Processing of Random Signals: Second Part of a

Processing of Random Signals: Second Part of a Course on Random Vibrations Rubens Sampaio

Roberta Lima

[email protected]

[email protected]

Departamento de Engenharia Mecânica

DINAME 2017 PUC-Rio: DEM

Organization: 1st part 1

Random process • statistics of first and second order − mean, variance, covariance and correlation • special types of random processes − Markov − Gaussian − second order − stationary − ergodic

2

Fourier analysis • Fourier series • Fourier transform • generalization: tempered distributions

3

Stationary random process • spectral density function • white noise PUC-Rio: DEM

Organization: 2nd part

4

Random vibrations of single-degree-of-freedom systems • random initial conditions • random excitation • random system

5

General case • filter, colored noise • delay • differentiation and integration (mean square sense)

6

Random vibrations of multi-degree-of-freedom systems • direct model method • normal model method

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Learning objectives

Signal analysis: main tools random processes − special types − statistics (in time and frequency) Fourier analysis

Signal processing: random vibrations basic formulas: relations between statistics of the inputs and outputs filter and colored noise

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Random vibrations of 1DOF systems Three main cases: 1

random initial conditions (random variables);

2

random excitation (the force is a random process);

3

random system (example: stiffness is a random variable).

In all the cases, the system response is a random process. Objectives: determine some properties of the system response − it is stationary? − it is ergodic? compute some statistics of the system response determine relations between statistics of the inputs and system response PUC-Rio: DEM

Random vibrations of 1DOF systems Random initial conditions Consider a system with

mass, m spring stiffness, k damping coefficient, b The equation of motion of the system is: m¨x(t) + b˙x(t) + kx(t) = 0 , with initial conditions x(0) = x0 x˙ (0) = v0 . PUC-Rio: DEM

Random vibrations of 1DOF systems Random initial conditions If 0 < b < 2mωn , where ωn =

−ζ ωn t

x(t) = e

q

k m

is the natural frequency:

  v0 + ζ ωn x0 x0 cos (ωd t) + sin (ωd t) ωd

where ζ is the damping factor and ωd damped frequency:

b 2mωn p ωd = ωn 1 − ζ 2

ζ=

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Random vibrations of 1DOF systems Random initial conditions The response x(t) can be written as x(t) = α1 (t) x0 + α2 (t) v0 ,  i h  α1 (t) = e−ζ ωn t cos (ωd t) + ζ ωn sin (ωd t) ωd  α (t) = 2

e−ζ ωn t ωd

sin (ωd t)

For ζ > 0:

lim α1 (t) = lim α2 (t) = 0 =⇒

t→∞

t→∞

lim x(t) = 0

t→∞

Regardless of the initial conditions. PUC-Rio: DEM

Random vibrations of 1DOF systems Random initial conditions If the initial conditions are random variables, X0 , V0 , the system response is a random process: X : T −→ RV(Ω,F, Pr) X (t) = α1 (t) X0 + α2 (t) V0 . Considering that X0 e V0 have expectations µX0 and µV0 and, variances σX20 and σV20 : X is a second order process. Question: is X a stationary second order process? PUC-Rio: DEM

Random vibrations of 1DOF systems Random initial conditions The expectation and variance of X are:

µX (t) = E[X (t)] = E[α1 (t) X0 + α2 (t) V0 ] = α1 (t) µX0 + α2 (t) µV0 ,

σX2 (t) = E[(X (t) − µX (t))2 ] = α12 (t) σX20 + 2α1 (t)α2 (t) (E[X0 V0 ] − µX0 µV0 ) + α22 (t) σV20 As µX (t) and σX2 (t) are not constants, X is not stationary. However, when t → ∞, X → 0. PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation Consider a system subject to a force f (t). mass, m spring stiffness, k damping coefficient, b The equation of motion of the system is: m¨x(t) + b˙x(t) + kx(t) = f (t) , with initial conditions (IC) x(0) = x0 x˙ (0) = v0 . PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation We decompose the problem into two initial value problems: 1st : system with no force and arbitrary IC m¨x1 (t) + b˙x1 (t) + kx1 (t) = 0 x1 (0) = x0 x˙ 1 (0) = v0 . 2nd : system with force and zero IC m¨x2 (t) + b˙x2 (t) + kx2 (t) = f (t) x2 (0) = 0 x˙ 2 (0) = 0. Theorem: x1 (t) + x2 (t) is solution of the original problem. PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation 1st : system with no force and arbitrary IC We saw previously that if 0 < b < 2mωn : x1 (t) = α1 (t) x0 + α2 (t) v0 ,  h i  α1 (t) = e−ζ ωn t cos (ωd t) + ζ ωn sin (ωd t) ωd  α (t) = 2

e−ζ ωn t ωd

sin (ωd t)

