General forced response

Impulse response •Dirac delta function •Unit impulse response

arbitrary response •Convolution integral

F(t)

t

cam

Material being compacted

Follower

Cam motion

Platform x(t)

m

F(t)

t

The principle of Superposition

02 =+ xx nω

Linear equations

21, xx are both solutions

2211 xaxax += is also a solution

12 fxx n =+ω 1xsolution

22 fxx n =+ω 2xsolution

212 ffxx n +=+ω 21 xx +solution

21, aa are constants

Impact excitation

Impact is a force applied for a very short period of time

Impulse due to F(t): ∫= dttFI )(

= area under F-t curve

Dirac delta function

Dirac delta function is a unit impulse function

Properties of dirac delta function

0)( =τ−δ t

1)( =τ−δ∫∞

∞−

dtt

)()()( τ=τ−δ⋅∫∞

∞−

gdtttg

F(t)

t τ

τ≠tfor

)()( τ−δ⋅= tItF

Impulse

Linear impulse-momentum equation

maF = dtmadtF ∫∫ = )(

vmvvmI ∆=−= )( 12

A mass m initially at rest subject to an impact of size I N.s

impact

0mvIdtF ==∫mIv /0 =

An impact causes initial velocity change of v0 = I/m while keeping initial displacement unchanged

Unit impulse response (1)

EOM )(tkxxcxm δ=++

Initial conditions 0)0()0( == xx

Just after subject to a unit impact (t = 0+)

Time response of a system initially at rest but subject to a unit impact (I = 1)

EOM 0=++ kxxcxm

Initial conditions mxx /1)0(;0)0( ==

m

c k

x

Impact response is just free vibration of a system with initial velocity 1/m

Unit impulse response (2)

EOM 0=++ kxxcxm

Initial conditions mxx /1)0(;0)0( ==

Unit impulse response is just free vibration of a system with initial velocity 1 / m

tem

txth dt

d

n ωω

== ζω− sin1)()(

For impact applied at time t = τ

)(sin1 )( τ−ωω

τ−ζω− tem d

t

d

n

0

h(t) =

for 0 < t < τ

for t > τ

Before impact

After impact

underdamped

Unit impulse response (3)

EOM 0=++ kxxcxm

Initial conditions mxx /1)0(;0)0( ==

tem

txth dt

d

n ωω

== ζω− sin1)()(underdamped

Impulse response

EOM δ=++ Ikxxcxm

Initial cond. 0)0(;0)0( == xx

Unit impulse response is just free vibration of a system with initial velocity I / m

)(sin)( thItem

Itx dt

d

n ⋅=ωω

= ζω−

For impact applied at time t = τ

)(sin)( τ−ωω

τ−ζω− tem

Id

t

d

n

0

h(t) =

for 0 < t < τ

for t > τ

Before impact

After impact

underdamped

0=++ kxxcxm

mIxx /)0(;0)0( ==

Example 1-1

Given: m = 1 kg, c = 0.5 kg/s, k = 4 N/m

)(1.0)(2.0)( τ−δ+δ= tttF

24 ==ωn 125.0)2/( =ω=ζ nmc

)(2.0)( ttF δ=For

tetx t ⋅−−

= − 2)2)(125.0(

21 125.012sin)125.012)(1(

2.0)(

;

)(1.0)( τ−δ= ttF

)984.1sin(1008.0)( 25.01 tetx t−=

)(984.1sin0504.0)( )(25.02 τ−= τ−− tetx t

For τ>t

Example 1-2

)984.1sin(1008.0 25.0 te t−

)(984.1sin0504.0)984.1sin(1008.0 )(25.025.0 τ−+ τ−−− tete tt τ>t

)()()( 21 txtxtx +=

= τ<< t0

[ ] )()(984.1sin0504.0)984.1sin(1008.0)( )(25.025.0 τττ −Φ⋅−+= −−− ttetetx tt

Heaviside step function

=−Φ )( τtτ<t,0

τ≥t,1)( τ−tHor

Example 1-3

)984.1sin(1008.0 25.0 te t−

)(984.1sin0504.0)984.1sin(1008.0 )(25.025.0 τ−+ τ−−− tete tt τ>t

)()()( 21 txtxtx +=

= τ<< t0

Impact forces 5.0=τ Response

)()()( 21 txtxtx +=

0.5 )()( 1 txtx =

Example 2-1

)4()()(4)(2)( −δ−δ=++ tttxtxtx

mm/s 1;mm 1 00 −== vx

Compute and plot the response

Arbitrary input

Vibration of systems subject to arbitrary nonperiodic inputs.

Analysis tools: • Impulse response function • Super position • Convolution integral

Convolution integral (1)

Excitation Response

Convolution integral (2)

+++= 321)( xxxtx

+−+−+−=

)()()()(

33

2211

tthItthItthItx

+−⋅∆+−⋅∆+−⋅∆=

)()()()(

33

2211

tthtFtthtFtthtFtx

0→∆t

ττ−τ= ∫ dthFtxt

0

)()()(

Excitation Response

Convolution integral (3)

=++ kxxcxm 11 0),( tttF <≤

12 ),( tttF ≥

For the non-continuous function,

ττ−τ∫ dthFt1

01 )()(

ττ−τ+ττ−τ ∫∫ dthFdthFt

t

t

1

1

)()()()( 20

1

=)(tx10 tt <≤

1tt ≥

when

when

Example 3-1

=++ kxxcxm 00,0 tt <<

00 , ttF ≥

000 == vx

10 <ζ< (underdamped)

ττ−τ= ∫ dthFtxt

0

)()()(

tem

txth dt

d

n ωω

== ζω− sin1)()(ττ−ω

ω= ∫ τ−ζω− dte

mFtx

t

td

t

d

n

0

)(sin1)( )(0

ττ−ωω

= ∫ τζωζω− dteemFtx

t

td

t

d

nn

0

)(sin)( 0

Example 3-2

00)(

200 ];)(cos[

1)( 0 tttte

kF

kFtx d

ttn ≥θ−−ωζ−

−= −ζω−

2

1

1tan

ζ−

ζ=θ −

Static displacement

1.0=ζ

16.3=ωn

300 =F

1000=k00 =t

rad/s

N

N/m

Example

Example 4-1

=++ kxxcxm 10 0, ttF ≤<

1,0 tt >

000 == vx

10 <ζ< (underdamped)

Homework

Calculate and plot the response of an undamped system to a step function with a finite rise time of t1 for the case m = 1 kg, k = 1 N/m, t1 = 4 s, and F0 = 20 N. This function is described by

=)(tF1

1

0 0, ttt

tF≤≤

10 , ttF >0F

1t