Dynamic response of oscillators to general excitations

© Eng. Vib, 3rd Ed.1/47 @ProfAdhikari

3 General forced response

• So far, all of the driving forces have been sine or cosine excitations

• In this chapter we examine the response to any form of excitation such as– Impulse– Sums of sines and cosines– Any integrable function

© Eng. Vib, 3rd Ed.2/47 @ProfAdhikari, #EG260

Linear Superposition allows us to break up complicated forces into sums of simpler forces, compute the response and add to get the total solution

© Eng. Vib, 3rd Ed.3/47 College of Engineering

3.1 Impulse Response Function

F(t)

+ -

Impulse excitation

is a small positive numberFigure 3.1


F(t)

+ -

i.e. area under pulse

From dynamics The impulse imparted to an object is equal to the change in the objects momentum


We use the properties of impulse to define the impulse function:

F(t)

t

Equal impulses

Dirac Delta function


The effect of an impulse on a spring-mass-damper is related to its change in momentum.

m

tF

m

FvmvF

tvtvmmvtF

ˆˆ

)]()([

00

00

change momentum=impulse

Thus the response to impulse with zero IC is equal to the free response with IC: x0=0 and v0 =FΔt/m

Just after impulse

Just before impulse


Recall that the free response to just non zero initial conditions is:

0

-0

For 0 this becomes:

( ) sinnt

dd

x

v ex t t


Next compute the response to x(0)=0 and v(0) =FΔt/m


So for an underdamped system the impulse response is (x0 = 0)

Response to an impulse at t = 0, and zero initial conditions

0 10 20 30 40-1

-0.5

0

0.5

1

Time

h(t)

m

k

x(t)

c

F̂


So, the response to ( ) is given by ( ).

( ) sin (3.8)

What is the response to a unit impulse applied

at a time different from zero?

The response

nt

dd

t h t

eh t t

m

to ( - ) is ( - ).

This is given on the following slide

t h t

The response to an impulse is thus defined in terms of the impulse response function, h(t).


( )

0

( )sin ( )

for the case that the impulse occurs at note that the effects of

non-zero initial conditions and other forcing terms must be super

imposed on this solution

n t

dd

t

h t et t

m

(see Equation (3.9))

0 10 20 30 40-1

0

1

h1

0 10 20 30 40-1

0

1

h2

0 10 20 30 40-1

0

1

Time

h1

+h

2

=0

=10

For example: If two pulses occur at two different times then

their impulse responses will superimpose


Consider the undamped impulse response

Setting 0 in the equation (3.8)

Response to unit impulse applied at ,

i.e. ( - ) is:

1 ( ) sin ( )n

n

t

t

h t tm


Example 3.1.1Design a camera mount with a vibration constraint

Consider example 2.1.3 of the security cameraagain only this time with an impulsive load


Using the stiffness and mass parameters of Example 2.1.3, does the system stay with in vibration limits if hit by a 1 kg bird traveling at 72 kmh?

The natural frequency of the camera system is

3

3

10 3

3

3

12

(7.1 x 10 N/m)(0.02 m)(0.02 m) 75.43 rad/s

4(3 kg)(0.55)

nc c

k Ebh

m m

From equations (3.7) and (3.8) with ζ = 0, the impulsive response is:

( ) sin sinbn n

c n c n

m vF tx t t t

m m

The magnitude of the response due to the impulse is thus

b

c n

m vX

m


Next compute the momentum of the bird to complete the magnitude calculation:

km 1000 m hour

1 kg 72 20 kg m/shour km 3600 sbm v

Next use this value in the expression for the maximum value:

20 kg m/s

0.088 m3 kg 75.45 rad/s

b

c n

m vX

m

This max value exceeds the camera tolerance


Example 3.1.2: two impacts, zero initial conditions (double hit).

1 2

0.25( )

0.25 0.25( )

( )

1.008 sin(1.984 ) 0

1.008 sin(1.984 ) 0.504 sin(1.984( ))

t

t t

x t x x

e t t

e t e t t


Example 3.1.3 two impacts and initial conditions

Note, no need to redo constants of integrationfor impulse excitation (others, yes)


Computation of the response to first impulse:

Treat (t) as x0 0 and v0 1, 0 t 4

xI (t) e nt v0

d

sind t

1

3e t sin 3t

0 t 4


Total Response for 0< t < 4

1( ) ( ) ( )

1(cos 3 sin 3 ),

3 for 0 4

h I

t

x t x t x t

e t t

t


Next compute the response to the second impulse:

Here the Heaviside step function is used to “turn on” the response to the impulse at t = 4 seconds.


