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<a class="moz-txt-link-abbreviated" href="mailto:sac-help-request@iris.washington.edu">sac-help-request@iris.washington.edu</a> wrote:<br>
<br>
The simple test of the FFT is to see the effect of an impulse:<br>
<br>
fg impulse delta 0.1 npts 1024<br>
w impulse.sac<br>
<br>
<br>
<b>1. Test of sac2000 FFT<br>
<br>
</b>Sac creates a time series with a centered impulse of height 1.0 and
not a unit area pulse.<br>
<br>
fft<br>
writesp am<br>
<br>
This creates a series of the amplitude spectrum, which has values of
0.1, e.g., 1 (from the time series) * DT from the fft<br>
<br>
ifft<br>
<br>
This creates a time series consisting of a centered impulse with height
1.0<br>
<br>
Thus sac2000 [8/8/2001 (Version 00.59.44)] is performing the following
operation to convert h(t) into H(f)<br>
N-1<br>
H(f) = SUM h(t) exp ( - j 2 pi f t ) dt<br>
k=0<br>
<br>
N-1<br>
h(t) = SUM H(f)exp(+j 2 pi f t ) df<br>
n=0<br>
<br>
where f = n df. t = k dt , and df = 1/Ndt<br>
<br>
<b>2. Constraints on testing the FFT using an analytic function</b><br>
<br>
If you wish to test an FFT, you cannot use a sine-wave since the value
of the FFT will be related to the total number of cycles in the time
window and because the Discrete Fourier Transform is an approximation
to the Fourier Transform that suffers from periodicity in both the time
and frequency domains.<br>
<br>
A properly designed test can use the analytical Gaussian <->
Gaussian transform pair, or perhaps some thing from Laplace transforms
using<br>
<br>
H(t) exp( - a t ) <-> 1/(s + a) where s = j 2 pi f<br>
<br>
You must always worry about the fact there there is a limited time
length and a limited frequency band. A perfect test of an analytical
transform pair is thwarted by the uncertainty principle<br>
<br>
<b>3. Other FFT's<br>
<br>
</b>If you use MATLAB or MATHCAD, you will always have to read the
documentation about the definition of the FFT used.<br>
I use the electrical engineering definition but incorporate the
physical dimensions to the integral. <br>
<br>
<blockquote
cite="mid:200803111913.m2BJDPOJ013487@norman.iris.washington.edu"
type="cite">
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Today's Topics:
1. About FFT scaling factor in SAC (wwxu)
----------------------------------------------------------------------
Message: 1
Date: Tue, 11 Mar 2008 07:53:16 GMT
From: "wwxu" <a class="moz-txt-link-rfc2396E" href="mailto:wwxu@mail.iggcas.ac.cn"><wwxu@mail.iggcas.ac.cn></a>
Subject: [SAC-HELP] About FFT scaling factor in SAC
To: <a class="moz-txt-link-abbreviated" href="mailto:sac-help@iris.washington.edu">sac-help@iris.washington.edu</a>
Message-ID: <a class="moz-txt-link-rfc2396E" href="mailto:20080311075316.5572.eqmail@mail.iggcas.ac.cn"><20080311075316.5572.eqmail@mail.iggcas.ac.cn></a>
Content-Type: text/plain; format=flowed; charset="gb2312"
Hi, ALL
We suppose that x[n] and y[k] are a DFT pairs, i.e.
y[k]=1/N*sum_n(x[n]*exp(-2*pi*j*k*n/N)) n=0,1,...,N-1
where sum_n means the sum for n=0,1,N-1(similar for sum_k),
and dt is the sampling interval, df is the sampling frequency,
N is the sampling number.
To obey Parseval's theorem, we get
sum_n(x[n]*x[n])*dt=sum_k(y[k]*y[k])*df
since sum_n(x[n]*x[n])=N*sum_k(y[k]*y[k])
and df=1/(N*dt), also consider the amplitude symmetry of the real series,
we should multiply a coefficent to y[k]:
sqrt(2)*N*dt
but in SAC software, this value is N*dt/2,
so do you have any idea?
Thanks.
------------------------------
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End of sac-help Digest, Vol 33, Issue 8
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</pre>
</blockquote>
<br>
<br>
<pre class="moz-signature" cols="72">--
Robert B. Herrmann
Otto W. Nuttli Professor of Geophysics
Department of Earth and Atmospheric Sciences
Saint Louis University
O'Neil Hall, Room 203
3642 Lindell Boulevard
St. Louis, MO 63108
TEL: 314 977 3120
FAX: 314 977 3117
Email: <a class="moz-txt-link-abbreviated" href="mailto:rbh@eas.slu.edu">rbh@eas.slu.edu</a>
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