Reply by glen herrmannsfeldt September 8, 20092009-09-08
Tim Wescott <tim@seemywebsite.com> wrote:
(snip)
 
<> This is a coincidence. Not a defining property.
 
< It's far more than a coincidence.  In fact I contend that there's very 
< little mathematical difference between a perfectly periodic function and 
< a function that's defined on a circle.

It gets more interesting in higher dimensions.  Much of solid state
physics can be done with periodic boundary conditions.  For a large
enough sample (1e18 atoms is not a big crystal, but plenty big enough),
and for bulk (not surface) properties, the analysis is much easier
with periodic boundary conditions.   That is, imagine a cube crystal
such that when an electron goes out one face it immediately enters
the opposite face.  You can then use the properties of periodic
basis functions for the electron wave function in the crystal.

Interestingly for this discussion, the actual crystal is not periodic,
but the periodic solution is used anyway.  With 1e18 atoms, the
crystal is 1e6 atoms on a side, plenty large N for a nice DFT.
 
< The DTFT of an infinitely long perfectly periodic sampled-time sequence 
< is a finite-length series of impulses, and the thing that you find 
< embedded in your integral equation that defines the strength of those 
< impulses is the summation that defines the DFT.

-- glen 
Reply by Tim Wescott September 8, 20092009-09-08
On Tue, 08 Sep 2009 10:55:14 -0700, Dale Dalrymple wrote:

> .On Sep 8, 9:28 am, Jerry Avins <j...@ieee.org> wrote: .> Rune Allnor > wrote: > .> > .> > From a previpouis post, > .> > .> >http://groups.google.no/group/comp.dsp/msg/4d0095164f78bcfc?hl=no .> > .> > pisted by Julius: > .> > .> > "To add to Rune's comment, the DFT operates on discrete, .> > > finite, PERIODIC time series and produces a discrete, finite, .> > > PERIODIC frequency series." > .> > .> > How is it possible to read the above and *not* get the .> > > impression that Jusius makes exatly that claim? .> > .> That's what a DFT does. That's not to say it's the only useful way to > .> use it. Ask the DFT what a particular term X[n] is and it will tell > you. > .> Ask what X[n+N] is, and it will tell you exactly the same thing, > whether > .> that makes sense or not. The periodicity is there whether we ignore > it > .> (the usual case) or not. > .> ... > .> Jerry > > The DFT defines no value for X[n+N]. Some people apply the DFT by > assuming that the defined samples into and out of the DFT are samples of > the Fourier Transform pair of some signal. For this assumption to be > true, the Fourier Transform pair of signals must have the characteristic > of periodic extension. That is not a characteristic of the DFT but the > requirement of an assumption made in the application of the DFT to some > subset (containing X[n+N]) of the domain of the FT. Many posters state > the assumption as if it were the DFT. What's the harm? The harm is that > we get people like the OP who apply the DFT as if it were the Fourier > Transform because that's what postings seem to mean since they leave out > the details of the assumptions made. Some people claim to know better > but still persist in making misleading postings. I think Rune is correct > in believing that there are also those who post here who don't know > better. The OP seems to have been one. At least the OP noticed the > difference and asked about it. > > Dale B. Dalrymple
And he spawned an enormous thread that he probably didn't expect at all! Hopefully he's not scared off by it. -- www.wescottdesign.com
Reply by Tim Wescott September 8, 20092009-09-08
On Tue, 08 Sep 2009 10:17:14 -0700, Rune Allnor wrote:

