Derivation of the Filter Coefficients for the Ramp Invariant Method... Applied to Base Excitation of a Single-degree-of-Freedom System

Derivation of the Filter Coefficients for the Ramp Invariant Method... Applied to Base Excitation of a Single-degree-of-Freedom System

Derivation of the Filter Coefficients for the Ramp Invariant Method as
Applied to Base Excitation of a Single-degree-of-Freedom System
Revision B
By Tom Irvine
Email: [email protected]
April 3, 2013
______________________________________________________________________________________
Introduction
Consider the single-degree-of-freedom system in Figure 1.
&&
x
m
c
k
&&
y
Figure 1.
where
m = Mass
c = viscous damping coefficient
k = Stiffness
x
y
= absolute displacement of the mass
= base input displacement
A free-body diagram is shown in Figure 2.
&&
x
m
k(y-x)
c ( y& − x& )
Figure 2.
1
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)
By convention,
(c / m) = 2ξωn
(6)
(k / m) = ωn 2
(7)
where ωn is the natural frequency in (radians/sec), and ξ is the damping ratio.
Substitute the convention terms into equation (5).
&u& + 2ξω n u& + ω n 2 u = −&y&
(8)
2
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.
Displacement
The damped natural frequency ωd is
ωd = ω n 1 - ξ 2
(9)
The displacement equation via a convolution integral is
u( t ) = −
1 t
y( τ)[ exp{− ξω n ( t - τ)}][ sin ωd ( t - τ) ]dτ
ωd ∫0
(10)
The corresponding impulse response function for the displacement is
ĥ d ( t ) =
1
[exp(− ξω n t )][sin ωd t ]
ωd
(11)
Further details regarding this derivation are given in Reference 2.
The corresponding Laplace transform is


1

H d (s) = 
 s 2 + 2ξω n s + ω n 2 
(12)
The Z-transform for the ramp invariant simulation is
 (z − 1)2   −1 
1
Ĥ d (z) = 
Z  L 
 s 2 s 2 + 2ξω n s + ω n 2
 Tz  
(
 

 

)
where T is the time step
(13)
3
Evaluate the inverse Laplace transform.


2

ωn
1 

2
2
=
− 2ξω n + ω n t + [2ξω n ]cos(ωd t ) +
− 1 + 2ξ sin (ωd t )

 s 2 s 2 + 2ξω n s + ω n 2  ω 4 
ωd

n 

L−1 
1
(
[
)
]
(14)
The Z-transform becomes
Ĥ d (z) =



2
ωn
1  (z − 1)2  


2
2
− 1 + 2ξ sin (ωd t ) 

 Z − 2ξω n + ω n t + exp(− ξω n t )[2ξω n ]cos(ωd t ) +
4  Tz  
ωn 

ωd
 
 

(15)
[
]
Let
2
α=
ωn
ωd
[− 1 + 2ξ2 ]
(16)
β = 2ξωn
(17)
4
Evaluate the Z-transform.
Ĥ d (z) =
- β  (z − 1)2   z 



4
ω n  Tz   z − 1
2
ω n  (z − 1)2   T z 
+



4  Tz   (z − 1)2 
ωn 


+

β z{z − exp[− ξω n T ]}{cos[ωd T ]}
1  (z − 1)2  



4
2
ω n  Tz   z − 2z{exp[− ξω n T ]}{cos[ωd T ]} + {exp[− 2ξω n T ]}
+

αz{exp[− ξω n T ]}{sin[ωd T ]}
1  (z − 1)2  



4
2
ω n  Tz   z − 2z{exp[− ξω n T ]}{cos[ωd T ]} + {exp[− 2ξω n T ]}
(18)
Ĥ d (z) =
1 
 z −1
2
(
− β )
 + ωn 

