ee 380 linear control systems lecture...
TRANSCRIPT
![Page 1: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/1.jpg)
EE 380 Fall 2014Lecture 17.
EE 380
Linear Control Systems
Lecture 17
Professor Jeffrey SchianoDepartment of Electrical Engineering
1
![Page 2: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/2.jpg)
EE 380 Fall 2014Lecture 17.
Lecture 17 Topics
• Basic Principles of Feedback– Steady-State Accuracy: Non-unity Feedback – Disturbance Rejection
2
![Page 3: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/3.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 4: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/4.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 5: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/5.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 6: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/6.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 7: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/7.jpg)
EE 380 Fall 2014Lecture 17.
Block Diagram Transformation
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
![Page 8: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/8.jpg)
EE 380 Fall 2014Lecture 17.
Block Diagram Transformation
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
![Page 9: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/9.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 10: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/10.jpg)
EE 380 Fall 2014Lecture 17.
Example 1 Solution
10
![Page 11: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/11.jpg)
EE 380 Fall 2014Lecture 17.
Example 1 Solution
11
![Page 12: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/12.jpg)
EE 380 Fall 2014Lecture 17.
Example 1 Solution
12
![Page 13: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/13.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 14: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/14.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 15: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/15.jpg)
EE 380 Fall 2014Lecture 17.
Determination of Y(s)
15
![Page 16: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/16.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 17: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/17.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 18: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/18.jpg)
EE 380 Fall 2014Lecture 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
LT
E
PK
1M
M
Ks
GKs
eK
EK
![Page 19: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/19.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 20: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/20.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 21: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/21.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 22: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/22.jpg)
EE 380 Fall 2014Lecture 17.
EE 380
Linear Control Systems
Lecture 17
Professor Jeffrey SchianoDepartment of Electrical Engineering
1
![Page 23: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/23.jpg)
EE 380 Fall 2014Lecture 17.
Lecture 17 Topics
• Basic Principles of Feedback– Steady-State Accuracy: Non-unity Feedback – Disturbance Rejection
2
![Page 24: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/24.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 25: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/25.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 26: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/26.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 27: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/27.jpg)
EE 380 Fall 2014Lecture 17.
Block Diagram Transformation
6
• Step 1: Add and subtract unity feedback paths
![Page 28: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/28.jpg)
EE 380 Fall 2014Lecture 17.
Block Diagram Transformation
7
• Step 2: Combine H(s) with a unity feedback path
![Page 29: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/29.jpg)
EE 380 Fall 2014Lecture 17.
Block Diagram Transformation
8
• Step 3: Replace the inner feedback loop with a single transfer function block; apply Lecture 16 results!
![Page 30: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/30.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 31: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/31.jpg)
EE 380 Fall 2014Lecture 17.
Example 1 Solution
10
![Page 32: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/32.jpg)
EE 380 Fall 2014Lecture 17.
Example 1 Solution
11
![Page 33: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/33.jpg)
EE 380 Fall 2014Lecture 17.
Example 1 Solution
12
![Page 34: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/34.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 35: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/35.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 36: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/36.jpg)
EE 380 Fall 2014Lecture 17.
Determination of Y(s)
15
![Page 37: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/37.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 38: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/38.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 39: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/39.jpg)
EE 380 Fall 2014Lecture 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
![Page 40: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/40.jpg)
EE 380 Fall 2014Lecture 17.
Example 2 Solution• Determine the transfer function from load torque to error
19
![Page 41: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/41.jpg)
EE 380 Fall 2014Lecture 17.
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
![Page 42: EE 380 Linear Control Systems Lecture 17courses.ee.psu.edu/schiano/ee380/Lectures/L17_EE380_f14.pdf · EE 380 Fall 2014 Lecture 17. Example 2 Solution • Because the mechanical time](https://reader034.vdocuments.us/reader034/viewer/2022052616/60a2c14920cdd1354e554872/html5/thumbnails/42.jpg)
EE 380 Fall 2014Lecture 17.
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