To get the total response add the partial solutions:

0 2 4 6 8 10

1

0.5

0.5

1

x t

t


3.2 Response to an Arbitrary Input

F(t)

tt1,t2 ,t3 ti

The response to general force, F(t), can be viewed as a series of impulses of magnitude F(ti)Δt

xi(t) = [F(ti)t ].h(t-ti) for t>ti

F(ti)

Impulses

Response at time t due to the ith impulse zero IC

ti

xi

t

(3.12)


Properties of convolution integrals: It is symmetric meaning:

0

0

0

Let , fixed so that =

and . Also : 0 : 0

( ) ( ) ( ) = ( ) ( )( )

= ( ) ( )

t

t

t

t t t

d d t t

x t F h t d F t h d

F t h d


The convolution integral, or Duhamel integral, for underdamped systems is:

0

0

1( ) ( ) sin ( )

1 ( ) sin (3.13)

n n

n

ttd

d

t

dd

x t e F e t dm

F t e dm

•The response to any integrable force can be computed with either of these forms•Which form to use depends on which is easiest to compute


Example 3.2.1: Step function input

Figure 3.6 Step function To solve apply (3.13):

0

0

0

0

0

0

1 1( ) (0) sin ( ) sin ( )

sin ( )

n n n n

n n

ttt t

d dtd d

ttdt

d

x t e e t d e F e t dm m

Fe e t d

m


Integrating (use a table, code or calculator) yields the solution:

0( )0 00 02

1

2

( ) cos ( ) , (3.15)1

tan (3.16) 1

n t td

F Fx t e t t t t

k k

Fig 3.7


1

1

00 1

1 2 1 2

0

0

1( ) sin

sin cos ,

( ) ( ) ,

( ) ( ) ( ) ( )

nn

h n nn

t

t t

t

h t tm

vx t x t t t

x F h t d t t t

F h t d F h t d

0

For an undamped system:

The homogeneous solution is

Good until the applied force acts at t1, then:

Example: undamped oscillator under IC and constant force

m

k

x(t)

F(t)t1 t2

F0

F(t)


1

1

1 2 0

0

012

1sin ( )

( 1)( 1) cos ( )

[1 cos ( )]

t

nnt

t

nn n t

nn

x F t dm

Ft

m

Ft t

m

For t1 t t2

Next compute the solution between t1 and t2


1 2

1 2

2

1

2

0

0

02 12

( ) ( ) ( ) ( ) ( ) ( )

1 cos ( )

[cos ( ) cos ( )]

t t t

t t

t

nn n t

n nn

x F h t d F h t d F h t d

Ft

m

Ft t t t

m

0 0

For t t2

Now compute the solution for time greater than t2


Total solution is superposition:

00 1

0 00 1 1 22

0 00 2 1 22

sin cos

( ) sin cos [1 cos ( )]

sin cos cos ( ) cos ( )

n nn

n n nn n

n n n nn n

vt x t t t

v Fx t t x t t t t t t

m

v Ft x t t t t t t t

m

0 1 2 0 01, 8, 2, 4, 0.1, 0

Check points: increases after application of . Undamped response around 0nm F t t x v

x F x

0 2 4 6 8 10-0.1

0

0.1

0.2

0.3

Time (s)

Dis

plac

emen

t x(t

)


Example 3.2.3: Static versus dynamic load

2

1( ) 1 cos

1ntd

d

m gx t e t

k

0 ( ) (1 cos )dd

m gx t t

k

This has max value of , twice the static load max 2 dm gxk


Numerical simulation and plotting• At the end of this chapter, numerical

simulation is used to solve the problems of this section.

• Numerical simulation is often easier then computing these integrals

• It is wise to check the two approaches against each other by plotting the analytical solution and numerical solution on the same graph


3.3 Response to an Arbitrary Periodic Input

• We have solutions to sine and cosine inputs.

