Modal Transient Analysis of a System Subjected to an Applied Force
via a Ramp Invariant Digital Recursive Filtering Relationship
Revision E
By Tom Irvine
Email: tomirvine@aol.com
February 9, 2012
_______________________________________________________________________________
This paper applies Smallwood’s methodology for base excitation in Reference 1 to the case of an
applied force. It also extends the method to multi-degree-of-freedom systems.
Ahlin & Brandt purported in Reference 2 to use the method given in the present paper, but they
omitted a derivation and the resulting filtering coefficients.
This paper fills the gap and gives accompanying Matlab scripts. The goals are to publicize the
method and to place the filter coefficients equations in the public domain.
Introduction
The purpose of this paper is to derive an efficient and accurate method for the modal transient
analysis of a dynamic system subjected to an external force or forces. The method will be a digital
recursive filtering relationship using the ramp invariant technique which models the slope between
adjacent points of the input force.
Assumptions
A modal analysis of the system has been previously performed. The system has been reduced to
uncoupled mass, damping and stiffness matrices per the method given in Reference 3, as well as in
common structural dynamics textbooks. The physical responses can then be recovered from the
modal responses after the transient analysis.
The initial conditions are zero. Note that the response to initial conditions can be solved exactly
using Laplace transforms, as needed.
Forcing Function
The forcing function may vary arbitrarily with time.
The forcing function may be either from measured data1 or from a synthesis.
1
Measured data should be collected using the proper sample rate and analog anti-aliasing filter.
1
The time step should be chosen so that there are at least ten points per cycle for the highest natural
frequency of interest. In other words, the sample rate should be at least ten times the highest
natural frequency of interest. This is an industry standard rule-of-thumb for time domain
calculations per Reference 4.
Smallwood wrote in his seminal paper that the ramp invariant method allows for analysis
frequencies approaching the Nyquist frequency, which in one-half the sample rate. But the x10
rule is still recommend in the present paper.
Note that the time step and corresponding sample rate are fixed for measured data as the data is
being collected. The time step can be shortened via interpolation or a cubic spline fit as a postprocessing step, but caution must be exercise in either case.
The time step can be set as small as needed for the case where a function is to be synthesized.
Alternatively, the number of modes included in analysis can be truncated if needed to meet the
rule-of-thumb. The highest frequency of interest is thus lowered as a trade-off.
Candidate Methods
The methods for the modal transient analysis are:
1. Convolution integral converted to a nested series for digital data
2. Convolution integral represented as digital cursive filtering relationship
3. Runge-Kutta fourth order method
4. Newmark-beta implicit numerical integration method
Convolution Integral
The Convolution method would yield an exact answer for the response if the system and the
forcing function could be analyzed in analog rather than digital form. This is impractical,
however.
The nested series representation of the Convolution integral is not commonly used because it is
numerically inefficient.2
The Convolution integral represented as digital cursive filtering relationship. The derivation is
performed using a Z-transform.
Convolution can also be performed in the frequency domain my by multiplying the Fourier
transform of the applied force by the frequency response function of the system. The inverse
Fourier transform of the product gives the response time history. A Fast Fourier transform (FFT)
and its corresponding inverse can be used to expedite the calculation. Note that there are some
potential error sources with this method such as leakage.
2
2
There are two Z-transform approaches: the impulse invariant and ramp invariant simulations. The
ramp invariant simulation is preferred because it adds a filtering term and has better accuracy at
frequencies approach the Nyquist frequency per Reference 1.
Note that either digital recursive filtering relationship requires a constant time step.
Newmark-beta Method
The Newmark-beta method is favored in structural dynamics books for multi-degree-of-freedom
systems, as given in References 5 and 6. It can accept a variable time step for the force input. Its
true strength is that it can be used for the direct integration of a system of uncoupled mass,
damping, and stiffness matrices. Obviously, it can be applied to an uncoupled system as well.
The Newmark-beta method is derived from the continuous time equation familiar to high school
physics students.
u  u t 
1 2
u t
2
(0)
It is typically implemented with   1 / 4 , thus assuming constant average acceleration over the
time step.
Runge-Kutta Method
The Runge-Kutta method extends the Taylor series method by estimating higher order derivatives
at points within the time step.
There are many types of Runge-Kutta algorithms. A common type which may be used for an
arbitrary forcing function is the Runge-Kutta fourth order method. This method is given in
Reference 7.
But the Runge-Kutta fourth order method may give unstable results3 for stiff systems with higher
natural frequencies.
The probability that instability will occur is difficult to predict. It depends on both the natural
frequency and the time step. The problem can be mitigated by using a smaller time step, but this
requires a longer analysis time.
If an instability occurs, then the analysis be repeated using a fewer number of modes for the case
of a multi-degree-of-freedom system.
3
“To infinity and beyond,” recalling Buzz Lightyear.
3
SDOF Equation of Motion
Consider a single-degree-of-freedom system.

