eth chair of structural mechanics structural identiﬁcation & … · outline the laplace...

ETH Chair of Structural Mechanics

Structural Identification & Health Monitoring

Lecture 05: Transform Domain Methods A: Laplace

Dr. V.K. Dertimanis & Prof. Dr. E.N. Chatzi

Outline

The Laplace transformTransfer functionsExercisesFurther Reading

ETH Chair of Structural Mechanics 04.03.2020 2

The Laplace transformDefinitions

x(t): continuous-time signal/function in [0,∞)

L{x(t)} ≡ X (s) =∫ ∞

0x(t)e−stdt

s = σ + jω: complex variable

Inverse Laplace transform:

L−1{X (s)} ≡ x(t) =1

2πj

∫ γ−jT

γ−jTX (s)estds


The Laplace transformDefinitions

- The Laplace transform is an integral transformation

- It maps the continuous-time domain into the Laplace domainand vice-versa1

- The functions x(t) and X (s) form a Laplace transform pair

1The Laplace transform can be also thought of as an operator that “transforms”analysis into algebra, constituting thus a very convenient tool for solving differentialequations


The Laplace transformCommon Laplace transforms

x(t), t ≥ 0 X (s)

δ(t) 1

δ(t − τ ) e−τs

u(t − τ ) 1s e−τs

t 1s2

e±at 1s∓a

sin(ωt) ωs2+ω2

cos(ωt) ss2+ω2

e−at sin(ωt) ω(s+a)2+ω2

e−at cos(ωt) s+a(s+a)2+ω2

1b−a (e−at − e−bt ) 1

(s+a)(s+b)


The Laplace transformFundamental properties

x(t): n-times differentiable

1. The Laplace operator is linear

2. Shifting:L{x(t − a)} = e−asX (s)

3. Translation:L{e−atx(t)} = X (s + a)




4. Scaling:

L{x(at)} =1|a|

X(s

a

)

5. Differentiation:L{x(t)} = sX (s)− x(0)

The Laplace transform of the n–th order derivative is

L{x (n)(t)} = snX (s)− sn−1x(0)− sn−2x(0)− x (n−1)(0)




6. Integration:

L{∫ t

0x(τ )dτ

}=

1sL{x(t)} =

F (s)s

7. Convolution:L{g(t) ∗ x(t)} = G(s)X (s)


Transfer functionsFrom differential equations

SISO differential equation:

x (n)(t)+an−1x (n−1)(t)+· · ·+a1x(t)+a0x(t) = b0u(t)+b1u(t)+· · ·+bmu(m)(t)

Zero I.C. on x (i)(t) and u(i)(t)

Step 1: apply L.T. on both sides

L{x (n)(t) + · · · + a0x(t)} = L{b0u(t) + · · · + bmu(m)(t)}



Step 2: use the linearity property

L{x (n)(t)} + · · · + a0L{x(t)} = b0L{u(t)} + · · · + bmL{u(m)(t)}

Step 3: use the differentiation property for zero I.C.

snL{x(t)} + · · · + a0L{x(t)} = b0L{u(t)} + · · · + bmsmL{u(t)}

Step 4: factorize

(sn + · · · + a0)L{x(t)} = (b0 + · · · + bmsm)L{u(t)}



Step 5: Set L{x(t)} = X (s) and L{u(t)} = U(s).

Step 6: Define

G(s) ≡ X (s)U(s)

=Laplace transform of outputLaplace transform of input

Step 7: G(s) −→ transfer function

G(s) =bmsm + bm−1sm−1 + · · · + b1s + b0

sn + an−1sn−1 + · · · + a1s + a0≡ B(s)

A(s)



G(s) is a rational function of the complex variable s

B(s) is the numerator polynomial

A(s) is the denominator or characteristic polynomial

Notice the analogy

x(t) =∫ t

0g(t − τ )u(τ )dτ

andX (s) = G(s)U(s)



Convolution in the time domain −→ multiplication in the Laplace domain

Steps for solving differential equations:

1. Calculate the Laplace transform of the input, U(s)

2. Multiply U(s) to the transfer function G(s), to obtain the Laplacetransform of the output, X (s) = G(s)U(s)

3. Apply the inverse Laplace transform, L−1{X (s)}, to obtain x(t)


Transfer functionsConnection to impulse response

When u(t) = δ(t) −→ U(s) = 1

X (s) = G(s)

x(t) = g(t) = L−1{X (s)} = L−1{G(s)}

Transfer function: the Laplace transform of the impulse response

Transfer function & impulse response −→ Laplace transform pair


Transfer functionsExample: spring-mass-damper

m

k c

u(t)

x(t)

