Download Structural Engineering - Indian Institute of Technology

25-05-2017
CE607: Random Vibration
By
Dr. A. Chakraborty
Department of Civil Engineering
Indian Institute of Technology Guwahati, India
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Syllabus
Concepts of probability, random variables, theory of
random process, stationary and nonstationary process,
Expected values, moments, spectral properties of random
process.
 Response of linear systems to random excitations, SDF
and MDF discrete systems, Continuous systems.
 Response of nonlinear systems to random excitations,
Fokker-plank equations, Markov vector approach, statistical
linearization and perturbation techniques. Level crossing,
Peaks envelops and first passage time, Monte-Carlo
simulation.

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Books (Text/Reference)
Texts
•D. Lutes and S. Sarkani, Random Vibrations: Analysis of Structural and
Mechanical Systems, Elsevier, Amsterdam, 2004.
•D.E. Newland, Random Vibration and Spectral Analysis, Longman, New York, 1984.
References
•A. Papoulis and S.U. Pillai, Probability, Random Variables and Stochastic
Processes, McGraw Hill, New Delhi, 2002.
•S.M. Ross, Stochastic Processes, Wiley, Delhi, 1996.
•Y.K. Lin, Probabilistic Theory of Structural Dynamics, McGraw Hill, New York, 1967.
•Y.K. Lin and G.Q. Cai, Probabilistic Structural Dynamics – Advanced Theory &
Applications, McGraw Hill, New York, 1995.
•N.C. Nigam, Introduction to Random Vibration, MIT Cambridge, 1983.
•N.C. Nigam and S. Narayanan, Applications of Random Vibrations, Narosa, New
Delhi, 1994.
•J.B. Roberts and P.D. Spanos, Random Vibration and Statistical Linearization,
Dover Publication, New York, 1999.
•R.A. Ibrahim, Parametric Random Vibration, Dover Publication, New York, 1985.
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Why do we study Random Vibration?
Example: Design against Earthquake
Sg
Design Philosophy:
• Identify loads acting
• Perform Analysis (Demand)
• Select suitable C/S and
material (Capacity)
IS Code
T
Demand = Capacity
OK
Obs: Structure fails even after designed
according to codal provisions
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Why do we study Random Vibration?
Uncertainty is a natural
phenomenon
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Uncertainty in Structural Engineering:
Aleatoric
Uncertainty
Epistemic
Sources of Uncertainty –
1. Loads (Earthquake, Wind, Wave, Road Roughness ….)
2. Structural Properties (Elastic Constants, BC, Damping…)
3. Modeling (Analytical, Experimental, Computational…)
4. Human Error
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Stochastic Structural Dynamics (RV):
Def: It is the branch of Structural Dynamics where
Uncertainties are modeled using Theory of Probability
(Stochastic Processes, statistics etc.)
How do we model Uncertainty?
1. Theory of Probability
2. Interval Algebra
3. Fuzzy Logic
4. AI
Etc.
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Theory of Probability:
Probability Can Defined by
1. Classical Approach
2. Relative Frequency Approach
3. Axiomatic Approach
Classical Definition
In a Random Experiment (where outcomes are Equally
likely, exhaustive, mutually exclusive), if ‘n’ no of outcome is
favorable to a event out of ‘N’ possibilities, the probability for
that event is defined as
P=n/N
Ex: Tossing a coin for head or tail
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Major Drawbacks in Classical Definition:
1.
2.
3.
4.
Outcomes need to be equally likely
What happens if the events are not equally likely?
No room for experimentation
Probability is a rational number
Relative Frequency Approach: if a random experiment
is performed and n no of outcome is favorable to an
event out of N possibilities
• experiment is required always
• what is that limit
• probability is again a rational number
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Axiomatic Definition:








Experiments, Trial, Outcomes – Notions
Axioms are not proofs
Samples Space – Toss a coin
Cardinality is 2
Sample Space – Finite or Infinite (Countable or Uncountable)
Event Space – B (let us consider finite ) is all the subset of
Ex. Tossing a coin –
B is the Sigma Algebra of subsets of
Note




P:B – [0 1] such that
Axiom 1 (Non negativity):
Axiom 2 (Normalization):
Axiom 3 (Additive):
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Theory of Probability (Axiomatic Definition):
Probability Space is the ordered triplet:
 Axiom 2:
 Axiom 3:


 What is conditional probability and Independence?

provided

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Theory of Probability (Axiomatic Definition):
Stochastic Independence – Two events A and B are
said to be independent if and only if


Notation:



