Modal Transient Analysis of a System Subjected to an Applied Force via a Ramp Invariant Digital Recursive Filtering Relationship

Revision F

By Tom IrvineEmail: 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 post-processing 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.

2 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

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+ 12

u t 2

(0)

It is typically implemented with β=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

m

k x xc

x

f(t)

SDOF Equation of Motion

Consider a single-degree-of-freedom system.

Figure 1.

The variables are

m is the massc is the viscous damping coefficientk is the stiffnessx is the absolute displacement of the mass

f(t) is the applied force

Note that the double-dot denotes acceleration.

The free-body diagram is

Figure 2.

m

k c

4

Summation of forces in the vertical direction

(1)

(2)

(3)

Divide through by m,

(4)

By convention,

( c/m)=2ξωn (5)

( k /m)=ωn2 (6)

where

n is the natural frequency in (radians/sec)

is the damping ratio.

By substitution,

(7)

Equation (7) does not have a closed-form solution for the general case in which 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.

(8)

5

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.

Displacement

The damped natural frequency ωd is

ωd=ωn √1- ξ2 (9)

The displacement equation via a convolution integral is

x ( t )= 1mωd

∫0

tf ( τ ) [ exp {−ξωn ( t- τ )} ] [ sin ωd( t- τ )] dτ

(10)

The corresponding impulse response function for the displacement is

hd ( t )= 1mωd

[exp (−ξωn t )] [sin ωd t ] (11)

Further details regarding this derivation are given in Reference 9.

The corresponding Laplace transform from Reference 10 is

Hd ( s )= 1m [ 1

s2+2ξωn s+ωn2 ]

(12)

The Z-transform for the ramp invariant simulation is

Hd ( z )= 1m [ ( z−1 )2

Tz ]Z { L−1[ 1s2 (s2+2ξωn s+ω

n2) ] }

where T is the time step

(13)

6

Evaluate the inverse Laplace transform per References 10 and 11.

L−1 [1s2 (s2+2 ξωn s+ωn2) ]=1

ωn 4 {−2 ξωn+ωn

2 t+ [2 ξωn ]cos (ωd t )+ωn 2

ωd [−1+2 ξ2 ]sin (ωd t )}

(14)

The Z-transform becomes

Hd ( z )=

1mωn

4 [( z−1 )2

Tz ]Z {−2 ξωn+ωn 2 t+exp (−ξωn t ){[2 ξωn ] cos (ωd t )+

ωn 2

ωd [−1+2ξ 2] sin (ωd t )} }

(15)

Let

α=ωn 2

ωd [−1+2ξ2 ]

(16)

β=2ξωn (17)

7

The Z-transform is evaluated using the method in Reference 12.

Hd ( z )=−βmωn

4 [( z−1 )2

Tz ] [zz−1 ]+

ωn 2

mωn 4 [( z−1 )2

Tz ] [T z( z−1 )2 ]

+1mωn

4 [( z−1 )2

Tz ] [βz {z−exp [−ξωn T ] } {cos [ωd T ]}z2−2 z {exp [−ξωnT ] }{cos [ωd T ] }+{exp [−2 ξωn T ] } ]

+1mωn

4 [( z−1 )2

Tz ] [αz {exp [−ξωnT ] }{sin [ ωd T ] }z2−2 z {exp [−ξωnT ] }{cos [ωd T ] }+{exp [−2 ξωn T ] } ]

(18)

Hd ( z )=1mωn

4 [ (−β )(z−1T )+ωn

2]+1

mωn 4 [ (z−1 )2

T ] [β { z−exp [−ξωn T ] }{cos [ ωd T ] }z2−2 z {exp [−ξωn T ] }{cos [ωd T ] }+ {exp [−2 ξωn T ] } ]

+1mωn

4 [ (z−1 )2

T ] [α {exp [−ξωn T ] } {sin [ωdT ] }z2−2 z {exp [−ξωn T ] }{cos [ωd T ] }+ {exp [−2 ξωn T ] } ]

(19)

