Identification Methods for Structural Systems · 2016. 3. 2. · Introduction to the Frequency...

Identification Methods for Structural Systems

Prof. Dr. Eleni Chatzi

Lecture 3 - 2 March, 2016

Institute of Structural Engineering Identification Methods for Structural Systems 1

Fundamentals

Overview

General Harmonic Response

Frequency Response Functions

The Fourier and Laplace Transforms for SDOFs

Bode plots for SDOFs


Introduction to the Frequency domain

General Harmonic Response of SDOF system

Let’s revisit the case of Forced Damped Vibration with aHarmonic Excitation F0cos(ωt)

mx + cx + kx = F0cosωt (1)

Then particular solution is of the type:

xp(t) = Ccosωt + Dsinωt

and its derivatives are:

xp(t) = ω(−Csinωt + Dcosωt)

xp(t) = ω2(Ccosωt − Dsinωt)



Let us plug these expressions in (??)[−mω2C + cωD + kC

]cosωt +

[−mω2D − cωC + kD

]sinωt = F0cosωt

If we equate the coefficients of similar terms:[−mω2C + cωD + kC

]= F0 &

[−mω2D − cωC + kD

]= 0

⇒ C =k −mω2

cωD & (k −mω2)

k −mω2

cωD + cωD = F0

which leads to the following solution for the two unknowncoefficients C , D.

D =(cω)

(k −mω2)2 + (cω)2F0 & C =

(k −mω2)

(k −mω2)2 + (cω)2F0



However, we know form trigonometry, that the expression

xp(t) = Ccosωt + Dsinωt

is equivalent to:

xp(t) = X0cos(ωt − φ) = X0cosωtcosφ+ X0sinωtsinφ

where X0 =√

C 2 + D2 & φ = arctan

[D

C

]Therefore,

X0 =F0√

(k −mω2)2 + (cω)2& φ = arctan

[cω

k −mω2

](2)



However we have previously defined that:

c = 2mωnζ, k = mω2n

Using the above relationships we can rewrite (??) as:

X0 =F0/k√(

1− ω2

ω2n

)2

+

(2ζ

ω

ωn

)2& φ = arctan

2ζω

ωn

1− ω2

ω2n


General Harmonic Response of SDOF system

Summarizing, the particular solution then becomes:

xp(t) = X0cos(ωt − φ)⇒ xp(t) =F0

kH(ω)cos(ωt − φ)

whereF0

k= δst is also known as static deflection

and H(ω) is what is known as:

Frequency Response Function

H(ω) =1√(

1− ω2

ω2n

)2

+ (2ζ ωωn)2


The half power Method

The half power Method for the estimation of Damping

The amplitude of the FRF at resonance is called the quality factor(Q) of the system

For ω ≈ ωn ⇒ H(ω) ≈ 1

2ζ= Q

Points R1, R2 where it holds thatH(ω) = Q/

√2 are called 1/2

power points. these are used forextracting the bandwidth of thesystem

t

( )F t

1τ

1F

2F

2τ

n

ωω

2R 1R

12

Qζ

=

2Q

( )H ω

1

n

ωω

2

n

ωω

1

m

k c ( )ty

( )tx t

( )F t

t∆

1t∆

0t∆ →



The Bandwidth is defined as: ∆ω = ω2 − ω1 where

ω1, ω2 are calculated from:

H(ω) =Q√

2=

1

ζ√

2⇒ 1√

(1−ω2

1,2

ω2n

)2 + (2ζω1,2

ωn)2

=1

ζ√

2

ω21,2

ω2n

= (1− 2ζ2 ∓ 2ζ√

1 + ζ2)ω2nζ<<→ ω2

1,2 = (1∓ 2ζ)ω2n

and ω22 − ω2

1 = 4ζω2n ⇒ (ω2 − ω1)(ω2 + ω1) = 4ζω2

n



Bandwidth - Identification of ζ

Since,ω1 + ω2

2= ωn, we have that

Bandwidth = ∆ω = ω2n = ω2 − ω1 ≈ 2ζωn

Therefore, from experimental response we can evaluate the dampingcoefficient:

ζ =ω2 − ω1

2ωn


Base Excited Systems

Structural Systems excited at the base - ground motion(earthquake)Two alternatives exist for formulating the Equation of Motion

Absolute Motion x(t):

mx + c(x − y) + k(x − y) = 0⇒mx + cx + kx = cy + ky

Relative Motion z(t) = x(t)− y(t)

m(z + y) + cz + kz = 0⇒mz + cz + kz = −my

n

ωω

2R 1R

12

Qζ

=

2Q

( )H iω

1

n

ωω

2

n

ωω

1

m

k c ( )ty

( )tx


Base Excited Systems-Absolute Motion


mx + cx + kx = cy + ky

Total solution xtot = xh + xp.