For ζ > 0:

lim α1 (t) = lim α2 (t) = 0 =⇒

t→∞

t→∞

lim x1 (t) = 0

t→∞

Regardless of the initial conditions. PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation 2nd : system with force and zero IC The response is a convolution between the force and h x2 (t) =

Z t

−∞

h(s)f (t − s) ds

where h is the response to a unit impulse applied at t = 0. h(t) =

1 −ζ ωn t e sin ωd t mωd

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Random vibrations of 1DOF systems Random excitation 2nd : system with force and zero IC Recalling:

⇐⇒

convolution in time

x2 (t) =

Z t

−∞

product in frequency

˜ ω ) f˜ (ω ) h(s)f (t − s) ds ⇐⇒ x˜ 2 (ω ) = h(

where h˜ is the response in frequency to a unit impulse. h(t) = ˜ ω) = h(

Z ∞

−∞

Z ∞

−∞

˜ ω ) e(i 2πω t) dω = h(

h(t) e(−i 2πω t) dt =

1 −ζ ωn t sin ωd t e mωd 1

m(ωn2 − ω 2 ) + 2iζ mωn ω

PUC-Rio: DEM

Random vibrations of 1DOF systems

Random excitation 2nd : system with force and zero IC

x2 (t) =

Z t

1 −ζ ωn (t) sin (ωd t) f (t − s) ds e m −∞ ωd

x˜ 2 (ω ) =

1 f˜ (ω ) m(ωn2 − ω 2 ) + 2iζ mωn ω

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Random vibrations of 1DOF systems Random excitation 2nd : system with force and zero IC If the force is a random process, F the system response is a random process: X2 : T −→ RV(Ω,F, Pr)

X2 (t) =

Z t

1 −ζ ωn t sin (ωd t)F(t − s) ds e −∞ mωd

X˜2 (ω ) =

1 m(ωn2 − ω 2 ) + 2iζ mωn ω PUC-Rio: DEM

˜ ω) F(

Random vibrations of 1DOF systems

Random excitation 2nd : system with force and zero IC If F is stationary with expectation µF and correlation rF : X2 is a second order process.

Question: is X2 a stationary second order process?

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Random vibrations of 1DOF systems Random excitation 2nd : system with force and zero IC Expectation of X2 :

µX2 (t) = E[X2 (t)] = E =

Z t

−∞

= µF = µF When t → ∞:

Z

t

−∞

h(s)F(t − s) ds



E [F(t − s)] h(s) ds Z t

Z−∞ t

h(s) ds h(s) e(−i 2π 0 s) ds

−∞

µF ˜ µX2 (t) −→ µX2 = µF h(0) = k PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation: 2nd : system with force and zero IC Correlation of X2 : rX2 (t1 , t2 )

= E[X2 (t1 )X2 (t2 )]  Z Z t 1 h(u)F(t1 − u) du = E

−∞

−∞

= = =

Z t1 Z t2

−∞ −∞

Z t1 Z t2

−∞ −∞

Z t1 Z t2

−∞ −∞

t2

 h(v)F(t2 − v) dv

h(u)h(v)E[F(t1 − u)F(t2 − v)] dv dv h(u)h(v)rF (t1 − u, t2 − v) dv dv h(u)h(v)rF (t2 − t1 − (v − u)) dv dv

When t1 , t2 → ∞: rX2 (t1 , t2 ) −→ rX2 (τ ) =

Z ∞Z ∞

−∞ −∞

h(u)h(v)rF (τ − (v − u)) dv dv

=⇒ X2 (t) goes to a stationary random process. PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation When t → ∞ e ζ > 0: 1st : system with no force and arbitrary IC X1 → 0 2nd : system with force and zero IC

µF ˜ = µX2 (t) −→ µX2 = µF h(0) k rX2 (t1 , t2 ) −→ rX2 (τ ) =

Z ∞Z ∞

−∞ −∞

h(u)h(v)rF (τ − (v − u)) dv dv

X1 (t) + X2 (t) goes to a stationary random process.

