EE 380 Linear Control Systems Lecture...

EE 380 Fall 2014Lecture 17.

EE 380

Linear Control Systems

Lecture 17

Professor Jeffrey SchianoDepartment of Electrical Engineering

1


Lecture 17 Topics

• Basic Principles of Feedback– Steady-State Accuracy: Non-unity Feedback – Disturbance Rejection

2


Advantages Afforded by Feedback • Achieve a stable closed-loop system when the plant is

unstable

• Obtain desired transient response characteristics

• Reduce the sensitivity of the closed-loop system to parameter uncertainty

• Minimize the steady-state error between the desired response and plant output

• Attain a desired response despite the presence of external disturbances

3


Steady-State Accuracy:Non-unity Feedback

• Consider the non-unity feedback system

• Results form Lecture 16 do not apply as the summer output is not the error signal

4

( ) ( ) ( )e t r t y t

( )R s

( )Y s

( )H s

( )G s


Steady-State Accuracy:Approach for Non-unity Feedback

• Transform the non-unity feedback system into a unity feedback system

• Apply the results from Lecture 16

5


Block Diagram Transformation

6

• Step 1: Add and subtract unity feedback paths

( )R s ( )Y s

( )H s

( )G s

1

( )R s

( )Y s

( )H s

( )G s



7

• Step 2: Combine H(s) with a unity feedback path

( )R s ( )Y s

( )H s

( )G s

1

( )R s ( )Y s

( ) 1H s

( )G s



8

• Step 3: Replace the inner feedback loop with a single transfer function block; apply Lecture 16 results!

( )R s ( )Y s

( ) 1H s

( )G s

( )R s

( )Y s

1 1G

G H


Example 1• Consider the non-unity feedback system

• Determine– The system type number– The error constants Kp and Kv

– Find the steady-state error for the inputs r(t) = uo(t)and r(t) = t uo(t)

9

( )R s

( )Y s

15s

100

10s s


Example 1 Solution

10


Example 1 Solution

11


Example 1 Solution

12


External Disturbance: Open-Loop Response

• Consider the effect of a disturbance d(t) on the plant output y(t) in the absence of feedback

13

( ) ( ) ( ) ( )Y s G s U s D s

( )U s ( )Y s( )G s

( )D s


External Disturbance:Closed-Loop Response

• Feedback provides a tool for reducing the effect of disturbances on the plant response

• Closed-loop system with disturbance input

• What is Y(s) in terms of R(s) and D(s)?

14

( )Y s( )pG s

( )D s

( )R s

( )cG s


Determination of Y(s)

15


Example 2• Consider a closed-loop system for regulating the angular

displacement of a DC motor shaft, where the ratio of the mechanical to electrical time constants is large

• Determine the steady-state error for the load torques TL(t) = TLo uo(t) and TL(t) = TLo t uo(t)

16

aV

1E

E

Ks

GK

s

eK

LT

eT 1M

M

Ks

bV

PKR

DC Motor

E


Example 2 Solution• Because the mechanical time constant is orders of

magnitude larger than the electrical time constant, it dominates the system response

• Equivalently, the bandwidth is limited by the mechanical system

• Within the bandwidth of the mechanical response, the following approximation holds

17

because 11

EE E

E

K K ss


Example 2 Solution• To determine the transfer function from load torque to output

– Set the reference input to zero– Approximate the electrical dynamics as a gain– Rearrange the block diagram

18

LT

E

PK

1M

M

Ks

GKs

eK

EK


Example 2 Solution• Determine the transfer function from load torque to error

19

2

( )1

11 1 1

1

1

G ML E P e

M G

G GM M ME P e L

M M M

M G M E P M e G M L

G M M

M e G M E PL

M M

K K sE T K K E K Es s K

K KK K KE K K K Ts s s s s

E s s K K K K K K s K K T

K KEK K K K K KT s s


Example 2 Solution• Determine the steady-state error for TL(t) = TLo uo(t)

• Determine the steady-state error for TL(t) = TLo tuo(t)

• Increasing Kp decreases the steady-state error when the load torque is constant– How does Increasing Kp affect the transient response?