8

Hd ( z )=1mωn

4 [(−β )(z−1T )+ωn

2 ]+1

mωn 4 [ (z−1 )2

T ] [ β { z−exp [−ξωnT ] }{cos [ ωd T ] }+α {exp [−ξωnT ] }{sin [ ωd T ] }z2−2 z {exp [−ξωnT ] }{cos [ωdT ] }+ {exp [−2 ξωn T ] } ]

(20)

Hd ( z )=1mωn

4 T[−βz+β+ωn

2 T ]

+1mωn

4 T[ ( z−1 )2] [βz+exp (−ξωn T ) {α sin (ωd T )−β cos (ωd T )}

z2−2 z {exp (−ξωn T )}{cos (ωd T ) }+{exp (−2 ξωn T ) } ](21)

Let

ψ=exp (−ξωn T ) {α sin (ωd T )−β cos (ωd T )} (22)

ρ=−2 {exp (−ξωn T )} {cos (ωd T )} (23)

λ=exp (−2 ξωn T ) (24)

η=β+ωn 2T

(25)

By substitution,

Hd ( z )= 1mωn

4 T[−βz+η ]+ 1

mωn 4 T

[ ( z−1 )2] [ βz+ψz2+ z ρ+λ ]

(26)

9

Hd ( z )= 1mωn

4 T[−βz+η ] [ z2+ z ρ+λ

z2+ z ρ+λ ]+ 1mωn

4 T[ z2−2 z+1 ] [ βz+ψ

z2+ z ρ+λ ]

(27)

Hd ( z )=1mωn

4 T[−βz ] [ z2+ z ρ+λ

z2+ z ρ+λ ]+1

mωn 4 T

[η ] [ z2+ z ρ+λz2+ z ρ+λ ]

+1mωn

4 T[z2 ] [ βz+ψ

z2+ z ρ+λ ]+1

mωn 4 T

[−2 z ] [ βz+ψz2+ z ρ+λ ]

+1mωn

4 T [ βz+ψz2+ z ρ+λ ]

(28)

H d( z )=1mωn

4 T [−βz3−β ρ z2 −βλ zz2+ z ρ+λ ]+1

mωn 4 T [η z2+ η ρz +η λ

z2+ z ρ+ λ ]+1

mωn 4T [ βz3+ψz2

z2+ z ρ+λ ]+1mωn

4T [−2 βz2−2 ψzz2+ z ρ+λ ]+1

mωn 4T [βz+ψ

z2+ z ρ+ λ ](29)

Hd ( z )=−βz3−β ρ z2 −βλ z+η z2+ η ρz +η λ+βz3+ψz2−2 βz2−2ψz+βz+ψmωn

4 T ( z2+ z ρ+ λ )

(30)

Hd ( z )=(−β ρ +ψ+η −2 β ) z2+(−βλ+ η ρ −2ψ+β ) z+(η λ+ψ )

mωn 4 T ( z2+ z ρ+λ )

(31)

10

Solve for the filter coefficient

s using the method in Reference 1.

b0 z2+b1 z+b2

z2+a1 z+a2

=[ (−β ρ +ψ+η −2 β ) z2+ (−βλ+ η ρ −2ψ+ β ) z+(η λ+ψ ) ] /[ mωn

4 T ]z2+ z ρ+ λ

(32

)

Solve for a1.

a1=ρ=−2 exp (−ξωn T )cos (ωd T )

(33)

Solve for a2.

a2= λ=exp (−2 ξωn T )

(34)

Note that the a1 and a2 coefficients are common for displacement, velocity and acceleration.

