Vibraciones Forzadas - Ejemplos y teoría.pdf

Ch. 3: Forced Vibration of 1-DOF System

3.0 Outline

Harmonic ExcitationFrequency Response FunctionApplicationsPeriodic ExcitationNon-periodic Excitation

3.0 Outline


3.1 Harmonic Excitation

Force input function of the harmonic excitation is theharmonic function, i.e. functions of sines and cosines. This type of excitation is common to many systeminvolving rotating and reciprocating motion. Moreover,many other forces can be represented as an infiniteseries of harmonic functions. By the principle ofsuperposition, the response is the sum of the individual harmonic response.

It is more convenient to use the frequency domaintechnique in solving the harmonic excitation problems.This is because the response to differentexcitation frequencies can be seen in one graph.




( )( )

0

20 0 0

0

20

Let us focus on the particular solution ofcos

normalize the equation of motion2 cos , /

Re

solve for from 2

and the solution

n n

i t

i tn n

mx cx kx F t

x x x f t f F m

f t f e

z t z z z f e

ω

ω

ω

ζω ω ω

ζω ω

+ + =

+ + = =

⎡ ⎤= ⎣ ⎦∴ + + =

( ) ( ) ( )

( ) ( ) ( )( ) ( )

( )

2 20

is the real part of ; Re

Assume the solution to have the same form as the forcing function same frequency as the input w/ different mag. and phase

2

i t

i t i tn n

z t x t z t

z t Z i e

i Z i e f e

fZ i

ω

ω ω

ω

ω ζωω ω ω

ω

= ⎡ ⎤⎣ ⎦

=

− + + =

=( )

( )

20 0

22 2

02

/2 1 / 2 /

1 / 2 /

n

n n n n

n n

fi i

Fk i

ωω ω ζωω ω ω ζω ω

ω ω ζω ω

=− + − +

=⎡ ⎤− +⎣ ⎦



( ) ( )

( )

( ) ( )

( ) ( ) ( )

( )( ) ( )

002

02

2

0

2 22

12

1 2

Re , /1 2

1If is the frequency response1 2

cos

1where magnitude1 2

2tan phas1

i t i t

i tn

i

Fz t e H i F ek r i r

Fx t e rk r i r

H i H i ek r i r

x t F H i t

H ik r r

rr

ω ω

ω

θ

ωζ

ω ωζ

ω ωζ

ω ω θ

ωζ

ζθ −

= =⎡ ⎤− +⎣ ⎦

⎡ ⎤⎢ ⎥∴ = =

⎡ ⎤− +⎢ ⎥⎣ ⎦⎣ ⎦

= =⎡ ⎤− +⎣ ⎦

∴ = +

= =− +

−= =

−

( ) ( )

e

The system modulates the harmonic input by

the magnitude and phase H i H iω ω



( ) ( )( ) ( ) ( ) ( )( )

1 2

0

1

total response homogeneous soln. particular soln.Recall the homogeneous solution of the underdamped system

cos or sin cos

cos cos

or

n n

n

n

t th d h d d

td

t

x Ce t x e A t A t

x t Ce t F H i t

x t e A

ζω ζω

ζω

ζω

ω φ ω ω

ω φ ω ω θ

− −

−

−

= +

= − = +

∴ = − + +

= ( ) ( ) ( )2 0

1 2

sin cos cos

The initial conditions will be used to determine , or ,They will be different from those of free responsebecause the transient term now is partly due to the excitatio

d dt A t F H i t

C A A

ω ω ω ω θ

φ

+ + +

n force

and partly due to the initial conditions



Ex. 1 Compute and plot the response of a spring-masssystem to a force of magnitude 23 N, drivingfrequency of twice the natural frequency and i.c.given by x0 = 0 m and v0 = 0.2 m/s. The massof the system is 10 kg and the spring stiffnessis 1000 N/m.



