introduction1 1.1 nastran model · nastran simulations are mainly performed in the frequency domain...
TRANSCRIPT
Introduction............................................................................................................................1
1 NASTRAN model and analytical model ..............................................................................2
1.1 NASTRAN model ........................................................................................................2
1.2 Analytical model...........................................................................................................3
1.3 Comparison of NASTRAN and analytical model ..........................................................7
2 Transformation to time domain............................................................................................9
2.1 Displacement to velocity and acceleration.....................................................................9
2.2 Mirroring ....................................................................................................................11
2.3 Inverse Fourier transformation ....................................................................................12
2.4 Tail .............................................................................................................................13
2.5 Further improvements .................................................................................................16
2.5.1 Interpolation.........................................................................................................16
2.5.2 Adding zero Hz....................................................................................................17
Conclusion and recommendations ........................................................................................18
References............................................................................................................................20
List of symbols.....................................................................................................................21
Appendix A Derivation of the analytical model parameters ..................................................22
Appendix B Fourier transformation .....................................................................................23
B.1 Basics.........................................................................................................................23
B.2 Discrete Fourier Transformation.................................................................................23
B.3 Fast Fourier transform................................................................................................24
B.4 Properties of the Fourier transform.............................................................................26
B.5 Symmetry...................................................................................................................26
B.6 Error sources in Fourier transform..............................................................................27
B.6.1 Signal leakage .....................................................................................................27
B.6.2 Aliasing...............................................................................................................28
Appendix C Convolution......................................................................................................30
C.1 Basic theory of convolution........................................................................................30
C.2 Properties of the convolution product .........................................................................31
Appendix D Testing methods ..............................................................................................32
D.1 Transfer function estimate method .............................................................................32
D.2 Analytical transfer function determination..................................................................33
Appendix E Deconvolution..................................................................................................35
Entire or partial reproduction of the content of this publication in any form, without
preliminary authorization by letter by DAF is illegal, except for restrictions registered by law.
This prohibition also includes entire or partial rewriting of the publication.
DCT 2006-117 1
Introduction
The drive-off behavior is an important characteristic for a truck. It influences both the comfort
and the performance of the truck. The drive-off behavior of a truck depends mainly on the
control strategy of the driveline (engine, clutch and gear box), the characteristics of the clutch
facing and the dynamic behavior of the vehicle structure (drive shafts, chassis, cabin
suspension).
In order to get the drive-off behavior on a high level, optimization and tuning of the control
strategy is a necessary part of the process. This optimization and tuning is mainly done on a
real vehicle, which takes a lot of time and effort. In order to speed up and improve the process
of optimizing and tuning and to be able to anticipate on future vehicles changes, it should be
possible to predict and calculate the drive off behavior up front. A necessary part of the
simulation models to do so is, besides the control strategy, the characteristic of the vehicle
dynamics.
Within DAF, NASTRAN models of the total vehicle dynamics are available. Recently these
models have been extended with a model of the driveline, which includes a clutch plate,
gearbox drive shaft and half-shafts. With these NASTRAN models any arbitrary transfer
function can be determined. These transfer functions are all determined in the frequency
domain.
Because the drive-off behavior is mainly a transient phenomenon, simulations in the time
domain with MATLAB/SIMULINK software are very useful. Therefore the available
NASTRAN model transfer functions will be transformed from the frequency domain into the
time domain. In this report the method of translating a FRF into a time domain impulse
response is discussed.
First an overview of the NASTRAN model is shown, together with an overview of an
analytical model that is used to approximate the NASTRAN driveline model. Next the
necessary steps to get a proper transformation of the frequency domain transfer function to the
time domain will be shown. The transformation routine that will be developed will be a
general usable tool. Finally a conclusion and recommendations will be given.
DCT 2006-117 2
1 NASTRAN model and analytical model
In order to get an optimal drive-off behavior, an optimization is necessary. The optimization
of the drive-off behavior is up to now mainly done by adjusting/tuning the vehicle and control
strategy parameters during tests on a real vehicle. This routine takes a lot of effort and time.
Therefore there is a need to make the drive off behavior more predictable by means of
simulations, which will speed up and improve this process/routine. For this kind of
simulations the vehicle/driveline dynamics are a major part of the simulation model. Within
DAF the software package NASTRAN is used to model the vehicle behavior/dynamics. The
NASTRAN simulations are mainly performed in the frequency domain and the output consist
of frequency response functions (FRF)
The manoeuvring and drive-off of a vehicle are mainly transient phenomena. Therefore a
simulation of this behavior in the time-domain using is more useful.
Therefore the output of the NASTRAN model (FRF’s) will be transformed from frequency to
time domain. This transformation will be discussed in chapter 2.
In this chapter a description of the NASTRAN vehicle model will be given.
Next a basic analytical model will be shown. This model can be used to approximate and
verify some transfer functions provided by the NASTRAN model. It will be used to describe
the time domain transformation in chapter 2.
1.1 NASTRAN model
To investigate the vehicle behavior of the vehicle, as was described before a NASTRAN
model is used. The NASTRAN model used was originally designed to perform fatigue and
comfort analyses. The original NASTRAN model is shown below.
Figure 1.1: Original NASTRAN model used for fatigue and comfort analyses.
The original NASTRAN model has been expanded with a model that describes the driveline.
The driveline consists of a clutch plate, gearbox, drive shaft, 2 cardanic joints, a differential
gear and 2 half-shafts.
The complete model can be used to determine the transfer functions between the clutch torque
and the vehicle and driveline displacements. These transfer functions are a necessary part in
the control loop to be optimized. Furthermore the influence of parameters like gear ratios and
shaft stiffness on the vehicle behavior can be investigated.
The NASTRAN driveline model is shown in the next figure.
DCT 2006-117 3
Figure 1.2: NASTRAN driveline model with all its components
The NASTRAN model determines the transfer functions by solving the general equation:
( ) ( ) ( ) ( )tftqKtqDtqM =++ ɺɺɺ (1.1)
Where ( )tq is the displacement column, ( )tf de force column, M de mass matrix, D the
damping matrix and K the stiffness matrix.
The general equation (1.1) is solved using:
[ ]aa FHFKDjMq ˆˆˆ
12 =+Ω+Ω−=−
(1.2)
Where ( ) tjeqtq Ω= ˆ and ( ) tj
a eFtf Ω= ˆ
[ ] 12 −+Ω+Ω−= KDjMH is called the transfer function matrix or the matrix of frequency
response functions (FRF matrix). All equations were found in [1].
1.2 Analytical model
The drive-behavior is also determined by the characteristics of the vehicle driveline. The
driveline can be seen as a two rotating mass spring-damper system.
A simple analytical model of the two mass spring-damper system can be used to approximate
and verify some of the NASTRAN vehicle model transfer functions.
In this analytical model the first mass represents the clutch plate and the primary part of the
gearbox. The second mass represents the vehicle with its mass. The stiffness k and the
damping b stand for the characteristics of the drive shaft and the two half-shafts.