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Random vibrations of 1DOF systems Random excitation Spectral density function: Fourier transform of rX (τ ). Considering that rX (τ ) ∈ L1 (R) or L2 (R): sX (ω ) = =

Z ∞

rX (τ ) e(−i 2πω τ ) dτ

−∞ Z ∞ Z ∞

Z ∞

−∞ −∞

−∞

 h(u)h(v)rF (τ − (v − u)) dv dv e(−i 2πω τ ) dτ

Let θ = τ + u − v and τ = θ − u + v sX (ω )

= = =

Z ∞ Z ∞ −∞ ∞

Z

−∞

∞

 rF (θ ) e(−i 2πω θ ) dθ dv dve(−i 2πω u) e(−i 2πω v) −∞ −∞   Z ∞  Z ∞ h(u)e(i 2πω u) du h(v)e(−i 2πω v) dv rF (θ ) e(−i 2πω θ ) dθ h(u)h(v)

Z

−∞

−∞

˜ ω )h( ˜ ω )rF (ω ) h(−

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Random vibrations of 1DOF systems Random excitation ¯˜ ω ): ˜ ω ) = h( Since h(− 2 sX (ω ) = ˜h(ω ) sF (ω ) Recalling: the area under sX (ω ) is equal to σX2 :

σX2

= rX (0) = =

Z ∞ −∞

Z ∞

−∞

(i 2πω 0)

sX ( ω ) e

dω =

˜h(ω ) 2 sF (ω ) dω PUC-Rio: DEM

Z ∞

−∞

sX ( ω ) d ω

Random vibrations of 1DOF systems Random excitation Cross-correlation: rF X (t1 , t2 ) = E[F(t1 )X (t2 )] Z t  1 = E F(t1 )h(s)F(t2 − s) ds = = = When t1 , t2 → ∞:

Z t1 −∞

Z−∞ t1 Z−∞ t1

−∞

h(s)E [F(t1 )F(t2 − s)] ds

h(s)rF (t1 , t2 − s) ds h(s)rF (t2 − t1 − s) ds

rF X (t1 , t2 ) −→ rX F (τ ) =

Z ∞

−∞

h(s)rF (τ − s) ds

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Random vibrations of 1DOF systems Random excitation Cross-correlations (time: convolution) rF X ( τ ) = rX F ( τ ) =

Z ∞

−∞ Z ∞ −∞

h(s)rF (τ −s) ds h(s)rF (τ +s) ds

Spectral densities (in frequency: product) ˜ ω )sF (ω ) sF X (ω ) = h(

∈C

˜ ω )¯sF (ω ) sX F (ω ) = h(

∈C

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Random vibrations of 1DOF systems Random excitation Example: deterministic system subject to a random force F(t). mass, m spring stiffness, k damping coefficient, b The equation of motion of the system is: mX¨ (t) + bX˙ (t) + kX (t) = F(t) , with deterministic initial conditions (IC) X (0) = x0 X˙ (0) = v0 . PUC-Rio: DEM

Random vibrations of 1DOF systems Random excitation Example: consider that F(t) is a white noise. Recalling: white noise is a Gaussian stationary random process with zero mean and sF ( ω ) = s0

r F ( τ ) = s 0 δτ

⇐⇒ (tempered distributions)

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Random vibrations of 1DOF systems Random excitation Example: sF ( ω ) = s0 2 2 1 ˜ sX (ω ) = h(ω ) sF (ω ) = m(ω 2 −ω 2 )+2iζ mω ω s0 n n

Considering m = 1 kg, k = 6 N/m, b = 1 kg/s and s0 = 1 0.2 0.15

sX (ω)

0.1 0.05 0 −5

0

ω PUC-Rio: DEM

5

Random vibrations of 1DOF systems

Random excitation Example: others statistics of the system response:

µF =0 k Variance: the area under sX (ω ) is equal to σX2 : Mean:

σX2

µX =

= rX (0) = =

Z ∞

Z ∞

−∞

(i 2πω 0)

sX ( ω ) e

dω =

˜h(ω ) 2 sF (ω ) dω = s0 π kb −∞

PUC-Rio: DEM

Z ∞

−∞

sX ( ω ) d ω

Random vibrations of 1DOF systems Random excitation We presented some analytical relations between statistics of the force and response: − mean − variance − correlation and cross-correlation − spectral density Limitation: the formulas are valid only for linear systems.

When the system is not linear, we should use computational techniques to estimate statistics of the system response. Monte Carlo method PUC-Rio: DEM

Random vibrations of 1DOF systems Monte Carlo method Fixed an error:

Sampaio, R. e Lima, R. Modelagem Estocástica e Geração de Amostras de Variáveis e Vetores Aleatórios. Notas de Matemática Aplicada, SBMAC, vol.70, 2012. PUC-Rio: DEM

Random vibrations of 1DOF systems In the Monte Carlo method is fundamental:

construct a probabilistic model to the entrance (in our example, the random force is modeled as a white noise); generate samples of the random object of the entrance; construct a statistical model to the response: − random variables; − random vectors; − random process.

