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General forced response
Impulse response
Dirac delta functionUnit impulse response
arbitrary responseConvolution integral
F(t)
t
cam
Material beingcompacted
Follower
Cam
motion
Platformx(t)
m
F(t)
t
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Impact excitation
Impact is a force applied for a very short
period of time
Impulse due toF(t): dttFI )(
area underF
-t
curve
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Dirac delta function
Dirac delta function is a unit impulse function
Properties of dirac delta function
0)( t
1)(
dtt
)()()(
gdtttg
F(t)
t
tfor
)()( tItF
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Impulse
Linear impulse-momentum equation
maF dtmadtF )(vmvvmI )( 12
A mass m initially at rest subject to an impact of sizeIN.s
impact
0mvIdtF mIv /0
An impact causes initial velocity change ofv0 =I/m while
keeping initial displacement unchanged
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Unit impulse response (1)
EOM )(tkxxcxm
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
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Unit impulse response (2)
EOM 0 kxxcxm
Initial conditions mxx /1)0(;0)0(
Unit impulse response is just free vibration of a systemwith initial velocity 1 / m
tem
txth dt
d
n
sin1)()(
For impact applied at time t
)(sin1 )(
te
md
t
d
n
0
h(t) =
for0 < t<
fort>
Before impact
After impact
underdamped
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Unit impulse response (3)
EOM 0 kxxcxm
Initial conditions mxx /1)0(;0)0(
tem
txth dt
d
n
sin1
)()(underdamped
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Example 1-1
Given: m 1 kg, c 0.5 kg/s, k 4 N/m
)(1.0)(2.0)( tttF
24 n 125.0)2/( nmc
)(2.0)( ttF For
tetx t
2)2)(125.0(2
1 125.012sin)125.012)(1(
2.0)(
;
)(1.0)( ttF
)984.1sin(1008.0)( 25.01 tetxt
)(984.1sin0504.0)( )(25.02 tetx t
For t
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Example 1-2
)984.1sin(1008.0 25.0 te t
)(984.1sin0504.0)984.1sin(1008.0 )(25.025.0 tete tt t
)()()( 21 txtxtx
t0
)()(984.1sin0504.0)984.1sin(1008.0)( )(25.025.0 ttetetx tt
Heaviside step function
)( tt,0
t,1)( tHor
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Example 1-3
)984.1sin(1008.0 25.0 te t
)(984.1sin0504.0)984.1sin(1008.0 )(25.025.0 tete tt t
)()()( 21 txtxtx
t0
Impact forces 5.0 Response
)()()( 21 txtxtx
0.5 )()( 1 txtx
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Example 2-1
)4()()(4)(2)( tttxtxtx
mm/s1;mm1 00 vx
Compute and plot the response
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Arbitrary input
Vibration of systems subject to arbitrary nonperiodic inputs.
Analysis tools: Impulse response function
Super position Convolution integral
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Convolution integral(1)Excitation Response
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Convolution integral(2)
321)( xxxtx
)(
)()()(
33
2211
tthI
tthItthItx
)(
)()()(
33
2211
tthtF
tthtFtthtFtx
0t
dthFtxt
0
)()()(
Excitation Response
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Convolution integral(3)
kxxcxm 11 0),( tttF
12 ),( tttF
For the non-continuous function,
dthFt1
0
1 )()(
dthFdthFt
t
t
1
1
)()()()( 20
1
)(tx
10 tt
1tt
when
when
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Example 3-1
kxxcxm
00,0 tt
00 , ttF
000 vx
10 (underdamped)
dthFtxt
0
)()()(
te
m
txth dt
d
n
sin1
)()(
dtemFtxt
t
d
t
d
n
0
)(sin1
)()(
0
dteemF
tx
t
t
d
t
d
nn
0
)(sin)( 0
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Example 3-2
00
)(
2
00 ];)(cos[
1
)( 0 tttte
k
F
k
Ftx d
ttn
2
1
1tan
Static displacement
1.0
16.3n
300 F
1000k
00 t
rad/s
N
N/m
Example
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Example 4-1
kxxcxm
10 0, ttF
1,0 tt000 vx
10 (underdamped)
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Homework
Calculate and plot the response of an undamped system to a step
function with a finite rise time oft1 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