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
Linear Superposition allows us to break up complicated forces into sums of simpler forces, compute the response and add to get the total solution
1 2
1 1 2 2
2
1 1
2
2 2
If , are solutions of a linear homogeneous equation, then
is also a solution.
If is the particular sol of and the particular sol of
n
n
x x
x a x a x
x x x f
x x x f
2
1 2 solves n 1 2
ax bx x x af bf
0 ˆ ( ) 2
0
t
F t F t
t
ˆ 2
F
3.1 Impulse Response Function
is a small positive numberFigure 3.1
F(t) Impulse excitation
impulse force = ( )
( ) ( ) ( ) N s
ˆ ˆ
2 2
F t dt F t I F t dt F t dt
F F
ˆ 2
F
From sophomore dynamics The impulse
imparted to an object is equal to the change in the objects momentum
F(t) i.e. area under
pulse
( ) 0, ( ) ˆ
If ˆ 1, this is the Dirac Delta (t)
F t t
F t dt F F
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.
impulse=momentum change
0 0
0 0
[ ( ) ( )]
ˆ ˆ
F t mv m v t v t F F t
F mv v
m m
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
2 2
0 0 0 - 1 0
0 0
The solution of:
0 (0) (0)
in underdamped case:
( ) n d nt sin( d tan d )
d n
mx cx kx x x x v
v x x x
x t e t
v x
0
- 0
For 0 this becomes:
( ) sin
nt
d d
x
x t v e t
Next compute the response to x(0)=0 and v(0) =FΔt/m
0
-
The solution of:
ˆ
0 (0) (0) /
in underdamped case from the previous slide is:
ˆ
( ) sin
nt
d d
mx cx kx x x x F t m F
m
x t Fe t
m
Response to an impulse at t = 0, and zero initial conditions
unit impulse response function
ˆ ˆ
( ) sin (response to ) (3.6) ( ) ˆ ( ), where ( ) sin (3.8)
n
n
t
d d
t
d d
x t Fe t F
m
x t Fh t h t e t
m
Fˆ
So for an underdamped system the impulse response is (x0 = 0)
c 0
10 20
30 40
-1
Response to an impulse at t = 0, and zero initial conditions
0 10 20 30 40
-1 -0.5
0 0.5
Time h(t
)
x(t) M
k c
The response to an impulse is thus defined in terms of the impulse response function, h(t).
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
d d
t h t
h t e t
m
to ( - ) is ( - ).
This is given on the following slide
t 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
d d
t
h t e
t t
m
(see Equation (3.9))
For example: If two pulses occur at two different times then their impulse responses will superimpose
0 1
0
2 0
3 0
4 0 -
1 0
1
h 1
0 10 20 30 40
-1 0 1 h2
0 10 20 30 40
-1 0 1
Time h1
+h 2
=0
=10
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 t
m
Example 3.1.2 Design a camera mount with a vibration constraint
Consider example 2.1.3 of the security camera again 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)
n
c c
k Ebh
m m
From equations (3.7) and (3.8) with ζ = 0, the impulsive response is:
( ) sin n b sin n
c n c n
F t m v
x t t t
m m
The magnitude of the response due to the impulse is thus
b
c n
X m v
m
Next compute the momentum of the bird to complete the magnitude calculation:
km 1000 m hour
1 kg 72 20 kg m/s
hour km 3600 s
m vb
Next use this value in the expression for the maximum value:
20 kg m/s
0.088 m 3 kg 75.45 rad/s
b c n
X m v
m
This max value exceeds the camera tolerance
Example 3.1.3: two impacts, zero initial conditions (double hit).
0.25 1
0.25( ) 2
1 kg, 0.5 kg/s, 4 N/m
ˆ 2 N s and ( ) 2 ( ) ( ) 2, 0.125
( ) 2 sin 1.008 sin(1.984 ), 0
( ) 0.504 sin(1.984( )),
n
n t
t d
d
t
m c k
F F t t t
x t e t e t t
m
x t e t t
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
0 0
0 0
0
2 4 ( ) ( 4), 1 mm, 1 mm/s Solve three simple problems and add the results.
Homogeneous solution ( 2rad/s, =0.5, = 3 rad/s)
( ) [ sin cos ]
[ 1 1s 3
n
n d
t n
h d d
d t
x x x t t x x
v x
x t e t x t
e
in 3t cos 3 ]t et cos 3t
Note, no need to redo constants of integration for impulse excitation (others, yes)
Computation of the response to first impulse:
0 0
0
Treat ( ) as 0 and 1, 0 4
( ) sin 1 sin 3
3 0 4
nt t
I d
d
t x v t
x t e v t e t
t
Total Response for 0< t < 4
1
( ) ( ) ( )
(cos 3 1 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:
4 2
4
Heaviside Step function
1 sin 3( 4), 4 3
sin 3( 4) ( 4)
3
t
t
x e t t
e t H t
Here the Heaviside step function is used to “turn on” the response to the impulse at t = 4 seconds.