Solve for b0.

b0= [−β ρ +ψ+η −2 β ] / [mωn 4 T ]

(35)

b0= [ψ+η −β ( ρ+2 ) ] / [mωn 4T ]

(36)

b0=exp (−ξωn T ) [ α sin (ωd T )−β cos (ωd T )]+β+ωn

2 T −β [−2 exp (−ξωn T )cos (ωd T ) +2 ]mωn

4 T

(37)

11

b0=exp (−ξωn T ) [ α sin (ωdT )−β cos (ωd T )]+β+ωn

2 T +2 β [ exp (−ξωnT ) cos (ωd T ) −1 ]mωn

4T

(38)

b0=

exp (−ξωn T ) [ωn 2

ωd [−1+2 ξ2 ]sin (ωd T )−2 ξωn cos (ωd T ) ]

mωn 4 T

+2ξωn+ωn

2T +4 ξωn [exp (−ξωn T )cos (ωd T ) −1 ]mωn

4 T

(39)

b0=

exp (−ξωn T ) [ωn 2

ωd [−1+2 ξ2 ]sin (ωd T ) ]+ωn

2 T +2ξωn [ exp(−ξωnT )cos (ωd T ) −1 ]mωn

4 T

(40)

b0=

exp (−ξωn T ) [ ωn

ωd [−1+2 ξ2] sin (ωdT )]+ωn

T +2 ξ [exp (−ξωn T )cos (ωd T ) −1 ]

mωn 3 T

(41)

b0=

2 ξ [exp (−ξωnT )cos (ωd T ) −1 ]+exp (−ξωnT )[ ωn

ωd [2ξ2−1 ] sin (ωdT )]+ωn

T

mωn 3 T

(42)

12

Solve for b1.

b1=−βλ+ η ρ −2 ψ+β

mωn 4 T

(43)

b1= η ρ −2 ψ+ β (1−λ )

mωn 4 T

(44)

b1=

−2 ( β+ωn 2T ) exp(−ξωn T )cos (ωd T )−2 exp (−ξωn T ) [α sin (ωd T )−β cos (ωd T ) ]

mωn 4T

+ β [1−exp (−2 ξωn T )]mωn

4 T

(45)

b1=−2 ωn

2 T exp (−ξωn T )cos (ωd T )−2exp (−ξωn T )α sin (ωd T )+β [ 1−exp (−2 ξωnT )]mωn

4 T

(46)

b1=−2 ωn

2 T exp (−ξωn T )cos (ωd T )−2exp (−ξωn T )α sin (ωd T )+β [ 1−exp (−2 ξωnT )]mωn

4 T

(47)

13

b1=

−2 ωn 2 T exp (−ξωnT )cos (ωd T )−2

ωn 2

ωd [−1+2ξ2 ]exp (−ξωn T )sin (ωd T )

mωn 4 T

+ 2ξωn [1−exp (−2 ξωnT )]mωn

4 T

(48)

b1=

−2 ωn T exp (−ξωn T )cos (ωd T )+2ξ [1−exp (−2 ξωn T ) ]−2

ωn

ωd [2 ξ2−1 ] exp (−ξωn T )sin (ωd T )

mωn 3 T

(49)

Solve for b2.

b2 = η λ+ψmωn

4T

(50)

b2 =( β+ωn

2 T ) exp (−2 ξωn T )+exp (−ξωnT ) {α sin (ωd T )−β cos (ωdT )}mωn

4 T

(51)

b2 =

(2ξωn+ωn 2 T )exp (−2 ξωn T )+exp (−ξωn T ){ωn

2

ωd [−1+2 ξ2] sin (ωd T )−2ξωn cos (ωd T )}

mωn 4 T

(52

)

14

b2 =

(2ξ+ωn T )exp (−2 ξωn T )+exp (−ξωn T ){ωn

ωd [2 ξ2−1 ]sin (ωd T )−2ξ cos (ωd T )}

mωn 3 T

(53

)

The digital recursive filtering relationship for the displacement is

x i = −a1 x i−1 −a2 x i−2

+ b0 f i +b1 f i−1 +b2 f i−2

(54)

