in article 1112443135.716309.30550@f14g2000cwb.googlegroups.com, Andor at an2or@mailcircuit.com wrote on 04/02/2005 06:58:> robert bristow-johnson wrote:...>> i am modifying some of your notation to keep it compatible to what you >> normally see in common DSP texts. one major difference is the numbering of >> indices, and i really think that the MATLAB/Fortran convention of numbering >> arrays (or matrices for that matter) is not equivalent elegance to what is >> normally done in DSP texts and i do not want to change that notation, so i'm >> changing yours so we can have compatible semantics. sorry, Andor, if you >> don't like it, but i really think we need more common elements in our >> notation and i refuse to yield to the MATLAB convention. > > Robert, allow me to say that sometimes you are pretty hard headed :-).no! what would ever give you that impression? i can't be my political POV. i'm such a softie! :^/ (that's my nose getting longer.)> If you wish, we can use zero-based C-style array indexing.otherwise we get this ugly exp(j 2 pi (n-1)(k-1)/N) kinda crap. you can call it "C-style", i might call it "Dykstra-style" indexing (check out http://www.cs.utexas.edu/users/EWD/ewd08xx/EWD831.PDF ). i really think that this indexing issue transcends "C" or whatever programming language.> I just wanted to make the point that two different bases (plural of > basis) are important.i got that.> Let's use {t[0], ..., t[N-1]} (t for time) as the names for the > standard basis set. > > >> are not the t[n] in this basis set simply equal to (... 0, 1, 0, 0, ...)? > > Well, if x is a vector, then (x[0], ... , x[N-1]) is just shorthand for > > x = sum_{n=0}^{N-1} x[n] t[n].but, for the "time domain" basis set, those t[n] basis vectors are column vectors (... 0, 1, 0, ...) where the "1" is in the nth row. right? (it has to be, or then i really dunno what the hell you mean.)> If we use another basis set, for example {f[0], ... , f[N-1]}, then we > can also write X = (X[0] , ... , X[N-1]) for > > X = sum_{k=0}^{N-1} X[k] f[k], > > in which case (.., 0, 1, 0, ..) means something different than in the > first.no, no,... (.., 0, 1, 0, ..) are NOT the basis vectors for the "frequency domain" basis. those are f[k] = (... , exp(j 2 pi n k/N), exp(j 2 pi (n+1) k/N), ...) where n is the row number (using Dykstra indexing) of the column vector.> Let's just call {t[0], ... , t[N-1]} the standard basis.i would call it the "time-domain basis".>>> x = (x[0], ... , x[N-1]) a vector from _X_, F the DFT operator, >> >> which is a square matrix with elements of exp(-j 2 pi n k/N) (where n & k >> are the column & row indices), right? i might reserve the symbol "X" for >> >> X = F x > > Well, for F to be unitary we need to define it with a 1/sqrt(N) > factor, but otherwise you are correct, the entries in the matrix are > 1/sqrt(N) exp(-j 2 pi n k/N).yeah, you're right, but i was trying to fudge what you were saying to make it most consistent with the DSP lit. it's just a scaling issue (and the loss of the semantic "unitary"). i really like that unitary transform property for the continuous Fourier Transform, not so much for the DFT (except maybe for implementation on a fixed-point machine). i know it means that repeatedly applying F, F, F, and F again makes your data set grow in magnitude by a factor of sqrt(N) each time, if you don't put the 1/sqrt(N) in there.>> >>> and S the cylic left-shift operator, i.e. >>> >>> S x = (x[1], x[2], .., x[N-1], x[0]). >> >> >> fine, this is an invertible (is the linear algebra term "complete"?) >> operator unlike the non-cyclic left-shift linear operator: >> >> T x = (x[1], x[2], .., x[N-1], 0) >> >> which destroys information. >> >> >>> Then the operator F S F* (where * denotes the adjoint of F) >> >> F* is a square matrix with elements of 1/N * exp(+j 2 pi n k/N) >> (where n & k are the row & column indices), right? > > Again, you need 1/sqrt(N) for unitarity.yeah, well.>> >>> is in diagonal form >> >> but not on the main diagonal, right? > > YES, in the main diagonal! That's the point! See below:no, it ain't! (perhaps i misunderstand what is meant by "in diagonal form".) doesn't that require being symmetric about the main diagonal? or is it just that each element in every diagonal is equal? / 0 1 0 0 \ | 0 0 1 0 | S = | 0 0 0 1 | \ 1 0 0 0 / is not symmetric about the main diagonal. maybe i misunderestimate (borrowed term from our illustrious Idiot-in-Chief) the meaning of "diagonal form".> The point really was the equation above that, which is the shift > theorem. Let's look at an example with N=4. > I'll use w := e^(-j 2 pi/N) to facilitate notation. Then > > / 1 1 1 1 \ > | 1 w w^2 w^3 | > F = 1/2| 1 w^2 w^4 w^6 |, > \ 1 w^3 w^6 w^9 / > > (note that eg w^4 = 1 etc.), > > / 1 1 1 1 \ > | 1 w w^-2 w^-3 | > F* = 1/2| 1 w^-2 w^-4 w^-6 |, > \ 1 w^-3 w^-6 w^-9 / >fine.> and > > / 0 1 0 0 \ > | 0 0 1 0 | > S = 1/2| 0 0 0 1 |. > \ 1 0 0 0 /are you sure there is a 1/sqrt(N) on this? (you're the expert, but i have my doubts about this statement.)