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© 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
© Eng. Vib, 3rd Ed.4/47 College of Engineering
F(t)
+ -
i.e. area under pulse
From dynamics The impulse imparted to an object is equal to the change in the objects momentum
© Eng. Vib, 3rd Ed.5/47 College of Engineering
We use the properties of impulse to define the impulse function:
F(t)
t
Equal impulses
Dirac Delta function
© Eng. Vib, 3rd Ed.6/47 College of Engineering
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
© Eng. Vib, 3rd Ed.7/47 College of Engineering
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
© Eng. Vib, 3rd Ed.8/47 College of Engineering
Next compute the response to x(0)=0 and v(0) =FΔt/m
© Eng. Vib, 3rd Ed.9/47 College of Engineering
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
© Eng. Vib, 3rd Ed.10/47 College of Engineering
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).
© Eng. Vib, 3rd Ed.11/47 College of Engineering
( )
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
© Eng. Vib, 3rd Ed.12/47 College of Engineering
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
© Eng. Vib, 3rd Ed.13/47 College of Engineering
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
© Eng. Vib, 3rd Ed.14/47 College of Engineering
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
© Eng. Vib, 3rd Ed.15/47 College of Engineering
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
© Eng. Vib, 3rd Ed.16/47 College of Engineering
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
© Eng. Vib, 3rd Ed.17/47 College of Engineering
Example 3.1.3 two impacts and initial conditions
Note, no need to redo constants of integrationfor impulse excitation (others, yes)
© Eng. Vib, 3rd Ed.18/47 College of Engineering
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
© Eng. Vib, 3rd Ed.19/47 College of Engineering
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
© Eng. Vib, 3rd Ed.20/47 College of Engineering
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.
© Eng. Vib, 3rd Ed.21/47 College of Engineering
To get the total response add the partial solutions:
0 2 4 6 8 10
1
0.5
0.5
1
x t
t
© Eng. Vib, 3rd Ed.22/47 College of Engineering
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)
© Eng. Vib, 3rd Ed.23/47 College of Engineering
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
© Eng. Vib, 3rd Ed.24/47 College of Engineering
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
© Eng. Vib, 3rd Ed.25/47 College of Engineering
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
© Eng. Vib, 3rd Ed.26/47 College of Engineering
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
© Eng. Vib, 3rd Ed.27/47 College of Engineering
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)
© Eng. Vib, 3rd Ed.28/47 College of Engineering
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
© Eng. Vib, 3rd Ed.29/47 College of Engineering
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
© Eng. Vib, 3rd Ed.30/47 College of Engineering
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
)
© Eng. Vib, 3rd Ed.31/47 College of Engineering
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
© Eng. Vib, 3rd Ed.32/47 College of Engineering
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
© Eng. Vib, 3rd Ed.33/47 College of Engineering
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
© Eng. Vib, 3rd Ed.34/47 College of Engineering
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)
© Eng. Vib, 3rd Ed.35/47 College of Engineering
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
© Eng. Vib, 3rd Ed.36/47 College of Engineering
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
© Eng. Vib, 3rd Ed.37/47 College of Engineering
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
© Eng. Vib, 3rd Ed.38/47 College of Engineering
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?
© Eng. Vib, 3rd Ed.39/47 College of Engineering
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)
© Eng. Vib, 3rd Ed.40/47 College of Engineering
3.4 Transform Methods
An alternative to solving the previous problems, similar to
section 2.3
© Eng. Vib, 3rd Ed.41/47 College of Engineering
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
© Eng. Vib, 3rd Ed.42/47 College of Engineering
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
© Eng. Vib, 3rd Ed.43/47 College of Engineering
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
© Eng. Vib, 3rd Ed.44/47 College of Engineering
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
© Eng. Vib, 3rd Ed.45/47 College of Engineering
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
© Eng. Vib, 3rd Ed.46/47 College of Engineering
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)
© Eng. Vib, 3rd Ed.47/47 College of Engineering
• 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
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