presentation slide based on ERROR ANALYSIS OF TIME RESPONSE

1. ERROR
ANALYSIS
Email : hasansaeedcontrol@gmail.com
URL: http://shasansaeed.yolasite.com/
1SYED HASAN SAEED

2. BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &
GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 2

3. STEADY STATE ERROR:
The steady state error is the difference between the
input 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 final
value theorem
SYED HASAN SAEED 3
)()(1
)(
)(
)()(1
1
)(
)(
sHsG
sR
sE
sHsGsR
sE




)(.lim)(lim
0
sEstee
st
ss



4. SYED HASAN SAEED 4
)(1
)(
.lim
1)(
)()(1
)(
.lim
0
0
sG
sR
se
sH
sHsG
sR
se
s
ss
s
ss







For unity feedback
Thus, the steady state error depends on the input and
open loop transfer function.

5. 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
KsHsG
e
sHsGs
se
s
p
p
s
ss
s
ss










Where is the Static position error constantpK
Steady state error

6. 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 sHsG
sRsLime
s
ss



v
s
ss
ss
ss
KsHssG
e
sHssGssHsGs
se
1
)()(
1
lim
)()(
1
lim
)()(1
1
.
1
.lim
0
020







Where )()(lim
0
sHssGK
s
v

 Static velocity error
coefficient

7. STATIC ACCELERATION ERROR CONSTANT (Ka):
The steady state error of the system with unit parabolic
input is given by
where,
SYED HASAN SAEED 7
a
s
ss
ss
ss
KsHsGs
e
sHsGsssHsGs
se
s
sR
1
)()(
1
lim
)()(
1
lim
)()(1
1
.
1
.lim
1
)(
20
22030
3








)()(lim 2
0
sHsGsK
s
a

 Static acceleration constant.

8. 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
sTsTK
sHsG
K
e
KK
e
KsHsGK
s
sR
ss
p
ss
s
p









1
1
1
1
1
1
)()(lim
1
)(
0 Hence , for type zero
system the static position
error constant Kp is finite.

9. 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
ba
ss
v
K
e
sTsT
sTsTK
ssHssGK
1
0
)....1)(1(
)...1)(1(
.lim)()(lim 21
00
 sse
a
ss
ba
ss
a
K
e
sTsT
sTsTK
ssHsGsK
1
0
)....1)(1(
)...1)(1(
.lim)()(lim 212
0
2
0






 sse

10. 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
)()(lim
0
sHsGK
s
p


0
1
1




p
ss
p
K
e
K
0 sse
)()(.lim
0
sHsGsK
s
v


KK
e
KK
v
ss
v
11


K
ess
1


11. 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
0
sHsGsK
s
a




a
ss
a
K
e
K
1
0
 sse

12. 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
K1
1

 
K
1
K
1

13. 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

14. SYED HASAN SAEED 14
)3(........
111
)(
)( 2
321
 s
K
s
KKsR
sE
)4().......(
1
)(
1
)(
1
)( 2
321
 sRs
K
ssR
K
sR
K
sEOr,
Take inverse Laplace of eqn.(4), the error is given by
)5(.......)(
1
)(
1
)(
1
)(
321


tr
K
tr
K
tr
K
te
Steady state error is given by
)(lim
0
ssEe
s
ss


Let s
sR
1
)( 

15. SYED HASAN SAEED 15
1
2
321
0
1
.......
1
.
11
..
11
.
1
.lim
K
e
s
s
Ks
s
KsK
se
ss
s
ss









Similarly, for other test signal we can find steady state
error.
.......,, 321 KKK are known as “Dynamic error
coefficients”

16. 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
)(


ss
sG
avp KKK ,,
0
)10)(1.01(
50
lim
)()(
0
)10)(1.01(
50
.lim
)()(.lim
5
)10)(1.01(
50
lim
)()(lim
2
0
2
0
0
0
0

















ss
s
sHsGsK
ss
s
sHsGsK
ss
sHsGK
s
a
s
s
v
s
s
p

17. EXAMPLE 2: The block diagram of electronic pacemaker is
shown in fig. determine the steady state error for unit
ramp input when K=400. Also, determine the value of K
for which the steady state error to a unit ramp will be
0.02.
Given that: K=400,
SYED HASAN SAEED 17
,
1
)( 2
s
sR  1)( sH
)20(
)()(


ss
K
sHsG

18. SYED HASAN SAEED 18
05.0
)20(
1
1
.
1
.lim
)()(1
)(
.lim 200







ss
Ks
s
sHsG
sR
se
ss
ss
Now, 02.0sse Given
1000
)20(
20
lim02.0
)20(
1
1
.
1
.lim
0
20









K
Kss
s
ss
Ks
se
s
s
ss

19. THANK YOU
SYED HASAN SAEED 19