THE TIME-DOMAIN RESPONSE OF A SINGLE-DEGREE-OF-FREEDOM
SYSTEM SUBJECTED TO AN IMPULSE FORCE
Revision C
By Tom Irvine
Email: [email protected]
December 19, 2013
Consider a single-degree-of-freedom system.
f(t)
y
m
c
k
Where
m
=
mass
c
k
y
f(t)
=
=
=
=
viscous damping coefficient
stiffness
displacement of the mass
applied force
Note that the double-dot denotes acceleration.
The free-body diagram is
f(t)
y
m
ky
cy
1
Summation of forces in the vertical direction
 Fmy
(1)
my   cy  ky  f ( t )
(2)
my  cy  ky  f ( t )
(3)
Divide through by m,
c
k
1
y    y    y   f ( t )
m
m
m
(4)
By convention,
(c / m)  2n
(5)
(k / m)  n 2
(6)
where
n

is the natural frequency in (radians/sec)
is the damping ratio
By substitution,
1
y  2n y  n2 y  f ( t )
m
(7)
Now apply an impulse force at t=0 via a pair of step functions.
f (t )  F u(t )  u(t  )
(8)
where  is a very small time step.
2
The governing equation becomes
y 2 n y   n2 y 
1
F u ( t )  u ( t  )
m
(9)
Now consider that the system undergoes oscillation with  < 1.
Take the Laplace transform of each side.


1

2
L y 2 n y   n y  L F u ( t )  u ( t  )
m

(10)
s 2 Y(s)  sy(0)  y(0)
 2 n sY (s)  2 n y(0)
F 1 1

 exp(  s)

m s s

2
  n Y(s) 
(11)
s
2  2 s   2 Y(s)  s  2 y(0)  y(0) 
n
n
n

s
2  2 s   2 Y(s) 
n
n

F 1 1

 exp(  s)

m s s

(12)
F 1 1

 exp(  s)  s  2n y(0)  y(0)

m s s

(13)
s 2  2n s  n  s  n 2  n 2  n
2
2

s 2  2n s  n  s  n 2  n 1   2
2
2
3

(14)
(15)
Let
d   n 1   2
(16)
Substitute equation (16) into (15).
s 2  2n s  n  s  n 2  d
2
s   
n
2

2   2 Y(s) 
d
(17)
F 1 1

 exp(  s)  s  2n y(0)  y(0)

m s s

(18)
Y(s) 




exp(  s)
F 
1
 F 






m s  s   2   2  m s  s   2   2 
n
n
d 
d 






s  2 n
1





y(0)  
y (0)

2
2
 s   n 2  d 
 s   n 2  d 
(19)
Divide the right-hand-side of equation (19) into three parts.
(20)
Y(s)  Y0 (s)  Y1 (s)  Y2 (s)
where

F 
Y0 (s) 

ms 




2
s  n 2  d 
1
(21)
4


exp(  s)
F 

Y1 (s)  


2
m s  s   2   
n
d 

(22)




s  2n
1




Y2 (s)  
y(0)  

y(0)
2
2
2
2
 s  n    
 s  n    
d 
d 


(23)
Consider Y0(s) from equation (21).

F 
Y0 (s) 

ms 




s  n 2  d2 
1
(24)
Expand into partial fractions using Reference 1.
 F 
 
 m s 

F


2
2
s 2  2 n s   n 
m n
1
1 
 
s 
 F 



2 
 m  n 



2
s  n 2  d 
s  n
 F  
 n


2  d
 m  n 







2
s  n 2  d 
d
(25)
5
The inverse Laplace transform per Reference 1 is
y 0 (t) 
F
2
m n
u(t) 


n
exp   n t  cosd t  
sin d t  , t  0
2
d


m n


F



n



y 0 (t) 
u ( t )  exp  n t  cosd t  
sin d t   , t  0
2
d


m n 



F
(26)
(27)
Now consider the term


exp(  s)
F 

Y1 (s)  


m s  s   2   2 
n
d 

(28)
Expand into partial fractions using Reference 1.
 F 
 
 m s 

F 1 
exp(  s)
 
2
2 
s 2  2 n s   n 
m n  s 
exp(  s)
 F 



2 
 m  n 


 exp(  s)
s  n 2  d2 
s  n
 F  
 n


2  d
 m  n 






 exp(  s)
s  n 2  d2 
d
(29)
6
Take the inverse Laplace transform from Appendices A, B & C.
y1 ( t )  
 F 
 exp   t    cos   t   
u t     
n

 d

2
2
m n
m

n 

F
 F  
 n


2  d
 m  n 

 exp    t    sin   t   
n
d


for t  
(30)
y1 ( t ) 




 
   n









u
t



exp



t


cos

t


  d
n
  d
2 
 
m  n 

F


 sin   t   

d



for t  
(31)
The inverse Laplace transform from Reference 2 for the natural response is
  y(0)  y(0) 
y 2 ( t )  y(0)exp  n t  cosd t    n
 exp  n t sin d t 
d