( )

( ) ( )( )( )

( )

32 2

31 2

31 2

2

/ 1000 /10 10 rad/s/ 2 0

2 10 20 rad/s1 1 0.333 10

1 2 1000 1 2

sin cos 23 0.333 10 cos

cos sin 23 0.333 10 sin

i.c. 0 0 23 0.333 10

n

n

n n

n n n n

k mc m

H ik r i r

x t A t A t t

x t A t A t t

x A

ωζ ωω

ωζ

ω ω ω

ω ω ω ω ω ω

−

−

−

= = =

= =

= × =

= = = − ×⎡ ⎤− + × −⎣ ⎦

= + − × ×

= − + × × ×

= = − × ×

( )( ) ( )

3 32

1 1

3

, 7.667 10

0 0.2 10 , 0.02

0.02sin10 7.667 10 cos10 cos 20 m

A

x A A

x t t t t

− −

−

= ×

= = × =

∴ = + × −





Ex. 2 Find the total response of a SDOF system withm = 10 kg, c = 20 Ns/m, k = 4000 N/m, x0 = 0.01 m,v0 = 0 m/s under an external force F(t) = 100cos10t.



( )

( ) ( )( ) ( ) ( ) ( )

( ) ( )( )

2 2

0

21 2

1

/ 20 rad/s/ 2 0.05/ 0.5

1 1 332.6 6 0.06661 2 4000 1 0.5 2 0.05 0.5

cos 33.26 3cos 10 0.0666

sin cos , 1 19.975 rad/s

s

n

n

n

n

n

p

th d d d n

t

k mc m

r

H i Ek r i r i

x t F H i t E t

x t e A t A t

x t e A

ζω

ζω

ωζ ω

ω ω

ωζ

ω ω θ

ω ω ω ω ζ−

−

= =

= =

= =

= = = − −⎡ ⎤− + × − + × ×⎣ ⎦

= + = − −

= + = − =

= ( ) ( ) ( )( ) ( ) ( )

( ) ( )

2 0

1 2 1 2

0

in cos cos

sin cos cos sin

sin

n n

d d

t tn d d d d d d

t A t F H i t

x t e A t A t e A t A t

F H i t

ζω ζω

ω ω ω ω θ

ζω ω ω ω ω ω ω

ω ω ω θ

− −

+ + +

= − + + −

− +





ωresponse finally becomes ω, and in phase



ωresponse finally becomes ω, and out of phase



F0ωnt/(2k)

( ) ( )

( ) 0

In case of 0 and , the guess solution of the form

cos sin is invalid. This is becauseit has the same form as the homogeneous solution.

The correct particular solution is

ni t

np

x t X i e A t B t

Fx t

ω

ζ ω ω

ω ω ω

ω

= =

= = +

= sin .2 n

t tk

ω


Beat when the driving frequency is close to natural freq.


( ) ( )0 00 2 2

2 2 20 0 1 0 0

2 20

The total solution can be arranged in the form

sin cos cos cos

2 sin tan sin sin2 2

If the system is at rest in

n n nn n

n n n nn

n n

v fx t t x t t t

x v x ft t tv

ω ω ω ωω ω ω

ω ω ω ω ω ωωω ω ω

−

= + + −−

+ ⎛ ⎞ − +⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) 02 2

02 2

the beginning,2 sin sin

2 2

The response oscillates with frequency inside2

2the slowly oscillated envelope sin2

The beat frequency is

n n

n

n

n

n

n

fx t t t

f t

ω ω ω ωω ω

ω ω

ω ωω ω

ω ω

− +⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠+

−⎛ ⎞⎜ ⎟− ⎝ ⎠

∴ −


Beat when the driving frequency is close to natural freq.



3.2 Frequency Response Function


( ) ( )2

The core of the particular solution to the harmonic function is1 ; frequency response function

1 2

It specifies how the system responds to harmonic excitation.As a standard, we normalize the

H ik r i r

ωζ

=− +

( ) 2

frequency response function1 and then study how it varies as the

1 2excitation frequency and system parameters , vary.It is indeed more convenient since we already normalized the frequen

n

G ir i r

ωζω ζ ω

=− +

( )( )

cy;/ . So we can now study its variation to and .

For the fixed damping ratio, we plot with varies.

has both magnitude and phase magnitude and phase plot.

Then we repeatedly evaluate

nr rG i r

G i

G i

ω ω ζω

ω

=

⇒

( ) by varying .ω ζ



Frequency response plot(Bode diagram)

( )( ) ( )2 22

1

1 2H i

r rω

ζ=

− +

12

2tan1

rrζθ − −⎛ ⎞= ⎜ ⎟−⎝ ⎠


Resonance is defined to be the vibration response atω=ωn, regardless whether the damping ratio is zero.At this point, the phase shift of the response is –π/2.