> Then a simple computation shows that > > / 1 0 0 0 \ > | 0 w^-1 0 0 | > F S F* = | 0 0 w^-2 0 | =: d_S, > \ 0 0 0 w^-3 / > > that is the diagonal matrix with the eigenvalues of S on its diagonal. > Multiplying each side k times by itself yields > > (F S F*)(F S F*)...(F S F*) = F S^k F* = d_S^k > > (where S^k is the linear operator that effects k cyclic rotations to > the left), and again by multiplying from the right with F we get, > > F S^k = d_S^k F (Eq. 1b). > > If you set X = F x, for any x in _X_, then the left hand side of (Eq. > 1b) is > > F S^k x = F S^k (x[0],...,x[N-1]) = F (x[k], x[k+1], ..., x[k-1]), (Eq. 2) > > that is we take the DFT of the cyclically rotated vector > (x[k], x[k+1], ..., x[k-1]). The right hand side of (Eq. 1b) gives > > d_S^k F x = d_S^k X (Eq. 3). > > Comparing components of (Eq.2) and (Eq.3) gives > > (F (x[k], x[k+1], ... , x[k-1]))[m] = w^(-k m) X[m], for 0 <= m <= N-1, > > which is the shift theorem as was stated originally. However, I prefer > the much more compact notation in (Eq. 1b). > > >>> Nowhere have I used modular addressing or periodic sequences or >>> anything like it. >> >> that, i disagree with. all of the linear algebra math is fine, but to stand >> back and look at the whole thing and say "Nowhere ... anything like" >> circular indexing or periodic extension, i really disagree. (but >> this is a disagreement of observation, i.e. what we, two different >> observers, say when we both observe the same phenomenon). >> >> on one hand, this operator S, clearly does a circular shift. it is as if >> your vectors were not a linear string of numbers (with two ends), but that >> string where the back end of the vector was bent around and attached to the >> front end (like a strip of paper). the boundary between x[N-1] and x[0] is >> qualitatively no different than the boundary between x[0] and x[1] or >> between any other x[n] and x[n+1]. that is, as i observe it, a periodic >> extension of x and all of the vectors in the _X_ space have that property. > > But we did not need to introduce periodic sequency, we could state > everything using just the linear operator S.sure, you wouldn't even have to go through all that rigmarole since it is obvious that / x[1] \ / 0 1 0 ... 0 \ / x[0] \ | x[2] | | 0 0 1 ... 0 | | x[1] | | x[3] | = | 0 0 0 ... 0 | | x[2] | | ... | | .. .. .. .. | | ... | | x[N-1]| | 0 0 0 ... 1 | | x[N-2]| \ x[0] / \ 1 0 0 ... 0 / \ x[N-1]/ and alls i'm saying is that this (along with the frequency-domain counterpart) wouldn't be happening if it weren't true that:>> ... this would not be the case if the elements of F and F* >> and d_S where not special (i wouldn't claim "unique", just "special"). they >> have to be these particular periodic functions as i have emphasized above. > > Yes, of course....> You can view it that way if you like, it is consistent. However, there > is no need to, as all the properties of F and S are just special cases, > and the theory of linear algebra completely explains what is going on > and why F and S have these properties.but the point i'm saying why, in this special case of the DFT, there *is* this inherent circularity or periodic extension is that there are all sorts of other linear operators (on the same hilbert space) that have no such property. suppose we have the same hilbert space _X_ and the same "time-domain" basis, {t[0], ... , t[N-1]} where t[n] = ( ..., 0, 1, 0, ...) (with the "1" in the nth element) x = x[0]*t[0] + x[1]*t[1] + ... + x[N-1]*t[N-1] and some other "polynomial-domain" basis, {p[0], ... , p[N-1]} where p[k] = (..., n^k, (n+1)^k, ...) (where n^k is in the nth element) x = X[0]*p[0] + X[1]*p[1] + ... + X[N-1]*p[N-1] there is a linear mapping, P, that will map x = (x[0], x[1], ... x[N-1]) to X = (X[0], X[1], ... X[N-1]) where N-1 x[n] = SUM{ n^k X[k] } k=0 or x = P* X we know that the P matrix is the inverse of P* and the elements of P* are obvious from above. isn't that sorta similar to your linear algebra expression of the DFT, except for at least two big differences? one is that the "polynomial-domain" basis, p[k], and the "frequency-domain" basis, f[k], are clearly different, and the other is that although P is linear (so all these nice Hilbert space properties apply), there is no equivalent *circular* "shift theorem" to this linear operator. (there might be some non-invertable linear shift theorem that corresponds multiplying by powers of n, but i wouldn't know what to do with any non-zero x[n] that falls off the edge in the linear shift.) not all linear operators periodically extend the data passed to it, but the DFT does (because of "special" properties it has due to the periodic nature of the elements of F or the periodic nature of the basis f). ...