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Final Project on the ECE 415 Control Systems course., Papers of Control Systems

A final project on compensator design for achieving desired specifications from a known plant of a system. The project employs techniques such as root locus, pole placement, step response and others to design lead and lag compensators. The compensators are designed to fulfill requirements such as overshoot, settling time, steady state errors fulfillment and others. three problems with their respective plants and required specifications, and the design of compensators to meet those specifications.

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ECE 415 Control Systems
Final Project
Submitted by:
Submitted on:
Abstract
The controller or compensator design is used to achieve desired specifications from a known plant
of a system. The compensator designed used techniques such as root locus, pole placement, step
response and other.
There are primarily two types of compensators designed in this project the lead compensator and
the lag compensator. The lead compensator has a pole further from the zero and thus adding phase
to the system. This makes the systemโ€™s root locus mover more towards the left half plane and gives
more stability.
The lag compensator is designed for fulfilling the margins in frequency domains such as phase
margin and gain margin. The zero is placed further in the system than the pole of the compensator
and tends to mover system more to the right.
The design of each of the compensator are carried out to fulfill requirements such as overshoot,
settling time, steady state errors fulfillment and others.
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ECE 415 Control Systems

Final Project

Submitted by:

Submitted on:

Abstract

The controller or compensator design is used to achieve desired specifications from a known plant

of a system. The compensator designed used techniques such as root locus, pole placement, step

response and other.

There are primarily two types of compensators designed in this project the lead compensator and

the lag compensator. The lead compensator has a pole further from the zero and thus adding phase

to the system. This makes the systemโ€™s root locus mover more towards the left half plane and gives

more stability.

The lag compensator is designed for fulfilling the margins in frequency domains such as phase

margin and gain margin. The zero is placed further in the system than the pole of the compensator

and tends to mover system more to the right.

The design of each of the compensator are carried out to fulfill requirements such as overshoot,

settling time, steady state errors fulfillment and others.

Design Problem Description

The standard design scheme of the negative feedback system incorporated with the compensator

is shown in figure 1. For each of the three systems in this project this particular scheme will be

employed to fulfill the requirements.

Figure 1 - Closed loop control system with compensator.

Problem 1

The Plant for this problem is ๐‘ƒ

!.#

$($&#)($&()

and is a stable plant with all the poles ๐‘  =

0 , โˆ’ 1 , โˆ’ 4 in the left half plane. This is a type 1 system with an integrator so the steady state error

for the step input for open loop response will be zero. The uncompensated closed loop system has

percentage overshoot of 27.7% while the settling time is about 11 seconds. The steady state error

when in a closed loop is about 27%. The uncompensated system velocity constant is about 1. The

required specifications are about ๐‘ƒ. ๐‘‚ โ‰ค 11% and๐‘‡ $

โ‰ค 2. 5 ๐‘ ๐‘’๐‘๐‘œ๐‘›๐‘‘๐‘ . So to make the system follow

such specifications a lead compensator will be employed to reduce the overshoot and the settling

time. The gain of the controller will computed to fulfill the velocity constant requirement of๐พ )

  1. The pole will be placed further than the zero of the system to move the system more towards

the left half plane.

Problem 2

The Plant for the second problem is ๐‘ƒ(๐‘ ) =

#!

$*

!

"

#".%&'

&

!

().*

&#+

and is a stable plant with all the poles

๐‘  = 0 , โˆ’ 0. 1016 ยฑ 3. 1536 ๐‘– the left half plane. This is a type 1 system with an integrator so the

steady state error for the step input for open loop response will be zero. The uncompensated closed

loop system has a very high percentage overshoot of 357240% while the system never properly

settles within a 1000 seconds frame. The steady state error when in a closed loop is 0. The

uncompensated system velocity constant is about 1. The required specifications are about ๐‘ƒ. ๐‘‚ โ‰ค

35% and๐‘‡ $

โ‰ค 0. 75 ๐‘ ๐‘’๐‘๐‘œ๐‘›๐‘‘๐‘ . So to make the system follow such specifications a lead compensator

will be employed to reduce the overshoot and the settling time. The gain of the controller will

computed to fulfill the velocity constant requirement of๐พ )

โ‰ฅ 15. The frequency domain technique

will be used to design two lead controller in series thus overall a second order compensator.

Problem 3

DC motor system control is the third problem to compensate. The input is in terms of voltage and

output in terms of position in radians. The estimated transfer function of the motor is

)($)

,($)

$(!.!!-.$&!./.0-)

. The analysis of the transfer function reveals a stable system with two poles

and no zero. The pole locations are๐‘  = 0 , โˆ’ 57. 7312. The step response of the closed loop of the

DC motor gives a peak time of 3.8 seconds. This time shows that the DC motor is not fast enough

Problem 2

The plant ๐‘ƒ(๐‘ ) =

#!

$*

!

"

#".%&'

&

!

().*

&#+

compensated by two lead controllers. The selected overshoot is

about 30% and settling time of about 0.45 seconds. This gives a damping ratio of about 0.

and natural frequency 24.839 radians/second. Taking a 20% increased value for the velocity

constant of about 18 and this gives a gain of: ๐พ 8

= ๐พ lim

$โ†’!