> On 8 Sep, 18:17, Tim Wescott <t...@seemywebsite.com> wrote: > > >> What my argument boils down to is that for the analysis to be >> meaningful in addition to being mathematically correct then (a) the >> numbers have to represent something that's useful to have broken down >> into frequency components (there's that danged pragmatism!) > > Exactky. Let go of the notion that 'the numbers have to *mean* > something' and you would be a far better engineer. And mathematician.
A better mathematician, perhaps, but if I can't assign a meaning to the numbers then what use are they to me as an engineer? And if I can't use them, then what's the business case for generating them?
>> >> But my overall explanation is that you are trying to make my point >> >> for me, by finding numerous examples of randomly applying the DFT to >> >> vectors of numbers in ways that are sometimes potentially meaningful >> >> and sometimes not. >> >> > The result of the DFT always 'means' the same thing: How a given data >> > set can be expressed in terms of a set of complex-valued basis >> > vectors in an N-dimensional vector space. >> >> Absolutely true, in a purely mathematical sense. > > Hold on to that thought! > >> &nbsp;But as I tried to point >> out in my 'fruity' example, the resulting set of basis vectors may have >> absolutely no meaning at all with respect to the problem at hand. &nbsp;So >> like any other mathematical analysis tool one must ask oneself (or >> one's collaborators) what level of relevance the tool has to the >> problem you're trying to solve. > > Exactly. Now, elaborate on that idea: > > The reason why people use the DFT is that it is the *only* flavour of > the Fourier Transforms that work on > > a) Finite amounts of data > b) Discrete data. > > Break any one of those (infinite amounts of data or continuous data) and > the digital computer is of little use to you. > > As you repeatedly point out, one often is interested in some other > flavour of the FT, but for *pragmatic* reasons one can only work with > the DFT. > >> >> But I can't for the life of me explain why you are doing this, and >> >> why you couch it in language that implies you are doing the >> >> opposite. >> >> > Again, you are a smart guy. You are amongst the very few people here >> > that >> >> > a) are smart enouhg to see that the DFT is nothing more >> > &nbsp; &nbsp;than a basis shift transform of the data actually presented to >> > &nbsp; &nbsp;the algorithm. >> > b) pragmatic enough to actually change your opinions once >> > &nbsp; &nbsp;you see that you have been wrong. >> >> > Over the past few weeks I have come to realize that people I used to >> > respect for their knowledge and skills, >> >> Perhaps you're just getting cranky because it's fall and the light's >> failing? > > Nope. Now is my best time of the year; reasonably light and cool (not > cold) weather. > >> > suffer from this DFT delusion. >> > This was too good an opportunity to test out if these people really >> > are as stupid as they appear. >> >> I think "delusion" is too strong a word. &nbsp;'Limited view', perhaps. >> >> I'll assert right here that it is correct to say that if you're taking >> the spectra of a physical signal in the sense of a Fourier transform >> then the DFT _itself_ is only _exact_ for exactly periodic (and >> sampled) signals (or ones on a circular domain -- if you can find >> them). > > This is where you - and just about everybody else - go wrong. Go back to > the statements you agreed with above, that the DFT is a basis transform > in N-dimensional vector space where N is finite. Since you agreed with > that statement, you will have to marry it with the statement you made > here, about periodic sequences. > > It's only one way to do that: By the pragmatic point of view, where one > would *want* to compute the DTFT, but is limited to actually compute the > DFT. > > What you end up with, is that the (continuous w domain) DTFT over one > period of an infinitely long, periodic sequence with period N *happens* > to be non-zero at exactly the same normalized frequencies where the > (discrete) coefficients of the DFT are located; AND that the continuous > w in detween these discrete non-vanishing DTFT coefficients is 0. > > This is a coincidence. Not a defining property.
It's far more than a coincidence. In fact I contend that there's very little mathematical difference between a perfectly periodic function and a function that's defined on a circle. The DTFT of an infinitely long perfectly periodic sampled-time sequence is a finite-length series of impulses, and the thing that you find embedded in your integral equation that defines the strength of those impulses is the summation that defines the DFT. So it's far more than a coincidence; it's that -- pretty much by definition of periodicity -- there is a 1:1 mapping between one period of a periodic sequence and the whole edge of a circle.
>> So while I have a good idea of where _I_ want to go, I'm still pretty >> unclear on where you want to lead me. > > You have already outlined the arguments and ideas I would like you to > see. The only thing left I might wish for, is that you contemplate them.
I will, but only if you promise to contemplate the mapping between a circle and the points within one period of a periodic function. -- www.wescottdesign.com
Reply by Dale Dalrymple September 8, 20092009-09-08
.On Sep 8, 9:28 am, Jerry Avins <j...@ieee.org> wrote:
.> Rune Allnor wrote:
.>
.> > From a previpouis post,
.>
.> >http://groups.google.no/group/comp.dsp/msg/4d0095164f78bcfc?hl=no
.>
.> > pisted by Julius:
.>
.> > "To add to Rune's comment, the DFT operates on discrete,
.> > finite, PERIODIC time series and produces a discrete, finite,
.> > PERIODIC frequency series."
.>
.> > How is it possible to read the above and *not* get the
.> > impression that Jusius makes exatly that claim?
.>
.> That's what a DFT does. That's not to say it's the only useful way
to
.> use it. Ask the DFT what a particular term X[n] is and it will tell
you.
.> Ask what X[n+N] is, and it will tell you exactly the same thing,
whether
.> that makes sense or not. The periodicity is there whether we ignore
it
.> (the usual case) or not.
.> ...
.> Jerry