> I wouldn't say stricken, but perhaps biased. I prefer the view that > gives my a better outlook on my surroundings. I think you do too, it is > just that your surroundings are different than mine :-).i really agree. the "surroundings" i behold are taking a real-time stream of data (audio, in my case, but that should not matter) or a really long (sound)file and processing it by sections. i think that this is very, very common in the actual application of the DFT. now, when we yank a section of data out of that stream, it is just a finite block (or "vector" if you prefer) of data. it might be appropriate (depending on what next we might do to the data) to consider that "yanking" operation to be windowing (even if it's just a rectangular window) and there are some effects due to that windowing. fine. no periodic extension yet. this block of data has two ends to it, a front end and a back end. then, suppose you pass this finite chunk (or "vector") of data to a DFT operation. BINGO! at that point, you better understand that your string of numbers (or "vector") has been rolled or bent around where the back end was just glued to the front end and there is NO qualitative difference between what connects x[N-1] to x[0] and what connects x[0] to x[1] or any other x[n] to x[n+1]. it's still describable as a finite set of numbers but it is equivalent in every respect to periodically extending that finite set in both directions forever. this is "mind-numbingly stupid brain-dead dogma". but if some practitioners (say someone doing fast convolution) forget that (or its equivalents, like they have to use modulo addressing), they will run into trouble. i guess dogma has its uses. we keep it in mind, we never question it, and sometimes it saves our asses. it can also be entertaining when it chases the karma. perhaps the karma runs over the dogma. ...> Forgive the long delay in answering.no need to. i don't think there is any obligation.> This morning I went to the flea > market with my family. We bought a new chair for our baby daughter.she your first? (getting any sleep? some people on comp.dsp remember when i was there in 1997.)> and a new bike for me!i'm still riding my old Gitane from the 1970's. good, lightweight, fast bike. looks like shit. "commuter" size tires (a little bigger than touring bikes and a lot thinner than the mountain bikes everyone here gets - it's a lot like driving SUVs in the city, i don't get it.) -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
shift theorem
Started by ●March 26, 2005
Reply by ●April 2, 20052005-04-02
Reply by ●April 2, 20052005-04-02
more mind-numbingly stupid brain dead dogma. in article 424d7f3c$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on 04/01/2005 12:04:> If by that you mean that a non-integer k when applied to the DFT would > result in only part of a waveform - well yes that *is* the point I was > making. The fact is is that not allowing t0 to be shifted off of a > sample point is a requirement for the DFT to have a period of N.it's not true at all, Jim, (assuming that you mean the input or output sequences of the DFT, since the periodicity is the property of sequences going in or out of the DFT) and i disposed of that in the post of 03/31/05 that you never responded to. the input or results of the DFT (or more specifically the iDFT in this case) are periodic with period N regardless whether the "d" in exp(-j*2*pi*d*k/N) * X[k] is an integer or not. it doesn't matter. it was a red herring from the beginning. BTW, the DFT has nothing to do with "sample point". there is no inherent assumption that either the input or output data (both which are periodic with period N) are associated with some continuous-time functions. you *can* make such an association with the sampling and reconstruction theorem (the "reconstruction" half is also the topic of interpolation), but that is another topic. so Jim, i already posted what the math is for a non-integer "d" in y[n] = iDFT{ exp(-j*2*pi*d*k/N) * X[k] } and how y[n] is related to x[n]. care to post the math that makes in not periodic? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●April 3, 20052005-04-03
robert bristow-johnson wrote:> > more mind-numbingly stupid brain dead dogma. >Well, I'm glad your proud of it. Your proofs that the ".." in y[..] can't be non-integer are boring. And besides, I started this by stating that *you* are incapable of having a non-integer n in the expression y[n] so there really is no need for you to prove it over and over. But here's the fallacy in your so called proof. You have after much hemming and hawing decide that you can accept a non-integer shift on the right side of the shift theorem equation but not on the left. Now the left side is just a definition which I have said from the beginning your too brain dead to see that it could be simply defined some other way. That is whenever you have a thorough (or is that throw) expression of the DFT you have an accompanying definition of n as a list of numbers ~ [0,1,2....n-1]. That list could be defined quite differently if you wanted it to be. You don't want it to be. That's fine that doesn't contradict anything I've said - I said from the beginning you would not accept a different definition and therefore you will always have a periodic DFT. But it is not the DFT that is producing the periodicity its the botux that you have injected into your brain that has paralyzed your sense of imagination. -jim ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Reply by ●April 4, 20052005-04-04