4
 T 

ωn 

β{z − exp[− ξω n T ]}{cos[ωd T ]}
1  (z − 1)2  

+


4  T   z 2 − 2z{exp[− ξω T ]}{cos[ω T ]} + {exp[− 2ξω T ]}
ωn 

n
d
n

+

α{exp[− ξω n T ]}{sin[ωd T ]}
1  (z − 1)2  



4  T   z 2 − 2z{exp[− ξω T ]}{cos[ω T ]} + {exp[− 2ξω T ]}
ωn 

n
d
n

(19)
5
Ĥ d (z) =
+
1 
 z −1
2
− β )
(
 + ωn 

4
 T 

ωn 
1  (z − 1)2   β{z − exp[− ξω n T ]}{cos[ωd T ]} + α{exp[− ξω n T]}{sin[ωd T ]}



4

z 2 − 2z{exp[− ξω n T ]}{cos[ωd T ]} + {exp[− 2ξω n T ]}
ω n  T  
(20)
Ĥ d (z) =
+
1 
2
− βz + β + ωn T


4
ωn T 
1
4
ωn T
exp(− ξω n T ){α sin (ωd T ) − β cos(ωd T )} 

[(z − 1)2 ] z 2 −βz2z+{exp
(− ξω T )}{cos(ω T )} + {exp(− 2ξω T )}

n
d
n

(21)
Let
ψ = exp(− ξω n T ){α sin (ωd T ) − β cos(ωd T )}
(22)
ρ = −2{exp(− ξω n T )}{cos(ωd T )}
(23)
λ = exp(− 2ξωn T )
(24)
2
η = β + ωn T
(25)
By substitution,
Ĥ d (z) =
Ĥ d (z) =
1
ωn T
1
4
4
[ − βz + η] +
ωn T
1
4
ωn T
 2
[(z − 1)2 ] z 2β+z z+ρψ+ λ 


[− βz + η ]  z 2 +zρ + λ  +
 z +zρ + λ 
6
1
4
ωn T
(26)

[z 2 − 2z + 1] z 2β+z z+ρψ+ λ 


(27)
4
ωn T
1
+
4
ωn T
1
+
4
ωn T
1
+
4
ωn T
+
 2
1
Ĥ d (z) =

[ − βz ]  z 2 +zρ + λ 
 z + zρ + λ 
 2

[η ]  z 2 +zρ + λ 
 z + zρ + λ 
[z 2 ] z 2β+z z+ρψ+ λ 


 βz + ψ


 z 2 + zρ + λ 
[− 2z] 
1  βz + ψ 


4
2
ω n T  z + zρ + λ 
(28)
1  − β z 3 − βρ z 2 − βλz 
1  η z 2 +η ρ z + η λ 

+


Ĥ d ( z ) =
4
 ω n4 T  z 2 +zρ + λ 
z 2 + zρ + λ
ω n T 
+
1  βz 3 + ψz 2 
1  − 2β z 2 − 2ψ z 
1  βz + ψ 

+

+


4  z 2 + zρ + λ 
4  z 2 + zρ + λ 
4  z 2 + zρ + λ 
ωn T 
 ωn T 
 ωn T 

(29)
Ĥ d (z) =
− βz 3 − βρ z 2 − βλz + η z 2 +η ρ z+ η λ + βz 3 + ψz 2 − 2β z 2 − 2ψz + βz + ψ
4
(
ω n T z 2 + zρ + λ
7
)
(30)
Ĥ d (z) =
(− βρ + ψ + η − 2β)z 2 + (− βλ +η ρ− 2ψ + β)z + (η λ + ψ )
4
(
ω n T z 2 + zρ + λ
)
(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
[(− βρ + ψ + η − 2β)z 2 + (− βλ +η ρ− 2ψ + β)z + (η λ + ψ )]/ mωn4T
z 2 + zρ + λ
(32)
Solve for a1.
a1 = ρ = −2 exp(− ξω n T ) cos(ωd T )
(33)
Solve for a2.
a 2 = λ = exp(− 2ξωn T )
(34)
Note that the a1 and a2 coefficients are common for displacement, velocity and acceleration.
Solve for b0.
b 0 = [− βρ + ψ + η − 2β] / ω n T 