The resonant frequency will give the peak amplitude for the response only when ζ=0. For ,the peak amplitude will be at , slightly before ωn.For , there is no peak but the max. value of theoutput is equal to the input for the dc signal (of course, for this normalized transfer function).


21 2nω ω ζ= −0 1/ 2ζ< <

1/ 2ζ ≥



Ex. 3 Consider the pivoted mechanism with k=4x103 N/m,l1=0.05 m, l2=0.07 m, l=0.10 m, and m=40 kg.The mass of the beam is 40kg which is pivotedat point O and assumed to be rigid. Calculate cso that the damping ratio of the system is 0.2.Also determine the amplitude of vibration of thesteady-state response if a 10 N force is appliedto the mass at a frequency of 10 rad/s.


( ) ( )

( )

12 2 1 1

2 212 1

2

12 2

0.1 10cos10 0.5 0.0049 59.050.004915.37, 0.2 , 627.3 Ns/m

2 0.5

O O

nn

l lM I Fl mgl Mg c l l k l l

l l l lml M M

t cc c

θ θ θ θ θ

θ

θ θ θ

ω ζω

ω

−⎛ ⎞⎡ ⎤= − − − −⎜ ⎟⎣ ⎦ ⎝ ⎠⎡ ⎤+ −⎛ ⎞= + +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦× = + +

= = = =× ×

=

∑

( ) ( )( )

10, 0.65061 0.02677 24.268

59.05 0.5767 0.26

0.02677 cos 10 0.424ss

r

H ii

t

ω

θ

=

= = − °+

= −



Ex. 4 A foot pedal for a musical instrument is modeledas in the figure. With k=2000 kg/s2, c=25 kg/s,m=25 kg, and F(t)=50cos2πt N, compute thesteady-state response assuming the system startsfrom rest. Use the small angle approximation.



( ) ( )

( ) ( )

2

2

0.15 0.05 0.05 0.05 0.1 0.15

5 1003.75 50cos 2 , positive CW6 3

Find the parameters2.98, 0.0373, 2 , 2.108

1 0.0087 177.41 2

since 0, the transie

O O

n

M I F k c m

t

r

H ik r i r

θ θ θ θ

θ θ θ π

ω ζ ω π

ωζ

ζ

⎡ ⎤= × − × − × = ×⎣ ⎦

+ + =

= = = =

= = − °− +

≠

∑

( ) ( ) ( )0

nt response will die out

cos 0.435cos 2 3.096ss F H i t tθ ω ω θ π= + = −



3.3 Applications

3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications


3.3 Applications

( ) ( )

( ) ( )

2

2 22

2 22

2

2

2

measured acc. 10 1actual acc. 9.81 1 2

1 2 0.962

From the problem statement, 628 rad/s, 1 628 rad/s

11

758 rad/s , 0.5621

5745.6 N/m, 8.49 Ns/

n

d n

d

dn

n

zy r r

r r

r

k cm m

k c

ω

ζ

ζ

ω ω ω ζωω ζ

ωω ζωζ

= = =− +

− + =

= = − =

= =−

= = = = =−

= = m


3.4 Periodic Excitation

A periodic function is any function that repeats itself in time, called period T.

It is more general than the harmonic function. Here, wewill find the response to the input that is a periodic function. The idea is to decompose that periodic inputinto the sum of many harmonics. The response, by thesuperposition principle of linear system, is then the sum of the responses of individual harmonic. The response of a harmonic function was studied in section 3.1


( ) ( )f t f t T= +


Fourier found the way to decompose the periodicfunction into sum of harmonic functions (sine & cosine)whose frequencies are multiples of the fundamental frequency. The fundamental frequency is the frequency of the periodic function.