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Random vibrations of 1DOF systems Construction of a statistical model

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Random vibrations of 1DOF systems Construction of a statistical model to a random variable. Given realizations x(1) , x(2) , . . . , x(m) of a r. v. X:

µˆ X =

1 m (j) ∑x m j=1

1

sample expectation:

2

sample variance:

3

sample standard deviation:

σˆ X

4

sample moment of order k:

1 m (j) k ∑ (x ) m j=1

5

histogram

σˆ2 X =

2 1 m (j) ∑ (x − µˆ X ) m − 1 j=1

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Random vibrations of 1DOF systems Construction of a statistical model to a random process. 1

Discretization of the continuous parameter t into n values: t

2

=⇒

t1 , t2 , . . . , tj , . . . , tn

∀tj ∈ T, ∀j ∈ N

The process X is represented by a random vector X ∈ Rn .   X (t1 )  ..   .     X =⇒ X =   X (tj )   ..   .  X (tn ) PUC-Rio: DEM

Random vibrations of 1DOF systems Construction of a statistical model to a random process. Given realizations X (1) , . . . , X (m) of X , some estimated statistics of first order:   µˆ X (t1 )   .. sample expectation: µˆ X (t) =   .

µˆ X (tn )

sample variance:

 σˆ X2 (t1 )   σˆ X2 (t) =  ...  σˆ X2 (tn )

sample standard deviation:



σˆ X (t)

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Random vibrations of 1DOF systems Construction of a statistical model to a random process. An estimated statistic of second order: sample correlation:

[ˆrX ] ∈ Rn×n

 rˆX (t1 , t1 ) · · · rˆX (t1 , tn )   .. .. .. [ˆrX ] =  . . . . rˆX (tn , t1 ) · · · rˆX (tn , tn ) 

Each element is:

rˆX (tj , tk ) =

1 m (j) ∑ X (tj ) X (j) (tk ) m j=1 PUC-Rio: DEM

Random vibrations of 1DOF systems Construction of a statistical model to a random process. Easy way to compute [ˆrX ]: 1

Build a matrix [Xs ] ∈ Rn×m that it column is a realization X (j) 

[Xs ] =  2

|

|

|

X (1)

X (2)

|

|

···

X (m)

|

The correlation matrix is: [ˆrX ] =

1 [Xs ] [Xs ]T m

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 

Random vibrations of 1DOF systems Construction of a statistical model to a random process. Another estimated statistics of second order: sample covariance:

[ˆcX ] ∈ Rn×n

 cˆ X (t1 , t1 ) · · · cˆ X (t1 , tn )   .. .. .. [ˆcX ] =  . . . . cˆ X (tn , t1 ) · · · cˆ X (tn , tn ) 

Each element is:

cˆ X (tj , tk ) =

1 m (j) ∑ [X (tj ) − µˆ X (tj )] [X (j) (tk ) − µˆ X (tk )] m j=1 PUC-Rio: DEM

Random vibrations of 1DOF systems Construction of a statistical model to a random process. Easy way to compute [ˆcX ]: 1

Build a matrix [Xs0 ] ∈ Rn×m that it column is a realization ˆX X (j) − µ 

| [Xs0 ] =  X (1) − µˆ X | 2

| ˆX X (2) − µ |

 | · · · X (m) − µˆ X  |

The covariance matrix is: [ˆcX ] =

1 [Xs0 ] [Xs0 ]T m PUC-Rio: DEM

Random vibrations of 1DOF systems

Random excitation Coming back to the example of 1DOF system forced with a white process. To make simulations, approximate the white noise by a low-pass process.

sF ( ω ) =

(

s0 , if |ω | ≤ ωc 0,

other cases

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Random vibrations of 1DOF systems Random excitation Example: Parameters used in the simulations: system: m = 1 kg, k = (4π )2 N/m, b = 2 kg/s excitation force: s0 = 5, ωc = 2π 50 rad/d −→ 50 Hz number of simulations: 5000 duration of each simulation: 20 seconds ∆ t = 0.01 s

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Random vibrations of 1DOF systems

Random excitation Example: spectral densities of the force and response 0.02 6

0.015

sˆF 4

0.01

2

0.005

0 0

20

40

frequency [Hz]

60

0 0

PUC-Rio: DEM

sX sˆX

20

40

frequency [Hz]

60

Random vibrations of 1DOF systems

Random excitation Example: expectation and variance of the random force 600

5

400 2 σ ˆF (t)

µ ˆF (t)

500

0

300 200 100

−5 0

10

t [s]

20

0 0

PUC-Rio: DEM

5

10

t [s]

15

20

Random vibrations of 1DOF systems

Random excitation Example: expectation and variance of the response 0.1

0.01 0.008

2 σ ˆX (t)

µ ˆX (t)

0.05

0

−0.05

−0.1 0

0.006 0.004 0.002

5

10

15

20

0 0

5

10

t [s]

t [s]

PUC-Rio: DEM

15

20

Random vibrations of 1DOF systems Random excitation

rˆX (t1 , t2 )