(35)
4
b 0 = [ψ + η − β(ρ + 2 )] / ω n T 


(36)
4
2
b0 =
exp (− ξω n T )[α sin (ω d T ) − β cos (ω d T )] + β + ω n T − β[− 2 exp (− ξω n T ) cos (ω d T )+ 2]
4
ωn T
(37)
8
2
b0 =
exp (− ξω n T )[α sin (ω d T ) − β cos (ω d T )] + β + ω n T + 2β[exp (− ξω n T ) cos (ω d T )− 1]
4
ωn T
(38)
b0 =
 2

ω
exp(− ξω n T ) n − 1 + 2ξ 2 sin (ωd T ) − 2ξω n cos(ωd T )
ω

 d

[
]
4
ωn T
2
+
2ξω n + ω n T + 4ξω n [exp(− ξω n T ) cos(ωd T )− 1]
4
ωn T
(39)
 2

ωn
2
2

exp(− ξω n T )
− 1 + 2ξ sin (ωd T ) + ω n T + 2ξω n [exp(− ξω n T ) cos(ωd T )− 1]
ω

 d

[
b0 =
]
4
ωn T
(40)


ωn
2

exp(− ξω n T )
− 1 + 2ξ sin (ωd T ) + ω n T + 2ξ[exp(− ξω n T ) cos(ωd T )− 1]
ω

 d

[
b0 =
]
3
ωn T
(41)
9


ω
2ξ[exp(− ξω n T ) cos(ωd T )− 1] + exp(− ξω n T ) n 2ξ 2 − 1 sin (ωd T ) + ω n T
ω

 d

[
b0 =
]
3
ωn T
(42)
Solve for b1.
b1 =
b1 =
− βλ +η ρ− 2ψ + β
(43)
4
ωn T
η ρ− 2ψ + β(1 − λ )
(44)
4
ωn T
b1 =
2
− 2 β + ω n T  exp(− ξω n T ) cos(ωd T ) − 2 exp(− ξω n T )[α sin (ωd T ) − β cos(ωd T )]


4
ωn T
+
β[1 − exp(− 2ξω n T )]
4
ωn T
(45)
2
b1 =
− 2ω n T exp (− ξω n T ) cos (ω d T ) − 2 exp(− ξω n T )α sin (ω d T ) + β[1 − exp(− 2ξω n T )]
4
ωn T
(46)
2
b1 =
− 2ω n T exp (− ξω n T ) cos (ω d T ) − 2 exp(− ξω n T )α sin (ω d T ) + β[1 − exp(− 2ξω n T )]
4
ωn T
(47)
10
b1 =
2
ωn
2
− 2ω n T exp(− ξω n T ) cos(ωd T ) − 2
− 1 + 2ξ 2 exp(− ξω n T )sin (ωd T )
ωd
4
ωn T
[
+
]
2ξω n [1 − exp(− 2ξω n T )]
4
ωn T
(48)
b1 =
[
]
ω
− 2ω n T exp(− ξω n T ) cos(ω d T ) + 2ξ[1 − exp(− 2ξω n T )] − 2 n 2ξ 2 − 1 exp(− ξω n T ) sin (ω d T )
ωd
3
ωn T
(49)
Solve for b2.
b2 =
ηλ + ψ
(50)
4
ωn T
 β + ω 2 T  exp(− 2ξω T ) + exp(− ξω T ){α sin (ω T ) − β cos(ω T )}
n
n
n
d
d

b2 = 
4
ωn T
b2 =
 2

 2ξω + ω 2 T  exp(− 2ξω T ) + exp(− ξω T ) ω n − 1 + 2ξ 2 sin (ω T ) − 2ξω cos(ω T )
n
n
n
n 
d
n
d 


 ωd

[
4
ωn T
11
]
(51)
(52)


 2ξ + ω T  exp(− 2ξω T ) + exp(− ξω T ) ω n 2ξ 2 − 1 sin (ω T ) − 2ξ cos(ω T )
n
n
n 
d
d 


 ωd

b2 =
3
ωn T
[
]
(53)
The digital recursive filtering relationship for the relative displacement is
ui =
− a1 u i −1
− b 0 &y& i
− a 2u i−2
− b1 &y& i −1 − b 2 &y& i − 2
(54)
12
The digital recursive relationship for the relative displacement is thus
ui =
+ 2 exp[− ξω n ∆t ]cos[ωd ∆t ]u i −1
− exp[− 2ξω n ∆t ]u i − 2