Fourier series


( ) ( )

( )

( )

( ) 0

00 0 0

1

00

00

Fourier series in real form:2cos sin ,

2Fourier coefficients:

2 cos , 0,1, 2,

2 sin , 1, 2,3,

Fourier series in complex form:

n nn

T

n

T

n

in tn

n

af t a n t b n tT

a f t n t dt nT

b f t n t dt nT

f t C e ω

πω ω ω

ω

ω

∞

=

∞

=−∞

= + + =

= =

= =

=

∑

∫

∫

…

…

( ) 0

0

0

2,

Fourier cofficients (complex):

1 , , 2, 1,0,1, 2,T

in tn

T

C f t e dt nT

ω

πω

−

=

= = − −

∑

∫ … …


Some properties of Fourier series


) ( )) ( )

) ( )

) ( ) ( ) 0

0

01

1 If is an even function, 0.

2 If is an odd function, 0.

3 is the average value of over one period.2

4 If is real, 2Re

n

n

in tk k n

n

f t b

f t aa f t

f t C C f t C C e ω∞

−=

=

=

⎛ ⎞= ⇒ = + ⎜ ⎟⎝ ⎠∑


Frequency spectrum tells how much each harmoniccontributes to the periodic function .

Plot of the amplitude of each harmonic vs. its frequencyis the (discrete) frequency spectrum.


( )f t

( )

2 20

0

In real form, the harmonic at has the amplitude

In complex form, the harmonic at has the amplitude 2 Re n n

n

n a b

n C

ω

ω

+




Superposition principle of linear system



Response to harmonic excitation




( )( )

( )

( )

( )

0

0

0

0

0

0 2

0 0

01

00

1

From section 3.1, 1where

1 2

2 Re

by superposition, 2Re

in tn

in tss n

n n

in tn

n

in tss n

n

mx cx kx F t C e

x C H in e

H inn nk i

mx cx kx F t C C e

Cx C H in ek

ω

ω

ω

ω

ω

ωω ωζω ω

ω

∞

=

∞

=

+ + = =

=

=⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟− +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞+ + = = + ⎜ ⎟⎝ ⎠⎛= +⎝

∑

∑ ⎞⎜ ⎟

⎠



response frequency spectrumexcitation frequency spectrum

system frequency response


Ex. Calculate the response of a damped system tothe periodic excitation f(t) depicted in the figureby means of the exponential form of the Fourierseries. The system damping ratio is 0.1 and thedriving frequency is ¼ of the system natural freq.




( )

( )

( )

( )

( )

0

0 0 0

0

/2

00 0 /2

odd

Expand as sum of harmonic series

1 1 2 ,

0, even1 1 2 , odd

2 2

in tn

n

T T Tin t in t in t

nT

nn

i nin t

n

f t

f t C e

C f t e dt Ae dt Ae dtT T T

niAC i An n

n

i A Af t e en n

ω

ω ω ω

ωω

πω

ππ

π π

∞

=−∞

− − −

=

=

⎡ ⎤= = + − =⎢ ⎥

⎣ ⎦=⎧

⎪⎡ ⎤= − − = ⎨⎣ ⎦ − =⎪⎩

∴ = − × = ×

∑

∫ ∫ ∫

∑

( )

( )( )

( ) ( ) ( )

( ) ( )

0 20odd

1,3,

02

2

1222 2

01,3,

4 1 sin

1 , 0.1, 1 2 4

11 / 4 0.05

1 0.05, tan1 0.251 0.25 0.05

4 1 sin

t

nn

nn n

n

n n

ss n nn

A n tn

n nG i rr i r

G in i n

nG Gnn n

Ax t G n t Gn

π

ωπ

ωωω ζζ ω ω

ω

ωπ

⎛ ⎞ ∞−⎜ ⎟⎝ ⎠

==

−

∞

=

=

= = = = =− +

=− +

−= =

−⎡ ⎤− +⎣ ⎦

∴ = +

∑ ∑

∑

…

…




Ex. The cam and follower impart a displacement y(t)in the form of a periodic sawtooth function to thelower end of the system. Derive an expressionfor the response x(t) by means of Fourier analysis.




