vibrationdata 1 unit 5 the fourier transform. vibrationdata 2 courtesy of professor alan m. nathan,...
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Unit 5
The Fourier TransformThe Fourier Transform
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Courtesy of Professor Alan M. Nathan,University of Illinois at Urbana-Champaign
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Unit 5
Many, many different equations for Fourier Transforms . . . . .
1. Mathematician Approach
2. Engineering Approach
Even within each approach there are many equations.
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Continuum of Signals
Random Stationary vibration signals can be placed along a continuum in terms of the their qualitative characteristics
A pure sine oscillation is at one end of the continuum
A form of broadband random vibration called white noise is at the other end
Reasonable examples of each extreme occur in the physical world.
Most signals, however, are somewhere in the middle of the continuum
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Frequency Example
-5
5
0
0 0.5 1.0 1.5 2.0
TIME (SEC)
AC
CE
L (
G)
TIME HISTORY EXAMPLE
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Resolving Spectral Components
The time history appears to be the sum of several sine functions
What are the frequencies and amplitudes of the components?
Resolving this question is the goal of this Unit
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Equation of Previous Signal
t22π2sin1.2t16π2sin1.5t10π2sin1.0ty
At the risk of short-circuiting the process, the equation of the signal in the previous figure is
The signal thus consists of three components with frequencies of 10, 16, and 22 Hz, respectively.
The respective amplitudes are 1.0, 1.5, and 1.2 G In addition, each component could have had a phase angle In this example, the phase angle was zero for each component Need some sort of "spectral function" to display the frequency and amplitude data Ideally, the spectral function would have the form shown in the next figure
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0.5
1.0
1.5
2.0
00 5 10 15 20 25
FREQUENCY (Hz)
AC
CE
L (G
)"SPECTRAL FUNCTION" OF TIME HISTORY EXAMPLE
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Textbook Fourier Transform
The Fourier transform X(f) for a continuous time series x(t) is defined as
where
- < f <
The Fourier transform is continuous over an infinite frequency range.
Note that X(f) has dimensions of [amplitude-time].
- dttf2πj-expx(t)=X(f)
1j
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Textbook Inverse Fourier Transform
The inverse transform is
Thus, a time history can be calculated from a Fourier transform and vice versa.
- dftf2πj+expX(f)=x(t)
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Magnitude and Phase
Note that X(f) is a complex function.
It may be represented in terms of real and imaginary components, or in terms of magnitude and phase.
The conversion to magnitude and phase is made as follows for a complex variable V.
bjaV
2b2aVMagnitude
)arctan(b/aVPhase
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Real Time History
Practical time histories are real
Note that the inverse Fourier transform calculates the original time history in a complex form
The inverse Fourier transform will be entirely real if the original time history was real, however.
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Fourier Transform of a Sine Function
The transform of a sine function is purely imaginary. The real component, which is zero, is not plotted.
tf2 πsinAx(t) ˆ
Im a g in a r y X ( f )
f̂f̂
ffδ
2A ˆ
ffδ
2A ˆ
f
f = 0
Dirac Delta function
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Characteristics of the Continuous Fourier Transform
The real Fourier transform is symmetric about the f = 0 line
The imaginary Fourier transform is antisymmetric about the f = 0 line.
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Characteristics of the ContinuousFourier Transform
This Unit so far has presented Fourier transforms from a mathematics point of view
A more practical approach is needed for engineers ………
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Discrete Fourier Transform
An accelerometer returns an analog signal
The analog signal could be displayed in a continuous form on a traditional oscilloscope
Current practice, however, is to digitize the signal, which allows for post-processing on a digital computer
Thus, the Fourier transform equation must be modified to accommodate digital data This is essentially the dividing line between mathematicians and engineers in regard
Fourier transformation methodology
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Discrete Fourier Transform Equations
The Fourier transform is
The corresponding inverse transform is
1N...,1,0,kfor,1N
0nnk
N2πjexpnx
N1
kF
1N...,1,0,nfor,1N
0knk
N2πjexpkFnx
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Discrete Fourier Transform Relations
Note that the frequency increment f is equal to the time domain period T as follows
The frequency is obtained from the index parameter k as follows
where k is the frequency domain index
T / 1 =fΔ
fΔk(k) frequency
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Nyquist Frequency
The Nyquist frequency is equal to one-half of the sampling rate
Shannon’s sampling theorem states that a sampled time signal must not contain components at frequencies above the Nyquist frequency
Otherwise an aliasing error occurs
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Half AmplitudeDiscrete
-1.0
-0.5
0
0.5
1.0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
N = 16384SR = 32 samples/sect = ( 1 / 32 ) secT = 512 secf = ( 1 / 512 ) Hz
Dashed line is line of symmetryat Nyquist frequency
FREQUENCY (Hz)
AC
CE
LE
RA
TIO
N (
G)
HALF-AMPLITUDE DISCRETE FOURIER TRANSFORM y(t) = 1 sin [ 2 ( 1 Hz) t ] G
This is the imaginary component of the transform. The real component is zero.