ωn
1 

2


−
2ξ[exp(− ξω n T ) cos(ωd T )− 1] + exp(− ξω n T )
2ξ − 1 sin (ωd T ) + ω n T  &y& i

3
ω


ωn T 
 d



[
]


ωn


2
−
− 2ω n T exp(− ξω n T ) cos(ωd T ) + 2ξ[1 − exp(− 2ξω n T )] − 2
2ξ − 1 exp(− ξω n T )sin (ωd T ) &y& i −1

3

ω n T 
ωd
[
1
]



1 
 ωn

2

−
2ξ − 1 sin (ωd T ) − 2ξ cos(ωd T ) &y& i − 2
 2ξ + ω n T  exp(− 2ξω n T ) + exp(− ξω n T )

3

ω n T 
 ωd



[
]
(55)
13
Relative Velocity
The impulse response function for the relative velocity response is
 − ξω n
ĥ v ( t ) = exp(-ξω n t ) 
 ωd


sin ωd t + cos ωd t 


(56)
The corresponding Laplace transform is
H v (s) =
s
(57)
s 2 + 2ξωn s + ω n 2
The Z-transform for the ramp invariant simulation is
 (z − 1)2   −1 
s
Ĥ v ( z) = 
Z  L 
2
2
 s s + 2ξω n s + ω n 2
 Tz  
 

 

)
(58)
 (z − 1)2   −1 
1
Ĥ v ( z ) = 
Z  L 
 s s 2 + 2ξω n s + ω n 2
 Tz  
 

 

(59)
(
(
)
Evaluate the inverse Laplace transform per References 3 and 4.

1
L−1 
 s s 2 + 2ξω n s + ω n 2
(



ξω
1
1
=
−
exp(− ξω n t ) cos(ωd t ) + n sin (ω d t )
ωd


 ω 2 ω 2
n
n


)
(60)
The Z-transform for the ramp invariant simulation is

 
 (z − 1)2   1
ξω n
1

−
(
−
ξ
ω
)
(
ω
)
+
(
ω
)
Ĥ v ( z ) = 
Z
exp
t
cos
t
sin
t
 
n
d
d 
ωd


 Tz   ω n2 ω n2


14
(61)
Ĥ v ( z ) =


 
ξω n
1  (z − 1)2  
sin (ωd t ) 

 Z  1 − exp(− ξω n t ) cos(ω d t ) +
2
ωd


ω n  Tz  


(62)
The Z-transform is evaluated using the method in Reference 5.
Ĥ v (z) =
1
2
ωn


ξω n


z
z
−
exp(
−
ξω
T
)
cos(
ω
T
)
+
z
exp
−
ξ
ω
t
sin
ω
t
[
]
[
(
)
(
)
]
n
d
n
d
 (z − 1)2   z
ωd

−



 Tz   z − 1
z 2 − 2 z exp(−ξω n T) cos(ωd T) + exp(− 2ξω n T )





(63)


ξω n

z
[
z
−
exp(
−
ξω
T
)
cos(
ω
T
)
]
+
z
[
exp
(
−
ξ
ω
t
)
sin
(
ω
t
)
]
n
d
n
d 
ωd
1  (z − 1)2   z

Ĥ v (z) =
−



2  z  z −1
2
z − 2 z exp(−ξω n T) cos(ωd T) + exp(− 2ξω n T )
ωn T 






(64)



[z − exp(−ξω n T) cos(ωd T)] + ξω n [exp(− ξω n t )sin (ωd t )]
ωd
1 

(
Ĥ v (z) =
z − 1) − (z − 1)2


2
z 2 − 2 z exp(−ξω n T ) cos(ωd T) + exp(− 2ξω n T )
ωn T 





(65)
15


 
ξω

z − exp(−ξω n T ) cos(ωd T ) − n sin (ωd t ) 
ωd

 
1 

 
2

Ĥ v (z) =
(z − 1) − (z − 1) 2
2 
z − 2 z exp(−ξω n T ) cos(ωd T ) + exp(− 2ξω n T ) 
ωn T






