Lecture 10-2
Linear Systems Analysis
in the Time Domain II
- Transient Response -
System Analysis Spring 2015
2
2
2 2
2
2 2
( ) 2
1 1
( ) ( ), ( )
2
n
n n
n
n n
G s s s
R s step input C s
s s s s
G(s) ( )
R s C s ( )
Second Order Systems
2 2
1
2
( ) 1 cos sin 1 cos( )
1 1
tan
1
n n
t t
d d d
c t e t t e t
2 3
1
2 2
2
n nK s K K
s s s
2
22 2 2
( ) / 1 1
1
( ) (1 )
n n
n n
s
s s
Damping ratio and Pole placement
i) ζ.>1 : poles are real and distinct (over damped)
ii) ζ.=1: poles are real and coincident (critically damped)
iii) 0 < ζ < 1 : pole are complex conjugates (under damped)
iv) ζ. = 0 : The pole are purely imaginary (undamped)
System Analysis Spring 2015
4
Step Response of Second-Order Systems
Step Response of Second-Order Systems
1. Over damped Case
1 2
1 2
( ) 1
p t p ty
stept C e
C e
2. Critically damped Case
3. Under damped Case
4. Undamped Case
1 2
( ) 1
pt pty
stept C e
C te
( ) 1
2cos( )
1
nt
step d
y t e t
1
tan
21
( ) 1 cos( )
step n
y t t
2
1
,
2 n n1
p p
1
,
2 np p
2
1
,
2 n n1
p p j
1
,
2 np p j
System Analysis Spring 2015
6
Under-damped Second-Order System
2
1
,
2 n n1
p p j
1
2d n
Damped Natural Frequency:
Influence of ω n and ζ on the pole locations
System Analysis Spring 2015
8
2 2
2 1
( ) 1 cos sin 1 sin( )
1 1
tan 1
n
n n
t
t t
step d d d
y t e t e t e t
Influence of ζ
T
r 0.10.9 1.0
t
T
pStep Response Based Second -Order System Specifications
T
s1) Peak Time: The time required to reach the first or maximum peak
2) Settling Time: The time required for the transeints’ damped oscillations to reach and stay within of the steady-state value.
3) Rise time : The time required to go from 0.1 to 0.9 of the final value
4) Percent Overshoot: % OS The amount that the waveform overshoots the steady-state at the peak time, expressed as a percentage of the steady-state value
( ) y t
2%
T
sT
pT
r%OS
System Analysis Spring 2015
10
2 2
2 2 2 2 2
( ) 2 ( ) (1 )
n n
n n n n
sY s s s s
1) Peak Time T
pStep Response Based Second -Order System Specifications
, 2 ,
d
t
2 2
2 2 2
1 1
( ) (1 )
n
n
n n
s
2
( )
2sin 1
1
nt n
y t e
n t
Set , y t ( ) 0
1
2 pd n
T
1 2 2
2 2
( ) 1 1 sin 1
1 1
n n
p p n
n
M y T e
2) % OS M
oStep Response Based Second -Order System Specifications
% OS = ( )
p steady state
100
steady state
y T y
y
1 exp
21
100 exp
2100
1
p s
o
s
percent overshoot
M y
M y
System Analysis Spring 2015
12 3) Settling time :
2
sin( )
1
nt
d
e y r e t
2% , 4 1
5% , 3 1
s
n
s
n
case T
case T
e
Step Response Based Second -Order System Specifications
2
0.2
1
nt
e
ln(0.2 1
2)
s
n
T
4) Rise time (01. to 0.9) :
Step Response Based Second -Order System Specifications
System Analysis Spring 2015
14
2
2 2
2 2
1
2 2
0, (0) 0
2 0
1
2 2
( ) (0) (0) 2 ( ) (0) ( ) 0
2 (0)
( ) 2
( ) (0) sin (0) cos (0) cos tan
1 1
n n
n n
n
n n
n
n n
t t
d d d
mx bx kx x
x x x
b b
m mk
s X s sx x sX s x X s
s x
X s s s
x t e
x t x t x e
t
1
2
Experimental Determination of Damping Ratio
m
b k
1
1
( 1) 1
( ( 1) )
n
n n
t
n T
t n T
n
x e
x e e
Experimental Determination of Damping Ratio
1 1
2 2
1
1
2
2 1
2 2 1
ln ln
1 1
ln ( 1)
1 ln 1
4 1 ln
1
n n
d n
n n
n
n
x x
x T n x
x n T
x
x
n x
x
n x
Logarithmic decrement
System Analysis Spring 2015
16
1
2 2
2 2
1
2
( ) (0) cos tan
1 1
1 1
1 , ,
tan 1 2
w tn
d
d n d
n
x t x e t
t t
n1 2
d n
Estimate of Response Time
End of Lecture 10-2