x
m
c
k
Figure 1.
The variables are
m
c
k
x
f(t)
is the mass
is the viscous damping coefficient
is the stiffness
is the absolute displacement of the mass
is the applied force
Note that the double-dot denotes acceleration.
The free-body diagram is
f(t)
m
kx
Figure 2.
4
Summation of forces in the vertical direction
F 

mx
(1)
   cx  kx  f ( t )
mx
(2)
  cx  kx  f ( t )
mx
(3)
 c
 k
 1
x    x    x    f ( t )
 m
 m
 m
(4)
Divide through by m,
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,
x  2   n x   n2 x 
1
f (t)
m
(7)
Equation (7) does not have a closed-form solution for the general case in which f(t) is an arbitrary
function. A convolution integral approach must be used to solve the equation.
The solution method proceeds by finding a solution to the homogeneous form of equation (7). In
other words, the solution is found for f(t)=0.
x  2  n x   n2 x  0
(8)
Note that equation (8) is essentially the same as equation (A-1) in Reference 8, except that
equation (8) is expressed in terms of absolute displacement.
5
Displacement
The damped natural frequency d is
d   n 1 -  2
(9)
The displacement equation via a convolution integral is
x( t ) =
t
1
f( ) exp  n ( t - ) sin d ( t - )d
md 0

(10)
The corresponding impulse response function for the displacement is
ĥ d ( t ) =
1
exp  n t sin d t 
md
(11)
Further details regarding this derivation are given in Reference 9.
The corresponding Laplace transform from Reference 10 is
H d (s) =

1
1


m  s 2  2 s   2 
n 
n

(12)
The Z-transform for the ramp invariant simulation is
Ĥ d (z) =

1  z  12  
1


 Z  L1 
m  Tz  
 s 2 s 2  2n s  n 2





 


where T is the time step
(13)
6
Evaluate the inverse Laplace transform per References 10 and 11.

1
L1 
 s 2 s 2  2n s  n 2



2

n
1 
2
2 sin  t 








2



t

2

cos

t


1

2


n
n
n
d
d 
 n4 


d




(14)
The Z-transform becomes
Ĥ d (z) =


2
 z  12  
n


2
2
 1  2  sin d t  

 Z  2n  n t  exp  n t 2n cosd t  
4  Tz  

 
mn 
d
 


1

(15)
Let

2
n
d
 1  22 
(16)
  2n
(17)
7
The Z-transform is evaluated using the method in Reference 12.
Ĥ d (z) 
-   z  12   z 



