steady-state analysis date: 28 th august 2008 prepared by: megat syahirul amin bin megat ali email:...
TRANSCRIPT
Steady-State Analysis
Date: 28th August 2008
Prepared by: Megat Syahirul Amin bin Megat Ali
Email: [email protected]
Introduction Steady-State Error for Unity Feedback
System Static Error Constants and System Type Steady-State Error for Non-Unity
Feedback Systems
Steady-state error, ess: The difference between the input and the output for a prescribed test input as time, t approaches ∞.
Step Input
Steady-state error, ess: The difference between the input and the output for a prescribed test input as time, t approaches ∞.
Ramp Input
Test Inputs: Used for steady-state error analysis and design.
Step Input: Represent a constant position. Useful in determining the ability of the control system to
position itself with respect to a stationary target. Ramp Input:
Represent constant velocity input to a position control system by their linearly increasing amplitude.
Parabolic Input: Represent constant acceleration inputs to position control. Used to represent accelerating targets.
To determine the steady-state error, we apply the Final Value Theorem:
The following system has an open-loop gain, G(s) and a unity feedback since H(s) is 1. Thus to find E(s),
Substituting the (2) into (1) yields,
)(lim)(0
ssFfs
)()()( sCsRsE
)()()( sGsRsC
)(1
)()(
sG
sRsE
…(1)
…(2)
By applying the Final Value Theorem, we have:
This allows the steady-state error to be determined for a given test input, R(s) and the transfer function, G(s) of the system.
)(1
)(lim
)(lim)(
0
0
sG
ssR
ssEe
s
s
For a unit step input:
The term:
The dc gain of the forward transfer function, as the frequency variable, s approaches zero.
To have zero steady-state error,
)(lim0
sGs
)(lim1
1
)(1
)/1(lim)(
0
0
sG
sG
sse
s
sstep
)(lim0
sGs
For a unit ramp input:
To have zero steady-state error,
If there are no integration in the forward path:
Then, the steady state error will be infinite.
)(lim
1
)(lim
1
)(1
)/1(lim)(
0
0
2
0
ssG
ssGs
sG
sse
s
s
sramp
)(lim0
ssGs
0)(lim0
ssGs
For a unit parabolic input:
To have zero steady-state error,
If there are one or no integration in the forward path:
Then, the steady state error will be infinite.
)(lim
1
)(lim
1
)(1
)/1(lim)(
2
0
2
0
2
3
0
sGs
sGss
sG
sse
s
s
sparabola
)(lim 2
0sGs
s
0)(lim 2
0
sGs
s
Example: Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t2u(t).
21
5
)(lim1
5)(
0
sGe
s
step
)(lim
5)(
0ssG
es
ramp
)(lim
5)(
2
0sGs
es
ramp
System Type: The value of n in the denominator or, the number of pure integrations in the forward path.
Therefore,i. If n = 0, system is Type 0ii. If n = 1, system is Type 1iii. If n = 2, system is Type 2
Example:i.
ii.
iii.
Problem: Determine the system type.
31
2
ss
sKsG
3
12
s
sKsG
1121
15.02
sssss
sKsG
Type 0
Type 1
Type 3
Static Error Constants: Limits that determine the steady-state errors.
Position constant:
Velocity constant:
Acceleration constant:
)(lim0
sGKs
p
)(lim0
ssGKs
v
)(lim 2
0sGsK
sa
Steady-state error for step function input, R(s):
Position error constant:
Thus,
sRsR )(lim11
lim0
0 sG
R
sG
ssRe
ss
ss
)(lim0
sGKs
p
pss K
Re
1
Steady-state error for step function input, R(s):
Position error constant:
Thus,
2sRsR )(lim1
lim0
0 ssG
R
sG
ssRe
ss
ss
)(lim0
ssGKs
v
vss K
Re
Steady-state error for step function input, R(s):
Position error constant:
Thus,
3sRsR )(lim1
lim2
00 sGs
R
sG
ssRe
ss
ss
)(lim 2
0sGsK
sa
ass K
Re
Relationships between input, system type, static error constants, and steady-state errors:
Example: Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t2u(t) by first evaluating the static error constants.
21
5
1)(
pstep K
Re
vramp K
Re )(
a
ramp K
Re )(
0 ,0 ,20 avp KKK
Example: Calculate the error constants and determine ess for a unit step, ramp and parabolic functions response of the following system.
515
;12
1
ss
sHss
sG
Example: Calculate the error constants and determine ess for a unit step, ramp and parabolic functions response of the following system.
For step input,
01
1lim
51215
1
1lim
1lim
0
0
0
s
s
sss
ssssss
sHsG
ssRe
512
15lim
)()(lim
0
0
sss
s
sHsGK
s
sp
Example: Calculate the error constants and determine ess for a unit step, ramp and parabolic functions response of the following system.
For ramp input,
12
51215
1
1lim
1lim
2
0
0
ssssss
sHsG
ssRe
s
sss
12
1
512
15lim
)(lim
0
0
ss
s
sHssGK
s
sv
Example: Calculate the error constants and determine ess for a unit step, ramp and parabolic functions response of the following system.
For parabolic input,
51215
1
1lim
1lim
3
0
0
ssssss
sHsG
ssRe
s
sss
0
512
15lim
lim
0
2
0
ss
ss
sHsGsK
s
sa
Problem: Calculate the error constants and determine ess for a unit step, ramp and parabolic functions response of the following system.
Chapter 5i. Dorf R.C., Bishop R.H. (2001). Modern Control
Systems (9th Ed), Prentice Hall. Chapter 7
i. Nise N.S. (2004). Control System Engineering (4th Ed), John Wiley & Sons.
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