7
(32)


  
y 2 ( t )  y(0) exp  n t cosd t    n  sin d t 


 d 
 1 
 y (0) 
 exp  n t sin d t 

d


(33)
The final displacement solution is
y( t )  y 0 ( t )  y1 (t )  y 2 ( t )
(34)


  
yt   y(0) exp   n t cosd t    n  sin d t 


 d 
 1 
 y (0) 
 exp   n t sin d t 
 d 



 


u ( t )  exp   n t  cosd t   n sin d t   ,

2
d


m  n 


F
0t
(35)
8


  
yt   y(0) exp   n t cosd t    n  sin d t 


 d 
 1 
 y(0) 
 exp   n t sin d t 
 d 



 
n








u
(
t
)

exp



t
cos

t

sin

t
n
d
d 
2



d
m  n 


F



n




 u t     exp   n t   cos d t    
sin  d t    ,

 d
2 
 

m  n 
F
t
(36)
The response for a Dirac Delta impulse is derived in Appendix D.
References
1. T. Irvine, Partial Fractions in Shock and Vibration Analysis, Revision G,
Vibrationdata, 2011.
2. T. Irvine, Table of Laplace Transforms, Revision I, Vibrationdata, 2011.
3. T. Irvine, Free Vibration of a Single-Degree-of-Freedom System, Revision B,
Vibrationdata, 2005.
9
APPENDIX A


F
 F 1

L1 
exp(  s) 
u t   
2 s
2


m

m

n
n


(A-1)
APPENDIX B
Consider the generic term


F(s)  




 exp(  s)
2
2
s     
s
(B-1)
Shift in the s-plane.
ŝ  s  


F(s)  


(B-2)

 ŝ     

F(s)  
 exp   ŝ   
2
2

 ŝ   

(B-3)

 ŝ     

F(s)  
exp   ŝexp  
2  2 


ŝ


(B-4)



  

exp   ŝexp    
exp   ŝexp  

2  2 


ŝ 2   2 
ŝ



(B-5)
ŝ
10
   
f ( t )  exp    exp   t  cos  t    
 exp    exp   t  sin   t   ,
  
t 
(B-6)


   
f ( t )  exp    exp   t  cos  t     
sin   t   

  


(B-7)
t
f(t) = 0
t<ρ
(B-8)
Note that the exp   t  term accounts for the phase shift.
Recall the term
 F 

Y12 (s)  

2 
 m  n 



 exp(  s)
2
2
s  n   d 
s  n
(B-9)
Compare the term in equation (B-9) to the generic term from (B-1).


F(s)  




 exp(  s)
2
2
s     
s
(B-10)
11
The inverse Laplace transform of equation (B-9) is
 F 
 exp    exp   t  cos   t   
y12 ( t )  
n
n

 d

2
 m n 
(B-11)
 F 
 exp   t    cos   t   
y12 ( t )  
n

 d

2
m

n 

(B-12)
12
APPENDIX C
Consider the generic term


F(s)  




 exp(  s)
2
2
s     

(C-1)
Shift in the s-plane.
ŝ  s  


F(s)  


(C-2)


 exp   ŝ   
2
2
ŝ   

(C-3)


 

F(s)  
 exp   ŝexp  
2
2


ŝ




(C-4)

 
f ( t )    exp    exp   t  sin   t  
 
t 
(C-5)
 
f ( t )    exp    t   sin   t  
 
t 
(C-6)
13
t<ρ
f(t) = 0
(C-7)
Note that the exp   t  term accounts for the phase shift.
 F  
 n
Y13 (s)  

2  d
 m  n 







 exp(  s)
2
s  n 2  d 
d
(C-8)
Compare with the generic term


F(s)  




 exp(  s)
2
2
s     
(C-9)


 

F(s)  
 exp   ŝexp  
2
2

 ŝ   

(C-10)

 
f ( t )    exp    exp   t  sin   t  
 
t 
(C-11)
 F  
 n
y13 ( t )   

2  d
 m  n 

 exp    t    sin   t   
n
d


t
(C-12)
14
APPENDIX D
Dirac Delta Impulse
1
y  2n y  n2 y  f ( t )
m
(D-1)
Let Î = total impulse
f ( t )  Î ( t )
(D-2)
.
y 2 n y   n2 y 
Î
( t )
m
(D-3)
s 2 Y(s)  sy(0)  y(0)
 2 n sY (s)  2 n y(0)
2
  n Y(s) 
Î
m
(D-4)
s