The method of determining the values of J1, J2, k and b can be found in appendix A.
The two mass spring-damper system is shown in the next figure.
DCT 2006-117 4
Figure 1.3: Schematic overview of the analytical two mass spring damper system
The next equations can be derived for the standard two mass spring-damper system.
=
⋅
−
−+
⋅
−
−+
⋅
020
01
2
1
2
1
2
1 M
kk
kk
bb
bb
J
J
α
α
α
α
α
αɺ
ɺ
ɺɺ
ɺɺ (1.3)
Using tje
ωαα ˆ= and tjeMM
ωˆ= This equation can also be written as:
=
⋅
−
−+
−
−+
−
0
ˆ
ˆ
ˆ
20
01
2
12 M
kk
kk
bb
bbj
J
J
α
αωω (1.4)
The equation for the first disc can be written as:
( ) ( ) MkbjJ ˆˆˆˆˆˆ212111
2 =−+−+− ααααωαω (1.5)
And the equation for second disc:
( ) ( ) 0ˆˆˆˆˆ121222
2 =−+−+− ααααωαω kbjJ (1.6)
( ) ( )122
2 ˆˆ αωαωω kbjkbjJ +=++− (1.7)
1
2
22ˆˆ α
ωωω
α
++−
+=
kbjJ
kbj (1.8)
Now the equation for the first disc can be rewritten by substituting equation (1.8)
( ) MkbjJ
JkbjJ ˆˆ
1
2
2
2
2
1
2 =
++−
−++− α
ωωω
ωω (1.9)
So the transfer function between the excitation moment and the displacement of the first disc
is:
k
M J1
b
1α 2α
J1
J2
DCT 2006-117 5
( )( )
++−
−++−
=
kbjJ
JkbjJ
H d
ωωω
ωω
ω
2
2
2
2
1
2
1
1 (1.10)
Using αωα j=ɺ the transfer function between excitation moment and rotational velocity of
the first disc can be determined:
( )
++−
−
−+
=
kbjJ
JkjbJj
H v
ωωω
ωω
ω
2
2
2
2
1
1
1 (1.11)
Next, using αωα 2−=ɺɺ the transfer function between the excitation moment and the
acceleration of the first disc can be written as:
( )( )
++−++
=
kbjJ
JkbjJ
H a
ωωω
ω
2
2
21
1
1 (1.12)
In the figure below an example of the transfer functions for the first disc of the two mass
spring-damper system same system is shown. The natural frequencies nω are indicated, where
1nω is an anti-resonance and 2nω a resonance. The static gains for the acceleration transfer
function are indicated in the figure as well.
10-2
10-1
100
101
102
-80
-60
-40
-20
0
20
frequency [Hz]
|H|
displacement transfer function
10-2
10-1
100
101
102
-200
-150
-100
-50
0
frequency [Hz]
an
gle
(H)
ωn1
=√ k/J2
ωn1
=√ k/J1
10-2
10-1
100
101
102
-60
-40
-20
0
20
frequency [Hz]
|H|
velocity transfer function
10-2
10-1
100
101
102
-100
-50
0
50
100
frequency [Hz]
an
gle
(H)
ωn1
=√ k/J2
ωn1
=√ k/J1
10-2
10-1
100
101
102
-40
-20
0
20
40
frequency [Hz]
|H|
acceleration transfer function
10-2
10-1
100
101
102
0
50
100
150
200
frequency [Hz]
an
gle
(H)
1/(J1+J2)
1/J1
Figure 1.4: Transfer functions for the first disc of the two mass spring-damper system.
DCT 2006-117 6
Next the transfer functions for the second mass are determined.
Equation (1.5) can be rewritten into:
( ) ( )211
2 ˆˆˆ αωαωω kbjMkbjJ +−=++− (1.13)
( )2
1
2
1
21ˆ
ˆˆ α
ωωω
ωωα
++−
+−
++−=
kbjJ
kbj
kbjJ
M (1.14)
Substituting (1.14) in to equation (1.6) gives:
( ) MkbjJ
kbj
kbjJ
JkbjJ ˆˆ
1
21
1
2
1
2
2
2
++−
+=
++−
−++−
ωωω
αωω
ωωω (1.15)
So the transfer function between the excitation moment and the displacement of the second
disc is:
( )( )
++−
−++−
++−
+
=
kbjJ
JkbjJ
kbjJ
kbj
H d
ωωω
ωω
ωωω
ω
1
2
1
2
2
2
1
2
2 (1.16)
Using αωα j=ɺ the transfer function between excitation moment and rotation velocity of the
second disc can be determined.
( )
++−
−
−+
++−
+
=
kbjJ
JkjbJj
kbjJ
kbj
H v
ωωω
ωω
ωωω
ω
1
2
1
2
2
1
2
2 (1.17)
Next, using αωα 2−=ɺɺ the transfer function between the excitation moment and the
acceleration of the first disc can be written as:
( )( )
++−++
++−
+
=
kbjJ
JkbjJ
kbjJ
kbj
H a
ωωω
ωωω
ω
1
2
12
1
2
2 (1.18)
In the next figure the transfer functions of the second disc on displacement, velocity and
acceleration level are shown.
DCT 2006-117 7
10-2
10-1
100
101
102
-150
-100
-50
0
50
frequency [Hz]
|H|
displacement transfer function
10-2
10-1
100
101
102
-350
-300
-250
-200
-150
frequency [Hz]
an
gle
(H)
ωn=√ k/J1
10-2
10-1
100
101
102
-100
-80
-60
-40
-20
0
frequency [Hz]
|H|
velocity transfer function
10-2
10-1
100
101
102
-300
-250
-200
-150
-100
-50
frequency [Hz]
an
gle
(H)
ωn=√ k/J1
10-2
10-1
100
101
102
-60
-40
-20
0
20
frequency [Hz]
|H|
acceleration transfer function
10-2
10-1
100
101
102
-200
-150
-100
-50
0
frequency [Hz]
an
gle
(H)
ωn=√ k/J1
1/(J1+J2)
Figure 1.5: Transfer functions for the second disc
1.3 Comparison of NASTRAN and analytical model
To compare the analytical model with the NASTRAN model, the transfer function between
clutch torque and the velocities of the first mass (clutch) and the second mass (end of half-
shaft) are shown in the figure below for both the models. For the analytical model the
parameters in appendix A were used. The transfer functions were calculated for third gear.
10-2
10-1
100
101
102
-60
-40
-20
0
20
frequency [Hz]
|H|
velocity transfer function first mass
analytical modelNastran model
10-2
10-1
100
101
102
-100
-50
0
50
100
frequency [Hz]
an
gle
(H)
10-2
10-1
100
101
102
-100
-80
-60
-40
-20
0
frequency [Hz]
|H|
velocity transfer function second mass
analytical modelNastran model
10-2
10-1
100
101
102
-300
-200
-100
0
frequency [Hz]
an
gle
(H)
Figure 1.6: Comparison of the analytical model with the NASTRAN model, showing the transfer functions
between clutch torque and the velocity of the first mass (clutch) and the velocity of the second mass (end of
half-shaft) for third gear.
