Fourier Transform unique?

On Sep 5, 2:19 pm, "westocl" <cwest...@hotmail.com> wrote:
>
> are these signals anomalous? I may have sampled them wrong. but they seem
> to have a strong sinusoidal group delay and the same FFT.
> signal 1
> -0.0240589395458152
> -0.0200948285137433
> -0.0214011982008984
....
>
> signal b
> 0.0100958116451182
> 0.00690202800032552
> 0.00368019172807294
....

this is funny on so many levels.

first, we're comparing signal "1" to, hmmmm, where's signal "2"? or
signal "a"?

well, west, we know what you mean. now the fact that you are asking
about the F.T. of two discrete signals means you mean the *discrete*
fourier transform (DFT). in that case, there aren't anomalous cases
like what Glen Hermannsfeldt pointed out: two continuous-time
functions that, although strictly not equal, *are* equal "almost
everywhere", so their integrals are equal. infinitely thin points on
the continuous-time axis don't matter to integrals. and the Fourier
Transform is also known as the Fourier Integral.

but with the Discrete Fourier Transform, there is no "almost
everywhere". two discrete functions are either identical, or they are
not. and if they are not identical, at every discrete value, their
DFTs will be different (assuming that machine precision and roundoff
are not an issue). that's well established.

r b-j

SteveSmith wrote:

>>
>
> Hi Fred,
> The "many:1" property of the phase doesn't stop the Fourier Transform
> from being a 1:1 mapping. Say you have a point in the frequency
> domain with a phase of P. This is ambiguous with a phase of P + 2
> pi, P + 4 pi, and so on. However, in the time domain this
> corresponds to shifting a cosine wave by 2 pi, 4 pi, etc., which
> leave it unchanged. So P in the frequency domain (with its 2 pi
> ambiguity) is a 1:1 match with a single waveform in the time domain
> (with its 2 pi ambiguity).
> Regards,
> Steve

Ah, yes. Good point. Thanks Steve!

On Sep 5, 5:00*pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
> > The answer depends upon how mathematically accurate you want to be.
> > For practical purposes, the time domain signal has to be the same.
> > Mathematically, you would say that the signals may at most be
> > different on a set of measure zero.
>
> For the continuous transform, any signal with a non-infinite
> (non delta function) discontinuity at a point can have the
> same transform. *Consider, for example,

The domain of the continuous Fourier transform is properly the set of
tempered distributions, which are defined only by their integrals
against smooth localized test functions and not by their values at
individual points. VIewed in this way, two functions that only differ
at isolated points are actually the same distribution, and the Fourier
transform is uniquely invertible.

In other words, in this and many other contexts, one nowadays
redefines the notion of "function" to include things like delta
functions and to exclude worries about differences on sets of measure
zero. It makes perfect sense, because in a physical context it is
never possible to measure the value of a function at an isolated point
-- one can only measure the average of the function in a small region
(and average == integral). And distributions eliminate a lot of
nitpicking headaches that have no practical significance.

Regards,
Steven G. Johnson

Steven G. Johnson wrote:
(snip)

> In other words, in this and many other contexts, one nowadays
> redefines the notion of "function" to include things like delta
> functions and to exclude worries about differences on sets of measure
> zero. It makes perfect sense, because in a physical context it is
> never possible to measure the value of a function at an isolated point
> -- one can only measure the average of the function in a small region
> (and average == integral). And distributions eliminate a lot of
> nitpicking headaches that have no practical significance.

Fourier transform of delta functions works fine, as the
area is not zero. Only discontinuities with zero area
are invisible to integral transforms.

-- glen

Fred Marshall wrote:
> SteveSmith wrote:
>
>> Hi Fred,
>> The "many:1" property of the phase doesn't stop the Fourier Transform
>> from being a 1:1 mapping. Say you have a point in the frequency
>> domain with a phase of P. This is ambiguous with a phase of P + 2
>> pi, P + 4 pi, and so on. However, in the time domain this
>> corresponds to shifting a cosine wave by 2 pi, 4 pi, etc., which
>> leave it unchanged. So P in the frequency domain (with its 2 pi
>> ambiguity) is a 1:1 match with a single waveform in the time domain
>> (with its 2 pi ambiguity).
>> Regards,
>> Steve
>
> Ah, yes. Good point. Thanks Steve!

Note to Rune: the confusion arose here from failing to distinguish the
real thing from its mathematical representation. :-)

Jerry
--
Engineering is the art of making what you want from things you can get.
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