2  2 s   2 Y(s)  s  2 y(0)  y (0) 
n
n
n
s   
n

2   2 Y(s) 
d
Î
m
(D-5)
Î
 s  2n y(0)  y(0)
m
(D-6)
15
Y(s) 


Î 
1



m  s   2   2 
n
d 





s  2 n
1





y(0)  
y (0)

2
2
 s   n 2  d 
 s   n 2  d 
(D-7)
The displacement is


  
yt   y(0) exp   n t cosd t    n  sin d t 


 d 
 1 
 y (0) 
 exp   n t sin d t 
 d 

Î  1 

 exp   n t sin d t 
m  d 
(D-8)
The displacement for zero initial conditions is
yt  
Î  1 
  exp  n t sin d t 
m  d 
(D-9)
The impulse response function is
h d (t) =
1
exp  n t sin d t 
md
16
(D-10)
The corresponding Laplace transform is
H d (s) =
 1 

1
1
1

 

m  s 2  2n s  n 2  m  s  n 2  d 2 
(D-11)


  
y t    n y(0) exp   n t cosd t    n  sin d t 


 d 
 y(0) exp   n t  d sin d t   n cosd t 
  
 y (0)  n  exp   n t sin d t 
 d 
 y (0) exp   n t  cosd t 

Î   n 

 exp   n t sin d t 
m  d 

Î
exp   n t  cosd t 
m
(D-12)
  2 2

n    sin  t 
y t    y(0) exp   n t 
d
d
 d




 y (0) exp   n t cosd t   n sin d t 
d






Î
exp   n t cosd t   n sin d t 
m
d


(D-13)
17
  2 2   2 
n
d  sin  t 
y t    y(0) exp   n t 
d
2


d



 y(0) exp   n t cosd t   n sin d t 
d






Î
exp   n t cosd t   n sin d t 
m
d


(D-14)
The velocity for zero initial conditions is
y t  



Î
exp  n t cosd t   n sin d t 
m
d


(D-15)
The impulse response function is
h v (t) =


1

exp(- n t ) cosd t   n sin d t 
d
m


(D-16)
The corresponding Laplace transform is
H v (s) =
 1 

1
s
s

 

m  s 2  2n s  n 2  m  s  n 2  d 2 
18
(D-17)
The acceleration for zero initial conditions is
yt  
Î
( t )  2n y ( t )   n 2 y( t )
m
(D-18)
yt  
Î
m




 1 
 n


sin d t    n 2 
 ( t )  2n exp   n t cosd t  
 exp   n t sin d t  
d




 d 


(D-19)
yt  
Î
m


2n 2  n 2  






(
t
)

exp



t

2

cos

t

sin d t 


n
n
d



d








(D-20)
yt  
Î
m





n 2

2  1 sin  t 







(
t
)

exp



t

2

cos

t

2



n
n
d
d
d









(D-21)
The Laplace transform is


Ya s  
2 2
2
2
Î 
  2n s  2  n   n 2   1 

1



2
2
m
s

2

s



n
n


(D-22)
Ya s  
 2n s   n 2 
Î 


1



2
2
m  s  2 s   
n 
n

(D-23)
19
Ya s  
2
2
 2n s   n 2 
Î 
 s  2n s   n


 2

2
2
2
m  s  2 s  
s  2n s   n 
n
n


(D-24)
Ya s  

Î 
s2


 2

2
m  s  2 s   
n 
n

(D-25)
The corresponding transfer function in terms of the Laplace transform is
H a (s) =
 1 

1
s2
s2

 

m  s 2  2n s  n 2  m  s  n 2  d 2 
20
(D-26)
APPENDIX E
Dirac Delta Impulse, Transmitted Force
Again, the impulse response function for displacement is
h d (t) =
1
exp  n t sin d t 
md
(E-1)
The impulse response function for velocity is
h v (t) =


1

exp(- n t ) cosd t   n sin d t 
d
m


(E-2)
The transmitted force f t (t) is
f t (t)  cy  ky
(E-3)
k 
c
f t (t) m  y  y 
m 
m
(E-4)
2
f t (t) m 2n y  n y


(E-5)
21
The impulse response function for transmitted force is



h ft ( t ) = 2n exp(- n t ) cosd t   n sin d t 
d


 2
 n exp  n t sin d t 
d
(E-6)


2 2 n 2
h ft ( t ) = exp(- n t )  2n cosd t  
sin d t 
d


 2

 exp  n t   n sin d t 
 d

(E-7)


  2 2 2  2 
n  sin  t 
h ft ( t ) = exp(- n t )  2n cosd t    n 
d
 d
d 






(E-8)
The final transmitted force impulse response function is

 1  2 2
2

h ft ( t ) = exp(- n t ) 2n cosd t    n 
 d




 sin  t 
d



(E-9)
Note that the transmitted force impulse response function is essentially the same as the
absolute acceleration impulse response function for the case of base excitation.
22