"Bernhard Holzmayer" <holzmayer.bernhard@deadspam.com> wrote in message news:39899646.CGuMU2cGCv@holzmayer.ifr.rt...> Greg Berchin wrote: > > > On Thu, 19 Feb 2004 08:21:03 +0100, Bernhard Holzmayer > > <holzmayer.bernhard@deadspam.com> wrote: > > > >>>Following is required for the signal then: > >>>1. Every second zero crossing (ZC) has opposite direction: > >>> neg->pos, pos->neg, neg->pos, ... > >>>2. Derivate dA/dt (slope) changes sign between consequent ZCs. > >>>3. Second derivate has ZCs between consequent ZCs > >>> of the signal. > > > > Do not forget that the zeroes of a function are not necessarily > > real. For zeroes in the complex plane, the concepts of "positive" > > and "negative" are not so well defined. > > > > Thanks for the hint. > > I thought that a function, if it is complex, it can be described by > a real part and an imaginary part, and that, because both are > orthogonal, they can be treated independently. > > But I guess, you're right, it's too much simplification. > > But now a question arises: if we talk of complex solutions or even > complex functions, would 0+5i or 5+0j be called a zero crossing? > Or only the value 0+0i ?Usually we are talking about functions of a complex variable ai+bj where i and j are unit vectors in a plane. And, usually i=1 and is what we call "real" so we forget it and j is in the orthogonal direction called "imaginary". Then, relative to that plane is an amplitude surface which can be positive or negative and you can think of it as real . So, the numbers above would be the coordinates on the plane and not the function that describes the amplitude surface. as an example..... If the surface is a magnitude then it never goes negative and can have distinct zeros anywhere on the plane - i.e. at any coordinate ri+sj. Fred
Band limited function zero crossings, how many?
Started by ●February 18, 2004
Reply by ●February 20, 20042004-02-20
Reply by ●February 20, 20042004-02-20
"Bob Cain" <arcane@arcanemethods.com> wrote in message news:c12t9o0ddg@enews3.newsguy.com...> > Do you know of applications wherein the prolate spheroid > decomposition has been used? > > > Bob > --Hello Bob, I have a sort of answer. IIRC, when various types of windows were being designed one optimal form that traded mainlobe to sidelobe width was discovered (Slepian et. al.) that used prolate spheroidal waves. Kaiser found a good approximation that while it doesn't use these functions, does use something that is close. He used a modified Bessel function (I.e., a Bessel function with a purely imaginary argument) and if you look at Slepian's paper (that Matt posted a link to), you can see how the prolate spheroidal functions are related to the Bessel functions. One way of viewing all of this stuff is we are just solving the wave equation for different shaped cavities for their eigenfunctions. If we pick a box, then we get sines and cosines along any of the three axes.. If we pick a cylinder, with get Bessel functions along the radial axis If we pick a sphere, we get spherical Bessel functions along the radial axis. And if we stretch the sphere along one axis, we get prolate spheroidal wave functions along the radial axis And if we compress the sphere along one axis, we get oblate spheroidal wave functions. In fact the sinc(x) function is a spherical Bessel function of order zero and we here all know it as the kernel function for uniform sampling. It is kinda cool how all these things relate. And sometimes these relations show up in not so obvious ways. Clay -- Clay S. Turner, V.P. Wireless Systems Engineering, Inc. Satellite Beach, Florida 32937 (321) 777-7889 www.wse.biz csturner@wse.biz
Reply by ●February 20, 20042004-02-20
Clay S. Turner wrote:> One way of viewing all of this stuff is we are just solving the wave > equation for different shaped cavities for their eigenfunctions. > > If we pick a box, then we get sines and cosines along any of the three > axes.. > > If we pick a cylinder, with get Bessel functions along the radial axis > > If we pick a sphere, we get spherical Bessel functions along the radial > axis. > > And if we stretch the sphere along one axis, we get prolate spheroidal wave > functions along the radial axis > > And if we compress the sphere along one axis, we get oblate spheroidal wave > functions. > > > In fact the sinc(x) function is a spherical Bessel function of order zero > and we here all know it as the kernel function for uniform sampling. It is > kinda cool how all these things relate. And sometimes these relations show > up in not so obvious ways.Fascinating, Clay. Many thanks. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein
Reply by ●February 20, 20042004-02-20