• What about periodic but non-harmonic inputs?

• We know that periodic functions can be represented by a series of sines and cosines (Fourier)

• Response is superposition of as many RHS terms as you think are necessary to represent the forcing function accurately

0 2 4 6-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (s)

Dis

plac

emen

t x(t

)

T

Figure 3.11


Recall the Fourier Series Definition:

0

1

20 0

2

0

2

0

Assume ( ) cos sin (3.20)2

2where

( ) (3.21) : twice the average

( ) cos (3.22) : Oscillations around average

( ) sin (3.

n n n nn

n

T

T

T

n nT

T

n nT

aF t a t b t

nn

T

a F t dt

a F t t dt

b F t t dt 23)


The terms of the Fourier series satisfy orthogonality conditions:

0

0

0

0sin sin (3.24)

2

0cos cos (3.25)

2

cos sin 0 (3.26)

T

T T

T

T T

T

T T

m nn t m tdt T m n

m nn t m tdt T m n

n t m tdt


Fourier Series ExampleF(t)

t1 t2=T

F0

1

01 1 2

2 1

0,

( ),

t t

F t Ft t t t t

t t

Step 1: find the F.S. and determine how many terms you need

0


Fourier Series Example

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Time (s)

For

ce F

(t)

F(t)2 coefficients10 coefficients100 coefficients


Having obtained the FS of input

• The next step is to find responses to each term of the FS

• And then, just add them up!

• Danger!!: Resonance occurs whenever a multiple of excitation frequency equals the natural frequency.

• You may excite at 100rad/s and observe resonance while natural frequency is 500rad/s!! Backwards?


Solution as a series of sines and cosines to

The solution can be written as a summation

Solutions calculated from equations of

motion (see section Example 3.3.2)


3.4 Transform Methods

An alternative to solving the previous problems, similar to

section 2.3


Laplace Transform•Laplace transformation

Laplace transforms are very useful because they change differential equations into simple algebraic equations

•Examples of Laplace transforms (see page 216) in book)

f(t) F(s)

Step function, u(t)

e-at

sin( t )

1/s

1/(s+a)

/ ( s2 + 2)

0( ) ( ) { ( )} (3.41)stF s f t e dt f tL


Laplace Transform•Example: Laplace transform of a step function u(t)

0

0

1{ ( )}

stst e

u t e dts s

L

•Example: Laplace transform of e-at

( )

0 0{ }at at st s a te e e dt e dtL

( )

0

1{ }

( ) ( )

s a tat e

es a s a

L

u(t)

t

e-at

t


Laplace Transforms of Derivatives

•Laplace transform of the derivative of a function

0

( ) ( ) stdf t df te dt

dt dtL

Integration by parts gives,

0 0

( )( ) ( )st stdf tf t e s f t e dt

dtL

( )(0) ( )

df tf s f t

dtL L


Laplace Transform Procedures

Steps in using the Laplace transformation to solve DE’s

•Find differential equations

•Find Laplace transform of equations

•Rearrange equations in terms of variable of interest

•Convert back into time domain to find resulting response (inverse transform using tables)

•Laplace transform of the integral of a function

01

( ) ( ) ( )tf t dt f t f t dt

sL L


Laplace Transform Shift Property

eat f t L F s a f t a (t a) L e asF s

(t) L 1 (t a) L e as

Note these shift properties in t and s spaces...

thus


Example 3.4.3: compute the forced response of a spring mass system to a step input using LT

Compare this to the solution given in (3.18)


• From Fourier series of non-periodic functions

• Allow period to go to infinity• Similar to Laplace Transform• Useful for random inputs

• Corresponding inverse transform

• Fourier transform of the unit impulse response is the frequency response function

Fourier Transform

X( ) x(t)e j t

dt

x(t) 12 X( )e j t

d

H ( ) h(t)e j t

dt0 1 2 3 4 5

-20

-10

0

10

20

Frequency (Hz)

N

orm

aliz

ed

H(

) (d

B)

0 1 2 3 4-0.1

-0.05

0

0.05

0.1

Time (s)

h(t

)

wn=2 and M=1

Fourier Transform

Education

Dynamic response of oscillators to general excitations