0 2 4 6 8 10
1 0.5 0.5
1
x t
t
To get the total response add the partial solutions:
4
initial condition
frist impulse second impulse
( ) ( 1 sin 3 cos 3 ) sin 3( 4) ( 4)
3 3
t
t e
x t e t t t H t
1
0
convolution integral
If (the time interval) ( ) [ ( ) ] ( )
0,
( ) ( ) ( )
th I
I
I i i
i i t
t t i
x t F t t h t t
t t
x t F h t d
3.2 Response to an Arbitrary Input
The response to general force, F(t), can be viewed as a series of impulses of magnitude F(ti)Δt
Response at time tdue to the ith impulse zero IC xi(t) = [F(t
i)Dt ].h(t-t
i) for t>t
i
F(t)
t t
1,t
2 ,t
3 ti
F(ti)
Impulses
ti
x
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
t t
d d
t
d d
x t e F e t d
m
F t e d
m
• 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
0
0 0
0 0
0 0
0, 0, 0 1 t t mx cx kx
F t t
x v
Example 3.2.1: Step function input
0
0
0
0 0
0
1 1
( ) (0) sin ( ) sin ( )
sin ( )
n n n n
n n
t t
t t
d t d
d d
t t t d d
x t e e t d e F e t d
m m
F e e t d
m
Figure 3.6 Step function To solve apply (3.13):
Integrating (use a table, code or calculator) yields the solution:
( 0)
0 0
0 0
2
1
2
( ) cos ( ) , (3.15)
1
tan (3.16) 1
n t t
d
F F
x t e t t t t
k k
( ) F t
Example: undamped oscillator under IC and constant force
1
1
0
0 1
1 2 1 2
0
0
( ) 1 sin
sin cos ,
( ) ( ) ,
( ) ( ) ( ) ( )
n n
h n n
n
t
t t
t
h t t
m
x v t x t t t
x F h t d t t t
F h t d F h t d
For an undamped system:
The homogeneous solution is
Good until the applied force acts at t1, then:
0
t1 t2 F0
F(t)
m x(t)
k
Next compute the solution between t
1and t
21 2
For t t t
1
1
1 2 0
0
0 2 1
1 sin ( )
( 1)( 1)
cos ( )
[1 cos ( )]
t
n t n
t n
n n t
n n
x F t d
m
F t
m
F t t
m
Now compute the solution for time greater than t
2For t t
21 2
1 2
2
1
2
0
0
0
2 1
2
( ) ( ) ( ) ( ) ( ) ( )
1 cos ( )
[cos ( ) cos ( )]
t t t
t t
t n
n n t
n n
n
x F h t d F h t d F h t d
F t
m
F t t t t
m
0 0
Total solution is superposition:
0
0 1
0 0
0 2 1 1 2
0 0
0 2 2 1 2
sin cos
( ) sin cos [1 cos ( )]
sin cos cos ( ) cos ( )
n n
n
n n n
n n
n n n n
n n
v t x t t t
v F
x t t x t t t t t t
m
v F
t x t t t t t t t
m
0 1, 8, 1 2, 2 4, 0 0.1, 0 0
Check points: increases after application of . Undamped response around 0
m F n t t x v
x F x
0 2 4 6 8 10
-0.1 0 0.1 0.2 0.3
Time (s) Displacement x(t)
0
0 0
m gd t mx cx kx
t
2
( ) 1 1 cos
1
nt d
d
x t m g e t
k
max 2 m gd
x k
Example 3.2.3: Static versus dynamic load
0 ( ) m gd (1 cos d )
x t t
k
This has max value of , twice the static load
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
2 n n2 ( ) where ( ) ( )
x
x
x F t F t F t T• 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) Displacement x(t)
T
Figure 3.11
Recall the Fourier Series Definition:
0
1
2
0 0
2 0 2
0
Assume ( ) cos sin (3.20)
2 where 2
( ) (3.21) : twice the average
( ) cos (3.22) : Oscillations around average ( ) sin (3.