Differential equation:

mx(t) + cx(t) + kx(t) = u(t)

L.T. (zero I.C.)→ Displacement TF

G(s) .=X (s)U(s)

=1

ms2 + cs + k

=1/m

s2 + 2ζnωns + ω2n



m

k c

u(t)

x(t)

Velocity transfer function

Gv (s) = sG(s) =s

ms2 + cs + k

Acceleration transfer function

Ga(s) = s2G(s) =s2

ms2 + cs + k


Transfer functionsPartial fraction expansion

TF

G(s) =bmsm + bm−1sm−1 + · · · + b1s + b0

sn + an−1sn−1 + · · · + a1s + a0

When n > m and roots of A(s) are distinct2

G(s) =R1

s − λ1+

R1

s − λ2+ · · · + Rn

s − λn=

n∑i=1

Ri

s − λi

Ri : residuesRi = G(s)(s − λi )|s=λi

2When n ≤ m polynomial long division must be appliedETH Chair of Structural Mechanics 04.03.2020 17


m

k c

u(t)

x(t)

Impulse response:

X (s) = G(s) =1/m

s2 + 2ζnωns + ω2n

Partial fraction expansion

1/ms2 + 2ζnωns + ω2

n=

R1

s − λ1+

R2

s − λ2

λ1,2 = −ζnωn ± ωd j (underdamped)



m

k c

u(t)

x(t)

Residues:

R1 =1/m

(s − λ1)(s − λ2)(s − λ1)

∣∣∣s=λ1

=1/m

λ1 − λ2= − 1

2mωdj

R2 = · · · = 12mωd

j



m

k c

u(t)

x(t)

Inverse Laplace transform

g(t) = L−1{G(s)} = R1L−1{ 1

s − λ1

}+ R2L−1

{ 1s − λ2

}−→

g(t) = R1eλ1t + R2eλ2t



m

k c

u(t)

x(t)

Replacing λ1, λ2, R1 and R2

g(t) = L−1{G(s)} = − 12mωd

je(−ωnζn+ωd j)t

+1

2mωdje(−ωnζn−ωd j)t



m

k c

u(t)

x(t)

Doing the algebra

g(t) =1

mωde−ωnζnt sin(ωd t)

(Compare to Lecture 2, p.31)



MDOF structures

Mx(t) + Cx(t) + Kx(t) = Pu(t)

Apply LT (zero I.C.) (Ms2 + Cs + K

)X(s) = PU(s)

Transfer functionG(s) =

[Ms2 + Cs + K

]−1P



G(s)→ a [n × nu] matrix

G(s) =

G11(s) G12(s) G13(s) . . . G1nu (s)G21(s) G22(s) G23(s) . . . G2nu (s)

......

.... . .

...Gn1(s) Gn2(s) Gn3(s) . . . Gnnu (s)

Gi,l (s)→ TF between the vibration response of the i-th DOF and the l-thinput excitation


Transfer functionsExample: two-story shear building

Equations of motion

m1r1(t) + (c1 + c2)r1(t) + (k1 + k2)r1(t)− k2r2(t)− c2r2(t) = −m1xg(t)

m2r2(t) + c2r2(t) + k2r2(t)− k2r1(t)− c2r1(t) = −m2xg(t)

Apply LT (zero I.C.)[m1s2 + (c1 + c2)s + (k1 + k2)

]R1(s)−

[c2s + k2

]R2(s) = −m1s2Xg(s)[

m2s2 + c2s + k2]R2(s)−

[c2s + k2

]R1(s) = −m2s2Xg(s)



Matrix notation[m1s2 + (c1 + c2)s + (k1 + k2) −(c2s + k2)

−(c2s + k2) m2s2 + c2s + k2

] [R1(s)R2(s)

]=

[−m1s2Xg(s)−m2s2Xg(s)

]

Algebraic system [α −β−β γ

] [R1(s)R2(s)

]=

[δ1

δ2

]s2Xg(s)

Apply Gaussian elimination to solve for R1(s) and R2(s)



Transfer functions

G1(s) ≡ R1(s)s2Xg(s)

=βδ1 + αδ2

αγ − β2 = −B1(s)A(s)

G2(s) ≡ R2(s)s2Xg(s)

=γδ1 + βδ2

αγ − β2 = −B2(s)A(s)

