1 Stochastic Structural Dynamics Lecture-19 Dr C S Manohar Department of Civil Engineering Professor of Structural Engineering Indian Institute of Science Bangalore 560 012 India [email protected] Failure of randomly vibrating systems-3
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111
Stochastic Structural Dynamics
Lecture-19
Dr C S ManoharDepartment of Civil Engineering
Professor of Structural EngineeringIndian Institute of ScienceBangalore 560 012 India
Failure of randomly vibrating systems-3
2
0 0
(0, , ) ( ) ,
,
T T
N T N T X t X t dt n t dt
n t X t X t
0 0
,0, 0 ,
, 0
T T
M T X t X t U X t dt m t dt
m t X t X t U X t
12 2 2
22 2 211 1 1
1exp 1
22 2 2 1 2 1pp erf
0
1, 12
T
x
y t erf dtT t
Level crossing
Number of peaks
pdf of peaks
Fractional occupation time
Recall
3
2
0 0
22 10 00
0
Recall0
0 ; 0
cos
; tan
x xx x x x
x t R t
x xR xx
R x t t
20 0
0 00
0 00
2 0; 0 ; 0
exp( ) cos sin
cos ; sin
exp( ) cos
d dd
d
d
x x x x x x x
x xx t t x t t
x xx R R
x t t R t
2
0 0
2 22
2 cos
0 ; 0
lim ( )cos
1;1 2
st dt
st
Px x x tm
x x x x
x t X DMF t
PX DMFk r r
Envelope and phaseprocesses
4
2
0
0
0
0
2
0 0; 0 0; 1
1 exp sin ( )
1 exp sin cos cos sin ( )
sin ( ) cos
1 exp cos ( )
1( ) exp sin
t
dd
t
d d d dd
d d
t
dd
t
d
x x x f t
x x
x t t t f d
t t t f d
A t t B t t
A t t f d
B t t
( )d f d
5
2 2
1
( ) cos
cos
sin
tan
dx t R t t t
A t R t t
B t R t t
R t A t B t
B tt
A t
6
22 2
2
22
2
22
2 222 2
KE+PE
d
n
x tR t x t
x tx t
mx tx t
kkx mx
k
Energy interpretation
7
whenever 0
( ) passes through extrema of ( ).If ( ) is a sample of a narrow band process,
( ) passes through all the peaks.Therefore one can expect similarities in theproperties of envelo
R t x t x t
R t x tx t
R t
pe and peaks of ( ).
x t
8
0
1 cos cos
( ) cos( ) 1 cos
m
m
m
m
x t A t t
E t A tE t A t
0 5 10 15 20 25-1.5
-1
-0.5
0
0.5
1
1.5
time t
x(t)
1 cos cosmx t A t t
( ) cos mE t A t
0 ( ) 1 cos mE t A t
0 ( ) 1 cos mE t A t
9
0 5 10 15 20 25-5
-4
-3
-2
-1
0
1
2
3
4
5
time t
x(t)
How do we generalize the notion of envelopeand phase to describe random processes?
10
-5 -4 -3 -2 -1 0 1 2 3 4 5-30
-20
-10
0
10
20
30
x(t)
xdot
(t)
1111
01
Let be a zero mean, stationary, Gaussian random process defined as
cos sin ;
Here ~ 0, , ~ 0, ,
0
n n n n nn
n n n n
n k
X t
X t a t b t n
a N b N
a a n
Fourier representation of a Gaussian random process
Assumptions
Recall
1
, 0 ,
0 , 1,2, ,
cos sin 0
n k
n k
n n n nn
k b b n k
a b n k
X t a t b t
1212
1 1
1 1
2
1
cos sin cos sin
cos sin cos sin
cos
n n n n n n n nn n
n n n n n n n nn m
XX n nn
X t X t a t b t a t b t
a t b t a t b t
R
is a WSS random process.
is Gaussian.
is a SSS process.