4
m n  Tz   z  1
2
 n  z  12   T z 




4  Tz   z  12 
m n 




 z  12  
zz  exp  n T cosd T 



4
2
m n  Tz   z  2zexp  n T cosd T   exp  2n T 


 z  12  
zexp  n T sin d T 



4
2
m n  Tz   z  2zexp  n T cosd T   exp  2n T 
1
1
(18)
Ĥ d (z) 

  z  1   n2 
4 
 T 

m n
1


 z  12  
z  exp  n T cosd T 



4  T   z 2  2zexp   T cos T   exp  2 T 
m n 

n
d
n 


 z  12  
exp  n T sin d T 



4  T   z 2  2zexp   T cos T   exp  2 T 
m n 

n
d
n 
1
1
(19)
8

  z  1   n2 
4 
 T 

mn
1
Ĥ d (z) 

 z  12   z  exp   T cos T   exp   T sin  T 
n
d
n
d 


4  T 
2

z  2zexp  n T cosd T   exp  2n T 
mn 

1
(20)
  z     2 T 
n 

1
Ĥ d (z) 
4 
m n T 

1
4
m n T
exp  n T  sin d T    cosd T  

z  12  z 2 z2zexp
  T cos T   exp  2 T 

n
d
n

(21)
Let
  exp  n T  sin d T    cosd T 
(22)
  2exp  n T cosd T 
(23)
  exp  2n T
(24)
2
    n T
(25)
By substitution,
Ĥ d (z) 
Ĥ d (z) 
1
4
mn T
1
4
mn T
 z   
 2
1
4
mn T
z  12  z 2zz  


 z    z 2 z    
 z z   
1
4
mn T
9
(26)

z 2  2z  1 z 2zz  


(27)




 2

4
 z   z 2 z   
4
   z 2 z   
1
Ĥ d (z) 
m n T
1
m n T
1
4
m n T
1
4
m n T
 z z   
 2

 z z   
z 2  z 2z z  


 z  


 z 2 z   
 2z 
 z   


4  z 2  z   
m n T 

1
(28)
Ĥ d (z) 
  z 3   z 2  z 
1   z 2   z    




4
 mn4 T  z 2 z   
z 2 z  
mn T 
1
  z 3  z 2 
1   2z 2  2z 
1  z   







4
2
4
2
4
2
mn T  z z    mn T  z z    mn T  z z   
1
(29)
Ĥ d (z) 
 z 3   z 2  z   z 2  z    z 3  z 2  2z 2  2z  z  
Ĥ d (z) 

4
m n T z 2 z  

       2z 2      2  z      
4

m n T z 2 z  

10
(30)
(31)
Solve for the filter coefficients using the method in Reference 1.
b 0 z 2  b1 z  b 2

z 2  a1z  a 2
       2z 2     2  z     / mn4T
z 2 z  
(32)
Solve for a1.
a1    2exp  n T cosd T 
(33)
Solve for a2.
a 2    exp  2n T
(34)
Note that the a1 and a2 coefficients are common for displacement, velocity and acceleration.
Solve for b0.
4
b 0         2 / mn T 


(35)
4
b 0        2 / mn T 


(36)
exp  n T  sin d T    cosd T     n T  2 exp  n T  cosd T  2
2
b0 
4
mn T
(37)
exp  n T  sin d T    cosd T     n T 2exp  n T  cosd T  1
2
b0 
4
mn T
(38)
11
b0 
 2

n
2

exp  n T 
 1  2  sin d T   2n cosd T 


 d



4
m n T
2n   n T 4n exp  n T  cosd T  1
2

4
m n T
(39)
 2


2
exp  n T  n  1  2 2 sin d T    n T  2n exp  n T  cosd T  1


 d

b0 
4
m n T


(40)


exp  n T  n

b0 
 d

 1  22 sin d T  n T 2exp  n Tcosd T 1

3
m n T
(41)


n
2

2  1 sin d T   n T
2exp  n T  cosd T  1  exp  n T 

 d


b0 

3
m n T
(42)
12
Solve for b1.
b1 
b1 
    2  
(43)
4
mn T
  2  1   
(44)
4
mn T
b1 
2
 2    n T  exp  n T  cosd T   2 exp  n T  sin d T    cosd T 