The DFT defines no value for X[n+N]. Some people apply the DFT by
assuming that the defined samples into and out of the DFT are samples
of the Fourier Transform pair of some signal. For this assumption to
be true, the Fourier Transform pair of signals must have the
characteristic of periodic extension. That is not a characteristic of
the DFT but the requirement of an assumption made in the application
of the DFT to some subset (containing X[n+N]) of the domain of the FT.
Many posters state the assumption as if it were the DFT. What's the
harm? The harm is that we get people like the OP who apply the DFT as
if it were the Fourier Transform because that's what postings seem to
mean since they leave out the details of the assumptions made. Some
people claim to know better but still persist in making misleading
postings. I think Rune is correct in believing that there are also
those who post here who don't know better. The OP seems to have been
one. At least the OP noticed the difference and asked about it.

Dale B. Dalrymple
Reply by Rune Allnor September 8, 20092009-09-08
On 8 Sep, 18:17, Tim Wescott <t...@seemywebsite.com> wrote:


> What my argument boils down to is that for the analysis to be meaningful > in addition to being mathematically correct then (a) the numbers have to > represent something that's useful to have broken down into frequency > components (there's that danged pragmatism!)
Exactky. Let go of the notion that 'the numbers have to *mean* something' and you would be a far better engineer. And mathematician.
> >> But my overall explanation is that you are trying to make my point for > >> me, by finding numerous examples of randomly applying the DFT to > >> vectors of numbers in ways that are sometimes potentially meaningful > >> and sometimes not. > > > The result of the DFT always 'means' the same thing: How a given data > > set can be expressed in terms of a set of complex-valued basis vectors > > in an N-dimensional vector space. > > Absolutely true, in a purely mathematical sense.
Hold on to that thought!
> &#2013266080;But as I tried to point > out in my 'fruity' example, the resulting set of basis vectors may have > absolutely no meaning at all with respect to the problem at hand. &#2013266080;So > like any other mathematical analysis tool one must ask oneself (or one's > collaborators) what level of relevance the tool has to the problem you're > trying to solve.
Exactly. Now, elaborate on that idea: The reason why people use the DFT is that it is the *only* flavour of the Fourier Transforms that work on a) Finite amounts of data b) Discrete data. Break any one of those (infinite amounts of data or continuous data) and the digital computer is of little use to you. As you repeatedly point out, one often is interested in some other flavour of the FT, but for *pragmatic* reasons one can only work with the DFT.
> >> But I can't for the life of me explain why you are doing this, and why > >> you couch it in language that implies you are doing the opposite. > > > Again, you are a smart guy. You are amongst the very few people here > > that > > > a) are smart enouhg to see that the DFT is nothing more > > &#2013266080; &#2013266080;than a basis shift transform of the data actually presented to the > > &#2013266080; &#2013266080;algorithm. > > b) pragmatic enough to actually change your opinions once > > &#2013266080; &#2013266080;you see that you have been wrong. > > > Over the past few weeks I have come to realize that people I used to > > respect for their knowledge and skills, > > Perhaps you're just getting cranky because it's fall and the light's > failing?
Nope. Now is my best time of the year; reasonably light and cool (not cold) weather.
> > suffer from this DFT delusion. > > This was too good an opportunity to test out if these people really are > > as stupid as they appear. > > I think "delusion" is too strong a word. &#2013266080;'Limited view', perhaps. &#2013266080; > > I'll assert right here that it is correct to say that if you're taking > the spectra of a physical signal in the sense of a Fourier transform then > the DFT _itself_ is only _exact_ for exactly periodic (and sampled) > signals (or ones on a circular domain -- if you can find them).
This is where you - and just about everybody else - go wrong. Go back to the statements you agreed with above, that the DFT is a basis transform in N-dimensional vector space where N is finite. Since you agreed with that statement, you will have to marry it with the statement you made here, about periodic sequences. It's only one way to do that: By the pragmatic point of view, where one would *want* to compute the DTFT, but is limited to actually compute the DFT. What you end up with, is that the (continuous w domain) DTFT over one period of an infinitely long, periodic sequence with period N *happens* to be non-zero at exactly the same normalized frequencies where the (discrete) coefficients of the DFT are located; AND that the continuous w in detween these discrete non-vanishing DTFT coefficients is 0. This is a coincidence. Not a defining property.
> So while I have a good idea of where _I_ want to go, I'm still pretty > unclear on where you want to lead me.
You have already outlined the arguments and ideas I would like you to see. The only thing left I might wish for, is that you contemplate them. Rune
Reply by Tim Wescott September 8, 20092009-09-08
On Tue, 08 Sep 2009 16:42:42 +0000, glen herrmannsfeldt wrote:

