Michael Soyka
Guest
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 Posted:
Sun Dec 11, 2005 12:16 am Post subject:
Re: Fourier Transform challenge |
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Raymond Toy wrote:
| Quote: | "John" == John <joehatesspam@nospam.spamshit> writes:
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
The use of the Laplace Transform is inappropriate here because the |
poster's function is defined for all real values whereas the Laplace
Transform is integrated over the non-negative reals alone. In effect
you are transforming the poster's function scaled by the Heaviside unit
step function, not the same thing.
If you set u equal to exp(-t) in the Fourier Transform integral, you
will get the Gamma function. No need for Maple here!
Mike |
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Guest
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| Posted:
Sun Dec 11, 2005 1:16 am Post subject:
Re: Fourier Transform challenge |
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Mike has found the answer in a handbook of mathematical functions;
magnitude-squared of gamma(1+iw) = pi * omega / sinh(pi*omega)
So this represents the Fourier Transform of exp(-t - exp(-t))
This is the answer I was looking for, and it's quite remarkable, I
think. Here's what I find very intruiging.
since sinh = (exp(x) - exp(-x))/2, as w becomes large, 1/sinh is
proportional to exp(-x), implying that the magnitude in dB decreases at
a constant rate as a function of LINEAR increase in frequency. The only
problem is the omega in the numerator which causes a slight deviation
from this rule (although presumably if one were to integrate the time
function such that f(t) = exp(-epx(-t)), that term would disappear).
If you were to go back to my Matlab code I posted previously, you will
see that the frequency-domain decay in dB appears to be a perfectly
straight line, as a function of LINEAR (not log) frequency. This is of
course very different than the usual relationship of a straight line on
a dB-per-log-frequency measure. Has anyone seen this before ?
I think there is a bigger picture at work here. There must be a rule
that states that if you have a time function which consists entirely of
exponentials, and you replace the t variable with yet another
exponential so that all terms are of the form exp(-exp(-t)), then the
frequency axis of the Fourier Transform also gets warped by an
exponential (or logarithmic ??) function. Anyone want to take a crack
at this theory? |
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