4
mn T

1  exp  2n T 
4
m n T
(45)
 2n T exp  n T  cosd T   2 exp  n T  sin d T   1  exp  2n T 
2
b1 
4
mn T
(46)
 2n T exp  n T  cosd T   2 exp  n T  sin d T   1  exp  2n T 
2
b1 
4
mn T
(47)
13
b1 


2
n
2
 2 n T exp  n T  cosd T   2
 1  2 2 exp  n T sin d T 
d
4
m n T

2n 1  exp  2n T 
4
m n T
(48)
b1 



 2 n T exp  n T  cosd T   21  exp  2n T   2 n 2 2  1 exp  n T sin d T 
d
3
m n T
(49)
Solve for b2.
b2 
  
(50)
4
mn T
    2 T  exp  2 T   exp   T  sin  T    cos T 
n
n
n
d
d

b2  
4
m n T
14
(51)
b2 
 2

 2   2 T  exp  2 T   exp   T   n  1  2 2 sin  T   2 cos T 
n
n
n
n 
d
n
d 


 d



4
m n T
(52)


 2   T  exp  2 T   exp   T   n 2 2  1 sin  T   2 cos T 
n
n
n 
d
d 


 d

b2 
3
m n T


(53)
The digital recursive filtering relationship for the displacement is
xi 
 a1 x i 1
 b0 f i
 a 2x i2
 b1 f i 1  b 2 f i  2
(54)
15
The digital recursive relationship for the displacement is thus
xi 
 2 exp  n t cosd t  x i 1
 exp  2n t x i  2




n


2




2exp  n T  cosd T  1  exp  n T 
2   1 sin d T    n T  f i

3



m n T 
 d




1



n


2


 2 n T exp  n T  cosd T   21  exp  2n T   2
2   1 exp  n T sin d T  f i 1

3

m n T 
d

1





 n

2



2   1 sin d T   2 cosd T  f i  2
 2   n T  exp  2n T   exp  n T 

3

 d

m n T 



1

(55)
16
Velocity
The impulse response function for the velocity response is
ĥ v ( t ) =

1
exp(- n t )  n
m
 d


sin d t  cos d t 


(56)
The corresponding Laplace transform is

1
s


H v (s) =
m  s 2  2 s   2 
n 
n

(57)
The Z-transform for the ramp invariant simulation is
1  z  12   1 
s
Ĥ v (z) =

Z  L 
2
2
m  Tz  
 s s  2n s  n 2


Ĥ v (z) =

1  z  12  
1


 Z  L1 
m  Tz  
 s s 2  2n s  n 2


 

 


(58)



 

(59)

Evaluate the inverse Laplace transform per References 10 and 11.

1
L1 
 s s 2  2n s   n 2





1
1


exp   n t  cosd t   n sin d t 
d


  n2  n2



(60)
The Z-transform for the ramp invariant simulation is
Ĥ v (z) =



1  z  12  
1
 1


exp  n t  cosd t   n sin d t  

Z 
2
2
m  Tz   
d




 n n
17
(61)


 z  12  
n



Ĥ v (z) =
sin d t  

 Z  1  exp   n t  cosd t  
2
d


m n  Tz  



1
(62)
The Z-transform is evaluated using the method in Reference 12.


n










z
z

exp(


T
)
cos(

T
)

z
exp



T
sin

T
n
d
n
d
d
1  z  12   z

Ĥ v (z) =




2
2
z  2 z exp( n T) cos(d T)  exp  2n T 
mn  Tz   z  1



(63)


n










z
z

exp(


T
)
cos(

T
)

z
exp



T
sin

T
n
d
n
d
d
1  z  12   z

Ĥ v (z) =





2
z 2  2 z exp( n T) cos(d T)  exp  2n T 
mn T  z   z  1



(64)