DCT 2006-117 8
For the first mass, the analytical model coincides well with the NASTRAN model, as can be
seen in figure 1.6. For the motion of the first mass, which is in fact the clutch, the simple two
mass spring damper system is a good approximation for the far more complicated NASTRAN
model.
For the second mass, the approximation of the NASTRAN model by the analytical model is
less accurate. The approximation of the complete vehicle structure by just one simple mass is
not accurate. Furthermore for both the first and second mass only one degree of freedom is
taken into account. The behavior of the vehicle (thus the second mass) cannot be described
using just one degree of freedom (for the first mass one degree of freedom is correct because
there is just one).
Therefore the conclusion can be drawn that in the investigation of the drive-off behavior the
NASTRAN model cannot be totally replaced by the analytical model. Parameters like the
vertical cabin acceleration, which are important parameters in the vehicle behavior, cannot be
determined with the analytical model. Therefore the NASTRAN model will still have to be
used and the NASTRAN model transfer functions will have to be transformed in order to be
able to use in MATLAB/SIMULINK.
DCT 2006-117 9
2 Transformation to time domain
The drive-off behavior is mainly a transient phenomenon and furthermore influenced by the
control strategy. Software like MATLAB/SIMULINK is therefore very useful to investigate
the drive-off problems. In order to use this software the NATRAN model output first has to be
transformed from the frequency domain to time domain. This can be done in several ways.
Making a fit of the frequency domain transfer function is a method that is often used. The fit
is then implemented in MATLAB/SIMULINK (using an s-block). Disadvantage of using a fit
is that it will practically always contain a small error. When the vehicle layout is changed a
new fit has to be made. The influence of the change in layout on the vehicle drive-off
behavior will be difficult to determine. The change in drive-off behavior could also be a result
of an error in the fit that was used to approximate the frequency domain transfer function.
Therefore another method will be used; the Fourier transformation method.
With this method the transfer function or frequency response function that was acquired using
the NASTRAN vehicle model will be converted to an impulse response function in the time
domain. The basic idea of the Fourier transformation is that any function ( )xf can be formed
as a summation of a series of cosine and sine terms of increasing frequency. This means that
any space or time varying data can be transformed into the frequency domain and vice versa.
More information on the Fourier transform theory can be found in appendix B.
In order to get a proper time domain transform, the following steps have to be followed:
2.1 Displacement to velocity and acceleration
The output of the NASTRAN model is a transfer function that describes the systems behavior
on displacement level. This means that all outputs are expressed as displacements. In the
research that is done in order to investigate the drive-off behavior the relation between the
clutch velocity and the clutch torque is an important parameter. To investigate this parameter
the transfer function on displacement level is converted to a transfer function on velocity
level.
Using equation (B.14):
)(2)(
uuFjdu
udFπ= (2.1)
With u the frequency f in Hz and F the NASTRAN transfer function ( )fH this gives
( ) ( )ffHjdf
fdHπ2= (2.2)
In the next figure a typical NASTRAN model output transfer function on velocity level is
shown.
DCT 2006-117 10
Figure 2.1: transfer function between clutch torque and clutch speed
Furthermore the vertical cabin acceleration is an important output concerning the drive off
comfort.
Therefore the NASTRAN model transfer function between clutch torque and cabin
displacement has to be converted to a transfer function between clutch torque and cabin
acceleration. Using equation (B.15):
( ) ( ) ( )fHfdf
fHd 2
2
2
2π−= (2.3)
In the figure below the NASTRAN model transfer functions between clutch torque and the
vertical cabin acceleration are shown for the first seven gears.
Figure 2.2: Transfer function between clutch torque and vertical cabin acceleration
DCT 2006-117 11
2.2 Mirroring
When a signal consists of real parts only, the Fourier transform of the signal is symmetrical
with respect to a folding frequency (see appendix B.5). This means that ( ) )Re(Re ff −= and
( ) ( ))ImIm ff −= with ( )fRe the real part of ( )fH and ( ))Im f the imaginary part of ( )fH .
This concept can also be used the other way around. In order to have an inverse Fourier
transform signal that is real, the original signal has to be symmetric.
Thus, before the NASTRAN model can be transformed to the time domain the NASTRAN
transfer function first has to be made symmetrical. This can be done by folding the original
transfer function data vector ( )fH around a folding frequency foldf (see figure B.3). The
folding frequency is chosen to be the maximum frequency of the NASTRAN model. This is
done in order to keep the information in the new symmetric transfer function the same as in
the original transfer function. Furthermore the mirrored part of the original transfer
function ( )fH is equal to the complex conjugated of ( )fH . This means that the real part
of ( )fH is line symmetrical and the imaginary part of ( )fH is point symmetrical with
respect to the folding frequency (see B.5).
The transfer function ( )fH can be made symmetrical in MATLAB using:
( ) ( ) ( ) ( )( )( )[ ] ( )( ) ( )( )( )[ ]( )endHimagfliplrendHimagjendHrealfliplrendHHrealfH s :2,0,:2,0:2,, −+=
With ( )fHH = the original NASTRAN model output transfer function (2.4)
The zero for the imaginary part and ( )endH for the real part indicates the values of ( )fH at
the folding frequency. This value is chosen to be zero for the imaginary part (otherwise the
method will not work) and ( )endH for the real part so that it does not influence the transfer
function.
In the figure below an example is shown of a transfer function and the corresponding
symmetric transfer function
frequency
|H|
transfer function
frequency
|H|
symmetric transfer function
frequency
ang
le(H
)
frequency
ang
le(H
)
Figure 2.3: Transfer function and the corresponding symmetric transfer function
DCT 2006-117 12
2.3 Inverse Fourier transformation
Now that the NASTRAN model transfer function is made symmetric, it can be transformed
from the frequency domain to the time domain. This can be done in MATLAB using the
inverse fast Fourier transform commando ifft.
( ) ( )( )fHifftth s= , with ( )fH s the symmetric transfer function (2.5)
This transformation will result in an impulse response function.
In the figure below the impulse response functions for the end of the right half-shaft are
shown on displacement, velocity and acceleration level. The behavior of the end of the right
(or left) half-shaft is the same as the behavior of the second mass of the analytical model.
0 5 10 15 201
1.5
2
2.5
3
3.5
4
4.5
5
5.5x 10
-3
time [s]
h(t
)
displacement impulse response function
0 5 10 15 20-8
-6
-4
-2
0
2
4x 10
-3
time [s]
h(t
)
velocity impulse response function
0 5 10 15 20-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
time [s]
h(t
)
acceleration impulse response function
Figure 2.4: Impulse response functions describing on displacement, velocity and acceleration level for the
end of the right half-shaft
The impulse response function that will be used to investigate the clutch judder behavior has
to be a finite impulse response. This means that the impulse response function has to be
damped out completely at the end of the time base. The finite impulse response function has
reached a constant value at the end of the time base, which is maintained till infinity.