in article 4250573f$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on 04/03/2005 16:51:> robert bristow-johnson wrote: >> >> more mind-numbingly stupid brain dead dogma. >> > > Well, I'm glad your proud of it.coming from you it appears to be an honor. why don't you follow in eBob's footsteps and put together a blacklist? i'd be happy to get on it.> Your proofs that the ".." in y[..] can't be non-integer are boring.what proofs? i didn't make any proofs about non-integral subscripts of a sequence of numbers. i'm just saying they're not defined (which they aren't). if you wanna define it, go ahead, tell us clearly what you're doing and and describe it mathematically (so you can't wiggle out of what might come of it). until then, what you have is a sequence of objects (in this case the objects are complex numbers), that means, by definition (of the term "sequence"), there is a one-to-one correspondence between the all of the objects in the sequence and the set of whole numbers. we can put whatever we want as a symbol in the subscript to differentiate one object from another, but given that correspondence it may as well be an integer.> And besides, I started this by stating that *you* are incapable of having > a non-integer n in the expression y[n]actually you were smarter than that when you started this, you're getting progressively dumber about it as time marches on. earlier you said ... in article 424766d4$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on 03/27/2005 21:06:> Now why is it inconceivable to shift the frame of reference for a DFT > by some non-integer amount? Why would the introduction of a linear phase > term make any difference to the DFT?that actually said something, i responded and said "i'm being very careful to not get sucked into a dispute of contradiction that happens here when we're not extremely careful about terminology and notation. if you want to talk about the subject of interpolation, fine, but let's call it what it is (a definition of a continous-whatever function from discrete-whatever samples)." and then i waited for you to actually make your case about would happen if a linear phase factor (i think that's what you meant when you said "term", your mathematical language really is sloppy, Jim) was "introduced" (i s'pose you meant multiply) to the DFT (and i think you meant iDFT even though they're both nearly the same). i waited for days (and posted several hints about that point in response to other people_ to see if you would actually get off your lazy-ass and show us some math (and i was prepared to respond to it). well you didn't. so, then about 4 days later, i wasted about 30 minutes and i did it. and i stuck to the rules and, rather than talk loosely of non-integral indices, i simply brought up another iDFT output sequence that would be the result of a linear phase shift (corresponding to an integer or non-integer amount) in the "frequency domain" or whatever you want to call it. i said: let DFT{ w[n] } = W[m] = Y[m] * exp(-j*2*pi*k*m/N) the inverse DFT: w[n] = N-1 SUM{y[i]*exp(-j*pi*(n-k-i)*(N-1)/N)*sin(pi*(n-k-i))/(N*sin(pi*(n-k-i)/N))} i=0 it turns out that *if* k is an integer then w[n] = y[n-k] or if you don't allow for periodically extended y[..] (which is what the original issue is about), then you have to say w[n] = y[ mod(n-k, N) ] for integer k. now if you want to use that to define y[n] for non-integer n (which is an abuse of notation of subscripts of a sequence of numbers), be my guest. i guess then N-1 y(n) = SUM{y[i]*exp(-j*pi*(n-i)*(N-1)/N)*sin(pi*(n-i))/(N*sin(pi*(n-i)/N))} i=0 so there you have your interpolated sequence (i'm trying to be less abusive of notation by showing y(..) as a continuous function with pareths, not a sequence with square brackets). your lazy ass didn't even have to get up and do any real math. someone else ended up doing it for you. funny thing is the very same expression that extends your sequence into non-integral "subscripts" also periodically extends it forever in both directions with period N. d(n) = exp(-j*pi*n*(N-1)/N)*sin(pi*n)/(N*sin(pi*n/N)) is a periodic function with period N. does that need to be proved to you? so it's kinda hard for you to claim your point (about being able to extend the indices to non-integers) without proving mine (about being able to extend the indices to outside the limits of 0 and N-1). and your other claim (if we can pin you down regarding its meaning) that "The fact is is that not allowing t0 to be shifted off of a sample point is a requirement for the DFT to have a period of N" is pure bullshit. it isn't true at all. there is no external or additional requirement or restriction that has to be made for the data of DFT or iDFT to have period N. it's built into the definition of the DFT and iDFT. it doesn't matter if you do the dirichlet thing to interpolate or not. the interpolation issue has nothing to do with the periodicity issue. at least it doesn't detract from it. getting back to the present...> so there really is no need for you to prove it over and over. > But here's the fallacy in your so called proof.i read with bated breath...