I used to think the operations are the same? Thank you.
What is the difference between FFT and IFFT?
Started by ●March 22, 2005
Reply by ●March 22, 20052005-03-22
>I used to think the operations are the same? > >Thank you. >Here is the equation for both forward and inverse transform: (Iam not used to typing math equations in text files, but never the less:) ---------------------------------------------- FFT: x(F)= summation( x(n)* exp[(-j*2*pi/N)kn] --------- 0->N-1 IFFT: x(n)= 1/N * summation( x(n)* exp[(j*2*pi/N)kn] --------- 0->n-1 ------------------------------------------------- They are the same from a computational point of view, where one utilizes the same function to perform both forward and inverse transform. But mathematically speaking they are different, only just though but from a signal processing view point, they are reaaaally different. One takes you from Time-> Frequency while the other takes you from Frequency->Time. So it depends on the context of the reference. Anyways,both from a theoretical/practical point of view, we can spread the normalization factor 1/N over the forward and inverse transform or move it to the forward completely etc. If you spread it across equally then you would have to multiply the result by 1/sq.rt(N) and take care of the negativity in the nth root of unity. --Bhooshan This message was sent using the Comp.DSP web interface on www.DSPRelated.com
Reply by ●March 22, 20052005-03-22
On Tue, 22 Mar 2005 19:28:16 +0800, "Sea Squid" <Sea.Squid@hotmail.com> wrote:>I used to think the operations are the same? > >Thank you.Sea Squid, uh oh. The FFT estimates the spectral content (the harmonic content) of a time-domain sequence of digital signal samples. The results of the FFT are frequency-domain samples. The IFFT is a process to convert frequency-domain samples back to time-domain samples. If you've had formal engineering (mathematical) training, then you must surely remember that the Fourier transform is *not* equal the Inverse Fourier transform. To think that they're equal is kinda like like saying that multiplication and division are equal. Good Luck, [-Rick-]
Reply by ●March 22, 20052005-03-22
> On Tue, 22 Mar 2005 19:28:16 +0800, "Sea Squid" > <Sea.Squid@hotmail.com> wrote: > >> I used to think the operations are the same?in article 4240c70e.40154093@news.sf.sbcglobal.net, Rick Lyons at r.lyons@_BOGUS_ieee.org wrote on 03/22/2005 20:46:> If you've had formal engineering (mathematical) training, > then you must surely remember that the Fourier transform is > *not* equal the Inverse Fourier transform. To think that > they're equal is kinda like like saying that multiplication > and division are equal.boy, Rick, we could get into a deep philosophical disagreement here. when expressed as: FT: +inf X(f) = integral{ x(t) * exp(-j*2*pi*f*t) dt} -inf iFT: +inf x(t) = integral{ X(f) * exp(+j*2*pi*f*t) df} -inf the Fourier Transform and Inverse Fourier Transform are exactly the same thing except the +j is replaced by -j. the funny thing is that -j and +j have equal claim to being the sqrt(-1). both square to be -1. it is only a matter of convention which separates the two. multiplication and division are also inverse operations, but they are qualitatively different. the FT and iFT are only quantitatively different and only because of convention. that is what allows the Duality Theorem to exist. the FT and iFT are closer to the 1/x function, which is its own inverse. a similar argument could be made for the DFT and iDFT if the convention was modified slightly to put a 1/sqrt(N) in front of both summations. they would be the same damn thing if you switch -j and +j. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●March 23, 20052005-03-23
Rick wrote: " If you've had formal engineering (mathematical) training, then you must surely remember that the Fourier transform is *not* equal the Inverse Fourier transform. To think that they're equal is kinda like like saying that multiplication and division are equal . " This is true for the DFT. However, some unitary transforms are also Hermitian, in which case the transform and its inverse are, in fact, the same. A non-trivial example is the Hadamard transform (with the proper normalization factor). Regards, Andor
Reply by ●March 23, 20052005-03-23
Haha, my example is not only non-trivial, it is also non-correct. The Hadamard transform, while unitary, is not Hermitian. Better examples of unitary and Hermitian operators are the Pauli operators. Unfortunately I can't think of an application of Pauli operators to DSP, which sort of nullifies my above comment (at least in this forum). Oh well.
Reply by ●March 23, 20052005-03-23
On Tue, 22 Mar 2005 21:39:14 -0500, robert bristow-johnson <rbj@audioimagination.com> wrote:>> On Tue, 22 Mar 2005 19:28:16 +0800, "Sea Squid" >> <Sea.Squid@hotmail.com> wrote: >> >>> I used to think the operations are the same? > >in article 4240c70e.40154093@news.sf.sbcglobal.net, Rick Lyons at >r.lyons@_BOGUS_ieee.org wrote on 03/22/2005 20:46: > >> If you've had formal engineering (mathematical) training, >> then you must surely remember that the Fourier transform is >> *not* equal the Inverse Fourier transform. To think that >> they're equal is kinda like like saying that multiplication >> and division are equal. > >boy, Rick, we could get into a deep philosophical disagreement here. when >expressed as: > > >FT: > +inf > X(f) = integral{ x(t) * exp(-j*2*pi*f*t) dt} > -inf > > >iFT: > +inf > x(t) = integral{ X(f) * exp(+j*2*pi*f*t) df} > -inf > > > >the Fourier Transform and Inverse Fourier Transform are exactly the same >thing except the +j is replaced by -j. the funny thing is that -j and +j >have equal claim to being the sqrt(-1). both square to be -1. > >it is only a matter of convention which separates the two. multiplication >and division are also inverse operations, but they are qualitatively >different. the FT and iFT are only quantitatively different and only >because of convention. that is what allows the Duality Theorem to exist. >the FT and iFT are closer to the 1/x function, which is its own inverse. > >a similar argument could be made for the DFT and iDFT if the convention was >modified slightly to put a 1/sqrt(N) in front of both summations. they >would be the same damn thing if you switch -j and +j. >Hi Robert B-J, well, ... humm. I was looking at Sea Squid's question from a strictly "practical", spectrum analysis, view point. But ... I agree with everything you wrote. I realize that the DFT and the IDFT differ only in their phase results, and that realization didn't occur to me when I replied to Sea Squid. Maybe his question was much "deeper" than I first thought. Good Robert. [-Rick-]
Reply by ●March 23, 20052005-03-23
Andor, The simple example of a self inverting transform is a Hartley transform. Clay "Andor" <an2or@mailcircuit.com> wrote in message news:1111567486.428107.161760@z14g2000cwz.googlegroups.com...> Haha, my example is not only non-trivial, it is also non-correct. The > Hadamard transform, while unitary, is not Hermitian. Better examples of > unitary and Hermitian operators are the Pauli operators. Unfortunately > I can't think of an application of Pauli operators to DSP, which sort > of nullifies my above comment (at least in this forum). Oh well. >
Reply by ●March 25, 20052005-03-25
Thanks Clay. I've always wondered, in what applications is the Hartley transform used? Regards, Andor
Reply by ●March 25, 20052005-03-25
Hello Andor, The Hartley is just a variation on the Fourier transform that is useful for real valued signals. A FHT is supposed to be pretty good, but FFT methods compete quite well. So I believe the Hartley transform is almost totally academic these days. There is a simple linear transformation between the Hartley and Fourier transforms, so one doesn't really have any info the other doesn't. Clay "Andor" <an2or@mailcircuit.com> wrote in message news:1111743674.514081.107790@z14g2000cwz.googlegroups.com...> Thanks Clay. I've always wondered, in what applications is the Hartley > transform used? > > Regards, > Andor >