15

The digital recursive relationship for the displacement is thus

x i =

+2exp [−ξωn Δt ] cos [ ωd Δt ] x i−1

−exp [−2ξωn Δt ] x i−2

+1mωn

3 T {2ξ [ exp (−ξωn T )cos (ωdT ) −1 ]+exp (−ξωn T )[ωn

ωd [2ξ2−1 ] sin (ωd T )]+ωn

T } f i

+1mωn

3 T {−2 ωn T exp(−ξωn T )cos (ωdT )+2ξ [1−exp (−2 ξωnT ) ]−2

ωn

ωd [2ξ2−1 ] exp (−ξωn T )sin (ωd T )} f i−1

+1mωn

3 T {(2ξ+ωn

T ) exp (−2 ξωn T )+exp(−ξωn T ){ωn

ωd [2ξ2−1 ] sin (ωd T )−2ξ cos (ωd T )}} f i−2

(55

)

16

Velocity

The impulse response function for the velocity response is

hv( t )= 1m [exp(−ξωn t ) ] [(−ξωn

ωd)sin ωd t+cos ωd t ]

(56)

The corresponding Laplace transform is

H v(s )= 1m [ s

s2+2 ξωns+ωn2 ]

(57)

The Z-transform for the ramp invariant simulation is

H v( z )= 1m [ ( z−1 )2

Tz ]Z { L−1 [ ss2 (s2+2 ξωn s+ω

n2 ) ] } (58)

H v( z )= 1m [ ( z−1 )2

Tz ]Z { L−1 [ 1s ( s2+2ξωn s+ω

n2) ] } (59)

Evaluate the inverse Laplace transform per References 10 and 11.

L−1 [ 1s (s2+2 ξωn s+ω

n2) ]= 1ωn 2−

1ωn 2 exp(−ξ ωn t ) [cos (ωd t )+

ξωn

ωdsin (ωd t ) ]

(60)

The Z-transform for the ramp invariant simulation is

H v( z )= 1m [ ( z−1 )2

Tz ]Z { 1ωn

2−1

ωn 2 exp (−ξ ωn t ) [cos (ωd t )+

ξωn

ωdsin (ωd t )] }

(61)

17

H v( z )= 1mωn

2 [ ( z−1)2

Tz ]Z { 1−exp (−ξ ωn t ) [cos (ωd t )+ξωn

ωdsin (ωd t ) ] }

(62)

The Z-transform is evaluated using the method in Reference 12.

H v( z )= 1mωn

2 [ ( z−1)2

Tz ] [ zz−1

−z [ z−exp(−ξωnT )cos(ωdT ) ]+ z

ξωn

ωd[exp (−ξ ωnT )sin (ωdT ) ]

z2−2 z exp(−ξωnT )cos(ωd T )+exp (−2 ξωnT ) ]

(63)

H v( z )= 1mωn

2 T [ ( z−1 )2

z ] [ zz−1

−z [ z−exp (−ξωn T )cos(ωd T ) ]+z

ξωn

ωd[exp (−ξ ωn T )sin (ωd T ) ]

z2−2 z exp(−ξωn T )cos(ωd T )+exp (−2 ξωn T ) ]

(64)

H v( z )= 1mωn

2T [( z−1 )−( z−1 )2[ z−exp(−ξωnT )cos(ωd T ) ]+

ξωn

ωd[exp (−ξ ωn T )sin (ωd T ) ]

z2−2 z exp(−ξωn T )cos(ωd T )+exp (−2 ξωn T ) ] (65)

18

H v( z )= 1mωn

2T [( z−1 )−( z−1 )2z− exp(−ξωnT )[cos (ωd T )−

ξωn

ωdsin (ωdT ) ]

z2−2 z exp(−ξωnT )cos(ωd T )+exp (−2 ξωnT ) ] (66)

Let

ψ=exp(−ξωn T )[cos (ωd T )−ξωn

ωdsin (ωd T ) ]

(67)

ρ=−2 {exp (−ξωn T )} {cos (ωd T )} (68)

λ=exp (−2 ξωn T ) (69)

H v( z )= 1mωn

2T [ ( z−1 )−( z−1 )2 z−ψz2+ρ z +λ ]

(70)

H v( z )= 1mωn

2T [ ( z−1 )−(z2−2 z+1 ) z−ψz2+ρ z +λ ]

(71)

H v( z )= 1mωn

2 T [ ( z−1 )−z (z2−2 z+1 )−ψ (z2−2 z+1 )

z2+ρ z + λ ]