As can be seen in figure 2.4, the impulse response function on acceleration level is the only
impulse response function that is finite. This is because the system is not constrained in space.
When the system is subjected to an impulse, the resulting displacement and velocity will keep
varying till infinity, but the acceleration will be finite. Therefore, the acceleration impulse
response function for this particular transfer function is the only function that can be used.
DCT 2006-117 13
Because the clutch facing properties are depending on slip speed, the acceleration impulse
response function has to be transformed to a velocity impulse response function.
Therefore, an integration step has to be applied. In MATLAB/SIMULINK this can be done by
adding an integrator to the model.
2.4 Tail
In the figure below a closer view on the acceleration impulse response function is given
0 5 10 15 20-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
time [s]
h
18 18.5 19 19.5 20
-0.1
-0.05
0
0.05
0.1
0.15
h
Figure 2.5: Acceleration impulse response function, describing the behavior of the clutch plate
The figure on the right side shows the close-up of the tail.
As can be seen in the figure, the impulse response does not totally damp out in time. There is
a little “tail” at the end of the impulse response function. This is not realistic and does not
correspond with what will happen in practice. Furthermore this phenomenon will give
problems when time response signals will be determined using this impulse response function.
In order to determine the influence of the tail a test is performed. In this test white noise is
applied to the impulse response and the transfer function of the impulse response filter is
determined using transfer function estimate. This transfer function is then compared to the
original transfer function. The testing method that was used is described in appendix D.
In the figure below the test of the impulse response function is shown.
10-3
10-2
10-1
100
101
-80
-60
-40
-20
0
20
40
|H|
bode plot original frf versus frf determined with impulse response
frequency [Hz]
original frfimpulse response frf
10-3
10-2
10-1
100
101
-200
-100
0
100
200
frequency [Hz]
ang
le(H
)
Figure 2.6: Testing the impulse response function
DCT 2006-117 14
Figure 2.6 shows that the transfer function of the tested impulse response function does not
coincide with the original transfer function. Especially in the low frequency area the
magnitude and phase of the tested impulse response function differ from the original.
The problem lies in the begin conditions of the filter that was used to test the impulse
response function. The filter uses convolution to determine the output signal (see appendix C).
The white noise input signal is convoluted with the impulse response function. In this
convolution the impulse response function is used as a periodic repeating time signal. This is
illustrated in the next figure.
periodic repeating of impulse reponse function
time [s]
h(t)
Figure 2.7: Periodic repeating of impulse response function
The idea of convolution is that the output signal at a certain point is equal to the multiplication
of the input signal with the value of the impulse response at that certain point summed with
the multiplications of the all the previous input signals with the previous values of the impulse
response function (see appendix E). Therefore the output signal values at the beginning of a
period are also determined by the values of the output signal of the previous period.
In the testing method that was used to determine figure 2.6 only one period was used.
Therefore the influence of the tail (which would be the end of the previous period, see figure
2.7) on the first output values is not taken into account. In others words, the begin conditions
are taken equal to zero, which is not the case because of the occurrence of the tail.
In order to take the begin conditions into account the testing method has to be changed. Now
a white noise input signal consisting of multiple periods is applied to the impulse response
function. The output signal will therefore also consist of multiple periods. For the transfer
function estimate only one period of the input and output signal is used, but not the first
period. This period will contain the influence of the previous period, thus the influence of the
tail. The transfer function corresponding to the impulse response function using this method is
shown in the figure 2.8, together with the original transfer function.
DCT 2006-117 15
10-3
10-2
10-1
100
101
-80
-60
-40
-20
0
20
40
|H|
bode plot original frf versus frf determined with impulse response
frequency [Hz]
original frfimpulse response frf
10-3
10-2
10-1
100
101
-50
0
50
100
150
200
frequency [Hz]
ang
le(H
)
Figure 2.8: Testing the impulse response function with begin conditions
As can be seen in figure 2.8, the transfer function determined with the impulse response
coincides well with the original transfer function. The impulse response function with the tail
is therefore a good representation of the dynamic behavior of the system.
However, the occurrence of the tail causes the problem that the begin conditions of the
impulse response are not equal to zero. Therefore the transient phase of the output signal
using the impulse response function is not correct. The big disadvantage of the tail is therefore
that the impulse response function can only be used for steady state situations or for multiple
period input signals.
Note that in the real world, the begin conditions will always be equal to zero. The system
cannot react and move before an excitation is applied to the system.
Another test to check the influence of the tail is an analytical transfer function determination.
This method is described in appendix D.1. With this method the influence that cutting of the
tail at the dynamic behavior is investigated. The result is the same as for the white noise
method; the impulse response with the tail is a good representation, the impulse response
without the tail is not. Furthermore in this test steady state signals are used to determine the
transfer function. This method again indicates that the impulse response is a good
representation, but only for steady state situations.
The cause of the tail lies in the limited frequency range that is analyzed in the NASTRAN
model. The transfer functions are only determined up to a maximum frequency, maxf . The
NASTRAN transfer function can be seen as a multiplication of the infinite transfer function
with a window function. This concept is illustrated in the figure next using the analytical
model.
10-1
100
101
102
10-2
100
102
frequency [Hz]
|H|
infinite model
10-1
100
101
102
0
50
100
150
200
frequency [Hz]
an
gle
(H)
10-1
100
101
102
0
0.2
0.4
0.6
0.8
1
frequency [Hz]
w
window function
10-1
100
101
102
10-2
100
102
frequency [Hz]
|H|
calculated finite model
10-1
100
101
102
0
50
100
150
200
frequency [Hz]
an
gle
(H)
Figure 2.9: Infinite transfer function (left), window function and the calculated finite model
(= NASTRAN model)
DCT 2006-117 16
In the frequency domain, the infinite transfer function is multiplied with the window function
to obtain the calculated finite transfer function. In time domain this is equal to a convolution
of the impulse response function of the infinite transfer function with the impulse response of
the window function (equation C.4). In the figure below the impulse response functions of the
infinite transfer function, the window function and the finite transfer function are shown.
0 2 4 6 8 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time [s]
h
impulse response infinite transfer function
0 2 4 6 8 10-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
time [s]
h
impulse reponse window function
0 2 4 6 8 10-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4impulse response finite transfer function
time [s]
h
Figure 2.10: Impulse response functions of the infinite transfer function, the window function and the
finite transfer function
As can be seen in figure 2.10, the impulse response function of the window function causes
the tail at the end of the impulse response of the finite transfer function. Furthermore also the
first part of the impulse response is influenced by the window function. This can also be seen
in figure 2.10: the amplitudes of the infinite and finite impulse responses are not the same.
The size of the tail is determined by the amount of information that is in the model just before
the maximum frequency where the model is cut off. The occurrence of (anti-) resonance peaks
and the presence of a slope in the transfer function just before the maximum frequency will
increase the size of the tail. The size tail will therefore increase for increasing gears (the
resonance peak moves towards the maximum frequency).