Example: correlation of the response

0.02 0

−0.02 4

4 2 t2 [s]

0 0

PUC-Rio: DEM

2 t1 [s]

Random vibrations of 1DOF systems

Random excitation

5

0.01

4

0.005

3

rˆX (τ )

t2 [s]

Example: correlation of the response

2

−0.005

1 0 0

0

2

t1 [s]

4

−0.01 0

5

10

τ [s]

PUC-Rio: DEM

15

20

Random vibrations of 1DOF systems Random system Consider a random system subject to a random force F(t). mass, m spring stiffness, K, is a r.v. damping coefficient, b The nonlinear equation of motion of the system is: mX¨ (t) + bX˙ (t) + KX (t) = f (t) , with initial conditions (IC) X (0) = x0 X˙ (0) = v0 . PUC-Rio: DEM

Random vibrations of 1DOF systems

Random system Problem: suppose that K is a discrete r.v. with Bernoulli mass function p(K = k1 ) = 1/2 p(K = k2 ) = 1/2

If F is a stationary random process, is X a stationary random process?

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Random vibrations of 1DOF systems Random system Problem: We analyze two initial value problems: 1st : when K = k1 mX¨1 (t) + bX˙1 (t) + k1 X1 (t) = F(t) X1 (0) = x0 X˙1 (0) = v0 . 2nd : when K = k2 mX¨2 (t) + bX˙2 (t) + k2 X2 (t) = F(t) X2 (0) = x0 X˙2 (0) = v0 . PUC-Rio: DEM

Random vibrations of 1DOF systems Random system Problem: Previously, we verified that: 1st : K = k1 when t −→ ∞, X1 goes to stationary random process. 2nd : when K = k2 when t −→ ∞, X2 goes to stationary random process. t −→ ∞, does X go to a stationary random process?

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Random vibrations of 1DOF systems Random system Problem: Idea: let’s make simulations and look at the statistics of X . Parameters used in the simulations: system: m = 1 kg, k1 = (4π )2 N/m, k2 = (10π )2 N/m, b = 2 kg/s excitation force: s0 = 5, ωc = 2π 10 rad/d −→ 10 Hz number of simulations: 5000 duration of each simulation: 10 seconds ∆ t = 0.05 s

PUC-Rio: DEM

Random vibrations of 1DOF systems

Random system Problem: expectation and variance of the response −3

6

0.1

2 σ ˆX (t)

µ ˆX (t)

0.05 0 −0.05 −0.1 0

5

t [s]

10

x 10

4 2 0 0

PUC-Rio: DEM

5 t [s]

10

Random vibrations of 1DOF systems Random system Problem: correlation of the response

−3

rˆX (t1 , t2 )

x 10 5 0 −5 4

4 2 t2 [s]

2 0 0 PUC-Rio: DEM

t1 [s]

Random vibrations of 1DOF systems Random system Example: correlation of the response −3

5

x 10

2.5

rˆX (t)

t2 [s]

2 1.5

0

1 0.5 0 0

1

2 t1 [s]

−5 0

5 τ [s]

t −→ ∞, X goes to a stationary random process How is the spectral density of X ? PUC-Rio: DEM

10

Random vibrations of 1DOF systems Random system Example: spectral density of the response −3

8

x 10

sˆX

6 4 2 0 0

5

f [Hz]

10

PUC-Rio: DEM

15

General case The basic idea of filter

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General case

Especial filter:

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General case

1/f noise:

What is a 1/f noise? How to transform samples of a white noise into samples of a 1/f noise? Idea of filter: which linear system will make the transformation?

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General case 1/f noise: 1/f noise is a random process X such that its spectral density function has the form: s0 = 1 and β = 2 30

s0 ωβ

sX (ω)

sX ( ω ) =

20 10 0 0

2

ω

4

for s0 , β > 0. Since lim sX = ∞, its correlation function is ω −→0

nonintegrable over (−∞, ∞): sX (0) =

Z ∞

−∞

rX (τ )dτ = ∞ PUC-Rio: DEM

6

General case

1/f noise:

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General case 1/f noise: Idea of filter: standard Langevin equation b X˙ (t) + k X (t) = F(ω ) F is a white noise with spectral density s0 . The spectral density of X is: sX ( ω ) =

s0 k + b2 ω 2

When k −→ 0, X is an approximation to the 1/f noise with β = 2.

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General case

1/f noise: simulations with the Langevin equation Parameters used in the simulations: system: k = 10(−4) N/m, b = 10(−4) kg/s excitation force: s0 = 20, ωc = 2π 500 rad/d −→ 500 Hz number of simulations: 500 duration of each simulation: 10 seconds ∆ t = 0.001 s

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General case 1/f noise: simulations with the Langevin equation Spectral densities of the force and response.