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Previous Fourier Transform Plot
Note that the sine wave has a frequency of 1 Hz
The total number of cycles is 512, with a resulting period of 512 seconds
Again, the Fourier transform of a sine wave is imaginary and antisymmetric.
The real component, which is zero, is not plotted.
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Spectrum Analyzer Approach
Spectrum analyzer devices typically represent the Fourier transform in terms of magnitude and phase rather than real and imaginary components
Furthermore, spectrum analyzers typically only show one-half the total frequency band due to the symmetry relationship
The spectrum analyzer amplitude may either represent the half-amplitude or the full-amplitude of the spectral components
Care must be taken to understand the particular convention of the spectrum analyzer
Note that the half-amplitude convention has been represented in the equations thus far
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Full Amplitude Fourier Transform
The full-amplitude Fourier transform magnitude would be calculated as
12N,...,1kfor
1N
0nnk
N2πjexpnx
N1magnitude2
0k for 1N
0nnx
N1magnitude
kG
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Full Amplitude Fourier Transform
Note that k = 0 is a special case
The Fourier transform at this frequency is already at full-amplitude
Consider a sine wave with an amplitude of 1 G and a frequency of 1 Hz
This signal would simply have a full-amplitude Fourier magnitude of 1 G at 1 Hz, as shown in the next figure
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0
0.5
1.0
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
N = 16384t = ( 1 / 32 ) secT = 512 secf = ( 1 / 512 ) Hz
FREQUENCY (Hz)
AC
CE
LE
RA
TIO
N (
G)
FULL-AMPLITUDE, ONE-SIDED DISCRETE FOURIER TRANSFORM OF y(t) = 1 sin [ 2 (1 Hz) t ] G
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Full Amplitude Fourier TransformTime History Example
-5
5
0
0 0.5 1.0 1.5 2.0
TIME (SEC)
AC
CE
L (G
)
TIME HISTORY EXAMPLE
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Full Amplitude One-Sided
0
0.5
1.0
1.5
2.0
5 10 15 20 25
N = 400t = 0.0025 secT = 2 secf = 0.5 Hz
FREQUENCY (Hz)
AC
CE
LE
RA
TIO
N (
G)
FULL-AMPLITUDE, ONE-SIDED DISCRETE FOURIER TRANSFORM OF TIME HISTORY IN FIGURE 1
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Unit 5 Exercise 1
A time history has a duration of 20 seconds.
What is the frequency resolution of the Fourier transform?
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Unit 5 Exercise 2
a) Generate sine time history using:
Amplitude = 7, Frequency = 2 Hz, Duration 100 sec, Sample rate = 100
b) Save sine function as ASCII text file.
c) Then take the Fourier transform of sine function:
vibrationdata > Fourier transform mean removal=yes, window=rectangular, min freq = 0 Hz, max freq = 10 Hz
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Unit 5 Exercise 3
Calculate Fourier transforms and identify dominant frequencies for these time histories from previous webinar units:
tuning_fork.txt - time (sec) & unscaled sound pressuredrop.txt - time (sec) & accel(G)channel.txt - time (sec) & accel(G)
Use: vibrationdata > Fourier transform mean removal=yes, window=rectangular, Min Freq=0 Hz
Data Max Freq (Hz)
tuning_fork 1000
drop 100
channel 1000