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Theorem of Total Probability:
Let
forms a partition of
B
Let B be a set
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Bayes’ Theorem:
Note:
 Prior Probability is
Posterior Probability is
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Summary
•
•
•
•
•
•
•
Why do we study Random Vibration
Different aspects of Random Vibration
Structural Dynamics
Theory of Probability
Three different definitions of Probability
Conditional Probability
Bayes’ Theorem
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Define Random Variable:
Random
Experiment
Ω
B
Sample Space
ω
Event Space
1
0
is a random variable
Example:
Toss an unbiased coin
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Define Random Variable:
Def: It is a function from Sample Space onto real line such that
 every outcome has a fixed probability

is an event

Example:
Throw a die and let
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Define Random Variable:
Probability Mass
Function or pmf
1/6
X(ω)
1.0
0.0
Probability Distribution
Function or PDF
X(ω)
Note: Probability at a particular outcome is
right continuous
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Random Variable:
 Map Sample Space on Real line
 Enable us to quantify uncertainty
 Provides mathematical framework to evaluate statistical
properties
pmf & PDF
Discrete
Random
Variable
pdf & PDF
Continuous
Note:
Once the RV is defined, the information
of sample space is not required
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Properties of Random Variable:
Probability is Right Continuous
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Probability Density Function:
As per definition
Properties:
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Heaveside’s Step Function:
a
Box Function:
a
b
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Dirac Delta Function:
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Bernoulli's Random Variable:
Let a random experiment has two outcomes i.e.
success & failure
1.0
p
PDF
x
p
1-p
pmf
x
Check Total Probability
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Binomial Random Variable B(N,p):
Random experiment has N repeated Bernoulli’s trial
Such that –
 trials are independent
 each trial has two outcomes
 probability corresponding to each outcome remains constant
 define X is the no of success in N trials
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Binomial Random Variable:
Note: Sequences are mutually exclusive
Ex: Plot pdf & PDF
of B(20,0.4)
Proof: Use Binomial Theorem
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Geometric Random Variable:
Random experiment are same as in Binomial RV, but stop
the experiment when first success is achieved
i.e. N trials are required for first success
Geometric
Progression is used
and hence the name
Example:
p=0.6
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Poisson Random Variable:
 Modeling isolated phenomenon in time/space continuum
 Impossible to put upper bound
 Actual no of occurrence is very small
Note:
Total probability is again 1.0
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Summary
•
•
•
•
•
•
•
Why do we study Random Vibration
Different aspects of Random Vibration
Structural Dynamics
Theory of Probability
Three different definitions of Probability
Conditional Probability
Bayes’ Theorem
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Gaussian Random Variable N(μ,σ):
 1  x   2 
1
p X ( x) 
exp  
 
2 
 2    
   x  
 σ is always positive
 It can shown that Gaussian
RV as a limit of Binomial RV
 The condition for this case is
as follows –
n  ,   np,  2  npq,  2  1,
np  npq  k  np  npq
Special Case: N(0,1)
Ckn p k q n k
 1  k  np  2 
1
 

exp  

 2  npq  
2npq


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Transformations of Random Variables:
Let X and Y are two RVs such that Y  g ( X )
Also, p X (x) is known. Find pY ( y )
g(x)
Equating the area
under the pdf of X and
Y i.e. total probability
x
P y  Y  y  dy   pY ( y)dy
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Transformations of Random Variables:
Let X and Y are two RVs such that Y  g ( X )
Also, p X (x) is known. Find pY ( y )
P y  Y  y  dy   pY ( y )dy
pY ( y )dy  p X ( x1 )dx1  p X ( x2 )dx2  p X ( x3 )dx3
g(x)
3
x
 pY ( y )  
i 1
p X ( xi )
dy
dx x  xi
In general,
n
pY ( y )  
i 1
p X ( xi )
dy
dx x  xi
xi  g 1 ( y )
Note: Modulus in slope evaluation
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X  N  ,  ,  x  
Y  exp( X ),0  y  
 x  log( y )
dy
 exp( x)  y
dx
Example:
pY ( y ) 
p X ( x)
exp( x)
2

1
1  log y    
pY ( y ) 
exp  
 

2 y
 
 2 
Y is known as Log-Normal RV
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Example:
Let us define T  d  H where ‘T’ is the dependent random
variable and ‘h’ is the independent random variable whose
pdf is given by Evaluate pdf of ‘T’ i.e. f T t 
f
h 
H
exp  h 2 2 


2
1
for
h0
Ans.: Using the transformation of random variables
h  t  d    g 1 t 
2

d t  d 
1 1
 1 t d  



f T t   
.
f
h

.
exp

.