30

40 30

sˆX

sˆF

20 10 0 0

20 10

200

400

f [Hz]

600

0

1

2

3

frequency [Hz]

PUC-Rio: DEM

4

General case 1/f noise: simulations with the Langevin equation Expectation and variance of the random force

4

3

50

x 10

2 (t) σF

µF (t)

2.5 0

2 1.5

−50 0

5

t [S]

10

1 0

PUC-Rio: DEM

5

t [S]

10

General case 1/f noise: simulations with the Langevin equation Expectation and variance of the response

1

15 2 (t) σX

µX (t)

0.5 0 −0.5 −1 0

5

t [S]

10

10 5 0 0

PUC-Rio: DEM

5

t [S]

10

General case 1/f noise: simulations with the Langevin equation Correlation of the response

rX (τ )

10 5 0 −5 −10

0

τ [s]

PUC-Rio: DEM

10

General case 1/f noise: simulations with the Langevin equation One realization of the force and of the response

5

1000

X (t)[s]

F(t)[s]

500 0

0

−500 −1000 0

5

t[s]

10

−5 0

5 t[s]

play

play

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10

General case

Delay Signal with delay present special features in its cross-correlation functions. Let us see two examples.

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General case Delay Example 1: given two ergodic random process X and Y and the realizations for t ∈ [0, ta ] X (t)

= a sin (ω t + θx ) + b

Y (t)

= c sin (ω t + θy ) + d sin (nω t + φ )

The cross-correlation rX Y can be computed using time average: rX Y

1 ta X (s)Y (s + τ ) ds ta −→∞ ta 0 ac cos (ω t − (θx − θy )) = 2

=

lim

Z

rX Y preservers the relative phases. PUC-Rio: DEM

General case Delay Example 1: The auto-correlations are: rX

=

rY

=

a2 cos (ω t) + b2 2 c2 d2 cos (ω t) + cos (nω t) 2 2

The phases θx , θy and φ do not appear in the correlations.

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General case Delay Example 2: consider a wheeled vehicle moving over a rough terrain

The correlation of X is known: rX (τ ). Let us study rX Y (τ )

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General case Delay Example 2: If the vehicle moves at a constant velocity v ans has length l: Y(t) = X (t − ∆)

with ∆ =

rX Y (τ ) = E[X (t)Y(t + τ )] = E[X (t)X (t + τ − ∆)] = rX (τ − ∆)

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l v

General case Delay Example 2: rX Y (τ ) is rX shifet by ∆

1

rX Y (τ )

rX (τ )

1 0.5 0 −0.5 −20

0

τ

20

0.5 0 −0.5 −20

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↔ ∆ 0

τ

20

General case Delay Example 2: Cross-density function: sX Y ( ω ) = =

Z ∞

Z−∞ ∞

−∞

rX Y (τ ) e(−i 2πω τ ) dτ rX (τ − ∆) e(−i 2πω τ ) dτ

Change of variable: u = τ − ∆: sX Y ( ω ) = = =

Z ∞

rX (u) e−i 2πω (u+∆) du −∞ Z ∞ −i 2πω ∆ e rX (u) e−i 2πω u −∞ e−i 2πω ∆ sX (ω ) PUC-Rio: DEM

du

General case

Delay Example 2: sX Y (ω ) = e−i 2πω ∆ sX (ω ) The frequency component ω in Y lags X by a phase angle 2πω ∆.

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General case Delay Example 2: Identification of the delay The relation ˜ ω )sX (ω ) sX Y (ω ) = h( can be used to identify the delay: ˜ ω ) = sX Y (ω ) = e−i 2πω ∆ h( sX (ω )

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General case Important concepts to deal with random vibrations: Given a second order random process: X : T −→ RV(Ω,F, Pr) we will formalize the definitions of:

continuity of X (in mean square sense) derivative of X (in mean square sense) integration of X (in mean square sense)

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General case To formalize the definitions, we need to understand what means: convergence of a sequence of random variables

Given a sequence of independent r.v. {Xn }n∈N . The concept of convergence of {Xn }n∈N is not unique. convergence in mean square (L2 ) convergence in mean (L1 ) convergence in probability convergence almost sure convergence in distribution PUC-Rio: DEM

General case Convergence in mean square L2 (Ω, Pr) is a Hilbert space, and the norm of X ∈ L2 (Ω, Pr) is: q kXk = E[X 2 ] .

The sequence {Xn }n∈N converges to the r.v. X in mean square if and only if: kXn − Xk −→ 0

when

n −→ +∞.

