A Digital Recursive Filtering Method for Calculating the Base Input

A Digital Recursive Filtering Method for Calculating the Base Input
for a Measured Response Acceleration
By Tom Irvine
Email: [email protected]
March 28, 2012
_____________________________________________________________________________________
Introduction
Consider a single-degree-of-freedom (SDOF) system subjected to base excitation, as shown in
Figure 1.
There are certain cases where the response of a system is known, but the base input acceleration
is unknown. An example would be a seismic sensor which behaved as an SDOF system. The
seismometer data would give the acceleration of the mass. A calculation would then be needed
to determine the base input which drove the mass to the measured response. This calculation
process is a form of deconvolution.

x
m
k
c
y
Figure 1.
where
m = Mass
c
k
x
y
=
=
=
=
viscous damping coefficient
Stiffness
absolute displacement of the mass
base input displacement
1
A free-body diagram is shown in Figure 2.

x
m
k(y-x)
c (y  x )
Figure 2.
Summation of forces in the vertical direction
F 

mx
(1)
  c (y  x )  k (y  x)
mx
(2)
Let
uxy
u  x  y
u  x  y
x  u  y
Substituting the relative displacement terms into equation (2) yields
m(u  y)  cu  ku
(3)
mu  cu  ku  my
(4)
Dividing through by mass yields
u  (c / m)u  (k / m)u  y
(5)
2
By convention,
(c / m)  2 n
(6)
(k / m)  n 2
(7)
where  n is the natural frequency in (radians/sec), and  is the damping ratio.
Substitute the convention terms into equation (5).
u  2 n u  n 2 u  y
(8)
Equation (8) does not have a closed-form solution for the general case in which y is an arbitrary
function. A convolution integral approach must be used to solve the equation. Note that the
impulse response function is embedded in the convolution integral.
Absolute Acceleration
The impulse response function for the acceleration response from Reference 2 is




n 2
ĥ a ( t ) = exp  n t 2 n cosd t  
1  2 2 sin d t 
d


(9)
The corresponding Laplace transform for Ha (s) (response/input) is
 2 s   2 
n
n

H a (s) = 
 s 2  2 n s  n 2 
(10)
The corresponding Laplace transform for H i (s) (input/response) is
H i (s) =
s 2  2 n s  n 2
(11)
2 n s  n 2
3
The Z-transform is found using the bilinear transform.
s
2 z 1
T z 1
(12)
 2 z  1 
 2 z  1 
2
 T z  1  2 n  T z  1  n



H i (z) = 
 2 z  1 
2 n 
 n 2

 T z  1
2
(13)
2
2

2

2
2
 T z  1  2 n  T z  1z  1  n z  1



H i (z) = 
2

2 n  z  1z  1  n 2 z  12
T

(14)
2z  12  4 n Tz  1z  1  T 2n 2 z  12
4Tn z  1z  1  T 2 n 2 z  12
(15)
H i (z) =
H i (z) =
H i (z) =






4 z 2  2z  1  4 n T z 2  1  T 2 n 2 z 2  2z  1




4Tn z 2  1  T 2 n 2 z 2  2z  1
4z 2  8z  4  4 n Tz 2  4 n T  T 2 n 2 z 2  2T 2 n 2 z  T 2 n 2
4Tn z 2  4Tn  T 2 n 2 z 2  2T 2 n 2 z  T 2 n 2
4
(16)
(17)
H i (z) =

( 4  4 n T  T 2 n 2 ) z 2  (  8  2T 2 n 2 ) z  4  4 n T  T 2 n 2

(18)
( 4Tn  T 2 n 2 ) z 2  ( 2T 2 n 2 ) z  4Tn  T 2 n 2

1 ( 4  4 n T  T 2 n 2 ) z 2  2(  4  T 2 n 2 ) z  4  4 n T  T 2 n 2
H i ( z) =
Tn
( 4  Tn ) z 2  ( 2Tn ) z  4  Tn

(19)
H i (z) =

( 4  4 n T  T 2 n 2 ) z 2  2(  4  T 2 n 2 ) z  4  4 n T  T 2 n 2
1
  4  Tn 
Tn ( 4  Tn )
2  2T n 

 z  

z  
 4  Tn 
 4  Tn 
(20)
Solve for the filter coefficients using the method in Reference 1.
c 0 z 2  c1 z  c 2
=
z 2  a1z  a 2

( 4  4 n T  T 2  n 2 ) z 2  2(  4  T 2 n 2 ) z  4  4 n T  T 2  n 2
1
  4  T n 
Tn ( 4  Tn )
2  2T n 

 z  

z  
4


T

4


T

n
n 


(21)
5
Solve for a1.
 2Tn
a1  
 4  Tn



(22)
Solve for a2.
  4  Tn
a 2  
 4  Tn



(23)
Solve for c0.
c0 
( 4  4 n T  T 2 n 2 )
(24)
Tn ( 4  Tn )
Solve for c1.
2(  4  T 2 n 2 )
c1 
Tn ( 4  Tn )
(25)
Solve for c2.
c2

4  4 n T  T 2 n 2 

(26)
Tn ( 4  Tn )
6
The digital recursive filtering relationship is
y i 
 a1 y i 1
c 0 x i
 a 2 y i  2
 c1 x i 1  c 2 x i  2
(27)
The digital recursive filtering relationship is
y i 
 2Tn
 
 4  Tn

 y i 1

  4  Tn
 
 4  Tn
 4  4 T  T 2  2 
n  
n

x
 Tn ( 4  Tn )  i



y i  2

 4  4 T  T 2  2 
 2(  4  T 2  2 ) 
n 
n
n


x i 1  

x i2

 Tn ( 4  Tn ) 
Tn ( 4  Tn ) 




(28)
References
1. David O. Smallwood, An Improved Recursive Formula for Calculating Shock Response
Spectra, Shock and Vibration Bulletin, No. 51, May 1981.
2. T. Irvine, The Impulse Response Function for Base Excitation, Vibrationdata, 2012.
3. T. Irvine, The Response of a Single-degree-of-freedom System Subjected to a Wavelet
Pulse Base Excitation, Vibrationdata, 2008.
7
Example
INPUT: WAVELET 80 Hz 7 half-sines
SDOF RESPONSE (fn=80 Hz, Q=10)
50
Response
Base Input
40
30
ACCEL (G)
20
10
0
-10
-20
-30
-40
-50
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
TIME (SEC)
Figure A-1.
An SDOF system is subjected to a wavelet pulse. Both the input and response are
shown in Figure A-1.
The response is calculated via Reference 3.
8
BASE INPUT
WAVELET 80 Hz 7 half-sines
15
Calculated
Original
10
ACCEL (G)
5
0
-5
-10
-15
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
TIME (SEC)
Figure A-2.
The Original and Calculated Base Input curves are nearly identical.
The calculation was performed via equation (28) given the response in Figure A-1.
9