( )

( )

( )

( ) ( )

( )

0

0

2 1

1 21 2 2

0

0

FBD and assume

, , 2

Write in the Fourier series expansion

2, , , 0

1 1

x x

nn

in tn

n

Tin t

n

y x

F ma mx k y x k x cx

k k cmx cx k k x k ym m

y t

Ay t C e y t B t t TT T

C y t e dt BeT T

ω

ω

ω ζω

πω∞

=−∞

− −

>

⎡ ⎤= = − − −⎣ ⎦

++ + + = = =

= = = + ≤ ≤

= =

∑

∑

∫

( )

0 0

0 0

2

2 2

22

00

0

1

Integration formula: and 1

2 1 , 02 22

2

T Tin t in t

ax axax ax

TT

in t in tT T

n

Adt te dtT T

e ee dx c xe dx ax ca a

B e A e iAC in t nT T T nin inT T

AC B

ω ω

π π

ππ ππ

−

− −

+

= + = − +

⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎛ ⎞⎢ ⎥= + − − = ≠⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠⎛ ⎞⎢ ⎥− −⎢ ⎥⎜ ⎟⎣ ⎦ ⎝ ⎠⎣ ⎦

= +

∫ ∫

∫ ∫



( ) ( )

( )

( )( )

( )( )

( )

0 01

01

21 2

2

0 01 2

22

0 01 2

2 Re cos sin2 2

1 sin2

1Frequency response 1 2

1

1 2

1

1 2

n

n

n

n

n n

n

n n

A iAy t B n t i n tn

A Ay t B n tn

H ik k r i r

H in nk k i

Hn nk k

ω ωπ

ωπ

ωζ

ωω ωζω ω

ω ωζω ω

∞

=

∞

=

⎡ ⎤∴ = + + +⎢ ⎥⎣ ⎦

= + −

=⎡ ⎤+ − +⎣ ⎦

=⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥+ − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

=⎛ ⎞⎛ ⎞⎜ ⎟+ − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑

∑

( ) ( )

0

12

20

2 20

11 2

2, tan

1

1 sin2

nn

n

ss n nn

n

Hn

k k AAx t B H n t Hk k n

ωζω

ωω

ωπ

−

∞

=

⎛ ⎞− ⎜ ⎟

⎝ ⎠=⎛ ⎞⎛ ⎞⎛ ⎞ − ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

⎛ ⎞= + − +⎜ ⎟+ ⎝ ⎠∑




















3.5 Non-periodic Excitation

Harmonic and steady-state excitation and response are conveniently described in the frequency domain. Fordeterministic non-periodic excitation and response, timedomain technique is more suitable.

We cannot find the repeated pattern that lasts forever (both in the past & future) for the non-periodic excitation.

System response to the unit impulse, called the impulse response, will be first studied. Then, this fundamentalresponse will be used to synthesize the response of theLTI system to arbitrary excitation.





Impulse

The unit impulse, or Dirac delta function, is defined as

This means that the unit impulse is zero everywhereexcept in the neighborhood of t=a. Since the areaunder the graph δ-t is 1, the value of is very large in the vicinity of t=a.The impulse of magnitude , which may represent alarge force acting over a short period, can be written as


( )

( )

0 for

1

t a t a

t a dt

δ

δ∞

−∞

− = ≠

− =∫

( )t aδ −

F̂

( ) ( )ˆF t F t aδ= −




The unit impulse has a useful property called the“sampling property”. Multiplying a continuous function

by , and integrating w.r.t. time:

which is just the value of f(t) at t=a. This is a way inevaluating integrals involving with impulse.


( )f t ( )t aδ −

( ) ( ) ( ) ( ) ( )f t t a dt f a t a dt f aδ δ∞ ∞

−∞ −∞

− = − =∫ ∫


Impulse response

The impulse response, h(t), is the response to the unitimpulse, δ(t), applied at t=0 with zero initial conditions.The impulse response is very important since it contains all the system characteristics and can be used to findthe response to arbitrary excitation of LTI system via the convolution integral theorem.

The impulse response of a 1 DOF MBK system mustsatisfy

subject to i.c.


( ) ( ) ( ) ( )mh t ch t kh t tδ+ + =

( ) ( )0 0, 0 0h h= =



( )

( ) ( )0 0

0

Get rid of the impulse function byintegrating over the duration 0, of the impulse

1

Take limit as 0 and apply the i.c.to evaluate the integral on the left hand side:

lim

mh ch kh dt t dt

m

ε ε

ε

ε

δ

ε

→

+ + = =

→

∫ ∫

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( )

000

00 00

00 00

lim 0 0, assuming is not continuous

lim lim 0 0, assuming is continuous

lim lim 0 0, assuming is continuous

0 1

h t dt mh t mh h t

ch t dt ch t ch h t

kh t dt gh t h t

mh

εε

ε

εε

ε ε

εε

ε ε

+

→

+

→ →

→ →

+

= = ≠

= = =

= =

∴ =

∫

∫

∫


Therefore, the effect of a unit impulse at t=0 is toproduce equivalent initial velocity (impulse-momentum)

Now, we are ready to find the impulse response. Theequivalent system is a homogeneous system with i.c.