> You have after > much hemming and hawing decide that you can accept a non-integer shift > on the right side of the shift theorem equation but not on the left.no, no, i got that right away (and said so, right away), but was waiting for you to actually make the point with some language of enough rigor so you couldn't wiggle out of the implications of it. you didn't bother to (despite a few hints for you to get on with it), so finally, after about 4 days, i spelled it out (as i did right here). and you still didn't get it, or if you did, you conveniently neglected to reply.> Now the left side is just a definition which I have said from the beginning > your too brain dead to see that it could be simply defined some other way.but it's a change of topic (which conveniently serves you in obscuring your bullshit). it's literally non-sequitur. that topic of mapping this sequence of numbers to a continuous function is most often called "interpolation". when that becomes the topic, then to talk intelligently about it we have to specify *how* that interpolation is defined. you never did that.> That is whenever you have a thorough (or is that throw) expression > of the DFT you have an accompanying definition of n as a list of numbers > ~ [0,1,2....n-1].you mean [0,1,2... N-1], don't you? "n" is one of the numbers in that set, no? if you're gonna do math, you gotta be careful about details, Jim.> That list could be defined quite differently if you > wanted it to be. You don't want it to be.no. it doesn't make sense in a subscript of a sequence of numbers. has nothing to do with what i want.> That's fine that doesn't > contradict anything I've said - I said from the beginning you would not > accept a different definition and therefore you will always have a > periodic DFT. But it is not the DFT that is producing the periodicity > its the botux that you have injected into your brain that has paralyzed > your sense of imagination.wow! THAT'S IT???! that's your case spelling out some fallacy?? how persuasive! i guess i'm taking the side of the "unimaginative" accountant rather than the "creative" accountant (e.g. Enron-style or WorldCom-style) so as not to try slip any bullshit past everybody. you're very good at invective, Jim, but you're awful sloppy with the math. actually, you're not really good at the math, are you? (since you don't do any of it.) and your reasoning in logical argument or debate leaves much to be desired. oh, and i use _botox_ in my brain. that's with two Oh's. botux is for losers. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●April 4, 20052005-04-04
Look Bob, you want to argue the merits of defining the function space differently than you do. I could care less. It's like arguing whether left and right have been defined correctly. If I claim that is an arbitrary designation - you would ask me to prove it mathematically. The point was, is and always will be that the DFT simply crunches numbers it doesn't make assumptions. When you say that it makes assumptions your talking thru your hat. -jim robert bristow-johnson wrote:> > in article 4250573f$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on > 04/03/2005 16:51: > > > robert bristow-johnson wrote: > >> > >> more mind-numbingly stupid brain dead dogma. > >> > > > > Well, I'm glad your proud of it. > > coming from you it appears to be an honor. why don't you follow in eBob's > footsteps and put together a blacklist? i'd be happy to get on it. > > > Your proofs that the ".." in y[..] can't be non-integer are boring. > > what proofs? i didn't make any proofs about non-integral subscripts of a > sequence of numbers. i'm just saying they're not defined (which they > aren't). if you wanna define it, go ahead, tell us clearly what you're > doing and and describe it mathematically (so you can't wiggle out of what > might come of it). > > until then, what you have is a sequence of objects (in this case the objects > are complex numbers), that means, by definition (of the term "sequence"), > there is a one-to-one correspondence between the all of the objects in the > sequence and the set of whole numbers. we can put whatever we want as a > symbol in the subscript to differentiate one object from another, but given > that correspondence it may as well be an integer. > > > And besides, I started this by stating that *you* are incapable of having > > a non-integer n in the expression y[n] > > actually you were smarter than that when you started this, you're getting > progressively dumber about it as time marches on. earlier you said ... > > in article 424766d4$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on > 03/27/2005 21:06: > > > Now why is it inconceivable to shift the frame of reference for a DFT > > by some non-integer amount? Why would the introduction of a linear phase > > term make any difference to the DFT? > > that actually said something, i responded and said "i'm being very careful > to not get sucked into a dispute of contradiction that happens here when > we're not extremely careful about terminology and notation. if you want to > talk about the subject of interpolation, fine, but let's call it what it is > (a definition of a