Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-I I

-Vib

ratio

ns

Response to Arbitrary Excitation

18 7 Response to arbitrary excitation

7 Response to arbitrary excitation

To be able to solve the response to an arbitrary excitation, first, the solution methods fora unit impulse and step response need to be known. They are treated first, after which thestandard solution to an arbitrary excitation is treated.

7-1 Unit impulse response

To calculate the response of a system to a unit impulse, first, the unit impulse function δneeds to be defined. This function is called the Dirac delta function and has the followingdefinition (see also figure 21):

δ(t− a) = 0 for t 6= a

δ(t− a) 6= 0 for t = a∞∫−∞

δ(t− a)dt = 1

δ has the unit of 1/s.

t

ε1

a

area = 1

ε

Figure 21: A visualisation of the Dirac delta function

From the definition, it follows that:∞∫−∞

f(t)δ(t− a)dt = f(a)∞∫−∞

δ(t− a)dt = f(a)

Let’s have a look at the impulse response of an underdamped system:

k

c x

m

Figure 22: A forced damped mass-spring system

The EOM is:mx+ cx+ kx = P δ(t)

Lecture Notes AE2135-II - Vibrations

7 Response to arbitrary excitation 19

and the IC are: {x(0) = 0x(0) = 0

To solve the impulse response problem, the impulse is converted to the initial conditions. Tothis end, integrate both sides from 0 to ε:

ε∫0

(mx+ cx+ kx)dt =ε∫

0

P δ(t)dt

This can be split into different integrals, looking at each individual part:

−→ limε→0

ε∫0

mxdt = limε→0

mx∣∣∣ε0

= limε→0

(mx(ε)−mx(0)) = mx(0+)

−→ limε→0

ε∫0

cxdt = limε→0

cx∣∣∣ε0

= limε→0

(cx(ε)− cx(0)) = cx(0+)

−→ limε→0

ε∫0

kxdt = limε→0

kxt∣∣∣ε0

= limε→0

(kx(ε)ε− kx(0)0) = 0

−→ limε→0

ε∫0

P δ(t)dt = P

x(0+) = 0 from the definition of impulse. The impulse can thus be written as an initialvelocity and the equation can be simplified to:

mx(0+) = P −→ x(0+) = P

m

So the EOM becomes:mx+ cx+ kx = 0

with IC: {x(0) = 0x(0) = P

m

Remember, for an underdamped system (see section 4):

x = Ae−ζωnt sin (ωd + ϕ)

with

A =

√x2

0ω2d + (x0 + ζωnx0)2

ωd

ϕ = arctan(

x0ωdx0 + ζωnx0

)

AE2135-II - Vibrations Lecture Notes

20 7 Response to arbitrary excitation

which for this case become:

ϕ = 0 + 2kπA = x0

ωd= P

mωd

−→ x(t) =

Pmωd

e−ζωnt sin (ωdt) t ≥ 00 t < 0

= P g(t)

where the displacement function to a unit impulse has been labelled g(x) for later conve-nience.

7-2 Unit step response

The unit step function H, or Heaviside step function, is defined as (see also figure 23):

H(t− a) ={

0 for t < a

1 for t ≥ a

H

t

1

a0

Figure 23: A visualisation of the Heaviside step function

The Heaviside step function is mathematically linked to the Dirac delta function:

δ(t− a) = dH(t− a)dt

H(t− a) =t∫

−∞

δ(τ − a)dτ

The response of an underdamped mass-spring system to a unit step at t = a can thus befound by integrating the response to a unit impulse at t = a. In this case a step starting at

Lecture Notes AE2135-II - Vibrations

7 Response to arbitrary excitation 21

t = 0 is regarded, so g(x) from section 7-1 can be inserted in the integral:

G(t) =t∫

−∞

g(τ)dτ

=t∫

−∞

1mωd

e−ζωnτ sin (ωdτ)dτ τ ≥ 0

=t∫

0

1mωd

e−ζωnτ sin (ωdτ)H(τ)dτ

where the lower limit has been set to zero because of the inclusion of the Heaviside functionand we are only interested in the displacement after t = 0. Substitute sin(ωdt) = eiωdt−e−iωdt

2i :

G(t) =t∫

0

12imωd

e−ζωnτ(eiωdτ − e−iωdτ

)H(τ)dτ

= 12imωd

[e−(ζωn−iωd)τ

−(ζωn − iωd)− e−(ζωn+iωd)τ

−(ζωn + iωd)