(72)

H v( z )= 1mωn

2T [ ( z−1 )−z3−2 z2+ z−(ψz2−2 ψz+ψ )

z2+ρ z +λ ]

(72)

H v( z )= 1mωn

2 T [ ( z−1 )− z3−2 z2+z−ψz2+2ψz−ψz2+ρ z +λ ]

(74)

19

H v( z )= 1mωn

2T [ ( z−1 )− z3+ (−2−ψ ) z2+ (1+2ψ ) z−ψz2+ρ z + λ ]

(75)

H v( z )= 1mωn

2T [ ( z−1 )+−z3+ (2+ψ ) z2−( 1+2ψ ) z+ψz2+ρ z + λ ]

(76)

H v( z )= 1mωn

2T [ ( z−1 ) z2+ ρ z +λz2+ ρ z +λ

+−z3+ (2+ψ ) z2−(1+2 ψ ) z+ψ

z2+ ρ z +λ ]

(77)

H v( z )= 1mωn

2T [ z3+ρ z2 +λz−z2−ρ z −λz2+ ρ z +λ

+−z3+(2+ψ ) z2− (1+2 ψ ) z+ψ

z2+ρ z +λ ]

(78)

H v( z )= 1mωn

2T [ z3+ ( ρ−1 ) z2 + ( λ−ρ ) z − λz2+ρ z + λ

+−z3+ (2+ψ ) z2−( 1+2ψ ) z+ψ

z2+ρ z + λ ]

(79)

H v( z )= 1mωn

2T [ (2+ψ+ρ−1 ) z2+ ( λ−ρ−1−2 ψ ) z+ψ−λz2+ρ z +λ ]

(80)

H v( z )= 1mωn

2 T [ (ψ+ρ+1) z2+( λ−ρ−2 ψ−1 ) z+ψ−λz2+ρ z +λ ]

(81)

H v( z )= 1mωn

2 T [ (ψ+ρ+1) z2+( λ−ρ−2 ψ−1 ) z+ψ−λz2+ρ z +λ ]

(82)

Solve for the filter coefficients using the method in Reference 1.

20

c0 z2+c1 z+c2

z2+a1 z+a2

= 1mωn

2T [ (ψ+ρ+1) z2+ ( λ−ρ−2ψ−1) z+ψ−λz2+ρ z+λ ]

(83)

Solve for a1.

a1=ρ=−2 exp (−ξωn T )cos (ωd T )

(84)

Solve for a2.

a2= λ=exp (−2 ξωn T )

(85)

Solve for c0.

c0=ψ+ ρ+1mωn

2 T

(86)

c0=

exp (−ξωnT )[cos( ωd T )−ξωn

ωdsin (ωd T )]−2 exp (−ξωnT )cos (ωd T )+1

mωn 2T

(87)

c0=

exp (−ξωnT )[−cos (ωd T )−ξωn

ωdsin (ωd T )]+1

mωn 2T

(88)

21

Solve for c1.

c1=λ−ρ−2 ψ−1

mωn 2T

(89)

c1=

exp (−2 ξωn T )+2 exp (−ξωn T )cos (ωd T )−2exp (−ξωnT )[cos (ωd T )−ξωn

ωdsin (ωd T ) ]−1

mωn 2T

(90)

c1=

exp (−2 ξωnT )+2exp (−ξωn T )[ ξωn

ωdsin (ωd T )]−1

mωn 2T

(91)

Solve for c2.

c2 = ψ− λmωn

2 T

(92)

c2 =

exp(−ξωn T )[cos (ωd T )−ξωn

ωdsin (ωd T ) ]−exp (−2 ξωn T )

mωn 2 T

(93)

The digital recursive filtering relationship for the velocity is

22

x i = −a1 x i−1 −a2 x i−2

+ c0 f i +c1 f i−1 +c2 f i−2

(94)