If the system is underdamped, the impulse response is

Note that the above i.c. is not the actual i.c.


( )0 1/h m+ =

( ) ( )0 0, 0 1/h h m= =

( )1 sin , 0

0, 0

ntd

d

e t tmh t

t

ζω ωω

−⎧ ≥⎪= ⎨⎪ <⎩



Impulse response of underdamped system


( )x t

Linear Time Invariant (LTI) system has the characteristic that the shape of the response will not be influenced bythe time the input is applied to the system. That is


LTI system( )f t

LTI system( )f t a− ( )x t a−

Hence if the impulse is applied at t=to, the response is

( )( ) ( )0

0 0

0

1 sin ,

0,

n t td

d

e t t t tmh t

t t

ζω ωω

− −⎧ − ≥⎪= ⎨⎪ <⎩


Total response of underdamped MBK with i.c. x(a)=x0and v(a)=v0 subject to the impulse force


( )F̂ t aδ −

( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

1 2

1 2

1 2

ˆ sin cos sin

ˆ sin cos ,

ˆsin cos

n n

n

n

h p

t a t ad d d

d

t ad d

d

t an d d

d

x t x x

Fe A t a A t a e t am

Fe A t a A t a t am

Fx t e A t a A t am

ζω ζω

ζω

ζω

ω ω ωω

ω ωω

ζω ω ωω

− − − −

− −

− −

= +

= − + − + −

⎧ ⎫⎛ ⎞⎪ ⎪= + − + − ≥⎜ ⎟⎨ ⎬⎪ ⎪⎝ ⎠⎩ ⎭

⎧ ⎫⎛ ⎞⎪ ⎪= − + − + −⎜ ⎟⎨ ⎬⎪ ⎪⎝ ⎠⎩ ⎭

+ ( ) ( ) ( )1 2

ˆcos sinn t a

d d d dd

Fe A t a A t am

ζω ω ω ω ωω

− − ⎧ ⎫⎛ ⎞⎪ ⎪+ − − −⎜ ⎟⎨ ⎬⎪ ⎪⎝ ⎠⎩ ⎭


Total response of underdamped MBK with i.c. x(a)=x0and v(a)=v0 subject to the impulse force


( )F̂ t aδ −

( ) ( )

( ) ( ) ( ) ( ) ( )

0 0 1 2

0 2

0 2 1

2 0 1 0 0

0 0 0

Apply i.c. and to solve for and :

ˆ

ˆ1 and

1 sin cos , n

n dd

nd

t an d d

d

x a x x a v A Ax A

Fv A Am

FA x A x vm

x t e x v t a x t a t aζω

ζω ωω

ζωω

ζω ω ωω

− −

= =

=

⎛ ⎞= − + +⎜ ⎟

⎝ ⎠⎛ ⎞

∴ = = + −⎜ ⎟⎝ ⎠

⎧ ⎫∴ = + − + − ≥⎨ ⎬

⎩ ⎭


Arbitrary Excitation

Ideally, arbitrary excitation can be expressed as linearcombinations of simpler excitations. The simplerexcitations are simple enough that the responseis readily available. This concept is exactly used byFourier.

Now, the idea is to regard the arbitrary excitation as asuperposition of impulses of varying magnitude andapplied at different times. It is used when the excitationcan be easily described in time domain.



Consider the excitation F(t). We can imagine that it is constructed from infinite impulses at different times.



Convolution integral theorm


( )( ) ( )

( )

Focus on the time interval , at whichthe impulse of magnitude is acting. This

shifted impulse can be written as .The response of the LTI system to this particularimpulse is

,

tF

F t

x t F

τ τ ττ τ

τ τδ τ

τ

< < + Δ

Δ

Δ −

Δ = ( ) ( )( ) ( ) ( )

( )( ) ( ) ( )

( ) ( ) ( )0

Since by sampling property ,

and the system is linear, the response to is

In the limit as 0, .t

h t

F t F t

F t

x t F h t

x t F h t d

τ

τ

τ τ τ

τ τδ τ

τ τ τ

τ τ τ τ

Δ −

= Δ −

= Δ −

Δ → = −

∑

∑

∫


Convolution integral theormThe response of the arbitrary excitation is thesuperposition of shifted impulse responses.


