Time Domain Specifications (Settling Time)
• Settling time (2%) criterion
• Time consumed in exponential decay up to 98% of the input.
t s  4T 
4
 n
T 
1
 n
• Settling time (5%) criterion
• Time consumed in exponential decay up to 95% of the input.
t s  3T 
3
 n
Summary of Time Domain Specifications
Rise Time
 
 
tr 

2
d
n 1  
Peak Time


tp 

d  1   2
n
Settling Time (2%)
t s  4T 
t s  3T 
4
 n
3
 n
Settling Time (4%)
Maximum Overshoot

Mp e

1 2
 100
Example#5
• Consider the system shown in following figure, where
damping ratio is 0.6 and natural undamped frequency is 5
rad/sec. Obtain the rise time tr, peak time tp, maximum
overshoot Mp, and settling time 2% and 5% criterion ts when
the system is subjected to a unit-step input.
Example#5
Rise Time
Peak Time

tp 
d
 
tr 
d
Settling Time (2%)
t s  4T 
t s  3T 
Maximum Overshoot
4
 n
3
 n
Settling Time (4%)

Mp e

1 2
 100
Example#5
Rise Time
 
tr 
d
tr 
3.141  
n 1  

2
2

1


  tan1( n
)  0.93 rad
 n
tr 
3.141  0.93
5 1  0.6 2
 0.55s
Example#5
Peak Time
Settling Time (2%)

tp 
d
3.141
tp 
 0.785s
4
ts 
4
 n
4
ts 
 1.33s
0.6  5
Settling Time (4%)
ts 
3
 n
3
ts 
 1s
0.6  5
Example#5
Maximum Overshoot

Mp e

Mp e

1 2
 100
3.1410.6
1 0.6 2
 100
M p  0.095  100
M p  9.5%
Example#5
Step Response
1.4
1.2
Mp
Amplitude
1
0.8
0.6
0.4
Rise Time
0.2
0
0
0.2
0.4
0.6
0.8
Time (sec)
1
1.2
1.4
1.6
Example#6
• For the system shown in Figure-(a), determine the values of gain K
and velocity-feedback constant Kh so that the maximum overshoot
in the unit-step response is 0.2 and the peak time is 1 sec. With
these values of K and Kh, obtain the rise time and settling time.
Assume that J=1 kg-m2 and B=1 N-m/rad/sec.
Example#6
Example#6
Since J  1 kgm 2 and
B  1 Nm/rad/sec
C( s )
K
 2
R( s ) s  (1  KK h )s  K
• Comparing above T.F with general 2nd order T.F
n2
C( s )
 2
R( s ) s  2 n s  n2
n  K
 
(1  KK h )
2 K
Example#6
n  K
• Maximum overshoot is 0.2.
 
(1  KK h )
2 K
• The peak time is 1 sec

tp 
d
1

ln( e

1 2
)  ln0.2
3.141
n 1   2
n 
3.141
1  0.456 2
n  3.53
Example#6
n  3.96
n  K
3.53  K
3.532  K
K  12.5
 
(1  KK h )
2 K
0.456  2 12.5  (1  12.5K h )
K h  0.178
Example#6
tr 
 
n  3.96
ts 
n 1   2
4
 n
t s  2.48s
t r  0.65s
ts 
3
 n
t s  1.86s
Example#7
When the system shown in Figure(a) is subjected to a unit-step input,
the system output responds as shown in Figure(b). Determine the
values of a and c from the response curve.
a
s( cs  1)
Example#8
Figure (a) shows a mechanical vibratory system. When 2 lb of force
(step input) is applied to the system, the mass oscillates, as shown in
Figure (b). Determine m, b, and k of the system from this response
curve.
Example#9
Given the system shown in following figure, find J and D to yield 20%
overshoot and a settling time of 2 seconds for a step input of torque
T(t).
Example#9
Example#9
Step Response of critically damped System ( 
 n2
C( s )

R( s ) s   n 2
Step Response
C( s ) 
 n2
ss   n 
• The partial fraction expansion of above equation is given as
n2
ss  n 
2
A
B
C
 

s s  n s  n 2
n
1
1
C( s )  

s s  n s  n 2
c(t )  1  e nt  n e nt t
c(t )  1  e nt 1  nt 
1
2
)