> Tim Wescott <tim@seemywebsite.com> wrote: (snip) > > < What you are doing wrong is trying to answer a continuous-time > question < with sampled-time analysis tools. Would you drive a nail > with a < screwdriver? Screw in a screw with a wrench? Tighten a nut > with a < hammer? Screws, nuts and nails are all fasteners; hammers, > screwdrivers < and wrenches are all tools to make fasteners work, yet > use the wrong tool < for a perfectly good fastener and you'll do no good > at all. Similarly, < the Laplace transform is the right tool for > analyzing continuous-time < ordinary linear differential equations, > which is what you're working with. > > This is true. Still, there are many cases where continuous data must be > analyzed with discrete techniques. For sufficiently large N, a DFT > becomes a sufficiently close approximation to the continuous transform. > > You can, for example, show that the Fourier Transform of a Guassian is a > Gaussian using the FFT. You do have to remember where zero is, and > supply enough points. With fast computers, one can apply the FFT to > very large data sets. (I am still interested in the FFT of an entire > CD, for example.) >
No argument -- but if the OP is really analyzing a lumped-element linear circuit, then Laplace analysis is the way to go, particularly for such a simple one. If nothing else, doing so gives you a symbolic answer, which makes it much easier to play "what if" games with component values. -- www.wescottdesign.com
Reply by Rune Allnor September 8, 20092009-09-08
On 8 Sep, 18:28, Jerry Avins <j...@ieee.org> wrote:
> >> What is claimed is that the operation on a finite > >> sequence gives the same result as operating on a repeated duplication of > >> that sequence, or on a circularly shifted version of it. You ably > >> demonstrated that in this thread. > > > I stand corrected. There are more people than Tim around > > that has the potential to read and understand an argument, > > and also change their minds. > > > Thanks, Jerry. My your post means that my faith in Homo > > Sapiens as an occasionally intelligent creature didn't > > shatter completely. Quite yet. > > I'm delighted to be an agent of hope.
Alas. You might have postponed my descent into the abyss, maybe. But not brought hope. Rune
Reply by glen herrmannsfeldt September 8, 20092009-09-08
Tim Wescott <tim@seemywebsite.com> wrote:
(snip)
 
< What you are doing wrong is trying to answer a continuous-time question 
< with sampled-time analysis tools.  Would you drive a nail with a 
< screwdriver?  Screw in a screw with a wrench?  Tighten a nut with a 
< hammer?  Screws, nuts and nails are all fasteners; hammers, screwdrivers 
< and wrenches are all tools to make fasteners work, yet use the wrong tool 
< for a perfectly good fastener and you'll do no good at all.  Similarly, 
< the Laplace transform is the right tool for analyzing continuous-time 
< ordinary linear differential equations, which is what you're working with.

This is true.  Still, there are many cases where continuous data
must be analyzed with discrete techniques.  For sufficiently large N,
a DFT becomes a sufficiently close approximation to the continuous
transform.  

You can, for example, show that the Fourier Transform of a Guassian
is a Gaussian using the FFT.  You do have to remember where zero is,
and supply enough points.  With fast computers, one can apply the
FFT to very large data sets.  (I am still interested in the FFT
of an entire CD, for example.)

