| Author |
Message |
Guest
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 Posted:
Thu Dec 01, 2005 4:00 pm Post subject:
Fourier Transform challenge |
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I am trying to take the Fourier transform of the following time
function;
f(t) = exp(-t)/(exp(exp(-t)), from t = -inf to t= +inf
This function approaches 0 at both time extremes, and seems
well-behaved; numerical evaluation of the Fourier Transform integral
seems to converge, and give very interesting results (it looks like an
asymmetric window function, and seems to have excellent sidelobe
rolloff characteristics). However, it seems
very difficult to arrive at a closed-form solution, due to the
(exp(exp(-t)) function. Can anyone help?
Thanks!
Bob Adams |
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John
Guest
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| Posted:
Thu Dec 01, 2005 5:16 pm Post subject:
Re: Fourier Transform challenge |
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you are welcome....
-------------------------------
| Quote: | This is great; thanks!
Bob Adams
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John
Guest
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| Posted:
Thu Dec 01, 2005 5:17 pm Post subject:
Re: Fourier Transform challenge |
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MATLAB:
fourier(exp(-x-exp(-x)))
ans =
gamma(1+i*w) |
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Guest
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| Posted:
Thu Dec 01, 2005 5:17 pm Post subject:
Re: Fourier Transform challenge |
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This is great; thanks!
Bob Adams |
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Jerry Wolf
Guest
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| Posted:
Thu Dec 01, 2005 5:17 pm Post subject:
Re: Fourier Transform challenge |
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(You might have pointed out that this can't be accomplished using
MATLAB itself, but rather MATLAB's Symbolic Math Toolbox.) |
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Guest
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| Posted:
Fri Dec 02, 2005 12:37 am Post subject:
Re: Fourier Transform challenge |
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Actually, this result is a bit confusing in that it's another integral
instead of a nice simple function of frequency. I'm trying to
understand why this function has such a marked frequency-domain lowpass
characteristic. To see why it's intriuging, plug the following into
Matlab.
Bob Adams
clear all;
close all;
numpoints = 2^16;
half = 2^15 + 1;
tstart = -30;
tend = 30;
tstep = (tend-tstart)/numpoints;
fs = 1/tstep;
t = tstart:tstep:tend;
fstep = fs/numpoints;
f = 0:fstep:fs/2;
w = 2.0*pi.*f;
f = (exp(-t))./exp(exp(-t));
F = fft(f);
fl = length(f);
Fyy = F.* conj(F);
plot(t,f);
figure;
plot(w,10*log10(Fyy(1:half) + 1e-64)); |
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naebad
Guest
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| Posted:
Fri Dec 02, 2005 1:17 am Post subject:
Re: Fourier Transform challenge |
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ie Maple.
Naebad |
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John
Guest
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| Posted:
Fri Dec 02, 2005 1:17 am Post subject:
Re: Fourier Transform challenge |
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Try this:
clc
close all
clear
T=-4:0.01:6;
y=exp(-T)./exp(exp(-T));
Y=abs(fft(y,2*length(y)));
plot(T,y)
title('Bobs signal')
figure
plot(fftshift(Y)./length(y))
title('FFT of Bobs signal')
figure
z=gausswin(length(y))
plot(T,z);
title('Gaussian Window')
figure
plot(fftshift(abs(fft(z,2*length(y))))./length(y))
title('FFT of Gaussian window')
I don't know.....but maybe litterature about Gaussian windows is relevant?
Cheers...
------------
| Quote: | Actually, this result is a bit confusing in that it's another integral
instead of a nice simple function of frequency. I'm trying to
understand why this function has such a marked frequency-domain lowpass
characteristic. To see why it's intriuging, plug the following into
Matlab.
Bob Adams
clear all;
close all;
numpoints = 2^16;
half = 2^15 + 1;
tstart = -30;
tend = 30;
tstep = (tend-tstart)/numpoints;
fs = 1/tstep;
t = tstart:tstep:tend;
fstep = fs/numpoints;
f = 0:fstep:fs/2;
w = 2.0*pi.*f;
f = (exp(-t))./exp(exp(-t));
F = fft(f);
fl = length(f);
Fyy = F.* conj(F);
plot(t,f);
figure;
plot(w,10*log10(Fyy(1:half) + 1e-64));
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Guest
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| Posted:
Fri Dec 02, 2005 5:17 pm Post subject:
Re: Fourier Transform challenge |
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Actually, in retrospect the gamma() solution is not that useful to me;
I need something that directly gives H(w) without need to evaluate an
integral.
It's strange that the function exp(-x)/exp(exp(-x)) would have such a
dramatic frequency-domain characteristic; it rolls off very steeply at
higher frequencies(as evidenced by taking the FFT of this function,
using a very dense time-sampling grid). It seems like there is a hole
to be filled here, and it smells kind of fundamental to me.
This formula popped up in my search to find a "logarithmic sampling
theorom"; that is, are there any functions that can be completely
reconstructed from samples that occur on log(integer) time points. The
function exp(-x)/exp(exp(-x)) seems to play the same role for
log-sampled signals that sin(x)/x plays in the reconstruction of
linearly-sampled signals. This interpretation comes from one of the
equations that equates the Reimann Zeta function with the series (SUM
of N^-s), which can be interpreted as a logarithmic time series if you
evaluate it on a vertical line in the complex plane. |
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Ron N.
Guest
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| Posted:
Sat Dec 03, 2005 1:16 am Post subject:
Re: Fourier Transform challenge |
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robert.w.adams@verizon.net wrote:
| Quote: | Actually, in retrospect the gamma() solution is not that useful to me;
I need something that directly gives H(w) without need to evaluate an
integral.