Both transfer functions share the same characteristic polynomial



Numerators / common denominator

B1(s) = m1m2s2 +[m2c1 + (m1 + m2)c2

]s + m2k1 + (m1 + m2)k2

B2(s) = m1m2s2 +[(m1 + m2)c2

]s + (m1 + m2)k2

A(s) = m1m2s4 +[m2c1 + (m1 + m2)c2

]s3 +

[m2k1 + (m1 + m2)k2 + c1c2

]s2

+[c1k2 + c2k1

]s + k1k2

Order of A(s): ALWAYS twice the DOFs



[2× 1] Transfer function matrix

G(s) =[−B1(s)

A(s) −B2(s)A(s)

]T= −

[B1(s)B2(s)

]A(s)

General rule: any [n × nu] transfer function matrix can be written as

G(s) ≡adj{[

Ms2 + Cs + K]−1P

}det{[

Ms2 + Cs + K]−1P

} =B(s)A(s)


Transfer functionsModal decomposition

Underdamped vibration modes −→ the 2n roots of A(s) arrive in ncomplex conjugate pairs

λi = −ζiωi + ωdi j, i = 1, . . . , n

TF

G(s) =n∑

i=1

Ri

s − λi+

Ri

s − λi

Ri −→ i-th residue matrix


Transfer functionsProportional damping

Modal space (x(t) = Φq(t), Φ is M-orthogonal)

Iq(t) +(αI + βΛ

)q(t) + Λq(t) = ΦT Pu(t)

Laplace transform (zero I.C.)[Is2 +

(αI + βΛ

)s + Λ

]Q(s) = ΦT PU(s)

Transfer function between Q(s) and U(s)

GQ (s) =[Is2 +

(αI + βΛ

)s + Λ

]−1ΦT P



By definitionX(s) = ΦQ(s)

Transfer function between X(s) and U(s)

G(s) = Φ[Is2 +

(αI + βΛ

)s + Λ

]−1ΦT P

[Is2 +(αI + βΛ

)s + Λ

]−→ diagonal with entries s2 + 2ζiωis + ω2

i



Expand

G(s) =[φ1 . . . φn

]

1s2+2ζ1ω1+ω2

1. . . 0

.... . .

...0 . . . 1

s2+2ζnωn+ω2n

φT1 P...

φTn P

Rewrite as

G(s) =n∑

i=1

φiφTi P

s2 + 2ζiωis + ω2i

φTi P .= πT : modal participation vector


Exercises

1. Consider the differential equation that describes the base excitation problem inrelative coordinates

r (t) + 2ζnωn r (t) + ω2nr (t) = −xg(t)

Calculate the transfer functions between

- the relative displacement of the mass and the displacement of the base- the relative acceleration of the mass and the displacement of the base- the relative acceleration of the mass and the acceleration of the base- the absolute acceleration of the mass and the acceleration of the base


Exercises

2. Consider the state-space model

ξ(t) = Acξ(t) + Bcu(t)

y(t) = Ccξ(t) + Dcu(t)

Calculate the transfer functions between

- the state vector and the input- the output vector and the input

Then apply the inverse Laplace transform to obtain the (matrix) impulse responseof the system

(Note: use the Neumann series expansion to obtain an expression for the matrixresolvent of Ac)


Exercises

3. Repeat the previous exercise for the state-space model in modal form

ζ(t) = Λcζ(t) + ΠTc u(t)

y(t) = Ψζ(t)

and obtain a modal decomposition of the output-to-input transfer function

4. Use the Laplace transform to solve the homogenous differential equation

x(t) + 2ζnωnx(t) + ω2nx(t) = 0

with x(0) = x0 and x(0) = v0


Exercises

5. Calculate the inverse Laplace transform of the following transfer functions

G(s) =as + b

s2 G(s) =s + 2

s2 + s + 1

G(s) =s2 + 2s + 3

(s + 1)3 G(s) =2s3 + 5s2 + 3s + 6s3 + 6s2 + 11s + 6


Further Reading

1. Debnath, L. & Bhatta, D. (2007), Integral transforms and their applications, 2nd edn,Chapman & Hall/CRC, Boca Raton, FL, USA.

2. Lancaster, P. & Salkauskas, K. (1996), Transform Methods in Applied Mathematics:An Introduction, John Wiley & Sons Ltd., New York, USA.

3. Ogata, K. (1997), Modern Control Engineering, 3rd edn, Prentice-Hall, Inc., NewJersey, USA. (Chapter 2)


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