

= 9𝑠2+3 𝑠+9

Solution-1

Final Design Check

𝐺𝑐 (𝑠)𝐺(𝑠)= 9𝑠 (𝑠+3)

Solution-1

𝐺 (𝑠 )=10

𝑠 (𝑠+1)

-4 -3 -2 -1 0 1-5

0

5

0.158

0.158

3.16

3.16

Root Locus

Real Axis

Imag

inar

y Ax

is

-4 -3 -2 -1 0 1-5

0

5

0.5

0.5

3

3

Root Locus

Real Axis

Imag

inar

y Ax

is

Final Design Check• The static velocity error constant for original system is

obtained as follows.

• The steady state error is then calculated as

𝑒𝑠𝑠= 1𝐾 𝑣

=110=0.1

𝐾 𝑣= lim𝑠→ 0

𝑠𝐺(𝑠)

𝐾 𝑣= lim𝑠→ 0

𝑠 [ 10𝑠(𝑠+1) ]=10

Solution-1

Final Design Check Solution-1

0 2 4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Ampl

itude

Actual SystemCompensated System

Final Design Check• The static velocity error constant for the compensated

system can be calculated as

• The steady state error is then calculated as

𝑒𝑠𝑠= 1𝐾 𝑣

=13=0.333

𝐾 𝑣= lim𝑠→0

𝑠𝐺𝑐 (𝑠 )𝐺(𝑠)

𝐾 𝑣=lim𝑠→ 0

𝑠 [ 9𝑠(𝑠+3) ]=3

Solution-1

Step-5 (Exampl-1)• Solution-2

Solution-2

-1

-1

-2

-2

90o

49.2o

-3

89.40

Step-5 (Exampl-1)• Solution-2

-1

-1

-2

-2

90o

49.2o

-3

89.40

Solution-2

𝐺𝑐 (𝑠)=1.03 𝑠+1.5𝑠+3.6

Step-5 (Example-1)• If no other requirements are imposed on the system, try to make

the value of α as large as possible. A larger value of α generally results in a larger value of Kv, which is desirable.

• Procedure to obtain a largest possible value for α.– First, draw a horizontal line passing through point P, the desired location for

one of the dominant closed-loop poles. This is shown as line PA in following figure.

– Draw also a line connecting point P and the origin O.

-1

-1

-2

-2

-3 O

AP

Solution-3

Step-5 (Example-1)• Bisect the angle between the lines PA and PO, as shown in following

figure.

-1

-1

-2

-2

-3 O

PA

2

2

Solution-3

Step-5 (Example-1)

• Draw two lines PC and PD that make angles with the the bisector PB.

• The intersections of PC and PD with the negative real axis give the necessary locations for the pole and zero of the lead network.

-1

-1

-2

-2

B

-3 O

PA

C

D

2d

2d

Solution-3

Step-5 (Example-1)• The lead compensator has zero at s=–1.9432 and pole at s=–4.6458.

• Thus, Gc(s) can be given as

-1

-1

-2

-2

B

-3 O

PA

C

D

2d

2d

=

Solution-3

Step-5 (Example-1)

• For this compensator value of is

• Also

=

1𝑇=1.9432 yields

→𝑇=0.514

1𝛼𝑇=4.6458 yields

→𝛼=0.418

Solution-3

Step-6 (Example-1)• Determine the value of Kc of the lead compensator from

the magnitude condition.

𝐺 (𝑠)𝐺𝑐 (𝑠 ) 𝐻 (𝑠)= 10𝐾 𝑐(𝑠+1.9432)𝑠(𝑠+1)(𝑠+4.6458)

| 10𝐾 𝑐 (𝑠+1.9432)𝑠(𝑠+1)(𝑠+4.6458)|𝑠=−1.5+ 𝑗 2.5981=1

Solution-3

Step-6 (Example-1)• The Kc is calculated as

• Hence, the lead compensator Gc(s) just designed is given by

𝐾 𝑐=1.2287

𝐺𝑐 (𝑠)=1.2287 𝑠+1.9432𝑠+4.6458

Solution-3

Final Design Check

Desired Closed Loop

Pole

Uncompensated System

Desired Closed Loop

Pole

Compensated System

Solution-3

Final Design Check• It is worthwhile to check the static velocity error

constant Kv for the system just designed.

• Steady state error is

94

𝐾 𝑣= lim𝑠→0

𝑠𝐺𝑐 (𝑠 )𝐺(𝑠)

𝐾 𝑣=lim𝑠→0

𝑠 [1.2287 𝑠+1.9432𝑠+4.6458

10𝑠(𝑠+1) ]=5.139

Solution-3

Final Design Check Solution-3

0 2 4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Ampl

itude

Actual SystemSolution-3

Final Design Check

0 2 4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Ampl

itude

Actual SystemSolution-1Solution-3

Solution-1

Solution-3

Mechanical Lead Compensator

• Figure shows the mechanical lead compensator.• Equations are obtained as

• Taking Laplace transform of these equations assuming zero initial conditions and eliminating Y(s), we obtain

Mechanical Lead Compensator

• By defining

• We obtain

Exampl-2• Design a mechanical lead compensator for following

system.

• The damping ratio of closed loop poles is 0.5 and natural undamped frequency 2 rad/sec. It is desired to modify the closed loop poles so that natural undamped frequency becomes 4 rad/sec without changing the damping ratio.

)2(4ss

Electrical Lead Compensator

𝑉 𝑜(𝑠)𝑉 𝑖 (𝑠) 𝑐

=𝑅2

𝑅1+𝑅2

𝑅1𝐶𝑠+1𝑅1𝑅2

𝑅1+𝑅2𝐶𝑠+1

𝑉 𝑜(𝑠)𝑉 𝑖 (𝑠 )

C 𝑎𝑇=𝑅1𝑅2𝐶𝑅1+𝑅2 1𝑎=

𝑅2

𝑅1+𝑅2

Example-3• Consider the model of space vehicle control system

depicted in following figure.

• Design an Electrical lead compensator such that the damping ratio and natural undamped frequency of dominant closed loop poles are 0.5 and 2 rad/sec.

END OF LECTURES-12-13-14-15

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