In practice, only the finite NASTRAN model transfer function and the window function are
known. To reconstruct the original infinite transfer function and its corresponding impulse
response function a deconvolution has been performed (see appendix E). With this
deconvolution an attempt has been done to filter out the influence of the window function in
the calculated NASTRAN transfer function. The results of this deconvolution, shown in
appendix E, are unsatisfying. With the deconvolution method that was used, it is not possible
to remove the tail without changing the dynamic behavior represented by the impulse
response function. Therefore the calculated impulse response function can still only be used
for steady state situations or for periodic input signals. The representation of the transient
phase using the impulse response function with the tail is not correct.
2.5 Further improvements
There are a few methods left to improve the results of the Fourier transformation if necessary.
2.5.1 Interpolation
The time base of the impulse response function is linearly proportional to the frequency step
of the original transfer function, as is shown in equation (B.8). If the impulse response
function is not yet damped out at the end of the time base, interpolation can be used. When
the original transfer function is interpolated, more data points can be created. The frequency
step then becomes smaller, creating a larger time base for the impulse response function.
DCT 2006-117 17
Decreasing the frequency step in the original NASTRAN model could also do this but this
would increase the calculation time NATRAN drastically.
Therefore, interpolation of the NASTRAN transfer function is a more suitable solution.
2.5.2 Adding zero Hz
When an analytical model is used the zero Hz frequency point information can be determined
easily. For NATRAN models this is more complicated. NATRAN uses the inverse of the
transfer function matrix to calculate the transfer function, as was shown in paragraph 1.1. This
method will not work for a frequency of zero. This problem can be avoided by choosing a
very small starting frequency for the NASTRAN model, which is in fact a simulation of the
zero frequency point. This method is especially attractive to use when a lot of information
about the dynamic behavior of the analyzed system is situated at low frequencies.
DCT 2006-117 18
Conclusion and recommendations
The drive-off behavior is an important characteristic for a truck. In order to get the drive-off
behavior on a high level, optimization is necessary. This optimization was mainly done on a
real vehicle, which takes a lot of time and effort. In order to speed up and improve the
process of optimizing and tuning and to be able to anticipate on future vehicles changes, it
should be possible to predict and calculate the drive off behavior up front. A necessary part
of the simulation models to do so is, besides the control strategy, the characteristic of the
vehicle dynamics.
The NASTRAN model outputs are transfer functions, which are all in the frequency domain.
Because the drive-off behavior is mainly a transient phenomenon and furthermore influenced
by the control strategy of the driveline, a simulation of the drive-off behavior is more useful.
In order to be able to do these time domain simulations, the NASTRAN model transfer
function have to be transformed from the frequency domain to time domain. This is done
using Fourier Transformation where the frequency domain transfer function will be
transformed into a time domain impulse response.
First a description of the NASTRAN model that was used was given together with an
analytical model that was used to approximate some of the NASTRAN model output transfer
functions. The analytical model can be used to approximate the transfer function between the
clutch torque and the clutch motion.
Next the steps used for the transformation from the frequency domain to the time domain
using Fourier transformation have been described. Furthermore the determined impulse
response has been tested in order to check the correctness.
The conclusion that can be drawn after following this procedure:
• The analytical model can be used to approximate the transfer function between clutch
torque and clutch motion. The approximation of the transfer function between clutch
toque and the motion of the half-shaft using the analytical model is less accurate and
therefore not useful. Because the analytical model can only be used to determine a
small part of the transfer functions the NASTRAN model will still have to be used.
• The impulse response function that was determined for the use in time domain can
only be used for steady state situations or for periodic input signals. This problem is
caused by the occurrence of a “tail” at the end of the impulse response. The tail is
caused by the limited frequency range that was analysed in the NASTRAN model. In
fact the real transfer function is multiplied with a window function. In order to filter
out the influence of this window function a deconvolution has been performed. The
results of this deconvolution however are unsatisfying.
Using these findings the next recommendations are done:
• Check the possibility to determine the impulse response function with the NATRAN
model by applying an impulse as input instead of a frequency excitation. The problem
with the tail will then probably not occur. Furthermore the transient behaviour will
surely be taken into account.
• Increase the maximum frequency for the NASTRAN model calculation. This will
decrease the effect of windowing and thus the size of the tail. Disadvantage of this
method is the increasing calculation time that is needed.
DCT 2006-117 19
• Look for other possible methods for deconvolution. With a well-operating
deconvolution method, the effect of windowing could possibly be filtered out.
DCT 2006-117 20
References
[1] Bram de kraker & Dick H. van Campen
Mechanical Vibrations
April 2001
[2] Yerin Yoo
Tutorial on Fourier Theory
Internet paper, March 2001
[3] Prof David Heeger
Signals, linear systems and convolution
Internet paper, September 2000
DCT 2006-117 21
List of symbols
engT = engine torque [Nm]
engω = engine speed [rad/s]
cT = clutch torque [Nm]
cω = clutch speed [rad/s]
deng ,ω = desired engine speed [rad/s]
cs = distance between clutch plates [mm]
M = mass matrix
D = damping matrix
K = stiffness matrix.
H = frequency domain transfer function
α = rotational displacement [rad]
αɺ = rotational speed [rad/s]
αɺɺ = rotational acceleration [rad/s2]
1J = inertia first mass of analytical model [kg.m2]
2J = inertia second mass of analytical model [kg.m2]
k = stiffness analytical model [N.m]
b = damping analytical model [N .m.s]
j = imaginary unit = 1−
ω = frequency [rad/s]
f = frequency [Hz]
( )th = impulse response function
sH = symmetric transfer function
N = number of samples
f∆ = frequency step
T = end time for impulse response function
t∆ = time step for impulse response function
sf = sample frequency
foldf = folding frequency used for symmetry
maxf = maximum frequency for frequency domain transfer function
( )tu = time domain input signal
( )ty = time domain output signal
DCT 2006-117 22
Appendix A Derivation of the analytical model parameters
The values of J1, J2, k and b of the two mass spring damper system can be derived using the
next input:
primJ = inertia input axle gearbox [kg.m2]
secJ = inertia output axle gearbox [kg.m2]
vm = vehicle mass [kg]
dynR = dynamic tyre radius [m]
gbi = gear ratio gearbox = gear dependent [-]
hai = gear ratio rear axle [-]
wheelJ = inertia rear wheels [kg.m2]
dsK = stiffness drive shaft [N.m]
hsK = stiffness half-shaft [N.m]
dsB = damping drive shaft [N.m.s]
hsB = damping half-shaft [N.m.s]
22
2
1
hagb
wheeldynv
ii
JRmJ
+= (A.1)
2
sec2
gb
primi
JJJ += (A.2)
21
ha
hs
hsi
KK = (A.3)
+
=
dshs
ds
KK
K11
1
1
1 (A.4)
2
1
gb
ds
i
Kk = (A.5)
21
ha
hs
hsi
BB = (A.6)
+
=
dshs
ds
BB
B11
1
1
1 (A.7)
The summation of dampers (A.7) is incorrect, but in this case where the damping of the drive
shaft is relatively large, the error will be minimal
2
1
gb
ds
i
Bb = (A.8)
DCT 2006-117 23
Appendix B Fourier transformation
In order to transform the transfer function provided by the NASTRAN model from the
frequency domain to the time domain a Fourier transformation is used. The basic ideas of the
Fourier transformation will be discussed in this paragraph.