Matt Timmermans wrote:> "Ron Hardin" <rhhardin@mindspring.com> wrote in message > news:4033F78F.68ED@mindspring.com... > >>And in fact, if my guess is right, any continuous function at all might > > appear in a given > >>time window and still be band limited (though huge outside that time > > window). > > Your guess is correct. There is a family of prolate spheroidal wave > functions, all limited to any given band, that is complete in any given time > interval. > >Yes there is such a family of PSWF's (one such family for any fixed finite frequency band and any fixed finite time interval). But I don't think that implies that the OP's guess is correct. In fact (see the first part of the paper of Slepian's you have mentioned) every band- limited function is smooth, so for instance the continuous but not differentiable function f(t) = |t| on the (time) interval [-1, 1] cannot be extended to a band-limited function on the real line (no matter what the band limit is). Don't let the (remarkable) orthogonality properties of the PSWF's mislead you. Fixing a frequency band [-W, W] we can (to use the notation of Slepian's paper) find an orthonormal basis r_n (n = 0,1,...) for the space of [-W, W]-bandlimited functions on the line, which are orthogonal on [-1, 1] and such that the L^2[-1, 1]-norm of r_n is the square root of \lambda_n, the n-th eigenvalue of the integral operator defining the PSWF's for this pair of bandwidth and time-width values. If we let s_n = r_n / (\lambda_n)^(1/2) then s_n (n = 0,1,...) is an orthonormal basis for L^2[-1, 1]. So any f in L^2[-1, 1] (including the absolute value function) has a series expansion on [-1, 1] in terms of the s_n: f = c_0 s_0 + c_1 s_1 + ..., and we know this will converge in L^2[-1, 1], and that the sum of the squares of the moduli of the c_n will converge. But in terms of the r_n we have f = d_0 r_0 + d_1 r_1 + ... (on [-1, 1]), where d_n = c_n / (\lambda_n)^(1/2). If this series converges in L^2(R) we can use it to construct a bandlimited extension of f to the whole real line, but (keep in mind that the \lambda_n converge to zero) there is no guarantee that it will. In fact, for functions like the absolute value function on [-1, 1] this series does not converge in L^2(R). I wonder just which functions on [-1, 1] have extensions in L^2(R) whose Fourier transforms are zero outside [-W, W] (for a given W). I also wonder if there is a bound (for W fixed, of course) on the number of zero-crossings in [-1, 1] such a function can have; that would address the OP's OQ (original question). My guess is that there is, but I need to think about this some more.... Bob Beaudoin
Reply by ●February 20, 20042004-02-20
Jerry Avins wrote:> Raymond Toy wrote: > >>>>>>> "Fred" == Fred Marshall <fmarshallx@remove_the_x.acm.org> writes: >> >> >> >> Fred> If you don't like to use infinite sincs then I bet you will >> find that >> Fred> suitably windowed sincs will support the same arguments. >> >> >> Is there not a theorem that says you can't have both a time-limited >> and a band-limited signal? Something about the time-bandwidth product >> being constant? >> >> Ray > > > Doesn't the development of that theorem assumes that the signals are > finite? If a clever mathematician can chop a sphere into an infinite > number of pieces, then assemble those pieces to make two spheres the > same size and density as the original, anything goes. If course, the > scheme to do this with gold spheres founders on the indivisibility of > gold atoms. > > JerryThose clever mathematicians actually have a way to chop a ball into five pieces that can be re-arranged into a ball of twice the radius. (Google for "Banach-Tarski paradox".) But the pieces have such irregular shapes that any attempt to apply this chopping to a golden sphere will indeed run afoul of the indivisibility of the gold atoms. Bob Beaudoin
Reply by ●February 20, 20042004-02-20
"Robert E. Beaudoin" <reb@sens.com> wrote in message news:42c6a$40369899$44a72252$11987@msgid.meganewsservers.com...> Don't let the (remarkable) orthogonality properties of the PSWF's > mislead you. [... just because you can get convergence in > L^2[-1,1] doesn't mean that the reconstruction will converge in > L^2(R) ...]I hadn't thought of that, but there remain reasonable interpretations of Ron's informally stated guess that are correct. You can find a bandlimited function on the whole real line that matches any other function to within any given bound on the error in L2[-1,1], for example. For practial purposes, however, it's certainly good to know that trying to converge in [-1,1] can produce a signal with infinite energy outside the interval.> I wonder just which functions on [-1, 1] have extensions in L^2(R) > whose Fourier transforms are zero outside [-W, W] (for a given W). > I also wonder if there is a bound (for W fixed, of course) on the > number of zero-crossings in [-1, 1] such a function can have; that > would address the OP's OQ (original question). My guess is that > there is, but I need to think about this some more....We can certainly rule out a bound on the number of zero crossings, given that the Nth PSWF *has* N zeros in the interval. Given how fast the concentrations fall off for successive PSWFs, though, I think you will fail to get convergence on L^2(R) for a lot of interesting functions -- even smooth ones.