The homogeneous solution is already explored.Assuming that the base excitation is of harmonic type, i.e.,y(t) = Y0cosωt, the particular solution will also be harmonic.

In accordance with the harmonic force excited system shown earlier,we now obtain:

xp(t) = Y0

√(k2 + (cω)2)

(k −mω2)2 + (cω)2cos(ωt − φ)⇒

xp(t) = Y0H(ω)cos(ωt − φ)




We therefore can define the following terms.

Gain Function or Transmissibility of Displacement:

Td = H(ω) =

√1 + (2ζr)2

(1− r2)2 + (2ζr)2

and Phase:

φ = tg−1

(2ζr3

1 + (4ζ2 − 1)r2

), r =

ω

ωn



Transmissibility of Displacement



Transmissibility of Displacement

This signifies how larger the maximum displacement of the system X0 is with

respect to the maximum amplitude of the input harmonic force Y0.Base excited systems: absolute motionBase excited systems: absolute motion

h l f i i 0 d l i f ll l f• The value of Td is unity at r=0 and close to unity for small values of r.

• For an undamped system ζ=0, Td ∞ at resonance (r=1).

• The value of Td is less than unity (Td <1) for values of r >√2 (for any amount of damping ζ)

• The value of Td is equal to unity (Td=1) for all values of ζ at r=√2

222

2

)2()1()2(1

rrr

YX

o

o

ξξ+−

+=


Base Excited Systems-Relative Motion

Relative Motion: z(t) = x(t)− y(t)⇒ mz + cz + kx = −myBase excited system: Relative motionBase‐excited system: Relative motion

d l f2rZ• In nondimensionless form,

• The gain function for the relative motion for the base‐excited system is

222 )2()1( rrr

YZ

o

o

ξ+−=

The gain function for the relative motion for the base excited system is shown in the figure:


Arbitrary Excitation

Response of SDOF System to Arbitrary ExcitationLet’s consider the impulse response:

Newton:∫ t+∆tt Fdt = (change in momentum) =∫ t+∆t

t mxdt =∫ t+∆tt m

x

dxdt ⇒ Fdt = mxt+∆t −mxt

Assuming mxt = 0⇒ xt+∆t =Fdt

m.

n

ωω

2R 1R

12

Qζ

=

2Q

( )H iω

1

n

ωω

2

n

ωω

1

m

k c ( )ty

( )tx t

( )F t

t t t+ ∆

Assume the unit impulse at t = 0, where

Fdt = F = 1⇒ F =1

∆t. F behaves like the Dirac δ

function∫ t+∆tt δ(t)dt = 1⇒ mx0 = 1⇒ x0 = 1 =

1

mand x0 = 0 (no move yet). Hence, the solution is the free

response with I.C. x0 = 0, x0 =1

m.

n

ωω

2R 1R

12

Qζ

=

2Q

( )H iω

1

n

ωω

2

n

ωω

1

m

k c ( )ty

( )tx t

( )F t

t∆

1t∆

0t∆ →



Unit Impulse Response for impulse at t = 0:

x(t) =

(x0cosωd t + (

x0

ωd+ x0ζωn)sinωd t

)e−ζωnt

with x0 = 0, x0 =1

m⇒ x(t) =

1

mωdsinωd te

−ζωnt , t > 0

For a unit impulse of magnitude Fdτ , applied at t = τ :

x(t) =Fdτ

mωdsinωd(t − τ)e−ζωn(t−τ), t > τ



For 2 unit impulses of magnitude F1dτ , F2dτapplied at t = τ1, t = τ2:

x(t) =F1dτ

mωdsinωd(t − τ1)e−ζωn(t−τ1)

+F2dτ

mωdsinωd(t − τ2)e−ζωn(t−τ2)

t

( )F t

1τ

1F

2F

2τ

n

ωω

2R 1R

12

Qζ

=

2Q

( )H iω

1

n

ωω

2

n

ωω

1

m

k c ( )ty

( )tx t

( )F t

t∆

1t∆

0t∆ →



Duhamels’s (or convolution) Integral Similarly, for n finiteimpulses:

x(t) =n∑

i=1

Fidτ

mωdsinωd(t − τi )e−ζωn(t−τi )

Adding up, for dτ → 0, we obtain the continuous expression:

x(t) =

∫ t

0

F(τ)

mωdsinωd(t − τ)e−ζωn(t−τ)dτ ⇒

x(t) =

∫ t

0F(τ)h(t − τ)dτ


From Time to Frequency Domain

SDOF systems: From Time to Frequency Domain

The Fourier Transform

Definition: F (ω) = F{f (t)} =∫∞−∞ f (t)e−iωtdt

Inverse: f (t) = F−1{F (ω)} =1

2π

∫∞−∞ F (ω)e iωtdω

Fourier Transform of a cosine



SDOF systems: From Time to Frequency Domain

Fourier Transform of a cosine - Note

The Fourier Transform (FT) of a cosine simply consists in twosymmetric spikes at values corresponding to the specific frequency ofthat cosine (symmetric transformation).