(66)
Let


ξω
ψ = exp(−ξω n T ) cos(ωd T ) − n sin (ωd t )
ωd




(67)
ρ = −2{exp(− ξω n T )}{cos(ωd T )}
(68)
λ = exp(− 2ξωn T )
(69)
Ĥ v (z) =
Ĥ v (z) =
1 
z−ψ 
(z − 1) − (z − 1)2

2
z 2 + ρ z + λ 
ω n T 
(70)
1 
z−ψ 
(z − 1) − z 2 − 2z + 1

2 
2
z + ρ z + λ 
ωn T 
(
)
(
) (
(71)
)
Ĥ v (z) =
1 
z z 2 − 2z + 1 − ψ z 2 − 2 z + 1 
(z − 1) −

2 
2

z +ρz+λ
ωn T 
Ĥ v (z) =
1 
z 3 − 2z 2 + z − ψz 2 − 2ψz + ψ
(z − 1) −
2
z2 + ρ z + λ
ω n T 
Ĥ v (z) =
(
(72)
)

1 
z 3 − 2z 2 + z − ψz 2 + 2ψz − ψ 
(z − 1) −

2 
2 +ρz+λ

z
ωn T 
16
(72)
(74)
Ĥ v (z) =
1 
z 3 + (− 2 − ψ )z 2 + (1 + 2ψ )z − ψ 
(z − 1) −

2

z2 + ρ z + λ
ω n T 
(75)
Ĥ v (z) =
1 
− z 3 + (2 + ψ )z 2 − (1 + 2ψ )z + ψ 
(z − 1) +

2

z2 + ρ z + λ
ω n T 
(76)
z 2 + ρ z + λ − z 3 + (2 + ψ )z 2 − (1 + 2ψ )z + ψ 
1 
Ĥ v (z) =
(z − 1)
+

2

z2 + ρ z + λ
z2 + ρ z + λ
ω n T 
(77)
1  z 3 + ρ z 2 + λz − z 2 − ρ z − λ − z 3 + (2 + ψ )z 2 − (1 + 2ψ )z + ψ 
Ĥ v (z) =

+

2

z2 + ρ z + λ
z2 + ρ z + λ
ω n T 
(78)
Ĥ v (z) =
1  z 3 + (ρ − 1) z 2 + (λ − ρ )z − λ − z 3 + (2 + ψ )z 2 − (1 + 2ψ )z + ψ 

+

2

z2 + ρ z + λ
z2 + ρ z + λ
ω n T 
(79)
Ĥ v (z) =
1  (2 + ψ + ρ − 1)z 2 + (λ − ρ − 1 − 2ψ )z + ψ − λ 


2 
2 +ρz+λ

z
ωn T 
(80)
1  (ψ + ρ + 1)z 2 + (λ − ρ − 2ψ − 1)z + ψ − λ 
Ĥ v (z) =


2

z2 + ρ z + λ
ω n T 
17
(81)
Ĥ v (z) =
1  (ψ + ρ + 1)z 2 + (λ − ρ − 2ψ − 1)z + ψ − λ 


2 
2

z +ρz+λ
ωn T 
(82)
Solve for the filter coefficients using the method in Reference 1.
c 0 z 2 + c1 z + c 2
z 2 + a1z + a 2
 (ψ + ρ + 1)z 2 + (λ − ρ − 2ψ − 1)z + ψ − λ 
=


2

z2 + ρ z + λ
ω n T 
1
(83)
Solve for a1.
a1 = ρ = −2 exp(− ξω n T ) cos(ωd T )
(84)
Solve for a2.
a 2 = λ = exp(− 2ξωn T )
(85)
Solve for c0.
c0 =
ψ + ρ +1
(86)
2
ωn T