๐‘ ๐‘ƒ(๐‘ ) = 18 รจ๐พ = 1. 8

The phase margin for the system to have is selected to be 100 times more than the damping ratio

of about 35.786 rad/s. The crossover frequency is calculated using the following formula:

C

6

The system is analyzed for the Phase margin deficiency and this about - 87.626 rad/s. Since this

compensation is very large so two lead controller will be required to fulfil. A safety factor of

about 15 rad/s is employed in calculating controller pole and zero. The gain of the controller is

calculated as:

D<E

= 35. 786 + O

F>. 070

7

P + 15 = 94. 6 ๐‘Ÿ๐‘Ž๐‘‘/๐‘  รจ๐›ผ =

3 $G6=(H

425

)

#&$G6=(H

425

)

๐บ๐‘Ž๐‘–๐‘› = โˆ’ 10 log [

\ = โˆ’ 27. 926 ๐‘‘๐ต

This DC gain is used to find the frequency point and is about 17rad/s. This gives ๐œ for the

controller: ๐œ =

#>โˆ— โˆš

K

= 1. 465. The first controller is computed as: ๐บ

C#

7 .0.>$&#.F

!.!!7.07$&#

The second controller phase margin deficiency is computed from the bode plot of plant with first

compensator. It comes out to be - 4.4205 rad/s

D<E

= 35. 786 + ( 4. 4205 ) + 15 = 55. 206 ๐‘Ÿ๐‘Ž๐‘‘/๐‘  รจ๐›ผ =

3 $G6=(H

425

)

#&$G6=(H

425

)

๐บ๐‘Ž๐‘–๐‘› = โˆ’ 10 log [

\ = โˆ’ 10. 08 ๐‘‘๐ต

This DC gain is used to find the frequency point and is about 43rad/s. This gives ๐œ for the

controller: ๐œ =

(.โˆ—โˆšK

The first controller is computed as: ๐บ

C#

!.!>(77$&#

!.!!>7F>$&#

and the gain from root locus is about 2.

C

C#

C 7

7

7

Problem 3

DC motor transfer function is ๐‘ƒ(๐‘ ) =

$(!.!!-.$&!./.0-)

Using the overshoot requirement the

damping ratio is range is calculated where the๐‘ƒ. ๐‘‚ โ‰ค 4 .6%.

3

+,

  • #.+

"

โ‰ค 0. 046 รจ๐œ โ‰ฅ 0. 7 , and selecting ๐œ = 0. 8

The natural frequency is calculated from peak time:๐‘‡ 1

L

5

/

M# 34

"

โ‰ค 0. 05 รจ

L

5

/

M# 34

"

โ‰ค 0. 05 รจ๐œ”

6

โ‰ฅ 97 , and selecting๐œ”

6

The second order system pole become ๐‘ = โˆ’๐œ๐œ” 6

6

C 1 โˆ’ ๐œ

7

รจ๐‘ = โˆ’ 80 ยฑ ๐‘— 60

Steady error requirement gives the gain where ๐‘’

$$

N

6

โ‰ค 0. 02 รจ๐พ

8

= lim

$โ†’!

N

!./.0-

2

1

โ‰ฅ 2 รจ๐พ โ‰ฅ 50 โˆ— 0. 5369

1

2

And selected gain is ๐พ = 1. 5 โˆ— 50 โˆ— 0. 5369 โˆ—

1

2

For the ease of design one zero is placed over the pole at - 57.312. This will help in reducing the

effect of angle compensation by both pole and zero. The angle deficiency to fulfill by the pole of

the compensator is calculated using trigonometry the angle compensation required by the pole for

dominant closed loop pole will be:

:;<=

= 180 โˆ’ 180 + tan

3 #

0!

F!

รจ ๐œƒ

:;<=

?

This lead angle will give pole location:

tan

:;<=

The steady state requirement error requirement for the ramp input gives the gain:

#0!

/>..#

Verification of Specifications

Problem 1

Controller designed ๐ถ

$&#

$&#0.#(

gives the step input as in figure 5. The overshoot of the

compensated system is about 10.9% which is less than required of 11% while the settling time has

gone below 2.5 second threshold to 2.09 seconds. The velocity constant comes out to be: ๐พ

8

lim

$โ†’!

๐‘ ๐ถ(๐‘ )๐‘ƒ(๐‘ ) = lim

$โ†’!

$&#

$&#0.#(

!.#

$($&#)($&()

= 2. 2 .Hence all the requirements of the design are

meet.

time is also reduced to 0.15 seconds. Whereas the required range was to be below 0.75 seconds.

The value of velocity constant comes out to be:

8

= ๐พ lim

$โ†’!

7

7

๐‘  [

7

+ 1 \

This is greater than the required value of 15. Thus all the requirements are fulfilled.

Problem 3

The compensator designed for the DC motor is: ๐ถ

$&/>..#

$&##

and the resulting response

for the compensated closed loop system is shown in figure below.

Figure 4 - DC Motor Response to Train of Steps

As seen the DC motor follows the train of steps successfully and with almost no delay. The step

response information for the compensated response is shown in the figure below.

Figure 5 - DC motor Compensated Step Response information

The peak time has gone below the threshold of 0.05 seconds to 0.042 seconds while the overshoot

is also below the threshold of 4.6% to about 3.48 %. This validates the designed controller to fulfil

the requirements.