Mathematically this can be written in the form of two deltafunctions δ(ω ± ω0) or δ(p ± p0) in Hz.

This reflects the fact that the frequency content of a perfect cosine(or sine) function contains a single frequency component.

see Laplace and Fourier transform pdfs uploaded on the website



The Laplace Transform

Definition: F (s) = L{f (t)} =∫∞

0 f (t)e−stdt, s = σ + iω

Inverse: f (t) = L−1{F (s)} =1

2πilimT→∞

∫ γ+iTγ−iT F (s)estds

Basic Laplace Transform Property

L[df

dt

]=

∫ ∞0

e−stdf (t)

dtdt

Applying integration by parts:∫udv = uv −

∫vdu, for u = e−st ,

v = f , we obtain:

L[df

dt

]= e−st f (t)

∣∣∞0−∫ ∞

0(−s)e−st f (t)dt

= −f (0) + s

∫ ∞0

f (t)e−stdt ⇒ L[df

dt

]= sF (s)− f (0)



Similarly,

L[d2f

dt2

]= s2F (s)− df

dt(0)− sf (0)

L[dnf

dtn

]= snF (s)− dn−1f

dtn−1(0)− ...− sn − 2

df

dt(0)− sn−1f (0)

Example - Obtaining the Transfer Function (TF):Apply the Laplace Transform on a 2nd order ODE:

d2y

dt2+ 2

dy

dt+ 3y = 4u

L→ s2Y (s)− dy

dt(0)− sY (0) + 2sY (s)− 2Y (0) + 3Y (s) = 4U(s)

0 I .C .−→ Y (s) =4

s2 + 2s + 3U(s) : Transfer Function



Transfer FunctionA transfer function (also known as the system function) is a mathematicalrepresentation, in terms of the system frequency, of the relation betweenthe input and output of a linear time-invariant system with zero initialconditions and zero-point equilibrium.A linear time -invariant (LTI) system is characterized by two properties:

Linearity which means that the relationship between the input and theoutput of the system is a linear map. If input x1(t) produces response y1(t)and input x2(t) produces response y1(t) then the scaled and summed inputα1x1(t) + α2x2(t) produces the scaled and summed responseα1y1(t) + α2x2(t) where α1, α2 are real scalars.

Time invariance which means that whether we apply an input to the

system now or T seconds from now, the output will be identical except for

a time delay of the T seconds. That is, if the output due to input x(t) is

y(t), then the output due to input x(t − T ) is y(t − T ). Hence, the

system is time invariant because the output does not depend on the

particular time the input is applied.



More Laplace Transform Properties

L{δ(t)} = 1, δ(t):Dirac

L{f (αt)} =1

αF( sα

)L{eαt f (t)} =

1

αF (s − α) (Frequency Shift)

L{f (t − α)H(t − α)} = e−αsF (s) (Time Shift) where H is the

Heaviside step function H(n) =

{0, n < 01, n ≥ 0

L{f ∗ g (t)} =∫ t

0 f (τ)g(t − τ)dτ = F (s)G (s) (Convolution)



Example: Equation of Motion

mx + cx + kx = f (t)L,0I .C .−→ ms2X (s) + csX (s) + kX (s) = F (s)⇒

X (s) =1

ms2 + cs + kF (s)⇒ H(s) =

1

ms2 + cs + kTransfer Function

Assuming s = iω we obtain the complex Frequency Response

Function (FRF):

H(iω) =1

mω2 + ciω + k

We can then use what we define as the Inverse Laplace Transformin order to determine x(t)



Basic Inverse Transform Properties (also look at given tables)

L−1

{1

s + α

}= e−αt

L−1

{1

(s + α)2

}= te−αt

L−1

{1

(s + α)(s + β)

}=

1

β − α[e−αt − e−βt

]Hence, x(t)L−1{X (s)} = L−1{ 1

ms2 + cs + kF (s)} =

L−1{ 1

m(s + α)(s + β)F (s)}

where,α, β =c ∓ i

√4mk − c2

2massuming an underdamped system



Also for f (t) = δ(t) (unit impulse) ⇒ F (s) = 1

Then, x(t) =1

m(β − α)

[e−αt − e−βt

]

Using k = mω2n, c = 2mωnζ, ωd = ωn

√1− ζ2

⇒ x(t) =1

mωde−ζωnt [sinωd t]

