error analysis
TRANSCRIPT
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 2
STEADY STATE ERROR:
The steady state error is the difference between theinput and output of the system during steady state.
For accuracy steady state error should be minimum.
We know that
The steady state error of the system is obtained by finalvalue theorem
SYED HASAN SAEED 3
)()(1
)()(
)()(1
1
)(
)(
sHsG
sRsE
sHsGsR
sE
)(.lim)(lim0
sEsteest
ss
SYED HASAN SAEED 4
)(1
)(.lim
1)(
)()(1
)(.lim
0
0
sG
sRse
sH
sHsG
sRse
sss
sss
For unity feedback
Thus, the steady state error depends on the input and open loop transfer function.
STATIC ERROR COEFFICIENTS
STATIC POSITION ERROR CONSTAN Kp: For unit step input R(s)=1/s
SYED HASAN SAEED 5
)()(lim
1
1
)()(lim1
1
)()(1
1.
1.lim
0
0
0
sHsGK
KsHsGe
sHsGsse
sp
ps
ss
sss
Where is the Static position error constantpK
Steady state error
STATIC VELOCITY ERROR CONSTANT (Kv):
Steady state error with a unit ramp input is given by
R(s)=1/s2
SYED HASAN SAEED 6
)()(1
1).(.
0 sHsGsRsLime
sss
vs
ss
ssss
KsHssGe
sHssGssHsGsse
1
)()(
1lim
)()(
1lim
)()(1
1.
1.lim
0
020
Where )()(lim0
sHssGKs
v
Static velocity error coefficient
STATIC ACCELERATION ERROR CONSTANT (Ka):
The steady state error of the system with unit parabolic input is given by
where,
SYED HASAN SAEED 7
as
ss
ssss
KsHsGse
sHsGsssHsGsse
ssR
1
)()(
1lim
)()(
1lim
)()(1
1.
1.lim
1)(
20
22030
3
)()(lim 2
0sHsGsK
sa
Static acceleration constant.
STEADY STATE ERROR FOR DIFFERENT TYPE OF SYSTEMS
TYPE ZERO SYSTEM WITH UNIT STEP INPUT:
Consider open loop transfer function
SYED HASAN SAEED 8
)1().........1)(1(
.).........1)(1()()( 21
ba
m sTsTs
sTsTKsHsG
Ke
KKe
KsHsGK
ssR
ss
p
ss
sp
1
1
1
1
1
1
)()(lim
1)(
0 Hence , for type zero system the static position error constant Kp is finite.
TYPE ZERO SYSTEM WITH UNIT RAMP INPUT:
TYPE ZERO SYSTEM WITH UNIT PARABOLIC INPUT:
For type ‘zero system’ the steady state error is infinite for ramp and parabolic inputs. Hence, the ramp and parabolic inputs are not acceptable.
SYED HASAN SAEED 9
v
ss
bass
v
Ke
sTsT
sTsTKssHssGK
1
0)....1)(1(
)...1)(1(.lim)()(lim 21
00
sse
a
ss
bass
a
Ke
sTsT
sTsTKssHsGsK
1
0)....1)(1(
)...1)(1(.lim)()(lim 212
0
2
0
sse
TYPE ‘ONE’ SYSTEM WITH UNIT STEP INPUT:
Put the value of G(s)H(s) from eqn.1
TYPE ‘ONE’ SYSTEM WITH UNIT RAMP INPUT:
Put the value of G(s)H(s) from eqn.1
SYED HASAN SAEED 10
)()(lim0
sHsGKs
p
01
1
p
ss
p
Ke
K
0 sse
)()(.lim0
sHsGsKs
v
KKe
KK
v
ss
v
11
Kess
1
SYED HASAN SAEED 11
TYPE ‘ONE’ SYSTEM WITH UNIT PARABOLIC INPUT:
Put the value of G(s)H(s) from eqn.1
Hence, it is clear that for type ‘one’ system step input and ramp inputs are acceptable and parabolic input is not acceptable.
)()(lim 2
0sHsGsK
sa
a
ss
a
Ke
K
1
0
sse
Similarly we can find for type ‘TWO’ system.
For type two system all three inputs (step, Ramp,Parabolic) are acceptable.
SYED HASAN SAEED 12
INPUT SIGNALS
TYPE ‘0’ SYSTEM
TYPE ‘1’ SYSTEM
TYPE ‘2’ SYSTEM
UNIT STEP INPUT 0 0
UNIT RAMP INPUT 0
UNIT PARABOLIC
INPUT
K1
1
K
1
K
1
DYNAMIC ERROR COEFFICIENT:
For the steady-state error, the static error coefficients gives the limited information.
The error function is given by
For unity feedback system
The eqn.(2) can be expressed in polynomial form (ascending power of ‘s’)
SYED HASAN SAEED 13
)1()()(1
1
)(
)(
sHsGsR
sE
)2()(1
1
)(
)(
sGsR
sE
SYED HASAN SAEED 14
)3(........111
)(
)( 2
321
sK
sKKsR
sE
)4().......(1
)(1
)(1
)( 2
321
sRsK
ssRK
sRK
sEOr,
Take inverse Laplace of eqn.(4), the error is given by
)5(.......)(1
)(1
)(1
)(321
trK
trK
trK
te
Steady state error is given by
)(lim0
ssEes
ss
Let s
sR1
)(
SYED HASAN SAEED 15
1
2
3210
1
.......1
.11
..11
.1
.lim
Ke
ss
Kss
KsKse
ss
sss
Similarly, for other test signal we can find steady state error.
.......,, 321 KKK are known as “Dynamic error coefficients”
EXAMPLE 1: The open loop transfer function of unity feedback system is given by
Determine the static error coefficients
SOLUTION:
SYED HASAN SAEED 16
)10)(1.01(
50)(
sssG
avp KKK ,,
0)10)(1.01(
50lim
)()(
0)10)(1.01(
50.lim
)()(.lim
5)10)(1.01(
50lim
)()(lim
2
0
2
0
0
0
0
sss
sHsGsK
sss
sHsGsK
ss
sHsGK
s
a
s
sv
s
sp
EXAMPLE 2: The block diagram of electronic pacemaker isshown in fig. determine the steady state error for unitramp input when K=400. Also, determine the value of Kfor which the steady state error to a unit ramp will be0.02.
Given that: K=400,
SYED HASAN SAEED 17
,1
)(2s
sR 1)( sH
)20()()(
ss
KsHsG
SYED HASAN SAEED 18
05.0
)20(1
1.
1.lim
)()(1
)(.lim
200
ss
Kss
sHsG
sRse
ssss
Now, 02.0sse Given
1000
)20(
20lim02.0
)20(1
1.
1.lim
0
20
K
Kss
s
ss
Ksse
s
sss
THANK YOU
SYED HASAN SAEED 19