z  exp( n T) cos(d T)  n exp  n T sin d T 
d
1 

z  1  z  12
Ĥ v (z) =

2 
2
z  2 z exp( n T) cos(d T)  exp  2n T 
mn T 



(65)
18


 


z  exp( n T) cos(d T)  n sin d T  
d

 
1 
Ĥ v (z) =
z  1  z  12

2
z 2  2 z exp( n T) cos(d T)  exp  2n T 
m n T 




(66)
Let



  exp( n T) cos(d T)  n sin d T 
d


(67)
  2exp  n T cosd T 
(68)
  exp  2n T
(69)

z 
z  1  z  12

2 
2   z  
z
mn T 

1
Ĥ v (z) =
Ĥ v (z) =
(70)



z 
z  1  z 2  2z  1

2 
2   z  
z
mn T 

Ĥ v (z) =
1

 
(71)


z z 2  2z  1   z 2  2z  1 
z  1 

2 
2 z

z
mn T 
1


z 3  2z 2  z  z 2  2z  
Ĥ v (z) =
z  1 
2
z2   z  
mn T 
1
Ĥ v (z) =
(72)



z 3  2z 2  z  z 2  2z   
z  1 

2 
2

z z
mn T 
1
19
(72)
(74)
Ĥ v (z) =

z 3   2   z 2  1  2 z   
z  1 

2 
2

z z
mn T 
(75)
Ĥ v (z) =

 z 3  2   z 2  1  2 z   
z  1 

2 
2

z z
mn T 
(76)
Ĥ v (z) =
1
1

z 2   z    z 3  2   z 2  1  2 z   

z  1

2 
2 z
2 z

z
z
mn T 
1
(77)
Ĥ v (z) =
 z 3   z 2  z  z 2   z    z 3  2   z 2  1  2 z   



2 
2 z
2 z

z
z
mn T 
(78)
Ĥ v (z) =
 z 3    1 z 2    z    z 3  2   z 2  1  2 z   



2 
2
2

z z
z z
mn T 
(79)
1
1
 2      1z 2      1  2 z     
Ĥ v (z) =


2

z2   z  
mn T 
1
Ĥ v (z) =
     1z 2      2  1z     


2 
2 z

z
mn T 
1
20
(80)
(81)
Ĥ v (z) =
     1z 2      2  1z     


2 
2

z z
mn T 
1
(82)
Solve for the filter coefficients using the method in Reference 1.
     1z 2      2  1z     
c 0 z 2  c1 z  c 2
1
=


2
z 2  a1z  a 2
z2   z  
mn T 

(83)
Solve for a1.
a1    2exp  n T cosd T 
(84)
Solve for a2.
a 2    exp  2n T
(85)
Solve for c0.
c0 
   1
(86)
2
m n T



exp( n T) cos(d T)  n sin d T   2 exp  n T  cosd T   1
d


c0 
2
mn T



exp( n T)  cos(d T)  n sin d T   1
d


c0 
2
mn T
21
(87)
(88)
Solve for c1.
c1 
    2  1
(89)
2
m n T



exp  2n T   2 exp  n T  cosd T   2 exp( n T) cos(d T)  n sin d T   1
d


c1 
2
mn T
(90)
 

exp  2n T   2 exp( n T)  n sin d T   1
 d

c1 
2
mn T
(91)
Solve for c2.
c2 

(92)
2
m n T



exp( n T) cos(d T)  n sin d T   exp  2n T 
d


c2 
2
mn T
(93)
The digital recursive filtering relationship for the velocity is
x i 
 a1 x i 1
c0 f i
 a 2 x i  2
 c1 f i 1  c 2 f i  2
(94)
22
The digital recursive relationship for the velocity is thus
x i 
 2 exp  n t cosd t  x i 1
 exp  2n t x i  2



 
n



exp(


T
)

cos(

T
)

sin

T
n
d
d   1 f i
2 


 
d
m n T 



 
 