]t0

The denominator has to be equal to add fractions. Multiplying both fractions with the otherdenominator yields a common denominator: −(ζωn − iωd) · −(ζωn + iωd) = ζ2ω2

n + ω2d = ω2

n.So:

G(t) = 12imωdω2

n

[(ζωn + iωd)e−(ζωn−iωd)τ + (ζωn − iωd)e−(ζωn+iωd)τ

]t0

= 12imωdω2

n

[(ζωn + iωd)

(e−(ζωn−iωd)t − 1

)+ (ζωn − iωd)

(e−(ζωn+iωd)t − 1

)]= 1

2imωdω2n

[−ζωne−ζωnteiωdt +���ζωn + iωd − iωde−ζωnteiωdt + ζωne

−ζωnte−iωdt����− ζωn

+ iωd − iωde−ζωnte−iωdt]

= 12imωdω2

n

[2iωd − e−ζωnt

(ζωn

(eiωdt − e−iωdt

)+ iωd

(eiωdt + e−iωdt

))]= 2iωd

2imωdω2n

[1− e−ζωnt

(ζωn2iωd

2i sin(ωdt) + 122 cos(ωdt)

)]= 1k

[1− e−ζωnt

(cos(ωdt) + ζωn

ωdsin(ωdt)

)]which is only valid for t ≥ 0.

7-3 Standard solution

The response to an arbitrary excitation can be found by dividing the excitation force intoinfinitesimally short impulses, see figure 24, and summing all the responses to these impulsesup:

AE2135-II - Vibrations Lecture Notes

22 7 Response to arbitrary excitation

t

Figure 24: An arbitrary load can be seen as a collection of impulses

P (τ) = f(τ)∆τF (τ) = P (τ)δ(t− τ) Impulsive force

= f(τ)∆τδ(t− τ)

linearsystem

Figure 25

x(τ) = f(τ)∆τg(t− τ)

x(t) = lim∆τ→0

∑τ

x(τ) = lim∆τ→0

∑τ

f(τ)∆τg(t− τ) =t∫

0

f(τ)g(t− τ)dτ

It does not matter which of the two functions (f or g) is function of τ which one is functionof (t− τ), as can be seen below:

t− τ = λ, so: τ = t− λ and: dτ = −dλ

also:

τ = 0 −→ λ = t

τ = t −→ λ = 0

x(t) = −0∫t

f(t− λ)g(λ)dλ =t∫

0

f(t− λ)g(λ)dλ

The response to an arbitrary excitation can be written as:

x =t∫

0

f(τ)g(t− τ)dτ

Lecture Notes AE2135-II - Vibrations

7 Response to arbitrary excitation 23

where g(t) is the impulse response function.

Integrals like these are conveniently solved in the Laplace domain, see also appendix F, wherethe integral vanishes into a straightforward multiplication and thus the solution is easilyfound. The challenge of this method usually only lies in transforming the solution back fromLaplace domain to the time domain. In the Laplace domain, the equation above becomes:

X = F (s)g(s)

so in case of the following initial conditions:

x(0) = x0

x(0) = x0

the equation of motion mx+ cx+ kx = f(t) becomes in the Laplace domain:

m(s2X − sx0 − x0

)+ c (sX − x0) + kX = F (s)

and solving the equation in the Laplace domain can be done by simply isolating X from theequation:

X = F (s)ms2 + cs+ k

+ (sm+ c)x0 +mx0ms2 + cs+ k

= F (s)m (s2 + 2ζωns+ ω2

n) + m [(s+ 2ζωn)x0 + x0]m (s2 + 2ζωns+ ω2

n)= F (s)g(s) +m [(s+ 2ζωn)x0 + x0] g(s)

where:g(s) = 1

m (s2 + 2ζωns+ ω2n)

This expression can be split up in partial fractions and transformed back to the time domainusing a table of standard Laplace transformations (Note: such a table will be provided at theexam).

7-4 Alternative solution

Instead of using the impulse response function, one can also use the step response functionin the manner as described below. This approach, though slightly more complex, is relevant,since it is often easier to get step response solutions from numerical codes than impulseresponse solutions.

∆f = ∆f∆τ ∆τH(t− τ)

−→ x(t) = lim∆τ→0

∑τ

∆f∆τ ∆τG(t− τ) =

t∫0

dfdτ G(t− τ)dτ

AE2135-II - Vibrations Lecture Notes

24 7 Response to arbitrary excitation

t

Figure 26: An arbitrary load can also be seen as a collection of steps

Lecture Notes AE2135-II - Vibrations