n n n n
n
n
T T
T
n T n
T
n T n
F t a a t b t
n n 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
0
sin sin (3.24)
2 0
cos cos (3.25)
2
cos sin 0 (3.26)
T
T T
T
T T
T
T T
m n n t m tdt
T m n
m n n t m tdt
T m n
n t m tdt
Fourier Series Example
1 0
1 1 2
2 1
0,
( ) ,
t t
F t F
t t t t t
t t
Step 1: find the F.S. and
determine how many terms you need
F(t) F0
0 t
1 t
2=T
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) Force F(t)
F(t)
2 coefficients 10 coefficients 100 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
2 n n2 ( )
x x x F t
The solution can be written as a summation
0
1 0
2 0 0
0 2
2 2
( ) ( ) ( ) ( )
where ( ) is a solution to
2 ( )
2 2
and ( ) and ( ) are a solutions to
2 cos( )
2 sin( )
p cn sn
n
n n
n
cn sn
n n n T
n n n T
x t x t x t x t
x t
a a
x x x x t
x t x t
x x x a n t
x x x b n t
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
( ) 0 ( ) st { ( )} (3.41) F s
f t e dt L f tLaplace transforms are very useful because they change differential equations into simple algebraic equations
• Examples of Laplace transforms (see page 244) in book)
f(t) F(s)
Step function, u(t) e-at
sin( t )
1/s 1/(s+a)
/ ( s2 + 2)
Laplace Transform
• Example: Laplace transform of a step function u(t)
0
0
{ ( )} 1
st
st e
u t e dt
s s
L
• Example: Laplace transform of e-at
( )
0 0
{ e
at}
e e dt
at st
e
s a tdt L
( )
0
{ } 1
( ) ( )
s a t
at e
e s a s a
L
u(t)
t
e-at
t
Laplace Transforms of Derivatives
• Laplace transform of the derivative of a function
0
( ) ( ) st
df t df t
dt dt e dt
L
Integration by parts gives,
0 0
( ) ( ) st ( ) st
df t f t e s f t e dt
dt
L
( ) (0) ( )
df t f s f t
dt
L L
Laplace Transform Procedures
• Laplace transform of the integral of a function
t f t dt( )
1s
f t( )
0 f t dt( )
L L
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 Shift Property
Note these shift properties in t and s spaces...
L
L
L L
( )
( ) 1 ( )
at
as
as
e f t F s a
f t a t a e F s
t t a e
thus
Example 3.4.3: compute the forced response of a spring mass system to a step input using LT
2
2 2 2
The equation of motion is ( ) ( ) ( )
Taking the Laplace Transform (zero initial conditions)
1 1 1 /
( ) ( ) ( )
( ) ( )
Taking the inverse Laplace Transform yiel
n
mx t kx t t
ms k X s X s m
s s ms k s s
2
ds:
1 / 1
( ) 1- cos n 1- cos n
n
x t m t t
k
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
( ) ( ) j t X x t e dt
1
( ) 2 ( ) j t x t
X e d( ) ( ) j t H h t e dt
Fourier Transform
0 1 2 3 4
-0.1 -0.05 0 0.05 0.1
Time (s) h(t
)
w n=2 and M=1
Fourier Transform
0 1 2 3 4 5
-20 -10 0 10 20
Frequency (Hz) Normalized H() (dB)
3.5 Random Vibrations
• So far our excitations have been harmonic, periodic, or at least known in advance
• These are examples of deterministic excitations, i.e., known in advance for all time
– That is given t we can predict the value of F(t) exactly
• Responses are deterministic as well
• Many physical excitations are nondeterministic, or random, i.e., can’t write explicit time descriptions
– Rockets – Earthquakes
– Aerodynamic forces – Rough roads and seas
• The responses x(t) are also nondeterministic
Random Vibrations
*ie given t we do not know x(t) exactly, but rather
we only know statistical properties of the response such as the average value
• Stationary signals are those whose statistical properties do not vary over time
• Functions are described in terms of probabilities – Mean values*
– Standard deviations
• Random outputs related to random input via system transfer function
Autocorrelation function and power spectral density
The autocorrelation function describes how a signal is changing in time or how correlated the signal is at two different points in time.
0
( ) lim 1 T ( ) ( )
xx T
R t x t x t d
T
The Power Spectral Density describes the power in a signal as a function of frequency and is the Fourier transform of the autocorrelation function.
( ) 1 ( )
2
j
xx xx
S R e d
Examples of signals
HARMONIC RANDOM
T
x(t) time
A
-A
Signal
x(t)
time Arms
T
Rxx(t) A2/2
-A2/2 t time shift
Autocorrelation
A2rms
t time shift Rxx(t)
Sxx(w)
1/T
Power Spectral Density
Frequency (Hz)
Sxx(w)
Frequency (Hz)
