presentation slide on TIME DOMAIN SPECIFICATIONS OF SECOND ORDER

1. TIME DOMAIN SPECIFICATIONS OF
SECOND ORDER SYSTEM
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1SYED HASAN SAEED

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

3. SYED HASAN SAEED 3
Consider a second order system with unit step input and
all initial conditions are zero. The response is shown in fig.

4. 1. DELAY TIME (td): The delay time is the time required
for the response to reach 50% of the final value in
first time.
2. RISE TIME (tr): It is time required for the response to
rise from 10% to 90% of its final value for over-
damped systems and 0 to 100% for under-damped
systems.
We know that:
SYED HASAN SAEED 4
  






2
1
2
2
1
tan
1sin
1
1)(







t
e
tc n
tn
Where,

5. Let response reaches 100% of desired value. Put c(t)=1
SYED HASAN SAEED 5
  
   01sin
1
1sin
1
11
2
2
2
2













t
e
t
e
n
t
n
t
n
n
0 tn
e 
Since,
)sin())1sin((
0))1sin((
2
2


nt
t
n
n


Or,
Put n=1

6. SYED HASAN SAEED 6
2
2
1
)1(







n
r
rn
t
t
2
2
1
1
1
tan









n
rt
Or,
Or,

7. 3. PEAK TIME (tp): The peak time is the time required
for the response to reach the first peak of the time
response or first peak overshoot.
For maximum
SYED HASAN SAEED 7
  






t
e
tc n
tn
2
2
1sin
1
1)(Since
  
   )1(0
1
1sin
11cos
1
)(
0
)(
2
2
22
2










tn
n
nn
t
n
n
et
t
e
dt
tdc
dt
tdc








8. Since,
Equation can be written as
Equation (2) becomes
SYED HASAN SAEED 8
0 tn
e 
     


sin1
1sin11cos
2
222

 tt nn
Put  cosand
       cos1sinsin1cos 22
 tt nn




cos
sin
))1cos((
))1sin((
2
2



t
t
n
n

9. SYED HASAN SAEED 9


nt
nt
pn
n


)1(
))1tan((
2
2
The time to various
peak
Where n=1,2,3,…….
Peak time to first overshoot, put n=1
2
1 



n
pt
First minimum (undershoot) occurs at n=2
2min
1
2


n
t

10. 4. MAXIMUM OVERSHOOT (MP):
Maximum overshoot occur at peak time, t=tp
in above equation
SYED HASAN SAEED 10
  






t
e
tc n
tn
2
2
1sin
1
1)(
2
1 



n
ptPut,























2
2
2
1
1
.1sin
1
1)(
2
n
n
n
n
e
tc

11. SYED HASAN SAEED 11
2
2
1
2
2
1
2
1
1
1
1)(
1sin
sin
1
1)(
)sin(
1
1)(
2
2
2




























e
tc
e
tc
e
tc
Put,
 sin)sin( 

12. SYED HASAN SAEED 12
2
2
2
1
1
1
11
1)(
1)(
















eM
eM
tcM
etc
p
p
p
100*%
2
1 



 eM p

13. 5. SETTLING TIME (ts):
The settling time is defined as the time required for the
transient response to reach and stay within the
prescribed percentage error.
SYED HASAN SAEED 13

14. SYED HASAN SAEED 14
Time consumed in exponential decay up to 98% of the
input. The settling time for a second order system is
approximately four times the time constant ‘T’.
6. STEADY STATE ERROR (ess): It is difference between
actual output and desired output as time ‘t’ tends to
infinity.
n
s Tt

4
4 
 )()( tctrLimite
t
ss 


15. EXAMPLE 1: The open loop transfer function of a servo
system with unity feedback is given by
Determine the damping ratio, undamped natural frequency
of oscillation. What is the percentage overshoot of the
response to a unit step input.
SOLUTION: Given that
Characteristic equation
SYED HASAN SAEED 15
)5)(2(
10
)(


ss
sG
1)(
)5)(2(
10
)(



sH
ss
sG
0)()(1  sHsG

16. SYED HASAN SAEED 16
0207
0
)5)(2(
10
1
2




ss
ss
Compare with 02 22
 nnss  We get
%92.1100*
7826.0
7472.4**2
sec/472.420
72
20
22
)7826.0(1
7826.0*
1
2


















eeM
rad
p
n
n
n
%92.1
7826.0
sec/472.4


p
n
M
rad



17. EXAMPLE 2: A feedback system is described by the
following transfer function
The damping factor of the system is 0.8. determine the
overshoot of the system and value of ‘K’.
SOLUTION: We know that
SYED HASAN SAEED 17
KssH
ss
sG



)(
164
12
)( 2
016)164(
16)164(
16
)(
)(
)()(1
)(
)(
)(
2
2





sKs
sKssR
sC
sHsG
sG
sR
sC
is the characteristic eqn.

18. Compare with
SYED HASAN SAEED 18
K
ss
n
n
nn
1642
16
02
2
22






.sec/4radn 
K1644*8.0*2  15.0K
%5.1
100*100*
22
)8.0(1
8.0
1

 



p
p
M
eeM




19. EXAMPLE 3: The open loop transfer function of a unity
feedback control system is given by
By what factor the amplifier gain ‘K’ should be multiplied so
that the damping ratio is increased from 0.3 to 0.9.
SOLUTION:
SYED HASAN SAEED 19
)1(
)(
sTs
K
sG


0
/
)(
)(
1.
)1(
1
)1(
)()(1
)(
)(
)(
2
2









T
K
T
s
s
T
K
T
s
s
TK
sR
sC
sTs
K
sTs
K
sHsG
sG
sR
sC
Characteristic Eq.

20. Compare the characteristic eq. with
Given that:
SYED HASAN SAEED 20
02 22
 nnss 
T
K
T
n
n


2
1
2

We get
TT
K 1
2 
T
K
n 
KT2
1
Or,
9.0
3.0
2
1




TK
TK
2
2
1
1
2
1
2
1





21. SYED HASAN SAEED 21
21
2
1
2
1
2
2
1
9
9
1
9.0
3.0
KK
K
K
K
K











Hence, the gain K1 at which 3.0 Should be multiplied
By 1/9 to increase the damping ratio from 0.3 to 0.9