X t
X t
X t
1313
1
1
1
2
Consider the psd function
1 cos2
1 cos2
Compare this with
cos
XX n n nn
XX n n nn
XX n n nn
XX n
S S
R S
R S
R
Fourier representation of a Gaussian random process (continued)
1
2By choosing , we see that the two ACF-s2
coincide.
nn
n nn
S
1414
0 5 10 15 20 25 30 35 40 45 500
0.005
0.01
0.015
frequency rad/s
psd
1n n
2Area 2n n nS
01
By discretizing the psd function as shown we can simulatesamples of ( ) using the Fourier representation
cos sin ; n n n n nn
X t
X t a t b t n
15
1
1 1
Let be a zero mean, stationary, random process defined as
cos
Here are deterministic constants and form
an iid sequence of random variables with
n n nn
n nn n
X t
X t A t
A
Alternative representation
1
12 2
1 0 0
a common PDFthat is uniformly distributed in 0 to 2 .
cos
cos cos sin sin
1 1cos cos sin sin2 2
0
n n nn
n n n n nn
n n n n n n nn
X t A t
A t t
A t d t d
16
1
1 1
2 2 2
1
2
1
cos cos sin sin
cos cos sin sin
cos cos sin sin
cos cos cos sin sin sin
cos
n n n n nn
n m n n n nn m
n n n n
n n n n n n nn
n nn
X t A t t
X t X t A A t t
t t
A t t t t
A
is a WSS random process.
is Gaussian (apply central limit theorem).
is a SSS process.
X t
X t
X t
17
01
1
cos sin ;
~ 0, , ~ 0, ,
0 , 0 ,
0 , 1,2, ,
cos sin
cos c
n n n n nn
n n n n
n k n k
n k
n n r r n n r rn
n n r
X t a t b t n
a N b N
a a n k b b n k
a b n k
X t a t t b t t
a t
Rice's definition of envelope and phase processes
1
1
os sin sin
sin cos cos sin
r n r rn
n n r r n r rn
t t t
b t t t t
Central frequencyr
18
1
1
1
1
cos cos sin sin
sin cos cos sin
cos sin cos
sin cos sin
cos sin
n n r r n r rn
n n r r n r rn
n n r n n r rn
n n r n n r rn
c r s r
c
X t a t t t t
b t t t t
a t b t t
a t b t t
I t t I t t
I
1
1
cos sin
sin cos
n n r n n rn
s n n r n n rn
t a t b t
I t a t b t
19
1
2 2 2
1
1
2 2 2
1
cos sin
sin cos
c n n r n n rn
c nn
s n n r n n rn
s nn
I t a t b t
I t X t
I t a t b t
I t X t
20
2 2 2
1
cos sin
cos ; sin
tan
cos
( ) Envelope process associated with ( )Phase process associated with ( )
Central frequency as
c r s r
c s
c s
s
c
r
r
X t I t t I t t
I t a t t I t a t t
a t I t I t
I tt
I t
X t a t t t
a t X tt X t
sociated with ( )X t
21
1
1 1
cos
Here are deterministic constants and form
an iid sequence of random variables with a common PDFthat is uniformly distributed in 0 to 2 .
c
n n nn
n nn n
n
X t A t
A
X t A
Alternative representation
1
1
1
1
os
cos cos sin sin
cos sin
cos
sin
n r n rn
n n r n r n r n rn
c r s r
c n n r nn
s n n r nn
t t
A t t t t
E t t E t t
E t A t
E t A t
22
2 2
1
cos sin
cos
sin
cos
tan
( ) Envelope process associated with ( )Phase process associated with ( )
Central frequency ass
c r s r
c
s
r
c s
s
c
r
X t E t t E t t
E t a t t
E t a t t
X t a t t t
a t E t E t
E tt
E t
a t X tt X t
ociated with ( )X t
23
2 2
1
Let ( ) and ( ) be two random processes.Define
( ) ( ) cos ( )sinLet A cos & sin
( ) ( ) cos
tan
A t B t
X t A t t B t tt R t t B t R t t
X t R t t t
R t A t B t
B tt
A t
Probability distributions of Envelope and phase processes
By definition( )=amplitude process, envelope, or amplitude modulation of ( )
phase process, or phase modulation of ( )carrier frequency or central frequency
R t X tt X t
24
1
cossin
2 2
2 222
ProblemGiven , ; to find , ; .