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What are most functions, other than evaluation of integrals
or series, or numerical approximations thereof? (other than
simple polynomials of course, but surely you can't be suggesting
that the FT of exp(-t)/(exp(exp(-t)) has a simple finite-degree
polynomial representation.) So what are your constraints
regarding the form or representation of H(w) which you desire?
--
rhn A.T nicholson d.O.t C-o-M |
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Michael Soyka
Guest
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| Posted:
Sat Dec 03, 2005 8:02 am Post subject:
Re: Fourier Transform challenge |
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John wrote:
| Quote: | MATLAB:
fourier(exp(-x-exp(-x)))
ans =
gamma(1+i*w)
This "solution" is actually meaningless because |
gamma(1+iw) = iw gamma(iw)
and the gamma function is analytic only in the right half plane (real
part of z positive).
That said, the original poster's function exp(-x-exp(-x)) is itself
integrable- its integal is
exp(-exp(-x))
as can be verified by differentiation.
Given this result, the Fourier Transform can be integrated "by parts".
Start by integrating only over the interval [-T,T]. Then, integrating
by parts and taking the limit as T becomes large we get
exp(-iwT-exp(-T)) - exp(iwT-exp(T))
FT{} = lim -------------------------------------
T->oo 1 + iw
Clearly the second term in the numerator converges to zero and so we are
left with
exp(-iwT-exp(-T))
FT{} = lim -------------------
T->oo 1 + iw
exp(-iwT)
= lim -----------
T->oo 1 + iw
This limit does not exist in the conventional sense but does exist as a
"distribution" or "generalized function" (the Dirac delta function is
the usual example- sorry to bring this subject up but I must). In this
sense, the limit exists and is zero by the Riemann-Lebesgue Lemma (see
Papoulis' book "The Fourier Integral and its Applications").
I suspect the limit being zero explains the excellent roll-off
characteristics seen by the original poster.
Mike |
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Michael Soyka
Guest
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| Posted:
Sat Dec 03, 2005 8:28 am Post subject:
Re: Fourier Transform challenge |
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Michael Soyka wrote:
| Quote: | John wrote:
MATLAB:
fourier(exp(-x-exp(-x)))
ans =
gamma(1+i*w)
This "solution" is actually meaningless because
gamma(1+iw) = iw gamma(iw)
and the gamma function is analytic only in the right half plane (real
part of z positive).
|
Actually, my statement above is ridiculous since 1+iw is obviously in
the right-half plane. Sorry John! I had also forgotten that the Gamma
function is analytic everywhere except at the negative integers.
And strike 3, that means the analysis below must be wrong somewhere even
though I did it twice before posting. Back to the clip board!
| Quote: |
That said, the original poster's function exp(-x-exp(-x)) is itself
integrable- its integal is
exp(-exp(-x))
as can be verified by differentiation.
Given this result, the Fourier Transform can be integrated "by parts".
Start by integrating only over the interval [-T,T]. Then, integrating
by parts and taking the limit as T becomes large we get
exp(-iwT-exp(-T)) - exp(iwT-exp(T))
FT{} = lim -------------------------------------
T->oo 1 + iw
Clearly the second term in the numerator converges to zero and so we are
left with
exp(-iwT-exp(-T))
FT{} = lim -------------------
T->oo 1 + iw
exp(-iwT)
= lim -----------
T->oo 1 + iw
This limit does not exist in the conventional sense but does exist as a
"distribution" or "generalized function" (the Dirac delta function is
the usual example- sorry to bring this subject up but I must). In this
sense, the limit exists and is zero by the Riemann-Lebesgue Lemma (see
Papoulis' book "The Fourier Integral and its Applications").
I suspect the limit being zero explains the excellent roll-off
characteristics seen by the original poster.
Mike |
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Raymond Toy
Guest
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| Posted:
Tue Dec 06, 2005 12:55 am Post subject:
Re: Fourier Transform challenge |
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| Quote: | "John" == John <joehatesspam@nospam.spamshit> writes:
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John> MATLAB:
John> fourier(exp(-x-exp(-x)))
John> ans =
John> gamma(1+i*w)
Are you sure it didn't say something like gamma(1+i*w,1)?
I have a table of Laplace transforms and it says the transform of
exp(-exp(-x)) is gammaincomplete(s,1) for Real(s) > 0, where
gammaincomplete(v,x) is the incomplete gamma function,
integral(z^(v-1)*exp(-z),z,x).
If this is correct, the Laplace transform of exp(-x-exp(-x)) is
gammaincomplete(s+1,1). I think that means the Fourier transform
would be gammaincomplete(1+i*w,1).
Ray |
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Guest
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| Posted:
Tue Dec 06, 2005 4:34 pm Post subject:
Re: Fourier Transform challenge |
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Thanks for all the interesting traffic on this. Michael, I like the
approach you took to get;
exp(-iwT)
lim -----------
T->oo 1 + iw
but I don't see how this limit could exist given that the numerator
oscillates with amplitude 1 as T approaches infinity, while the
denomiator is fixed.
Regards
Bob Adams |
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Guest
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| Posted:
Tue Dec 06, 2005 4:54 pm Post subject:
Re: Fourier Transform challenge |
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Another interesting formula; the gamma function can be written as;
gamma(z) = integral from t=0 to 1 of ((-ln(t))^(z-1)) dt (from
mathworld)
so if we plug in z = 1 + iw we get
integral from t=0 to 1 of (-ln(t)) ^ iw dt
Not sure what this means but it looks interesting!
Bob Adams |
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