ξω
exp(−ξω n T) cos(ωd T) − n sin (ωd t ) − 2 exp(− ξω n T ) cos(ωd T ) + 1
ωd




c0 =
2
ωn T


ξω
exp(−ξω n T) − cos(ωd T) − n sin (ωd t ) + 1
ωd




c0 =
2
ωn T
18
(87)
(88)
Solve for c1.
c1 =
λ − ρ − 2ψ − 1
(89)
2
ωn T


ξω
exp(− 2ξω n T ) + 2 exp(− ξω n T ) cos(ωd T ) − 2 exp(−ξω n T) cos(ωd T) − n sin (ωd t ) − 1
ωd




c1 =
2
ωn T
(90)
 ξω

exp(− 2ξω n T ) + 2 exp(−ξω n T)  n sin (ωd t ) − 1
 ωd



c1 =
2
ωn T
(91)
Solve for c2.
c2 =
ψ−λ
(92)
2
ωn T


ξω
exp(−ξω n T) cos(ωd T) − n sin (ωd t ) − exp(− 2ξω n T )
ωd




c2 =
2
ωn T
19
(93)
The digital recursive filtering relationship for the relative velocity is
u& i =
− a1 u& i −1
−c 0 &y& i
− a 2 u& i − 2
− c1 &y& i −1
− c 2 &y& i − 2
(94)
u& i =
+ 2 exp[− ξω n ∆t ]cos[ωd ∆t ]u& i −1
− exp[− 2ξω n ∆t ]u& i − 2
+

 
ξω n
1 

exp(
−
ξω
T
)
−
cos(
ω
T
)
−
sin
(
ω
T
)
n
d
d  + 1 &y& i
2 
ω

 
d
ω n T 

 
+

 
 ξω

n sin (ω T ) − 1 &y&

exp
(
−
2
ξω
T
)
+
2
exp(
−
ξω
T
)
 i −1
n
n
d
2 
 ωd
 
ω n T 

 
+




ξω
1 

exp(−ξω n T ) cos(ωd T ) − n sin (ωd T ) − exp(− 2ξω n T ) &y& i − 2

2
ωd


ω n T 



1
(95)
20
Absolute Acceleration
The impulse response function for the absolute acceleration response is


ω 2
ĥ a ( t ) = exp(− ξω n t ) 2ξω n cos(ω d t ) + n 1 − 2ξ 2 sin (ωd t )
ωd


(
)
(96)
The corresponding Laplace transform is
 2ξω s + ω 2 
n
n

H a (s) = 
 s 2 + 2ξω n s + ω n 2 
(97)
The Z-transform is
 (z − 1)2   −1  1  2ξω s + ω 2  
n
n

Ĥ a (z) = 
Z  L  


2
2
2
Tz
 s s + 2ξω s + ω  

 
n
n






1
L−1 
s2

 2ξω s + ω 2 


1
n
n

 = L−1  1 −

 s 2 + 2ξω s + ω 2 
2 s 2 + 2ξω s + ω 2 

s
n 
n 
n
n


1
L−1 
s2

 2ξω s + ω 2 


1
n
n

 = L−1  1 −

 s 2 + 2ξω s + ω 2 
2 (s + ξω )2 + ω 2 

s
d 
n 

n
n

21
(98)
(99)
(100)
1
L−1 
s2

 2ξω s + ω 2 
n
n

 = L−1  t −  1
 
 s 2 + 2ξω s + ω 2 
  ωd
n 
n



 exp(−ξω n t ) sin (ωd t )


(101)
Evaluate the Z-transform.
 (z − 1)2    1
Ĥ a (z) = 
 Z  t − 
 Tz    ωd


 exp(−ξω n t ) sin (ωd t ) 


(102)


 1 

 exp[− ξω n T ]sin[ωd T ]


 (z − 1)2   zT
ωd 


Ĥ a (z) = 
−


2
2
 Tz   (z − 1)
z − 2z{exp[− ξω n T ]cos[ωd T ]} + {exp[− 2ξω n T ]} 




Ĥ a (z) = 1 − (z − 1)2
Ĥ a (z) = 1 −
 1 

 exp[− ξω n T ]sin[ωd T ]
 ωd T 
z 2 − 2z{exp[− ξω n T ]cos[ωd T ]} + {exp[− 2ξω n T ]}
(z 2 − 2z + 1) ω1d T  exp[− ξωn T]sin[ωd T]
z 2 − 2z{exp[− ξω n T ]cos[ωd T ]} + {exp[− 2ξω n T ]}
(
(103)
(104)
(105)
)
 1 
 exp[− ξω n T ]sin[ωd T ]
z 2 − 2z exp[− ξω n T ]cos[ωd T ] + exp[− 2ξω n T ] − z 2 − 2z + 1 
ωd T 

