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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
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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
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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
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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)
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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)
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The Laplace transformFundamental properties
x(t): n-times differentiable
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)
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The Laplace transformFundamental properties
x(t): n-times differentiable
6. Integration:
L{∫ t
0x(τ )dτ
}=
1sL{x(t)} =
F (s)s
7. Convolution:L{g(t) ∗ x(t)} = G(s)X (s)
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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)}
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Transfer functionsFrom differential equations
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)}
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Transfer functionsFrom differential equations
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)
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Transfer functionsFrom differential equations
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)
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Transfer functionsFrom differential equations
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)
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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
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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
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Transfer functionsExample: spring-mass-damper
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
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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
Transfer functionsExample: spring-mass-damper
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)
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Transfer functionsExample: spring-mass-damper
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
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Transfer functionsExample: spring-mass-damper
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
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Transfer functionsExample: spring-mass-damper
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
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Transfer functionsExample: spring-mass-damper
m
k c
u(t)
x(t)
Doing the algebra
g(t) =1
mωde−ωnζnt sin(ωd t)
(Compare to Lecture 2, p.31)
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Transfer functionsFrom differential equations
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
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Transfer functionsFrom differential equations
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
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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)
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Transfer functionsExample: two-story shear building
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)
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Transfer functionsExample: two-story shear building
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
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Transfer functionsExample: two-story shear building
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
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Transfer functionsExample: two-story shear building
[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)
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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
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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
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Transfer functionsProportional damping
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
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Transfer functionsProportional damping
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
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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
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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)
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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
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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
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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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