cossincos sinsin cos
, ,
Let and be jointly Gaussian
21 1, exp2 12 1
AB R
a rR ABb r
abAB
a b a baba b ab
p a b t p r tA RB R
RJ R
R
p r rp a b
A B
r aba bp a brr
25
cossin
2
22 2 2 2
2 22
2
0
0
, ,
,2 1
2 sin cos1 cos sinexp2 1
0 ;0 2
, ;0
, ;0 2
a rR ABb r
R
a b ab
ab
a b a bab
R R
R
p r rp a b
rp rr
r rr rr
r
p r p r d r
p p r dr
26
22 22
2 22 2
02 2 2 22
0
2
2 2
2
2exp
4 1 2 11
0 .Bessel's function of the first kind
1
cos sin2
a bab
a ba bR
a b ab a b aba b ab
ab
a bb
rrp r r I r
r rr
rI
rp
2
;0 2sin cosab
a a b
r
is generalized Rayleigh random variable is a generalized random variable.
R
27
2
00
0
exp cos 2
modified Bessel's function of argument and order 0.
b d I b
I b b
Note
28
2 2 2 2
2 2 2
2
2 2
2 2 2
2 20 0
2
2 2
0,
1 cos sin, exp2 2
exp ;0 ;0 22 2
, exp2 2
ex [Rayleigh RV]p ;0 2 2
Simi
ab a b
R
R R
R
r
r r rp r
r r r
r rp r p r d d
r rp r r
Special case
[Un1lar iforly we ge m RVt ;0 ]2 2
,R R
p
p r p r p R
29
Envelope and peaks share common properties.This is not surprising since the envelope passes throughall the peaks.The heuristic approach based on which we derived
Rayleigh model for the peak di
Remarks
stribution stands justifiedsince using a more rigorous arguement we have arrivedat Rayleigh model for the envelope.
and are independent of the central frequency R rp r p
30
1 2 1 2
1 1 1 1 1
2 2 2 2 2
1 1 1
1 1 1
2 2 2
2 2 2
1 1 1
1 1 1
2 2 2
2 2 2
, , ,&
cos sin
cos sin
cos
sin
cos
sin
cossincossin
R t R t t t
X t A t t B t t
X t A t t B t t
A t R t t
B t R t t
A t R t t
B t R t t
A RB RA RB R
Joint pdf of1 1 1
1 1 11
2 1 2
2 2 2
1 1
1 2 1 2
2 2 2
1
1 1 2 1 2
2 2 2
1 2
cos sin 0 0sin cos
0 0 cos sin0 0 sin cos
cos 0 0cos 0 cos sin
0 sin cos
sin 0 0sin 0 cos sin
0 sin cos
RR
JR
R
RR
R
R RR
R R
31
1 1 12 2 21 2 1 2 1 1 2 21 1 12 2 2
1 2 1 1 2 2
coscos1 2 1 2 1 2 1 2 1 2
sinsin
2 2
1 2 1 2 1 1 1 1 2 2 2 2 1 20 0
, , , , , ,
, cos , sin , cos , sin
If ( ) is a stationary Gaussian random proces
a ra rR R A B A Bb rb r
R R A B A B
p r r r r p a a b b
p r r r r p r r r r d d
X t
1 2
22 2 2 21 2 1 2
1 2 0 13 14 1 2
2 2 2 2
13
14
4 2 213 14
s with zero mean it can beshown that (exercise)
, exp2R R
r r r rp r r I r r
A B X t
A t A t B t B t
A t B t B t A t
32
1
&
cos
sin
cos sin
sin cos
cos sin 0 0sin cos 0 0
sin sin cos cos sincos cos sin sin cos
cos 0 0cos sin cos c
A t A t
A t R t t
B t R t t
A t R t t R t t t
B t R t t R t t t
RR
JR R R R
R R R
RR R
Joint pdf of
2 2 2 2 2
sin 0 0os sin sin sin cos sin
cos sin sin cos cos sin cos
cos sin
R R R RR R R R
R R R
33
2 cossincos sinsin cos
, , , ; , , , ;
0
0 2
a rb rRR AABBa r rb r r
p r r t r p a a b b t
rr
34
2 2 2 2
2 2 21
Let and be zero mean, Gaussian, stationary random processes
, , , are independent
( ) ( ) cos ( )sin
( ) sin ( )cos ( ) cos ( )sin
X
A t B t
A t B t X t
A t A t B t B t
A t B t
X t A t t B t t
X t A t t A t t B t t B t t
B
2 2 2 2 21
sin cos
XX
A t A B t
X t
35
2 2 2 2 2
2 2 22 211
2
0
2 2
1 2 22 12
, , , ; exp22
0 ; ;0 2 ;
, ; , , , ;
exp ; 0 ;22
RRXX
RR RR
XX
r r r rp r r t
r r
p r r t p r r t d d
r r r r r
Exercise: verify if the following are true.