-- glen
Reply by Jerry Avins September 8, 20092009-09-08
Rune Allnor wrote:
> On 8 Sep, 17:32, Jerry Avins <j...@ieee.org> wrote: >> Rune Allnor wrote: >> >> ... >> >> >> >> >> >>> Doesn't matter. I used the numbers as arguments to the DFT. >>> Your point seems to be that in case 1, my use of the DFT >>> is wrong because no more numbers exist in the data set than >>> those I feed to the DFT. In case 2 you seem to think that >>> using the DFT is correct, because of the data items I *don't* >>> feed to the DFT. In this last case, that argument turns >>> straight and all of a sudden the DFT is wrong because of the >>> data points I *don't* feed to the DFT. >>> Would you ever present that sort of argument to a paying >>> customer? If so, would you be surprised if the same paying >>> customer called in people to express a 2nd, a 3rd and >>> maybe even a 4th opinion? >>>> But my overall explanation is that you are trying to make my point for >>>> me, by finding numerous examples of randomly applying the DFT to vectors >>>> of numbers in ways that are sometimes potentially meaningful and >>>> sometimes not. >>> The result of the DFT always 'means' the same thing: How >>> a given data set can be expressed in terms of a set of >>> complex-valued basis vectors in an N-dimensional vector space. >>>> But I can't for the life of me explain why you are doing this, and why >>>> you couch it in language that implies you are doing the opposite. >>> Again, you are a smart guy. You are amongst the very few >>> people here that >>> a) are smart enough to see that the DFT is nothing more >>> than a basis shift transform of the data actually >>> presented to the algorithm. >>> b) pragmatic enough to actually change your opinions once >>> you see that you have been wrong. >>> Over the past few weeks I have come to realize that people >>> I used to respect for their knowledge and skills, suffer >>> from this DFT delution. This was too good an opportunity to >>> test out if these people really are as stupid as they appear. >> I don't think anyone here claims that a DFT is legitimate only with >> periodic data. > > From a previpouis post, > > http://groups.google.no/group/comp.dsp/msg/4d0095164f78bcfc?hl=no > > pisted by Julius: > > "To add to Rune's comment, the DFT operates on discrete, > finite, PERIODIC time series and produces a discrete, finite, > PERIODIC frequency series." > > How is it possible to read the above and *not* get the > impression that Jusius makes exatly that claim?
That's what a DFT does. That's not to say it's the only useful way to use it. Ask the DFT what a particular term X[n] is and it will tell you. Ask what X[n+N] is, and it will tell you exactly the same thing, whether that makes sense or not. The periodicity is there whether we ignore it (the usual case) or not.
>> What is claimed is that the operation on a finite >> sequence gives the same result as operating on a repeated duplication of >> that sequence, or on a circularly shifted version of it. You ably >> demonstrated that in this thread. > > I stand corrected. There are more people than Tim around > that has the potential to read and understand an argument, > and also change their minds. > > Thanks, Jerry. My your post means that my faith in Homo > Sapiens as an occasionally intelligent creature didn't > shatter completely. Quite yet.
I'm delighted to be an agent of hope. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Tim Wescott September 8, 20092009-09-08
On Mon, 07 Sep 2009 13:38:39 +0200, Eugen Artus wrote:

> Hi! > > Suppose there is resistor R and a capacitor C connected in series. A > voltage U_0 is applied at time t=0. The current in the time domain is > then > > I(t) = U_0/R * e^(-t/RC) (1) > > Taking a Laplace transform and substituting s by j*2*\pi*f, I get the > frequency domain current: > > I(f) = U_0*C/(j*2*\pi*f*R*C+1) (2) > > Now the problem: I defined values for U_0, R and C, calculated the time > domain signal by Eq. 1 and transformed it by DFT (using the Python > function numpy.fft.rfft). Then, I normalized the result by multiplying > with 2/N, except the first and the last sample, which were multiplied > with 1/N. N is the number of samples of the time signal. > > A comparison with the signal predicted by Eq. 2 shows a significant > difference. What am I doing wrong?
(sorry for getting off on the tangent without answering your real question). What you are doing wrong is trying to answer a continuous-time question with sampled-time analysis tools. Would you drive a nail with a screwdriver? Screw in a screw with a wrench? Tighten a nut with a hammer? Screws, nuts and nails are all fasteners; hammers, screwdrivers and wrenches are all tools to make fasteners work, yet use the wrong tool for a perfectly good fastener and you'll do no good at all. Similarly, the Laplace transform is the right tool for analyzing continuous-time ordinary linear differential equations, which is what you're working with. Keep your analysis in the Laplace domain, and things should work well. If you're trying to analyze what happens if you sample the result of the continuous-time operation then say so, and someone will point out how to do that part. -- www.wescottdesign.com