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General case

Continuity (in mean square sense) Definition: in mean square sense, continuity of X with respect to its parameter means that:

lim E[(X (t + ε ) − X (t))2 ] = 0

ε −→0

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General case Derivative (in mean square sense) Definition: in mean square sense, X is differentiable with respect to its parameter if: X (t + ε ) − X (t) =0 ε −→0 ε lim

If the limit exists, it is called derivative: dX (t) = X˙ (t) dt

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General case Derivative (in mean square sense) Theorem: X is differentiable in mean square sense if and only if the following limit exists and is continuous:    X (t + ε ′ ) − X (t) X (t + ε ) − X (t) lim E ε ,ε ′ −→0 ε ε′ This limit can be rewritten as: n h 1 rX (t+ε ,t+ε ′ )−rX (t+ε ,t) lim ε ε′ ′ ε ,ε −→0 oi ′ 2 X (t,t) − rX (t,t+εε)−r = ∂∂t∂ s rX (t, s) ′ m rX (t, s) has a continuous and mixed derivative on the diagonal t = s. PUC-Rio: DEM

General case

Derivative (in mean square sense) Theorem: if X is a second order and derivable random process:

µX˙ (t)

=

d dt µX (t)

rX˙ X (t1 , t2 ) =

d dt1 rX (t1 , t2 )

rX X˙ (t1 , t2 ) =

d dt2 rX (t1 , t2 )

,t2 )−rX (t1 ,t2 +ε )+rX (t1 ,t2 ) rX˙ X˙ (t1 , t2 ) = limε ,ε ′ −→0 rX (t1 +ε ,t2 +ε )−rX (t1 +εεε ′ ′

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′

General case Derivative (in mean square sense) Theorem: if X is a stationary second order and derivable random process:

µX˙ (t)

= 0

σX2˙ (t)

= 0 = rX˙ X (0)

rX˙ X (τ )

=

d d τ rX ( τ )

rX X˙ (t1 , t2 ) = − ddτ rX (τ ) rX˙ X˙ (τ )

2

= − ddτ 2 rX (τ )

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General case Integration (in mean square sense) Given a discretization of the continuous parameter t into n + 1 values in the interval [a, b], such that:

t

=⇒

a = t1 < . . . < tj < . . . < tn < tn+1 = b

∀tj ∈ T, ∀j ∈ N

We approximate the integral of a random process X over [a, b]: Y=

Z b

n

X (t) dt

a

=⇒ Yn = ∑ X (tj ) × (tj+1 − tj ) j=1

In the sense of convergence in mean square, the integral exits if: lim Yn = Y

n−→∞

=⇒

lim kXn − Xk −→ 0

n−→∞

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General case

Integration (in mean square sense) Theorem: In the sense of convergence of mean square, a necessary and sufficient condition to the existence of the integral of X over [a, b] is:

Z b Z b a a rX (t1 , t2 ) dt1 dt2 < ∞

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Random vibrations of multi-DOF systems Consider a multi-degree-of-freedom system: masses m1 , . . . , mn springs, k1 , . . . , kn dampers, b1 , . . . , bn

 X1 (t)   X (t) =  ...  Xn (t) 

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 F1 (t)   F (t) =  ...  Fn (t) 

Random vibrations of multi-DOF systems Equation of motion of the system in time: [m]X¨ (t) + [b]X˙ (t) + [k]X (t) = F (t)

In the frequency: (−ω 2 [m] + iω [b] + [k])X˜ (ω ) = F˜ (ω ) ˜ ω )]F˜ (ω ) X˜ (ω ) = [h( ˜ ω )] is the transfer function matrix. where [h(

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Random vibrations of multi-DOF systems

Impulse response and transfer function matrices:   ˜ h11 (ω ) · · · h˜ 1n (ω ) .. .. ..  [h˜ (ω )] =  . . . h˜ n1 (ω )

···

h˜ nn (ω )



[h(t)] = 

h11 (t) .. . hn1 (t)

··· .. . ···

h1n (t) .. . hnn (t)

 

The response for each degree-of-freedom Xi is the sum of the response for all forces: n

Xi (t) = ∑

Z ∞

j=1 −∞

Fj (t − θ )hij (θ )dθ

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Random vibrations of multi-DOF systems

Two methods: 1

direct model method

2

normal model method

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Random vibrations of multi-DOF systems Direct model method Considering that each Fj (t) is a second order stationary random process: When t→∞: ˜ µX (t) → µX = [h(0)] µF

where:  µX1   µX =  ...  µXn 

 µF1   µF =  ...  µFn 

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Random vibrations of multi-DOF systems Direct model method Correlation and spectral density matrices of F (t): 

[rF (τ )] = 

rF1 F1 (τ ) . . . rFn F1 (τ )

··· .. . ···

rF1 Fn (τ ) . . . rFn Fn (τ )