Interpretationfor the wholerange of time; t



( ) ( )To obtain from , we need to carry outtwo operation; shifting and folding. This is another interpretationfor the specific time t. The figures show the steps in evaluating the convolution.

If we

h t hτ τ−

( ) ( ) ( )( ) ( ) ( )

( ) ( )

0

0

define a new variable , then and . With the change of the integration limits,

That is the convolution is symmetric in and .

To decide which formula to u

t

t

t td d

x t F t h d F t h d

F t h t

λ τ τ λτ λ

λ λ λ τ τ τ

= − = −= −

= − − = −∫ ∫

( ) ( )

( ) ( )

se depends on the nature of and .

It is obvious that if the excitation or the impulse response is too complicated, we may be unable to evaluate the closed formsolution of the convolution inte

F t h t

F t h t

gral. The excitation may not at allbe written as functions of time. In these cases, the integration mustbe carried out numerically.



Interpretation for the specific time; t


Ex. Determine the response of the underdamped MBKto the unit step input.


u(t)1

0 t



( ) ( ) ( )

( ) ( ) ( )

( ) ( )( ) ( ) ( )

( )

( ) ( )

0

0

and is the system impulse responseshifted by and mirrored about the vertical axis.If 0, 0 because of no overlap

If 0,

0, 0

, 0

Let

t

t

x t F h t d

F u h tt

t F h t

t F h t h t

x t t

x t h t d t

t

τ τ τ

τ τ τ

τ τ

τ τ τ

τ τ

= −

= −

< − =

> − = −

∴ = <

∴ = − >

−

∫

∫

( ) ( )

( )

0 0

. Hence

1 sin

Substitute sin and use 2

1 1 cos sin , 0

n

d d

n

t t

dd

i i axax

d

t nd d

d

d d

x t h d e dm

e e ee dx ci a

x t e t t tk

ζω λ

ω λ ω λ

ζω

τ λ τ λ

λ λ ω λ λω

ω λ

ζωω ωω

−

−

−

= = −

∴ = =

−= = +

⎡ ⎤⎛ ⎞∴ = − + >⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

∫ ∫

∫


Ex. Find the undamped response for the sinusoidalpulse force shown using zero i.c.




( ) ( ) ( )

( )

( )

( )

( ) ( )( )

0

0 00 0

2sin sin2

1 sin

is the system impulse response shifted by and mirrored about the vertical axis.If 0, 0 because of no overlap

0, 0

t

nn

x t F h t d

F F FT T

hm

h t t

t F h t

x t t

τ τ τ

π πτ τ τ

τ ω τω

τ

τ τ

= −

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

=

−

< − =

∴ = <

∫



( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

0 00

0

00 0

002

1If 0 , sin sin

sin sin

1using the relation sin sin cos cos and some arrangements2

sin sin , 0 where1

nn

t t

nn

n

t T F h t F tT m

Fx t F h t d t dm T

Fx t t r t t Tk r

πτ τ τ ω τω

πτ τ τ τ ω τ τω

α β α β α β

ω ω

⎛ ⎞< < − = × −⎜ ⎟

⎝ ⎠⎛ ⎞

= − = −⎜ ⎟⎝ ⎠

= − − +⎡ ⎤⎣ ⎦

∴ = − < <−

∫ ∫

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) [ ] ( ) ( ){ }

0

0

2

0

00 0

00 0 02

, ,

If ,

sin sin sin sin , 1

superposition of the out-of-phase shifted sine trains

nn

Tt t

t T

n n

r k mT

t T x t F h t d F h t d F t h d

Fx t t r t t T r t T t Tk r

π ωω ωω

τ τ τ τ τ τ τ τ τ

ω ω ω ω

−

= = =

> = − = − = −

∴ = − − − − − >⎡ ⎤⎣ ⎦−

∫ ∫ ∫

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