R S Langley, 1986, On various definitions of the envelope of arandom process, Journal of Sound and Vibration, 105(3), 503-512.
Determine p , ;t
36
0
21
1 222
Knowing , ; the number of crossing of level by the envelope process ( ) can be characterized.
, , ;
exp22
Average rate of crossing of the level
RR
R RR
XX
p r r tR t
n t rp r t dr
Remarks
with positive slope by ( )
: ( ) is non-Gaussian Crossing of a level for narrow band processes
occur in clumps
R tR t
Note
37
0 5 10 15 20 25-1.5
-1
-0.5
0
0.5
1
1.5
time t
x(t)
Clumping
38
1
Average clump size
, 1, 2
X X X
R
n tcs
n t
Poisson model for number of level crossing is more appropriate for ( ) than for ( )R t X t
39
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10
0
10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10
0
10
x(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10
0
10
time s
Time required by ( ) to reach level for the first timeX t
40
The time required by ( ) to cross level for the first timeA real valued random variable taking values in 0 to Given the complete description of ( ),
can we characterize ?
This is known a
f
f
T
X t
X tT
s the First passage problem Barrier crossing problem Outcrossing problem
41
•The threshold level is high (so that crossing is a rare event)•Crossing times are mutually independent• ( ,0, )is a Poisson random variable
,0, x
,
p
0,
e!
k
N T
N
TP N T
T
kk
AssumptionsPoisson model for
2
2
rate of crossing of level ,
If ( ) is a stationary gaussian random process with zero mean
1, exp2
x
x x
T
n t
X t
n t
42
2
2
2
2 2
2
,0, exp!
1, exp2
1exp2 1,0, exp exp ;
! 2
0,1, 2, ,
k
x
x x
k
x
x x x
x x
TP N T k T
k
n t
TTP N T k
k
k
43
2
2
2
2
2 2
2
no points in 0 to
,0, 0
1exp exp2 2
1
11 exp exp2 2
1 1exp exp exp2 2 2 2
f
f
f
f
x
x x
T f
x
x x
TT
x x
x x x x
P T t P t
P N T
t
P t P T t
t
dP tp t
dtt
2
0 t
44
2
2
0
1 exp
exp 0
1exp2 2
1exp
f
f
T
T
x
x x
f
P t t
p t t t
T t t dt
is exponentially distributedfT
45
1 2
Let ( ) be a nonstationary, zero mean, Gaussian random processwith autocovariance function , .Determine the average rate of crossing of level bythe process ( ).
XX
X tR t t
X t
Example
2 2
2 222
, , ;
We need the jpdf of and .
1 1 2, ; exp2 12 1
,
XX
x x x xx x
n t X t X t x p x t dx
X t X t
x x rxxp x x trr
x x
46
22
2 2
2
2 2
1, 1 [exp2 2 1
exp 1 ]2 2 1
The quantities , ,& are time varying.If 0 and ,& are time invariant, the
above express
x
x x
x x x
x x
x x
n t rr
r rerfr
rr
Note
ion reduces to the expression forthe case when ( ) is stationary.This is what is expected.
X t
47
0
1 exp ,f
t
T XP t n d
48
21
1 220 2
21
1 222
( )
, , ; exp22
1 exp exp22
f
R RRX
X
TX
X
R t
n t rp r t dr
tP t
First passage time for envelope process Recall
49
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10
0
10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10
0
10
x(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10
0
10
time s
The maximum value of ( ) in interval 0 toX t T
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