B.1 Basics
The basic idea of the Fourier transformation is that any function ( )xf can be formed as a
summation of a series of cosine and sine terms of increasing frequency. This means that any
space or time varying data can be transformed can be transformed into the frequency domain
and vice versa [5].
For a one-dimensional function ( )xf the Fourier transform is defined by:
( ) ( ) dxexfuFjuxπ2−
∞
∞−∫= (B.1)
And the inverse Fourier transform:
( ) ( ) dueuFxfjuxπ2
∫∞
∞−= (B.2)
Where 1−=j and u is called the frequency variable.
Using Euler’s Formula:
θθθ sincos jei += (B.3)
Equation 1.1 can be rewritten into;
( ) ( )( )∫∞
∞−−= dxuxjuxxfuF ππ 2sin2cos (B.4)
In this equation the summation of sine and cosine terms is clearly visible.
B.2 Discrete Fourier Transformation
In practice, the function ( )xf is not continuous but discrete. The transfer function provided
by the NASTRAN model for example is a discrete function. The difference between a
continuous function and a discrete function is shown in figure B.1.
DCT 2006-117 24
0 5 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time [s]
x(t
)
continuous signal
0 5 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1discrete signal
Figure B.1: Continuous function versus discrete function
Therefore the Fourier transform is changed into the Discrete Fourier Transform [5]
( ) ( )∑−
=
−=
1
0
2N
x
N
jux
exfuF
π
(B.5)
And the inverse Discrete Fourier transform
( ) ( )∑−
=
=1
0
21 N
u
N
jux
euFN
xf
π
(B.6)
Where N is the number of samples of ( )xf or ( )uF .
The discrete Fourier transform takes ( )2NO time to process for N samples.
B.3 Fast Fourier transform
A proper decomposition of equation 1.5 can make the number of multiplications and addition
operations proportional to ( )NN 2log instead of 2N [8].
There are number of different ways that this algorithm can be implemented. The most
common used algorithm is the Cooley-Tukey algorithm.
In MATLAB the Fast Fourier transform Y of a vector x can be calculated using
Y = fft(x)
And the inverse Fast Fourier transform
y = ifft(X)
DCT 2006-117 25
In the figure below an example is given of a transfer function in frequency domain and the
corresponding transfer function in time domain. The time domain function is called the
impulse response function of the system.
frequency
H(f
)
transfer function in frequency domain
timeh(t
)
transfer function in time domain
∆ f
fmax T
∆ t
Figure B.2: A transfer function in frequency domain and the corresponding transfer function in time
domain.
When the original function consists of N sampled points:
Nff /max=∆ (B.7)
fT
∆=
1 (B.8)
sfN
Tt
1==∆ , with sf the sample frequency (B.9)
maxmax
/1
/1f
f
ff
T
N
tf s =
∆
∆==
∆= (B.10)
In case the impulse response function was acquired using the mirrored frequency domain
transfer function:
frfimpulse NN 2= (B.11)
max22
fT
N
T
Nf
frfimpulse
s === (B.12)
DCT 2006-117 26
B.4 Properties of the Fourier transform
Linearity: The Fourier transform is a linear operation so that the Fourier transform of the sum
of two functions is given by the sum of the individual Fourier transforms. Therefore:
( ) ( ) ( ) )(ubGuaFxbgxaf +=+ℑ (B.13)
Complex conjugate: The Fourier transform of the Complex Conjugate of a function is given
by:
( )( ) ( )uFxf** =ℑ where ( )xf
* is the complex conjugate of ( )xf . (B.14)
Forward and inverse:
( )( ) ( )xfuF −=ℑ (B.15)
Differentials: The Fourier transform of the derivative of a function is given by
( ) ( )uuFjdx
xdfπ2=
ℑ (B.16)
And the second derivative is given by
( ) ( ) ( )uFudx
xfd 2
2
2
2π−=
ℑ (B.17)
B.5 Symmetry
In practice for most cases a sampled signal consists of real function values only.
In that case it can be easily shown that there is an additional property for the Fourier
transform which states [5]:
[ ] [ ]pXpX NN −=+2
*
2 , where *X is the complex conjugate of X . (B.18)
Thus, for real values of the sampled signal, the Fourier transform has a special symmetry with
respect to the frequency-line2Nf
foldf = , also called the folding frequency. Consequently, of
all the N complex numbers X only half contain essential information.
The theory of symmetry is shown in the next figure.
DCT 2006-117 27
n
Re
[cn]
n
Im[c
n]
ffold
=fs/2
Figure B.3: Discrete frequency spectrum of complex Fourier series
B.6 Error sources in Fourier transform
B.6.1 Signal leakage
From a signal ( )tx , which is defined for infinite time, only a part is used while sampling.
In this process, called windowing, only the data points from 0 to T in time are taken into
account.
There are two situations where the windowing effect does not have any influence:
• In case of a transient signal (for example an impulse response) which is (practically)
damped out before the end of the window
• In case of a periodic signal when the window-length t is exactly a multiple of the
period time
If the windowed part is not exactly a multiple of the basic period of the harmonic signal,
discontinuities will arise on the edges of the window. This is illustrated in figure 1.4.
This phenomenon can be avoided using a different window function in stead of the
rectangular window function. An example is the Hanning-window, which reduces the edge
discontinuities considerably.
In case of transient signals signal leakage can be avoided by multiplying the response with a
so-called exponential window. This window forces the response signal to be practically zero
before the end of the measurement time. However this method introduces some additional
(numerical) damping in the response. If a system is judged on damping level using these
measurements this additional damping should be taken into account.
DCT 2006-117 28
time [s]
x(t
)
0 T
Figure B.4: Time domain illustration of signal leakage
B.6.2 Aliasing
Aliasing is an effect which is closely related to the sampling of the original (analogous) signal
on the discrete time points Tnti ∆= .
The sampling frequency sf is defined as:
[ ]HzT
f s ∆=
1 (B.19)
A sampled signal can be written as a so-called pulse train with variable intensity:
( ) ( )[ ] ( )∑∞
∞−
∆−∆= TnttxTtxB δ (B.20)
Using [1] this relation can be transformed with Fourier transformation into:
( ) ( )∑∞
∞−
−= sB fnfXfX (B.21)
The Fourier transform of the sampled signal can be seen as the infinite sum of the exact
Fourier transform each time shifted over multiples of Nf .
2 situations can be distinguished:
Situation 1: 2/max sff ≤
In the interval [ ]2/2/ ss fff ≤≤− the function ( )fX B is exactly equal to ( )fX .
DCT 2006-117 29
So the approximation ( )fX B is perfect in this interval.