Ĥ a (z) =
z 2 − 2z exp[− ξω n T ]cos[ωd T ] + exp[− 2ξω n T ]
(106)
22
Ĥ a (z) =
(
)
 1 
 exp[− ξω n T ]sin [ωd T ]
z 2 − 2z exp[− ξω n T ]cos[ωd T ] + exp[− 2ξω n T ] + − z 2 + 2z − 1 
ω
T
 d 
z 2 − 2z exp[− ξω n T ]cos[ωd T ] + exp[− 2ξω n T ]
(107)
Ĥ a (z) =


 1 
 exp(− ξω n T )sin (ωd T )
1 − 


ωd T 



2
z 

2
 z − 2z exp[− ξω n T ]cos(ωd T ) + exp[− 2ξω n T ]







 1 
 sin (ωd T ) 
 exp(− ξω n T ) − cos(ωd T ) + 


 ωd T 


+ 2z 

2
 z − 2z exp(− ξω n T ) cos(ωd T ) + exp(− 2ξω n T ) 






 1 
 exp(− ξω n T )sin (ωd T ) 
 exp(− 2ξω n T ) − 


 ωd T 
+

2
 z − 2z exp(− ξω n T ) cos(ωd T ) + exp(− 2ξω n T )




(108)
23
c 0 z 2 + c1 z + c 2
=
z 2 + a 1z + a 2


 1 
 exp(− ξω n T )sin (ωd T )
1 − 


ωd T 



2
z 

2
 z − 2z exp[− ξω n T ]cos(ωd T ) + exp[− 2ξω n T ]







 1 
 sin (ωd T ) 
 exp(− ξω n T ) − cos(ωd T ) + 


 ωd T 


+ 2z 

2
 z − 2z exp(− ξω n T ) cos(ωd T ) + exp(− 2ξω n T ) 






 1 
 exp(− ξω n T ) sin (ωd T ) 
 exp(− 2ξω n T ) − 


 ωd T 
+

2
 z − 2z exp(− ξω n T ) cos(ωd T ) + exp(− 2ξω n T ) 




(109)
Solve for a1.
a1 − 2 exp(− ξω n T ) cos(ωd T )
(110)
Solve for a2.
a 2 = exp(− 2ξωn T )
(111)
Solve for c0.
c0 =

1   1 
 exp(− ξω n T ) sin (ωd T )
1 − 
T   ωd T 

(112)
Solve for c1.



 1 
 sin (ωd T )
c1 = 2exp(− ξω n T ) − cos(ωd T ) + 


 ωd T 


24
(113)
Solve for c2.


 1 
 exp(− ξω n T )sin (ωd T )
c 2 = exp(− 2ξω n T ) − 


 ωd T 
(114)
The digital recursive filtering relationship for absolute acceleration is
&x& i =
− a1 &x& i −1
−c 0 &y& i
− a 2 &x& i − 2
− c1 &y& i −1
− c 2 &y& i − 2
(115)
&x& i =
+ 2 exp[− ξω n ∆t ]cos[ωd ∆t ]&x& i −1
− exp[− 2ξω n ∆t ]&x& i − 2
  1 

 exp(− ξω n T ) sin (ωd T )  &y& i
+  1 − 
  ωd T 




 1 
 sin (ωd T ) &y& i −1
+  2 exp(− ξω n T ) − cos(ωd T ) + 


 ωd T 




 1 
 exp(− ξω n T )sin (ωd T ) &y& i − 2
+  exp(− 2ξω n T ) − 


 ωd T 
(116)
Relative Acceleration
The impulse response function for the relative acceleration response is
25