Reply by ●February 21, 20042004-02-21
Clay S. Turner wrote: (snip)> If we pick a cylinder, with get Bessel functions along the radial axis> If we pick a sphere, we get spherical Bessel functions along the radial > axis.(snip)> In fact the sinc(x) function is a spherical Bessel function of order zero > and we here all know it as the kernel function for uniform sampling. It is > kinda cool how all these things relate. And sometimes these relations show > up in not so obvious ways.Yes, sinc() is a spherical Bessel function, but it is also the Fourier transform of Rect(). But then at least the Fourier series came from an eigenfunction differential equation, so it makes some sense. I am not sure, though, that there is a connection between them. -- glen
Reply by ●February 21, 20042004-02-21
"Robert E. Beaudoin" <reb@sens.com> wrote in message news:42c6a$40369899$44a72252$11987@msgid.meganewsservers.com...> I wonder just which functions on [-1, 1] have extensions in L^2(R) > whose Fourier transforms are zero outside [-W, W] (for a given W).Bob, The Fourier Series, as I recall, minimizes the least square error for a given number of sin cos basis functions. If that's what you meant by L^2 here. Fred
Reply by ●February 21, 20042004-02-21
Matt Timmermans wrote:> I hadn't thought of that, but there remain reasonable interpretations of > Ron's informally stated guess that are correct. You can find a bandlimited > function on the whole real line that matches any other function to within > any given bound on the error in L2[-1,1], for example.Convergence of a series of bandlimited functions to any arbitrary L2[-1,1] function in the L2 norm does not imply that they (the limit of the bandlimited function and a representative of this limit in L2) have the same number of zero crossings. Counter example: consider f \in L2[-1,1], defined as f(x) = chi_Q(x) - 1/2, where chi_Q is the characteristic function of the rational numbers on the interval [-1,1] (this is my favourite counter-example function :). f is in the same L2 equivalence class as the constant function g \in L2[-1,1], defined as g(x) = -1/2. So, when you have a series of L2 convergent functions h_i, with h_i -> f in L2 for i->\infty, ie. || h_i - f ||_{L2} -> 0 you also have ||h_i - g ||_{L2} -> 0 for i->\infty (h_i converges to both f and g in L2) Now f has denumerable infinite many zero crossings in [-1,1], whereas g has none. What does that say about the number of zero crossings of the limit of h_i? Regards, Andor
Reply by ●February 21, 20042004-02-21
"Andor" <an2or@mailcircuit.com> wrote in message news:ce45f9ed.0402210421.790394b5@posting.google.com...> Convergence of a series of bandlimited functions to any arbitrary > L2[-1,1] function in the L2 norm does not imply that they (the limit > of the bandlimited function and a representative of this limit in L2) > have the same number of zero crossings.Yes, I know, but the family of functions we are using as a basis here includes a single member, bandlimited, with whatever number of zero crossings you want, so there is no need even to converge to settle the question of whether or not that number is bounded.