The digital recursive relationship for the velocity is thus

x i =

+2 exp [−ξωn Δt ] cos [ ωd Δt ] x i−1

−exp [−2 ξωn Δt ] x i−2

+1mωn

2 T {exp (−ξωn T )[−cos (ωd T )−

ξωn

ωdsin (ωd T ) ]+1} f i

+1mωn

2 T {exp(−2 ξωn T )+2exp(−ξωn T )[ ξωn

ωdsin (ωd T )]−1} f i−1

+1mωn

2 T {exp(−ξωn T )[cos(ωd T )−

ξωn

ωdsin ( ωd T )]−exp (−2 ξωn T )} f i−2

(95)

Acceleration

The acceleration response for a unit impulse is

y (t )= 1

m { δ( t )+exp (−ξ ωn t )[−2ξωn cos (ωd t )+ω

n2

ωd[ (2 ξ )2−1 ]sin (ωd t ) ] }

(96)

The corresponding Laplace transform is

Ha (s )= 1m [ s2

s2+2 ξωn s+ωn2 ]

(97)

23

The Z-transform is

H a( z )= 1m [ ( z−1 )2

Tz ]Z { L−1 [ s2

s2 (s2+2 ξωn s+ωn2 ) ] }

(98)

H a( z )= 1m [ ( z−1 )2

Tz ]Z { L−1 [ s2

s2 (s2+2 ξωn s+ωn2 ) ] }

(99)

Ha ( z )= 1m [ ( z−1 )2

Tz ]Z { L−1 [ 1s2+2 ξωn s+ω

n2 ] }

(100)

Evaluate the inverse Laplace transform per Reference 10 and 11.

L−1 [ 1s2+2 ξωn s+ω

n2 ]=L−1 [ 1(s+ξ ωn)2+ωd

2 ]

(101)

L−1 [ 1s2+2ξωn s+ω

n2 ]= 1ωd exp (−ξωnt ) sin (ωd

t )

(102)

The Z-transform for the ramp invariant simulation is

Ha ( z )= 1m [ ( z−1 )2

Tz ]Z { 1ωd

exp (−ξωn t )sin (ωd t ) }

(103)

Ha ( z )= 1m [ ( z−1 )2

Tz ]Z { 1ωd

exp (−ξωn t )sin (ωd t ) }

(104)

24

The Z-transform is evaluated using the method in Reference 12.

Ha( z )= 1mωd

[ ( z−1 )2

Tz ] [ z [exp (−ξ ωn t )sin (ωd t ) ]z2−2 z exp (−ξωnT )cos(ωdT )+exp (−2 ξωn T ) ]

(105)

Ha ( z )= 1mωd

T [ ( z−1 )2 [exp (−ξ ωn t )sin (ωd t ) ]z2−2 z exp (−ξωnT )cos (ωdT )+exp (−2 ξωn T ) ]

(106)

Ha ( z )= exp (−ξ ωn t ) sin (ωd t )

mωd T [ ( z−1 )2

z2−2 z exp(−ξωn T )cos (ωd T )+exp (−2 ξωnT ) ]

(107)

Ha ( z )= exp (−ξ ωn t ) sin (ωd t )

mωd T [ z2−2 z+1

z2−2 z exp(−ξωnT )cos (ωd T )+exp (−2 ξωnT ) ]

(108)

Solve for the filter coefficients using the method in Reference 1.

d0 z2+d1 z+d2

z2+a1 z+a2

=exp (−ξ ωn t )sin (ωd t )

mωd T [ z2−2 z+1

z2−2 z exp(−ξωn T )cos (ωdT )+exp (−2 ξωnT ) ]

(109)

Solve for a1.

a1=ρ=−2 exp (−ξωn T )cos (ωd T )

(110)

Solve for a2.

a2= λ=exp (−2 ξωn T )

(111)

25

Solve for b0.

d0=exp (−ξ ωn t )sin (ωd t )

mωd T

(112)

Solve for d1.

d1=−2 d0

(113)

Solve for b2.

d2=d0

(114)

The digital recursive filtering relationship for the acceleration is

x i = −a1 x i−1 −a2 x i−2 + d0 f i +d1 f i−1 +d2 f i−2 (115)

x i = +2 exp [−ξωn Δt ] cos [ωd Δt ] x i−1 − exp [−2ξωn Δt ] x i−2

+ξωn

exp (−ξ ωn t )sin (ωd t )mωd

2T

{ f i −2 f i−1 + f i−2 }

(116)

26

There are three response parameters: displacement, velocity and acceleration.