20

0 0lim lim Lo Lo

ss Ls sL L E P

T TE Ee s T sT T s K K

20 0 0lim lim limLo Lo

ss Ls s sL L L

T TE E Ee s T sT T s T s


Example 2 Solution• The poles of the closed-loop system satisfy

• It follows that

• Increasing Kp decreases z and increases the peak overshoot– Tradeoff between steady-state accuracy and transient

response

21

2 2 21 2 0M e G M E Pn n

M M

K K K K K Ks s s s

1 12 2

G M E Pn

M

M e M e

n M M G M E P

K K K K

K K K KK K K K


EE 380

Linear Control Systems

Lecture 17

Professor Jeffrey SchianoDepartment of Electrical Engineering

1


Lecture 17 Topics

• Basic Principles of Feedback– Steady-State Accuracy: Non-unity Feedback – Disturbance Rejection

2


Advantages Afforded by Feedback • Achieve a stable closed-loop system when the plant is

unstable

• Obtain desired transient response characteristics

• Reduce the sensitivity of the closed-loop system to parameter uncertainty

• Minimize the steady-state error between the desired response and plant output

• Attain a desired response despite the presence of external disturbances

3


Steady-State Accuracy:Non-unity Feedback

• Consider the non-unity feedback system

• Results form Lecture 16 do not apply as the summer output is not the error signal

4


Steady-State Accuracy:Approach for Non-unity Feedback

• Transform the non-unity feedback system into a unity feedback system

• Apply the results from Lecture 16

5



6

• Step 1: Add and subtract unity feedback paths



7

• Step 2: Combine H(s) with a unity feedback path



8

• Step 3: Replace the inner feedback loop with a single transfer function block; apply Lecture 16 results!


Example 1• Consider the non-unity feedback system

• Determine– The system type number– The error constants Kp and Kv

– Find the steady-state error for the inputs r(t) = uo(t)and r(t) = t uo(t)

9


Example 1 Solution

10


Example 1 Solution

11


Example 1 Solution

12


External Disturbance: Open-Loop Response

• Consider the effect of a disturbance d(t) on the plant output y(t) in the absence of feedback

13


External Disturbance:Closed-Loop Response

• Feedback provides a tool for reducing the effect of disturbances on the plant response

• Closed-loop system with disturbance input

• What is Y(s) in terms of R(s) and D(s)?

14


Determination of Y(s)

15


Example 2• Consider a closed-loop system for regulating the angular

displacement of a DC motor shaft, where the ratio of the mechanical to electrical time constants is large

• Determine the steady-state error for the load torques TL(t) = TLo uo(t) and TL(t) = TLo t uo(t)

16


Example 2 Solution• Because the mechanical time constant is orders of

magnitude larger than the electrical time constant, it dominates the system response

• Equivalently, the bandwidth is limited by the mechanical system

• Within the bandwidth of the mechanical response, the following approximation holds

17


Example 2 Solution• To determine the transfer function from load torque to output

– Set the reference input to zero– Approximate the electrical dynamics as a gain– Rearrange the block diagram

18


Example 2 Solution• Determine the transfer function from load torque to error

19


Example 2 Solution• Determine the steady-state error for TL(t) = TLo uo(t)

• Determine the steady-state error for TL(t) = TLo tuo(t)

• Increasing Kp decreases the steady-state error when the load torque is constant– How does Increasing Kp affect the transient response?

20


Example 2 Solution• The poles of the closed-loop system satisfy

• It follows that

• Increasing Kp decreases z and increases the peak overshoot– Tradeoff between steady-state accuracy and transient

response

21

EE 380 Linear Control Systems Lecture...

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