continous-whatever function from discrete-whatever > samples)." > > and then i waited for you to actually make your case about would happen if a > linear phase factor (i think that's what you meant when you said "term", > your mathematical language really is sloppy, Jim) was "introduced" (i s'pose > you meant multiply) to the DFT (and i think you meant iDFT even though > they're both nearly the same). i waited for days (and posted several hints > about that point in response to other people_ to see if you would actually > get off your lazy-ass and show us some math (and i was prepared to respond > to it). well you didn't. > > so, then about 4 days later, i wasted about 30 minutes and i did it. and i > stuck to the rules and, rather than talk loosely of non-integral indices, i > simply brought up another iDFT output sequence that would be the result of a > linear phase shift (corresponding to an integer or non-integer amount) in > the "frequency domain" or whatever you want to call it. i said: > > let DFT{ w[n] } = W[m] = Y[m] * exp(-j*2*pi*k*m/N) > > the inverse DFT: > > w[n] = > > N-1 > SUM{y[i]*exp(-j*pi*(n-k-i)*(N-1)/N)*sin(pi*(n-k-i))/(N*sin(pi*(n-k-i)/N))} > i=0 > > it turns out that *if* k is an integer then > > w[n] = y[n-k] > > or if you don't allow for periodically extended y[..] (which is what the > original issue is about), then you have to say > > w[n] = y[ mod(n-k, N) ] > > for integer k. > > now if you want to use that to define y[n] for non-integer n (which is an > abuse of notation of subscripts of a sequence of numbers), be my guest. i > guess then > > N-1 > y(n) = SUM{y[i]*exp(-j*pi*(n-i)*(N-1)/N)*sin(pi*(n-i))/(N*sin(pi*(n-i)/N))} > i=0 > > so there you have your interpolated sequence (i'm trying to be less abusive > of notation by showing y(..) as a continuous function with pareths, not a > sequence with square brackets). your lazy ass didn't even have to get up > and do any real math. someone else ended up doing it for you. > > funny thing is the very same expression that extends your sequence into > non-integral "subscripts" also periodically extends it forever in both > directions with period N. > > d(n) = exp(-j*pi*n*(N-1)/N)*sin(pi*n)/(N*sin(pi*n/N)) > > is a periodic function with period N. does that need to be proved to you? > > so it's kinda hard for you to claim your point (about being able to extend > the indices to non-integers) without proving mine (about being able to > extend the indices to outside the limits of 0 and N-1). > > and your other claim (if we can pin you down regarding its meaning) that > "The fact is is that not allowing t0 to be shifted off of a sample point is > a requirement for the DFT to have a period of N" is pure bullshit. it isn't > true at all. there is no external or additional requirement or restriction > that has to be made for the data of DFT or iDFT to have period N. it's > built into the definition of the DFT and iDFT. > > it doesn't matter if you do the dirichlet thing to interpolate or not. the > interpolation issue has nothing to do with the periodicity issue. at least > it doesn't detract from it. > > getting back to the present... > > > so there really is no need for you to prove it over and over. > > But here's the fallacy in your so called proof. > > i read with bated breath... > > > You have after > > much hemming and hawing decide that you can accept a non-integer shift > > on the right side of the shift theorem equation but not on the left. > > no, no, i got that right away (and said so, right away), but was waiting for > you to actually make the point with some language of enough rigor so you > couldn't wiggle out of the implications of it. you didn't bother to > (despite a few hints for you to get on with it), so finally, after about 4 > days, i spelled it out (as i did right here). > > and you still didn't get it, or if you did, you conveniently neglected to > reply. > > > Now the left side is just a definition which I have said from the beginning > > your too brain dead to see that it could be simply defined some other way. > > but it's a change of topic (which conveniently serves you in obscuring your > bullshit). it's literally non-sequitur. that topic of mapping this > sequence of numbers to a continuous function is most often called > "interpolation". when that becomes the topic, then to talk intelligently > about it we have to specify *how* that interpolation is defined. you never > did that. > > > That is whenever you have a thorough (or is that throw) expression > > of the DFT you have an accompanying definition of n as a list of numbers > > ~ [0,1,2....n-1]. > > you mean [0,1,2... N-1], don't you? "n" is one of the numbers in that set, > no? if you're gonna do math, you gotta be careful about details, Jim. > > > That list could be defined quite differently if you > > wanted it to be. You don't want it to be. > > no. it doesn't make sense in a subscript of a sequence of numbers. has > nothing to do with what i want. > > > That's fine that doesn't > > contradict anything I've said - I said from the beginning you would not > > accept a different definition and therefore