This situation is shown in figure 1.5.
f max
-f max
Xb(f)
0 -f s f
s / 2 f
s 2 f
s
X(f)
Figure B.5: Situation without Aliasing
Situation 2: 2/max sff >
This situation is shown in figure 1.6
Xb(f)
0 f s / 2 f
s 2 f
s
X(f)
-f s
-fmax
fmax
Figure B.6: Situation with aliasing
The basic Fourier transform is indicated by the thin solid line and the shifted functions
( )sfnfX − by dotted Lines. The final result ( )fX B is indicated by the thick solid line.
In the interval [ ]2/2/ ss fff ≤≤− this function ( )fX B is not coinciding anymore with the
original ( )fX . This clearly shows the Aliasing effect.
To avoid the Aliasing effect there is one action that should always be taken care of:
,2/ maxfff sfold >= or max2
1f
T>
∆ (B.22)
The frequency T∆2/1 is called the Nyquist frequency. In case of a known maxf the FFT
parameters have to be chosen such that this criterion is fulfilled.
DCT 2006-117 30
Appendix C Convolution
C.1 Basic theory of convolution
When the transfer function provided by the NASTRAN model is transferred from the
frequency domain to the time domain, the result will be an impulse response function.
This function describes the way the system reacts to a unit impulse. Just like the NATRAN
model transfer function the impulse response function describes the dynamic behaviour of the
system. Once the impulse response function is determined, the system response to any input
signal can be predicted. In order to make this prediction a convolution product is needed
Figure C.1: A linear system
The output y(t) can be determined by applying a convolution product of the input u(t) and the
systems impulse response function h(t) [9].
( ) ( ) )(thtuty ⊗= , where ⊗ is the sign used for the convolution product. (C.1)
The concept of convolution is shown in figure 1.7. When the impulse response function of a
system is known, the output can be determined. First the input signal u(t) is transformed into
the sum of a set of impulses. The response to each input pulse can be calculated using the
impulse response function h(t). The output signal y(t) is equal to the sum of the responses of
all input pulse in the input signal u(t).
Figure C.2: Characterizing a linear system using its impulse response.
u(t) y(t)
Linear
System
h(t)
DCT 2006-117 31
The convolution integral is defined as:
( ) ( ) ( ) ( )∫∞
∞−−=⊗= dssthsuthtuty )( (C.2)
And the discrete form:
( ) ( ) ( )∑ −+=j
jkhjuky 1 (C.3)
t
sig
na
l x(t
)
t
sig
na
l h
(t)
s
Figure C.3: Graphical presentation of the convolution integral
The length of the output vector y in discrete form:
( )( ) ( )( ) ( )( ) 1−+= thlengthtulengthtylength
C.2 Properties of the convolution product
Some general properties of the convolution product
Commutative: xyyx ⊗=⊗
Associative: ( ) ( )zyxzyx ⊗⊗=⊗⊗ (C.4)
Distributive: ( ) ( ) ( ) zyxzyzx ⊗+=⊗+⊗
Some properties of the convolution product with respect to the Fourier transform.
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )tytxfYfX
tytxfYfX
fYfXtytx
fYfXtytx
⊗→
→⊗
→⊗
⊗→
−
−
ℑ
ℑ
ℑ
ℑ
1
1 (C.5)
DCT 2006-117 32
Appendix D Testing methods
In order to validate the correctness of the impulse response that was determined two testing
methods are used.
D.1 Transfer function estimate method
The first method is the transfer function estimate method. In order to use this method, the
impulse response function first has to be implemented in a MATLAB or
MATLAB/SIMULINK model. This can be done by using the filter function. In MATLAB the
filter commando is used, in MATLAB/SIMULINK a Finite Impulse Response filter block
(FIR) is used.
Now the impulse response function can be compared to the original transfer function by
applying a white noise signal to the impulse response filter. The SIMULINK model where
white noise is used to test the impulse response function is shown in the figure below.
Figure D.1: MATLAB/SIMULINK model used to test the impulse response function
In MATLAB the same routine can be done using the following commandos.
x = randn(size(h));
a = zeros(size(h));
a(1) = 1;
y = filter(h, a, x)
with h the impulse response function h(t).
taking the begin conditions into account:
creating input signal with multiple periods (in this case 5):
ne = 5;
xe = [];
for i=1:ne,
xe = [xe x];
end
xe = [xe x];
DCT 2006-117 33
ye = filter(h, a, xe)
y= ye((ne*length(h)+1):(ne*length(h)+length(h)));
The impulse response filter can now be analyzed and compared in the frequency domain to
the original frequency domain transfer function by using the Transfer function estimate (tfe)
commando.
[ ] ( )fsnfftyxtfefH ,,,, =
Where:
H = estimated transfer function
f = corresponding frequency vector
x = white noise input signal
nfft = number of points used to estimate the transfer function
fs = sampling frequency of the original model
D.2 Analytical transfer function determination
One of the disadvantages of the transfer function estimate method is that a possible
occurrence of the tail will not be taken into account, unless multiple-period input signals are
used, as was explained before. The tfe function does not take the begin conditions into
account. This is undesired when the impulse response function has to be checked.
Therefore also another method is used to check the functionality of the acquired impulse
response function. In this method a sine signals with varying frequencies are put on the FIR
filter in stead of white noise. The magnitude and the phase of the transfer function are then
determined analytically.
The magnitude can be determined by dividing the maximum amplitude of the output signal by
the maximum amplitude of the input signal. The maximum amplitude is determined in the
steady state area, after the transient phenomenon in the first seconds of the output signal.
To determine the phase the maximum amplitude of the input and output signal are scaled to
one and then the two signals are summed. The maximum amplitude of the resulting signal can
be used as a standard for the phase. The relation between the amplitude of the summed input
and output signal and the phase are shown in figure D.1.
To determine the sign of the phase, the sign of both the input and the output signal are
checked. When the output signals sign changes before the sign of the input signal does, the
phase will be positive and vice versa. The main disadvantage of this method is that less
transfer function information can be acquired compared to the transfer function estimate
method.
DCT 2006-117 34
0 0.5 1 1.5 2-0.5
0
0.5
1
1.5
2
2.5
3
3.5
amplitude(input + output)
ph
ase
[ra
d]
Figure D.2: Relation between the maximum amplitude of the summed scaled input and output signal and
the phase.
In the figure below an example is shown of the analytical transfer function determination
method. In this example the influence of the tail is shown. In the left figure the tail is taken
into account, in the right figure the tail is cut of.
10-2
10-1
100
101
102
10-2
100
102
frequency [Hz]
|H|
transfer function with tail
10-2
10-1
100
101
102
-50
0
50
100
150
200
frequency [Hz]
an
gle
(H)
10-2
10-1
100
101
102
10-2
100
102
frequency [Hz]
|H|
transfer function without tail
10-2
10-1
100
101
102
-50
0
50
100
150
200
frequency [Hz]
an
gle
(H)
Figure D.3: Bode plot of the NASTRAN transfer function versus the transfer function acquired using the
impulse response function with (left) and without (right) the tail.