Obviously this agrees with the time domain derived SDOF impulseresponse


Linear Systems

The Frequency Response Function (FRF)Assume a linear system characterized by its Transfer Function:

Y(s) = G(s)U(s), where s ∈ C (Laplace Domain)

Evaluated at the imaginary axis s = iω, the TF yields the FRF, G(iω):

Case of a SISO System

d2y

dt2+ 4

dy

dt+ 3y =

du

dt+ 2u

L, 0 ICs−→

(s2 + 4s + 3)Y (s) = (st2)U(s)⇒ G (s) =s + 2

(s + 3)(s + 1)

Hence the FRF is, G (iω) =iω + 2

(iω + 3)(iω + 1)


The FRF

As shown FRF describes the response of the system for sinusoidalinputs. It is also the FT of the system’s impulse response.

The FRF can be visualized using Bode Plots

Assume G (iω) =1

iω + 1⇒ G (iω) =

1

1 + ω2− iω

1 + ω2

Bode Magnitude Plot

For ω = 1⇒ |G (iω)| =

∣∣∣∣1− i

2

∣∣∣∣ =

1√2⇒ −20log |G | = −3.02dB

( )ϕ ω

45o−

90o−

1 100.1

5.7o

5.7o

( ) logscaleω

45 / decadeo−

3dB

( ) logscaleω

( ) ( )20log G i dBω

20 / decadedB−

30−

10−

0

0.1 1 10

20−


The FRF

Bode Magnitude Plot - Asymptotes

For ω << 1⇒ 1

iω + 1≈ 1 and −20log |G | = −20log(1) = 0

For ω >> 1⇒ 1

iω + 1≈ 1

iωand

∣∣∣∣ 1

iω

∣∣∣∣ =1

ω

Then assuming ω2 = 10ω1 ⇒

20log

(1

ω2

)= 20log

(1

10ω1

)= 20log

(1

10

)+ 20log

(1

ω1

)Hence we can approximate the decline per “decade”

20log

∣∣∣∣ 1

iω2 + 1

∣∣∣∣− 20log

∣∣∣∣ 1

iω1 + 1

∣∣∣∣ ≈ 20log

(1

ω2

)− 20log

(1

ω1

)⇒

−20(dB/decade)


The FRF

G (iω) =1

1 + ω2− iω

1 + ω2⇒ φ(ω) = tan−1(−ω)

Bode Phase Plot

For ω = 1⇒ φ(ω) = tan−1(−1) = −45o

For ω =1

10⇒ φ(ω) = tan−1(

1

10) = −5o

For ω = 10⇒ φ(ω) = tan−1(−10) = −90o

( )ϕ ω

45o−

90o−

1 100.1

5.7o

5.7o

( ) logscaleω

45 / decadeo−

3dB

( ) logscaleω


20 / decadedB−

30−

10−

0

0.1 1 10


The FRF

Generalizing:

Suppose we had G (iω) =α

α + iω=

1

i(ωα

)+ 1

Then, the same plots apply for ω =ω

α

3dB

( ) logscaleω


20 / decadedB−

30−

10−

0

0.1α α 10α

20−

( )ϕ ω

90o−

α 10α 0.1α

5.7o

5.7o

( ) logscaleω

45 / decadeo−


The FRF

Returning to the SISO Example

We had G (iω) =iω + 2

(iω + 3)(iω + 1)⇒

20log |G (iω)| = 20log |iω + 2|+ 20log

∣∣∣∣ 1

(iω + 3)

∣∣∣∣+ 20log

∣∣∣∣ 1

(iω + 1)

∣∣∣∣= 20log

∣∣∣2i ω2

+ 1∣∣∣+ 20log

∣∣∣∣∣∣ 1/3

(iω

3+ 1)

∣∣∣∣∣∣+ 20log

∣∣∣∣ 1

(iω + 1)

∣∣∣∣⇒Superposition of Plots

20log |G (iω)| =

20log

(2

3

)+ 20log

∣∣∣i ω2

+ 1∣∣∣+ 20log

∣∣∣∣∣∣ 1

(iω

3+ 1)

∣∣∣∣∣∣+ 20log

∣∣∣∣ 1

(iω + 1)

∣∣∣∣Institute of Structural Engineering Identification Methods for Structural Systems 35

The FRF

Plot Superposition

3dB

( ) logscaleω

20 / decadedB 10

20

0.2 2 20

0

( )20log 1 2 dBω +

+ + 220log 3

3dB

( ) logscaleω

( )120log 13

dBω +

20 / decadedB−

30−

10−

0

0.3 3

30

20−

+

3dB

( ) logscaleω

( )120log 1

dBiω +

20 / decadedB−

30−

10−

0

0.1 1 10

20−

+


Identification Methods for Structural Systems · 2016. 3. 2. · Introduction to the Frequency...

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