you will always have a > > periodic DFT. But it is not the DFT that is producing the periodicity > > its the botux that you have injected into your brain that has paralyzed > > your sense of imagination. > > wow! THAT'S IT???! that's your case spelling out some fallacy?? how > persuasive! > > i guess i'm taking the side of the "unimaginative" accountant rather than > the "creative" accountant (e.g. Enron-style or WorldCom-style) so as not to > try slip any bullshit past everybody. > > you're very good at invective, Jim, but you're awful sloppy with the math. > > actually, you're not really good at the math, are you? (since you don't do > any of it.) and your reasoning in logical argument or debate leaves much to > be desired. > > oh, and i use _botox_ in my brain. that's with two Oh's. botux is for > losers. > > -- > > r b-j rbj@audioimagination.com > > "Imagination is more important than knowledge."----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Reply by ●April 4, 20052005-04-04
robert bristow-johnson wrote:> :^/ (that's my nose getting longer.):-)> >>> Then the operator F S F* (where * denotes the adjoint of F) > >>> is in diagonal form > >> > >> but not on the main diagonal, right? > > > > YES, in the main diagonal! That's the point! See below: > > no, it ain't! (perhaps i misunderstand what is meant by "in diagonal > form".) doesn't that require being symmetric about the maindiagonal? or> is it just that each element in every diagonal is equal? > > > / 0 1 0 0 \ > | 0 0 1 0 | > S = | 0 0 0 1 | > \ 1 0 0 0 / >S is not diagonal, F S F* is diagonal. ...> > / 0 1 0 0 \ > > | 0 0 1 0 | > > S = 1/2| 0 0 0 1 |. > > \ 1 0 0 0 / > > > are you sure there is a 1/sqrt(N) on this? (you're the expert, but ihave> my doubts about this statement.)Sorry, you are right, the 1/2 is just a leftover from the cut'n'paste and should be deleted. [snipping periodicity and linearity and unitarity and linear algebra and everything connected with it] ...> > This morning I went to the flea > > market with my family. We bought a new chair for our baby daughter. > > she your first? (getting any sleep? some people on comp.dspremember when> i was there in 1997.)Yes, my first. Maybe if I sent some new pictures to Jerry he might be kind enough to post them again?> > and a new bike for me! > > i'm still riding my old Gitane from the 1970's. good, lightweight,fast> bike. looks like shit. "commuter" size tires (a little bigger thantouring> bikes and a lot thinner than the mountain bikes everyone here gets -it's a> lot like driving SUVs in the city, i don't get it.)Yeah, those MTB are silly in the city -- too slow and too heavy. I bought an old race bike (light with extremely thin tires) which has break-handles for upright and bent-over position. The price was right (it was the cheapest on the market) and since summer is coming again and my last bike was stolen, this seemed the right moment for a good bargain.> > -- > > r b-j rbj@audioimagination.com > > "Imagination is more important than knowledge."One of these days I might also install signature line. My favourite up to now is the following quote from Alan Turing (about programing the first programmable computer): "One can do as one wishes. One must just make up one's mind". Regards, Andor
Reply by ●April 4, 20052005-04-04
in article 42512592$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on 04/04/2005 07:31:> Look Bob, you want to argue the merits of defining the function space > differently than you do.whatever that means. care to be specific?> I could care less. It's like arguing whether > left and right have been defined correctly.yup. on the left side of DFT{ y[n] } = Y[m] or DFT{ y[n-k] } = Y[m] * exp(-j*2*pi*k*m/N) there is no defined (as far as the DFT is concerned) meaning in y[..] when the stuff in the subscript is non-integral. the right side expression of the 2nd equation DOES have meaning for non-integral k, but when you iDFT it you don't get ANY y[n]. there is no subscript of n that will get any version of y[n] to correspond to the right side with non-integral k. that, if you're gonna have any rigor, is the way it is. but what *is* the iDFT of the right side when k is either integral or non-integral? it is N-1 SUM{y[i]*exp(-j*pi*(n-k-i)*(N-1)/N)*sin(pi*(n-k-i))/(N*sin(pi*(n-k-i)/N))} i=0 and, gee, it's also periodic with period N.> If I claim that is an > arbitrary designation - you would ask me to prove it mathematically.no, you claimed that it wasn't so if the delay, k, was non-integral and i asked you to prove that, you never got off your butt to do it, i finally checked and typed in the (standard dirichlet) expression and said, "see, it IS periodic" and you kept saying it's brain-dead. so is the moral of the story is that the person who doesn't and can't prove the point correct over the person that does prove his or her point?