exp  2n T   2 exp( n T)  n sin d T   1 f i 1

2
 d
 
m n T 





n






exp(


T
)
cos(

T
)

sin

T

exp

2

T
n
d
d
n  f i2
2 
d


m n T 

1
1
1
(95)
Acceleration
The acceleration response for a unit impulse is
yt  
1
m





n 2

2  1 sin  t 







(
t
)

exp



t

2

cos

t

2



n
n
d
d
d









(96)
The corresponding Laplace transform is
H a (s) =

1
s2


m  s 2  2 s   n 2 
n


(97)
The Z-transform is
Ĥ a (z) =
1  z  12   1 
s2

Z
L

 
m  Tz  
 s 2 s 2  2 s  n 2
n



23
 




(98)
1  z  12   1 
s2
Ĥ a (z) =

Z  L  2 2
m  Tz  
 s s  2 s  n 2
n



 




 
1  z  12   1 
1


Ĥ a (z) =

Z  L
m  Tz  
 s 2  2 s  n 2  
n



(99)
(100)
Evaluate the inverse Laplace transform per Reference 10 and 11.




1
1

  L1 
L1 
2
2
2
2


 s  2 s  n 
n


 s  n   d 
 


1
1

L1 
exp  n t sin d t
2
2
 s  2 s   n  
n


d
(101)
(102)
The Z-transform for the ramp invariant simulation is
 
(103)
 
(104)
Ĥ a (z) =

1  z  12   1



Z
exp


t
sin

t

 

n
d
m  Tz   

 d
Ĥ a (z) =

1  z  12   1



Z
exp


t
sin

t

 

n
d
m  Tz   

 d
The Z-transform is evaluated using the method in Reference 12.
Ĥ a (z) =

 z  12  
z exp  n t sin d t 


 2
md  Tz   z  2 z exp( n T) cos(d T)  exp  2n T 
1
24
(105)
Ĥ a (z) =


z  12 exp  n t sin d t 


2
md T  z  2 z exp( n T) cos(d T)  exp  2n T 
(106)
Ĥ a (z) =

exp  n t sin d t  
z  12


 z 2  2 z exp( n T) cos(d T)  exp  2n T 
md T
(107)
Ĥ a (z) =

exp  n t sin d t  
z 2  2z  1


2


z

2
z
exp(


T
)
cos(

T
)

exp

2

T

md T
n
d
n 

(108)
1
Solve for the filter coefficients using the method in Reference 1.

d 0 z 2  d1 z  d 2
exp  n t sin d t  
z 2  2z  1



z 2  a 1z  a 2
 z 2  2 z exp( n T) cos(d T)  exp  2n T 
md T
(109)
Solve for a1.
a1    2exp  n T cosd T 
(110)
Solve for a2.
a 2    exp  2n T
(111)
Solve for b0.
d0 
exp   n t sin d t 
(112)
md T
25
Solve for d1.
d1  2d 0
(113)
Solve for b2.
d2  d0
(114)
The digital recursive filtering relationship for the acceleration is
x i   a1 x i 1
 a 2 x i  2
d0 f i
x i   2 exp  n t cosd t x i 1

