General Forced Response

7/28/2019 General Forced Response

1/21

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


2/21


3/21

Impact excitation

Impact is a force applied for a very short

period of time

Impulse due toF(t): dttFI )(

area underF

-t

curve


4/21

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


5/21

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


6/21

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


7/21


EOM 0 kxxcxm


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


8/21


EOM 0 kxxcxm


tem

txth dt

d

n

sin1

)()(underdamped


9/21


10/21

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


11/21

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


12/21

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


13/21

Example 2-1

)4()()(4)(2)( tttxtxtx

mm/s1;mm1 00 vx

Compute and plot the response


14/21

Arbitrary input

Vibration of systems subject to arbitrary nonperiodic inputs.

Analysis tools: Impulse response function

Super position Convolution integral


15/21

Convolution integral(1)Excitation Response


16/21

Convolution integral(2)

321)( xxxtx

)(

)()()(

33

2211

tthI

tthItthItx

)(

)()()(

33

2211

tthtF

tthtFtthtFtx

0t

dthFtxt

0

)()()(

Excitation Response


17/21

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


18/21

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


19/21

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


20/21

Example 4-1

kxxcxm

10 0, ttF

1,0 tt000 vx

10 (underdamped)


21/21

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

Documents

General Forced Response