If two are known, the third can be calculated via equation (7).

27

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 = 0.97 inch minimum = -0.971 inch overall = 0.601 inch RMS Velocity Response

maximum = 61.6 in/sec minimum = -61.6 in/sec overall = 38.1 in/sec RMS Acceleration Response

maximum = 9.98 G minimum = -9.97 G overall = 6.17 G RMS

28

-1.0

-0.5

0

0.5

1.0

0 0.5 1.0 1.5 2.0

TIME (SEC)

FOR

CE

(lbf

)

APPLIED FORCE

Figure 3.

29

-1.2

-1.0

-0.8

-0.6

-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

LaplaceRamp Invariant

TIME (SEC)

DIS

P (i

nch)

SDOF DISPLACEMENT RESPONSE fn=10 Hz Q=10

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.

30

-100

-80

-60

-40

-20

0

20

40

60

80

100

0 0.5 1.0 1.5 2.0

LaplaceRamp Invariant

TIME (SEC)

VE

LOC

ITY

(inc

h/se

c)

SDOF VELOCITY RESPONSE fn=10 Hz Q=10

Figure 5.

There is excellent agreement between the two velocity curves as shown in Figure 5.

31

-14

-12

-10

-8

-6

-4

-2

0

2

4

6

8

10

12

14

0 0.5 1.0 1.5 2.0

LaplaceRamp Invariant

TIME (SEC)

AC

CE

L (G

)

SDOF ACCELERATION RESPONSE fn=10 Hz Q=10

Figure 6.

There is excellent agreement between the two acceleration curves as shown in Figure 6.

32

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.

33

k1

k2

x1

m2

m1

k3x2

f2(t)

f1(t)

APPENDIX A

MDOF Example

The example from Reference 2 is repeated here.

Figure A-1.

Assume 5% damping for each mode. Assume zero initial conditions.

34

The parameters are

Table A-1. Parameters

Variable Value Unitm1 3.0 lbf sec^2/in

m2 2.0 lbf sec^2/in

k1 400,000 lbf/in

k 2 300,000 lbf/in

k 3 100,000 lbf/in

B1 100 lbf

B2 200 lbf

55 Hz 100 Hz

The mass matrix is

M=[m1 00 m2 ]=[3 0

0 2 ] (A-1)

The stiffness matrix is

K=[k1+k2 −k2

−k2 k2+k3 ] =[700 ,000 −300 ,000−300 ,000 400 ,000 ]

(A-2)

Each of the two forcing functions is synthesized using Matlab script: generate.m, as a pre-processing step.

The modal transient analysis is performed using Matlab script: mdof_modal_arbit_force_ri.m

35

>> 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: stiff_case5 Natural Frequencies No. f(Hz)1. 48.552 2. 92.839 Modes Shapes (column format)

ModeShapes =

0.3797 -0.4349 0.5326 0.4651

36

Enter duration(sec) 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...

37

-0.003

-0.002

-0.001

0

0.001

0.002

0.003

0 0.05 0.10 0.15 0.20 0.25 0.30

dof 2dof 1

TIME (SEC)

DIS

P (I

NC

H)

DISPLACEMENT

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.

38

-1.5

-1.0

-0.5

0

0.5

1.0

1.5

0 0.05 0.10 0.15 0.20 0.25 0.30

dof 2dof 1

TIME (SEC)

VE

L (IN

CH

/SE

C)

VELOCITY

Figure A-3.

-1.5

-1.0

-0.5

0

0.5

1.0

1.5

0 0.05 0.10 0.15 0.20 0.25 0.30

dof 2dof 1

TIME (SEC)

AC

CE

L (G

)

ACCELERATION

39

Figure A-4.

40