 d1 f i 1  d 2 f i  2
(115)
 exp  2n t x i  2
n exp   n t sin d t 
2
md T
 f i  2 f i 1  f i  2 
(116)
There are three response parameters: displacement, velocity and acceleration.
If two are known, the third can be calculated via equation (7).
26
SDOF Example
An SDOF system has a natural frequency of 10 Hz with an amplification of Q=10. Its mass is 1
lbm. It is subjected to 1 lbf sinusoidal excitation at is natural frequency as shown in Figure 3.
The analysis is performed using the ramp invariant digital recursive filters derived in this paper as
implemented in Matlab script: arbit_force.m.
This is a simple problem which does not demonstrate the full power of the ramp invariant filtering
method for calculating the response to a force which varies arbitrarily with time.
But the simple example is useful for checking the accuracy of the method. The peak results agree
with the expected values from the corresponding frequency response functions.
The descriptive statistics for the response from the Matlab script are:
Displacement Response
maximum =
minimum =
overall =
0.97 inch
-0.971 inch
0.601 inch RMS
Velocity Response
maximum =
minimum =
overall =
61.6 in/sec
-61.6 in/sec
38.1 in/sec RMS
Acceleration Response
maximum =
minimum =
overall =
9.98 G
-9.97 G
6.17 G RMS
27
APPLIED FORCE
1.0
FORCE (lbf)
0.5
0
-0.5
-1.0
0
0.5
1.0
TIME (SEC)
Figure 3.
28
1.5
2.0
SDOF DISPLACEMENT RESPONSE fn=10 Hz Q=10
1.2
Laplace
Ramp Invariant
1.0
0.8
0.6
DISP (inch)
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1.0
-1.2
0
0.5
1.0
1.5
2.0
TIME (SEC)
Figure 4.
The resulting displacement, velocity and acceleration responses for the ramp invariant method are
shown in Figures 4, 5 and 6, respectively. The exact results from the Laplace transform
calculation are superimposed for comparison, as calculated from the formulas Reference 13 as
implemented via Matlab script: sdof_sine_force.m.
There is excellent agreement between the two displacement curves as shown in Figure 4.
29
SDOF VELOCITY RESPONSE fn=10 Hz Q=10
100
Laplace
Ramp Invariant
80
60
VELOCITY (inch/sec)
40
20
0
-20
-40
-60
-80
-100
0
0.5
1.0
1.5
2.0
TIME (SEC)
Figure 5.
There is very good agreement between the two velocity curves as shown in Figure 4.
The Ramp Invariant curve is slightly ahead of the Laplace curve in terms of phase.
difference will be investigated in the next revision of this paper.
30
This
SDOF ACCELERATION RESPONSE fn=10 Hz Q=10
14
Laplace
Ramp Invariant
12
10
8
6
ACCEL (G)
4
2
0
-2
-4
-6
-8
-10
-12
-14
0
0.5
1.0
1.5
2.0
TIME (SEC)
Figure 6.
There is excellent agreement between the two acceleration curves as shown in Figure 6.
31
MDOF Application
The ramp invariant digital recursive filtering relationship can be readily used as the numerical
engine in an MDOF modal transient analysis for each of the respective response parameters.
An example is shown in Appendix A.
References
1. David O. Smallwood, An Improved Recursive Formula for Calculating Shock Response
Spectra, Shock and Vibration Bulletin, No. 51, May 1981.
2. A. Brandt & K. Ahlin, A Digital Filter Method for Forced Response Computation, Society
for Experimental Mechanics ( SEM ) Proceedings, IMAC-XXI.
3. T. Irvine, The Generalized Coordinate Method for Discrete Systems, Revision F,
Vibrationdata, 2012.
4. Himelblau, Piersol, et al., IES Recommended Practice 012.1: Handbook for Dynamic Data
Acquisition and Analysis, Institute of Environmental Sciences and Technology, Mount
Prospect, Illinois.
5. Rao V. Dukkipati, Vehicle Dynamics, CRC Press, Narosa Publishing House, New York.
2000.
6. K. Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, Englewood
Cliffs, New Jersey, 1982.
7. W. Thomson, Theory of Vibrations with Applications, Second Edition, Prentice-Hall, New
Jersey, 1981.
8. T. Irvine, An Introduction to the Shock Response Spectrum, Revision R, Vibrationdata,
2010.
9. T. Irvine, The Impulse Response Function, Vibrationdata, 2010.
10. T. Irvine, Table of Laplace Transforms, Revision J, Vibrationdata, 2011.
11. T. Irvine, Partial Fractions in Shock and Vibration Analysis, Revision I, Vibrationdata,
2012.
12. R. Dorf, Modern Control Systems, Addison-Wesley, Reading, Massachusetts, 1980.
13. T. Irvine, The Time-domain Response of a Single-degree-of-freedom System Subjected to
a Sinusoidal Force, Revision B, Vibrationdata, 2010.
32
APPENDIX A
MDOF Example
The example from Reference 2 is repeated here.
f2(t)
k3
x2
m2
k2
f1(t)
x1
m1
k1
Figure A-1.
Assume 5% damping for each mode. Assume zero initial conditions.
33
The parameters are
Table A-1. Parameters
Variable
Value
Unit
m1
m2
k1
k2
k3
3.0
lbf sec^2/in
2.0
lbf sec^2/in
400,000
lbf/in
300,000
lbf/in
100,000
lbf/in
B1
100
lbf
B2
200
lbf