[sF (ω )] = 





sF1 F1 (ω ) . . . sFn F1 (ω )

··· .. . ···

sF1 Fn (ω ) . . . sFn Fn (ω )



··· .. . ···

sX1 Xn (ω ) . .. sXn Xn (ω )



Correlation and spectral density matrices of X (t): 

[rX (τ )] =  When t→∞:

rX1 X1 (τ ) . .. rXn X1 (τ )

··· .. . ···

rX1 Xn (τ ) . .. rXn Xn (τ )



[sX (ω )] = 

sX1 X1 (ω ) . .. sXn X1 (ω ) T

˜ ω )][sF (ω )][h( ˜ ω )] [sX (ω )] = [h( PUC-Rio: DEM





Random vibrations of multi-DOF systems Example: consider the system with two degrees of freedom

[sF (ω )] =

F1 is a white noise. Compute sX1 X1 and sX2 X2 .

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S0 0

0 0



Random vibrations of multi-DOF systems Example: Equation of motion of the system in time: [m]X¨ (t) + [b]X˙ (t) + [k]X (t) = F (t) where:

[m] =



[k] =

m1 0 0 m2 



k1 + k2 −k2 −k2 k2

[b] =





b1 + b2 −b2 −b2 b2 + b3

F (t) =

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F1 0





Random vibrations of multi-DOF systems Example: T

˜ ω )][sF (ω )][h( ˜ ω )] [sX (ω )] = [h(

For m1 = m2 = 1 kg, k1 = 3 N/m, k2 = 5 N/m, b1 = 0.1 kg/s, b2 = 1 10−4 kg/s, b3 = 0.10 kg/s, and s0 = 5 =⇒ ω1 = 0.18 Hz and ω2 = 0.54 Hz

200

200

sX2 X2

250

sX1 X1

250

150

150

100

100

50 0 0

50 0.5

1

1.5

Frequency [Hz]

2

0 0

0.5

1

1.5

Frequency [Hz]

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2

Random vibrations of multi-DOF systems Normal model method Natural frequencies and modes of the deterministic system: ¨ + [k]x(t) = 0 [m]x(t) Solution: x(t) = v eiω t (−ω 2 [m] + [k])v = 0

Natural frequencies Eigenvectors

⇒

det(−ω 2 [m] + [k]) = 0

[v] =

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"

| v1 |

| v2 |

···

| vn |

#

Random vibrations of multi-DOF systems Normal model method Considering proportional damping: [b] = α [m] + β [k] We diagonalize the matrices: 

[m⋆ ] = [v]T [m][v] = 

m⋆1 .. . 0

··· .. . ···

0 .. . m⋆n

  

[b⋆ ] = [v]T [b][v] = 



[k⋆ ] = [v]T [k][v] =  b⋆1 .. . 0

··· .. . ···

0 .. . b⋆n

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 

ω1 2 .. . 0

··· .. . ···

0 .. . ωn 2



,

Random vibrations of multi-DOF systems Normal model method Change of variable: x(t) = [v] y(t) The equation of motion become: ¨ + [b⋆ ]y(t) ˙ + [k⋆ ]y(t) = q(t) [m⋆ ]y(t) where:

q(t) = [v]T f(t)

n uncoupled equations PUC-Rio: DEM

Random vibrations of multi-DOF systems Normal model method ˜ and [h] become: With the change of variable, [h]  ˜⋆ h1 (ω ) · · · .. ⋆ ..  [h˜ (ω )] = . . 0

···

0 .. . ⋆ h˜ n (ω )

 



[h⋆ (t)] = 

where: ⋆ h˜ j (ω ) =

1 −m⋆j ω 2 + ib⋆j ω

+ kj⋆

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h⋆1 (t) .. . 0

··· .. . ···

0 .. . h⋆n (t)

 

Random vibrations of multi-DOF systems Normal model method Previous result: spectral density matrix of X ˜ ω )][sF (ω )][h( ˜ ω )]T [sX (ω )] = [h(

In the new variables: ⋆

⋆

[sX (ω )] = [v][h˜ (ω )][sQ (ω )][h˜ (ω )][v]T

where

[sQ (ω )] = [v]T [sF (ω )][v]

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Additional material: journal articles

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Additional material: short courses

Short course in Uncertainties 2016 (Parts 1 and 2)

Short course in LNCC 2017 (Workshop)

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Additional material: published books 2012

SBMAC vol. 66

2012

SBMAC vol. 70 PUC-Rio: DEM

2014

Next events

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Processing of Random Signals: Second Part of a Course on Random Vibrations Rubens Sampaio

Roberta Lima

[email protected]

[email protected]

Departamento de Engenharia Mecânica

DINAME 2017 PUC-Rio: DEM