ω 2
ĥ ra ( t ) =  − δ( t ) + exp(− ξω n t )2ξω n cos(ωd t ) + n 1 − 2ξ 2 sin (ω d t )
ωd



(
)



(117)
The corresponding Laplace transform is
H ra (s ) = −
s2
(118)
s 2 + 2ξω n s + ω n 2
The Z-transform is
 
 (z − 1)2   −1  1 
s2



Ĥ a (z) = 
Z  L

 
2
2
2
Tz


 
s
s
+
2
ξω
s
+
ω
n  
n
 


(119)
 
 (z − 1)2   −1 
1

Ĥ a (z) = 
Z  L 
 s 2 + 2ξωn s + ω n 2  
 Tz  

(120)
Evaluate the inverse Laplace transform.



1
−
1


=L
L
 (s + ξω )2 + ω 2 
 s 2 + 2ξω s + ω n 2 
n
n


d 


−1 
1


1
1
=
L−1 
exp(− ξω n t )sin ωd t
2
2
 s + 2ξω s + ω n  ω

n

d
(121)
( )
(122)

 (z − 1)2   1



Ĥ ra (z) = 
exp(− ξω n T )sin  ωd T  
Z 


 Tz   ω
 d

(123)
The Z-transform for the ramp invariant simulation is
26

 (z − 1)2   1



Ĥ ra (z) = 
exp(− ξω n T )sin  ωd T  
Z 


 Tz   ω
 d

(124)
The Z-transform is evaluated as.
Ĥ ra (z) =

z [exp(− ξω n T )sin (ωd T )]
1  (z − 1)2  



2
ωd  Tz   z − 2 z exp(−ξω n T) cos(ωd T) + exp(− 2ξωn T ) 

(
z − 1)2 [exp(− ξω n T )sin (ωd T )]
1 


Ĥ ra (z) =
2
ωd T  z − 2 z exp(−ξωn T) cos(ωd T) + exp(− 2ξωn T ) 
(125)
(126)

exp(− ξωn T ) sin (ωd T ) 
(z − 1)2


2 − 2 z exp(−ξω T) cos(ω T) + exp(− 2ξω T ) 

z
ωd T
n
d
n 

(127)

exp(− ξωn T ) sin (ωd T ) 
z 2 − 2z + 1


Ĥ ra (z) =
 z 2 − 2 z exp(−ξωn T) cos(ωd T) + exp(− 2ξωn T ) 
ωd T
(128)
Ĥ ra (z) =
Solve for the filter coefficients.

d 0 z 2 + d1 z + d 2
exp(− ξω n T ) sin (ωd T ) 
z 2 − 2z + 1


=
2 − 2 z exp(−ξω T) cos(ω T) + exp(− 2ξω T ) 

z 2 + a 1z + a 2
z
ω T
n
d
n 

d
(129)
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Solve for a1.
a1 = ρ = −2 exp(− ξω n T ) cos(ωd T )
(130)
Solve for a2.
a 2 = λ = exp(− 2ξωn T )
(131)
Solve for d0.
d0 =
exp(− ξω n T ) sin (ωd T )
(132)
ωd T
Solve for d1.
d1 = − 2d 0
(133)
Solve for d2.
d2 = d0
(134)
The digital recursive filtering relationship for the relative acceleration is
&u& i =
− a1 &u& i −1
− a 2 &u& i − 2
+ d 0 &y& i
&u& i = + 2 exp[− ξω n T ]cos[ωd T ]&u& i −1
+
+ d1 &y& i −1 + d 2 &y& i − 2
(135)
− exp[− 2ξω n T ]&u& i − 2
exp(− ξω n T ) sin (ωd T )
ωd T
{ &y& i − 2 &y& i −1 + &y& i − 2 }
(136)
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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, Table of Laplace Transforms, Revision J, Vibrationdata, 2011.
4. T. Irvine, Partial Fractions in Shock and Vibration Analysis, Revision I, Vibrationdata,
2012.
5.
R. Dorf, Modern Control Systems, Addison-Wesley, Reading, Massachusetts, 1980.
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