 

H
dt   
 2
2



 


td
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Example:
P
PL3

48 EI
EI
L
Qs.: Find out pdf of the deflection at the free end if
pP ( p)  N (10,2)
Remember:
n
pY ( y )  
i 1
p X ( xi )
dy
dx x  xi
d
L3
Ans.: dP  48EI
pP ( p) 
p ( ) 
48EI
p  3  *
L
 1  p  10  2 
1
exp  
 
2 2
 2  2  
48 EI
L3
2
*



1
1   10 
 
exp  
2 2
 2  2  
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Different Types of Random Variables:
Discrete
 Bernoulli
 Binomial
 Poison
 Etc.
Defined by pmf or PDF
Continuous
 Gaussian (Normal)
 Log-Normal
Defined by pdf or PDF
 Extreme Value
 Etc.
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Moment of a Random Variable:
Discrete RV
Continuous RV
Note:
E is called the expectation operator
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Algebra of Variance:
Proof:
Note: X & Y are independent
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Algebra of Variance:
Proofs are straight forward.
Try yourself.
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Algebra of Variance:
Some Important Properties
Prove that –
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Moment Generating Function:

Let us define: X ( s)  E exp( sX )   exp( sx ) p X ( x)dx

X (s) is the Laplace transformation of the pdf
( sX ) 2 ( sX )3
exp( sX )  1  sX 

 ...
2!
3!
 
 
3
s2
s
X ( s)  Eexp( sX )  1  sE X   E X 2  E X 3  ...
2!
3!
 d 2 X 
 d 3X 
 dX 
2
3
EX   
,E X  
,E X  
2 
3 

ds
ds
ds

 s 0

 s 0

 s 0
 
 
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Characteristic Function:

Let us define:  X ( )  Eexp( iX )   exp( ix) p X ( x)dx
Therefore:
1
p X ( x) 
2



X
( ) exp( ix)dx

 X ( ) is the Fourier transformation of the pdf
Using similar steps in Moment Generating Function,
one can show that
 
E Xn
 1 d n X 
 n
n 
i
d


  0
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Example:
P( X  0)  p, P( X  1)  q  1  p
Bernoulli’s RV
2
E X    xi P( X  xi )  0 * p  1* (1  p)  1  p  
i 1
    x P( X  x )  0 * p  1 * (1  p)  1  p
  E X    (1  p)  (1  p)  p(1  p)
2
EX
2
2
i 1
2
X
2
i
i
2
2
2
2
 X    exp( i 0). p  exp( i1).(1  p)  p  (1  p) exp( i )
 
E X n  (1  p )
All moments are equal
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Summary:
 Complete description of a random variable
 Probability Space
 pdf, PDF
 moments of all order
 Moment Generation Function & Characteristic
Function
 Transformation of Random Variables
 Algebra of Variance
 Solution of Static case with Uncertainty
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Markov Inequality:
Let a random variable has positive sample space i.e.
P( X  0)  0
Then for any
a0
p X (x)
area
E[ X ]
P( X  a) 
a
a
x
Hint:

a

0
0
a
E[ X ]   xpX ( x)dx  xpX ( x)dx   xpX ( x)dx
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Chebychev Inequality:
Let the RV X has mean and standard deviation
P X    k  
Then
( , )
p X (x)
2
k2
2


X


Note
is positive, also
X   
2

 k   X    k
2
Then, by Markov
Inequality
 k



x
k
E X   
P X     k 
k2

2
2
2

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Multi-dimensional RV:
Consider two RVs X and Y, the joint PDF and pdf is defined
as –
PXY x, y   P X  x  Y  y   P X  x, Y  y 
2
 PXY x, y 
p XY x, y  
xy
Note: Comma denotes
intersection
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Properties of Joint Distribution:
PXY x1 , y1   P X  x1  Y  y1 
PXY x,    P X  x  Y     P X  x 
PXY , y   P X    Y  y   PY  y 
Called
Marginal PDF
PXY ,    P X    Y     P   1.0
PXY x,   P , y   0
Px1  X  x2  y1  Y  y2   ?
x1 , y1
Top view of (X,Y) plane
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Independence of RVs:
P A  B 
Remember: P A | B  
PB  0
P B 
A  B  P A  B   P A.PB 
Extend the same for two RVs
X  Y  P X  x  Y  y   P X  x .PY  y 
PXY x, y   PX x .PY  y 
p XY x, y   p X x .PY  y 
Note: if two RVs are independent, then complete
description of joint PDF and pdf are through their
individual PDFs and pdfs respectively
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Conditional Distribution:
PY  y  X  x  PXY  x, y 
PY  y | X  x  

