Ch. 3: Forced Vibration of 1-dof system



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x t

e

x

v

t

a

x

t

a

t

a

ζω

ζω



ω

ω

ζω



ω

ζω

ω



ω

ω



=

=



=



= −

+

+







=

=



+ −





=



+

+







Ch. 3: Forced Vibration of 1-DOF System

Arbitrary Excitation

Ideally, arbitrary excitation can be expressed as linear

combinations of simpler excitations.  The simpler

excitations are simple enough that the response

is readily available.  This concept is exactly used by

Fourier.

Now, the idea is to regard the arbitrary excitation as a

superposition of impulses of varying magnitude and

applied at different times.  It is used when the excitation

can be easily described in time domain.

3.5 Non-periodic Excitation



Ch. 3: Forced Vibration of 1-DOF System

Consider the excitation F(t).  We can imagine that it is 

constructed from infinite impulses at different times.

3.5 Non-periodic Excitation



Ch. 3: Forced Vibration of 1-DOF System

Convolution integral theorm

3.5 Non-periodic Excitation

( )


( )

(

)



( )

Focus on the time interval 

,  at which

the impulse of magnitude 

 is acting.  This

shifted impulse can be written as 

.

The response of the LTI system to this particular



impulse is

,

t



F

F

t

x t

F

τ

τ



τ

τ 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 →

=





Ch. 3: Forced Vibration of 1-DOF System

Convolution integral theorm

The response of the arbitrary excitation is the

superposition of shifted impulse responses.

3.5 Non-periodic Excitation

Interpretation

for the whole

range of time; t



Ch. 3: Forced Vibration of 1-DOF System

3.5 Non-periodic Excitation

(

)

( )



To obtain 

 from 


, we need to carry out

two operation; shifting and folding.  This is another interpretation

for the specific time t.  The figures show the steps in evaluating the convolution.

If we


h t

h

τ

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

t

d

d

x t

F t

h

d

F t

h

d

F t

h t

λ

τ



τ

λ

τ



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 form

solution of the convolution inte

F t

h t

F t

h t

gral.  The excitation may not at all

be written as functions of time.  In these cases, the integration must

be carried out numerically.



Ch. 3: Forced Vibration of 1-DOF System

3.5 Non-periodic Excitation

Interpretation for the specific time; t


Ch. 3: Forced Vibration of 1-DOF System

Ex.  


Determine the response of the underdamped MBK

to the unit step input.

3.5 Non-periodic Excitation

u(t)


1

0

t



Ch. 3: Forced Vibration of 1-DOF System

3.5 Non-periodic Excitation

( )

( ) (


)

( )


( )

(

)



( ) (

)

( ) (



)

(

)



( )

( )


(

)

0



0

 and 


 is the system impulse response

shifted 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 t

t

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

d

d

i

i

ax

ax

d

t

n

d

d

d

d

d

x t

h

d

e

d

m

e

e

e

e dx

c

i

a

x t

e

t

t

t

k

ζω λ


ω λ

ω λ


ζω

τ λ


τ

λ

λ λ



ω λ λ

ω

ω λ



ζω

ω

ω



ω



=

= −



=

=



=

=



+





=

+



>









Ch. 3: Forced Vibration of 1-DOF System

Ex.  


Find the undamped response for the sinusoidal

pulse force shown using zero i.c.

3.5 Non-periodic Excitation


Ch. 3: Forced Vibration of 1-DOF System

3.5 Non-periodic Excitation

( )

( ) (


)

( )


( )

(

)



( ) (

)

( )



0

0

0



0

0

2



sin

sin


2

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

n

n

x t

F

h t

d

F

F

F

T

T

h

m

h t

t

t

F

h t

x t

t

τ

τ τ



π

π

τ



τ

τ

τ



ω τ

ω

τ



τ

τ

=





=



=







=



<

=



=

<



Ch. 3: Forced Vibration of 1-DOF System

3.5 Non-periodic Excitation

( ) (

)

(



)

( )


( ) (

)

(



)

(

)



(

)

( )



(

)

(



)

0

0



0

0

0



0

0

0



0

2

1



If 0

,  


sin

sin


sin

sin


1

using the relation  sin

sin

cos


cos

 and some arrangements

2

sin


sin

,  0


  where

1

n



n

t

t

n

n

n

t

T

F

h t

F

t

T

m

F

x t

F

h t

d

t

d

m

T

F

x t

t

r

t

t

T

k

r

π

τ



τ

τ

ω



τ

ω

π



τ

τ τ


τ

ω

τ τ



ω

α

β



α β

α β


ω

ω





< <

=



×





=



=





=



+





=



< <



( )


( ) (

)

( ) (



)

(

) ( )



( )

(

)



[

]

(



)

(

)



{

}

0



0

2

0



0

0

0



0

0

0



0

2

  ,  ,  



If ,  

sin


sin

sin


sin

,  


1

superposition of the out-of-phase shifted sine trains



n

n

T

t

t

t T

n

n

r

k

m

T

t

T

x t

F

h t

d

F

h t

d

F t

h

d

F

x t

t

r

t

t T

r

t T

t

T

k

r

π

ω



ω

ω

ω



τ

τ τ


τ

τ τ


τ

τ τ


ω

ω

ω



ω

=



=

=

>



=

=



=



=





>











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