DCT 2006-117 35
Appendix E Deconvolution
As was explained before, the calculated impulse response has a little tail at the end. This tail
was caused by the limited frequency area that was analyzed in NASTRAN. The calculated
NASTRAN transfer function is a multiplication of the infinite transfer function with a
window function (see paragraph 2.4). In time domain this is equal to a convolution of the
impulse response of the infinite model with the impulse response of the window function. In
this case, the impulse response of the finite model and the window function are known and the
impulse response of the infinite model is unknown. In order to determine this infinite model
deconvolution is applied. Note that the result will in fact not be a real infinite model, but a
model with a longer frequency range. The length of the frequency range is determined by the
length of the window function, which can be chosen freely.
First the convolution equation (C.2) is written in matrix form
=
+−
nn
n
n
n
n
n
n
y
y
y
y
u
u
u
hhhh
hhhh
hhhh
hhhh
hhhh
⋮⋮
⋮
⋮
⋮
⋮
⋮
⋯
3
2
1
1
1
123
123
123
123
123
0000
0000
0000
0000
0000
or yuH = (E.1)
In this case ih is the impulse response coefficient of the window function,
iu the impulse
response coefficient of the infinite model and iy the impulse response coefficient of the finite
model (=NASTRAN model).
In order to decrease the calculation time, only a part of the window function impulse response
is used. Therefore first the zero-frequency component of the impulse response function is
shifted to the center of the spectrum. This can be done using fftshift in MATLAB.
Next a part at the beginning and the end of the impulse response function is disposed. This
part can be disposed because the impulse response is almost equal to zero. This is illustrated
in the next figure.
0 1 2 3 4 5 6 7 8 9 10-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
time [s]
h(t
)
shifted impulse response
disposed disposed used
hmax
hmax
Figure E.1: Shifted impulse response
DCT 2006-117 36
The same routine is applied to the impulse response function of the NASTRAN model.
Equation (E.1) can now be rewritten into:
=
+−
max
1
max
1
1
max1max
max1max
max1max
max1max
max1max
0000
0000
0000
0000
0000
y
y
y
u
u
u
hhh
hhh
hhh
hhh
hhh
n
n
⋮
⋮
⋮
⋮
⋯⋮
⋯⋮
⋯⋮
⋯⋮
⋯⋯
(E.2)
This matrix form equation cannot be solved. There are 12 −n unknown variables of u and
only n equations. In order to be able to solve this problem the size of H has to be reduced
from ( ) nn ×−12 to nn × . This can be done in several ways. Here, two ways are shown:
1. use only the right half side of the matrix H. the matrix then changes to:
max1max
max1
max1
max
max
0
00
000
0000
hhh
hh
hh
h
h
⋯⋮
⋯⋮
⋯
⋮
(E.3)
In this way no begin conditions are taken into account.
2. use the center part of the matrix H. the matrix then changes into:
1max
1max
max1max
max1
max1
00
0
0
00
hh
hh
hhh
hh
hh
⋯
⋯⋮
⋯⋮
⋯⋮
⋯
(E.4)
Now a part of the begin conditions are taken into account.
In order to determine U, the impulse response of the infinite model, the inverse of H has to be
determined.
YHU1−= (E.5)
Because H is a large sparse singular matrix, singular value decomposition is used to determine
the inverse of H. In MATLAB:
[ ] ( )kHsvdsVSU ,,, = (E.6)
( ) ( )( )( )( ) ( )( )':1,:*:1,:1/1*:1,:1kUkkSdiagdiagkVH =− (E.7)
DCT 2006-117 37
With k the number of largest singular values and associated singular vectors of the matrix H ,
U the kn × orthonormal columns, S the kk × diagonal and V the kn × orthonormal columns.
'** VSU is the closest rank k approximation to H .
In order to determine the optimal value for k , the number of singular values, the so-called
L-curve plot is used. In this plot the norm of the solution is plotted against the norm of the
error for different values of k . The optimal value for k lies at the point where both the norm of
the error and the solution are small. This is illustrated in the next figure.
10-2
10-1
100
10-1
100
101
12
34567891010
50
100110
150210220
350
450490
||U
est||
||H.Uest - Y'||
L-curve
optimum
Figure E.2: L-curve used to determine the optimal number of singular values
Now, both equation (E.3) and equation (E.4) are used to calculate the impulse response
functions of the infinite model. This calculation was done using the analytical model.
The window function is chosen to be such that the “infinite” model will be four times longer
then the original model. The impulse responses are now calculated using the deconvolution
method. Next the actual impulse response function of the infinite model is determined by
increasing maximum frequency of the analytical model. The impulse response functions are
then tested by applying white noise (see appendix D) and compared with the original transfer
function. In the next figure the calculated impulse response for the infinite model using the
right side part of the window impulse response matrix (equation (E.3)) is shown together with
the real impulse response of the infinite model. The third gear was used for the calculation
0 5 10 15 20 25 30 35 40-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5impulse response with deconvolution using right side part
time [s]
h(t
)
with deconvolutionreal
10-1
100
101
102
-60
-40
-20
0
20
40
|H| [d
B]
bode plot input frf versus numerical experiment
frequency [Hz]
original tftf infinite impulse responsetf finite impulse response
10-1
100
101
102
-200
-100
0
100
200
frequency [Hz]
an
gle
(H)
[de
g]
Figure E.3: Impulse response calculated with deconvolution (left) and tested using white noise
DCT 2006-117 38
The left figure shows that the impulse response function calculated with the deconvolution
method differs from the real impulse response function, determined by increasing the
maximum frequency of the analytical model. The right figure shows the influence of this
error. The new infinite impulse response is not an improvement compared to the original
impulse response of the finite model.
The only advantage that can be seen is the increase of high frequency information. The
maximum frequency of the model has increased.
Next the center part of the window impulse response matrix (equation (E.4)) is used.
The calculated impulse response is again tested with white noise.
0 5 10 15 20 25 30 35 40-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5impulse response with deconvolution using center part
time [s]
h(t
)
with deconvolutionreal
10-1
100
101
102
-60
-40
-20
0
20
40
|H| [d
B]
bode plot input frf versus numerical experiment
frequency [Hz]
10-1
100
101
102
-200
-100
0
100
200
frequency [Hz]
an
gle
(H)
[de
g]
original tftf infinite impulse responsetf finite impulse response
Figure E.4: Impulse response calculated with deconvolution using the center part of the window impulse
response matrix (left) and tested using white noise
The calculated impulse response function using the center part of the matrix is even worse, as
can be seen in figure E.4. The tail is larger then for the impulse response function using the
right side part of the matrix. The advantage of gaining more high frequency information is
gone, because the information is not correct.
Conclusion is that the deconvolution method does not give the desired result. The method
using the right side part of the window impulse response matrix works better that the method
using the center part of the same matrix, but the results are for both methods unsatisfying.
The main problem is that the original impulse response matrix H is not square and therefore
only a part of this matrix can be used. This will always give errors in the calculated impulse
response.