P X  x 
PX  x 
x y
p u , v dudv


 x, y  
PXY
 PY  y | X  x  
PX x 
XY
  
x 
  p u, v dudv
XY
  
x
dPY  y | X  x 
 pY  y | X  x  

dy
 p u, y du
XY

x 
  p u, v dudv
XY
  
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Joint Distribution (Contd.):
p XY x, y 
Prove that p  y | X  x  
p X x 

 
  x
Moment M nk  E X nY k 
n
y k p XY x, y dxdy
  
E g  X , Y  
  
  g  X , Y  p x, y dxdy
XY
  
Evaluate the following moments –
M 01, M 10
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Joint Distribution (Contd.):
n
k
 nk  E  X   x  Y   y  
 20   X2
Variance
 02  
11  E  X   X 
. Y  Y    XY Covariance
 XY
Correlation Coefficient
rXY 
 X . X
2
Y
Note: Covariance and Correlation
Coefficient are always between two RVs
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Correlation Coefficient:
Prove that –
1  rXY  1
Y  aX  b
Proof: Let us define
EY   EaX  b  a X  b


  E Y  Y   a 2 X2
2
Y
2
Note: if X and Y
are independent
 XY  E X   X Y  Y   a X2
rXY 
a
2
X
 X a 2 X2
rXY  0
Hint: Use definition of
joint moment to prove it
 1
Note:
rXY  0 does not mean X  Y
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2D Gaussian RVs:
Let X and Y are jointly Gaussian
p XY x, y  
X* 
x  X
X *Y* 
X
1
2 X  Y
; Y* 
1  r 
2
XY


1
2
2
exp 
X *  Y*  X *Y* 
2
 2 1  rXY





y  Y
Y
2rXY x   X  y  Y 
Ex: Plot the joint density
function in MATLAB
 XY
   x  ;  y  
 X    X2
X 
   N   
 Y  rXY X  Y
Y 
rXY X  Y  

2
 Y  
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Transformation of pdf in 2D:
Let X and Y are two RVs where jpdf is given. Also, let
U  gX ,Y 
V  h X , Y 
Similar as single RV. First find the roots xi , yi i 1
n
u
x
J
v
x
u
y
v
y
or
x
1
J  u
v
u
x
v
v
v
 p XY x, y 
pUV u, v    

J
i 1 
 x  xi , y  yi

n
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Example:
Find
pUV u, v 
uv
x
2
u v
y
2
0 1 0 
U  X  Y X 
   N   


V  X  Y Y 
0 0 1 
J
1
1
1 1
Note: inverse of
Jacobian is not 0.5
 p XY x, y  
pUV u, v   

J

 x  u  v , y  u v
2

 2
J 1  2
2

1
 1 2 2

. exp  u  v 
4
 4

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Example:

Given R  X  Y
2

2 1/ 2
X
  tan  
Y 
1
Find pR r , 
0  r  
0    2
J
1
cos 

sin 
0 1 0 
X 
   N   


Y 
0 0 1 


1
 1 2

p XY  x, y  
. exp   . x  y 2 
2
 2

   x  
   y  
 r sin 
r
r cos 
Note: Inverse
Jacobian is easier to
evaluate
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Example (Contd.):
 p XY x, y 
 r2 
r
pR r ,   

exp  

J

 x r cos 2
 2
y  r sin
Consider Marginal Distribution
2
0  r  
 r2 
pR r    pR r ,  d  r exp  
 2
0
Rayleigh

0    2
1
p     pR r ,  dr 
2
0
Uniform
pR r ,   pR r . p  
R
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Multi-dimensional RV:
Let X be a n dimensional joint random variable

Joint PDF

Joint pdf
 n

PX x  P   X i  xi 
 i 1

 n PX1 , X 2 ,.... X n x1 , x2 ,...xn 
pX x 
x1x2 ...xn
    g xp xd x

Eg X 
X
Expectation

Note: Multi-dimensional mathematical
operations required
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Summary:
 Markov and Chebychev Inequalities are valid for
any distribution
 Multi-dimensional RVs are defined
 Conditional probability, independence, jpdf, JPDF
and moments are defined
 Marginal pdf/PDF and correlation coefficient are
obtained
 Transformation of jpdf is developed
 Example shows different distributions can be
obtained by non-linear transformations
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Thank You!!!
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