> The point was, is and always will be that the DFT simply crunches > numbers it doesn't make assumptions.yup, and the way it crunches the numbers is to fit the data passed to it to a set of basis functions, all that are periodic with period N. that, in and of itself, periodically extends that data for all purposes, practically and theoretically, in both directions with period N. shifting or convoluting the data? you have to either apply the modulo operator y[ mod(n,N) ] or simply understand that y[n +/- N] is the same as y[n]. both are equivalent. when you say "y[ mod(n,N) ]" (which you have to when there is shifting of convoluting), you are saying the same thing as "y[n] is periodically extended with period N". this periodic extension concept is necessary for, at least those two operations and does not hurt in any other operation, including your interpolation of y[n] to non-integral n, given the assumptions of interpolation you have made. the DFT is inanimate, so *it* cannot make assumptions, but when using it *we* have to recognize that periodic extension, lest when we do certain operations with the DFT data, we will run into trouble.> When you say that it makes assumptions your talking thru your hat.i am using anthropomorphic language to try to describe something that is, admittedly, inanimate. it's kinda like some lawyer in the Supreme Court saying that the right to free speech in the 1st Amendment of the U.S. Constitution "stands resolute". the 1st amendment, or the Constitution, or any other law of any country, does not "stand resolute" at all. *people* do when they use or interpret such a law. when people interpret some inanimate law in a resolute way, it makes perfect sense when they say the law "stands resolute". in the same manner i will continue to repeat, as do some pretty heavyweight DSP text authors, that the DFT periodically extends the data passed to it or (by anthropomorhizing) the DFT assumes the N samples passed to it are N contiguous samples of a periodic sequence with period N and returns as results N samples of one period of another periodic sequence of period N. the DFT is one and the same as the so-called Discrete Fourier Series. the math supports it consistently (it *has* to since the periodicity is built into the definition of the DFT). it's just the way it is, despite the periodic denial we get here periodically. somehow, i doubt that's what you mean by "talking thru your hat". -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●April 4, 20052005-04-04
robert bristow-johnson wrote:> > > I could care less. It's like arguing whether > > left and right have been defined correctly. > > yup. on the left side ofMan you can't possibly be this dense. I'm talking about the assignment of the words "left" and "right".> > > If I claim that is an > > arbitrary designation - you would ask me to prove it mathematically. > > no, you claimed that it wasn't so if the delay, k, blah blah blah.....The concept of left and right have nothing to do with delay they are just arbitrary definitions. It's like I'm talking to brick. -jim ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Reply by ●April 4, 20052005-04-04
in article 425152c6$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on 04/04/2005 10:44:> robert bristow-johnson wrote: > >> >>> I could care less. It's like arguing whether >>> left and right have been defined correctly. >> >> yup. on the left side of > > Man you can't possibly be this dense. I'm talking about the assignment > of the words "left" and "right". > > >> >>> If I claim that is an >>> arbitrary designation - you would ask me to prove it mathematically. >> >> no, you claimed that it wasn't so if the delay, k, blah blah blah..... > > The concept of left and right have nothing to do with delay they are > just arbitrary definitions. It's like I'm talking to brick.your credibility is slipping away, Jim. (actually, with me it as evaporated completely, but i can't speak for the others.) remember saying this?: in article 4250573f$1_2@127.0.0.1, jim at "N0sp"@m.sjedging@mwt.net wrote on 04/03/2005 16:51:> But here's the fallacy in your so called proof. You have after > much hemming and hawing decide that you can accept a non-integer shift > on the right side of the shift theorem equation but not on the left. Now > the left side is just a definitionit was in (late) response to this: in article BE71C701.5C91%rbj@audioimagination.com, robert bristow-johnson at rbj@audioimagination.com wrote on 03/31/2005 15:25:>> "jim" <"N0sp"@m.sjedging@mwt.net> wrote in message >> news:424c297d$1_1@127.0.0.1......>>> Here's the shift theorem as presented by Robert. >>> >>> DFT{ y[n-k] } = exp(-j*2*pi*k*m/N) * Y[m] > > cause is on the left and effect is on the right. > >>> Will you permit k to be a non-integer? > > be specific, Jim. "k" where? on the right of the = sign, there is some > meaning. on the left there is no meaning for non-integer k. so the > specific question to ask is what is the inverse DFT of the expression on the > right when k is non-integer? will you answer that? then we can argue about > your next question (when properly phrased):this is on the record, Jim. you can't change what you said in the past (you can correct it if you want). people on this newsgroup are smart enough to see through attempts to obfuscate. that includes invective. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