55
Hz

100
Hz
The mass matrix is
0  3 0 
m
M 1
  0 2
0
m

2 

(A-1)
The stiffness matrix is
k  k 2
K 1
  k2
 k 2   700,000  300,000

k 2  k 3   300,000 400,000 
(A-2)
Each of the two forcing functions is synthesized using Matlab script: generate.m, as a preprocessing step.
The modal transient analysis is performed using Matlab script: mdof_modal_arbit_force_ri.m
34
>> mdof_modal_arbit_force_ri
mdof_modal_arbit_force_ri.m ver 1.3 February 4, 2012
by Tom Irvine Email: tomirvine@aol.com
This program calculates the response of an MDOF
system to arbitrary force excitation via the ramp
invariant digital recursive filtering relationship.
The system is decoupled using normal modes as an
intermediate step.
Enter the units system
1=English 2=metric
1
Assume symmetric mass and stiffness matrices.
Select input mass unit
1=lbm 2=lbf sec^2/in
2
stiffness unit = lbf/in
Select file input method
1=file preloaded into Matlab
2=Excel file
1
Mass Matrix
Enter the matrix name:
mass_case5
Select damping input method
1=uniform damping ratio
2=damping ratio vector
1
Enter damping ratio
0.05
Stiffness Matrix
Enter the matrix name:
Natural Frequencies
No.
f(Hz)
1.
48.552
2.
92.839
stiff_case5
Modes Shapes (column format)
ModeShapes =
0.3797
0.5326
-0.4349
0.4651
Enter duration(sec)
35
0.3
Enter sample rate (samples/sec)
(recommend
1857)
2000
Each force file must have two columns: time(sec) & force(lbf)
Enter the number of force files
2
Note: the first dof is 1
Enter force file 1
Enter the matrix name:
sine1
Enter the number of dofs at which this force is applied
1
Enter the dof number for this force
1
Enter force file 2
Enter the matrix name:
sine2
Enter the number of dofs at which this force is applied
1
Enter the dof number for this force
2
begin interpolation
end interpolation
Calculating response...
36
DISPLACEMENT
0.003
dof 2
dof 1
0.002
DISP (INCH)
0.001
0
-0.001
-0.002
-0.003
0
0.05
0.10
0.15
0.20
0.25
0.30
TIME (SEC)
Figure A-2.
The displacement, velocity and acceleration responses are shown in Figures A-2, A-3 and A-4,
respectively.
The results appear to be the same as the Laplace transform results in Reference 3. A formal
comparison will be given in the next revision of this paper.
37
VELOCITY
1.5
dof 2
dof 1
VEL (INCH/SEC)
1.0
0.5
0
-0.5
-1.0
-1.5
0
0.05
0.10
0.15
0.20
0.25
0.30
TIME (SEC)
Figure A-3.
ACCELERATION
1.5
dof 2
dof 1
1.0
ACCEL (G)
0.5
0
-0.5
-1.0
-1.5
0
0.05
0.10
0.15
TIME (SEC)
38
0.20
0.25
0.30
Figure A-4.
39