robert bristow-johnson wrote:> ... > from what they taught me, sin(x)/x clearly has a "singularity" at > 0 but it > is a "removable singularity". sin(x)/x and *all* its derivatives > exist > everywhere and it is analytic everywhere. in fact, a decent > McLaurin series can be determined for it. > > r b-jIf I consider sin(x) to just be a handy notation for x-(x^3)/3!+(x^5)/5! ..., the question of singularity of sin(x)/x is obviously the same as the question if f(x) = x/x has a singularity at x=0. Bernhard

# what is the z-transform of sinc function?

Started by ●February 28, 2004

Reply by ●March 3, 20042004-03-03

Reply by ●March 3, 20042004-03-03

>>>Given the fraction A/B, A and B each being differentiable expressions, >>>A/B had a computable value where I come from even when each is zero.The words as written apply equally to sin(x) / x and sin(x) / x^2 but as you later notice they do not work for the second expression. You got the facts correct for the specific examples but did not communicate the rule required to address further examples. This is a pretty standard introductory calculus exercise and the text books all give the full statement which is a bit more elaborate than your statement. The comment was that you dropped the extra conditions to make your rule work for all examples. Or perhaps you have a nonstandard definition of computable value for what most would call an indeterminant form.>>>With sin(x)/x^2, there's a different result, which shows simply that not >>>all zeros are the same. >> >> >> You just provided a counterexample to your claim so the conditions are >> a bit stronger than your statement. > >Gordon, > >I'm surprised! I didn't claim that 0/0 was known without separate >specification, but only that sin(x)/x, x being zero, is. The claim is >no stronger than the what can be supported. > >Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >

Reply by ●March 3, 20042004-03-03

Gordon Sande wrote:>>>>Given the fraction A/B, A and B each being differentiable expressions, >>>>A/B had a computable value where I come from even when each is zero. > > > > The words as written apply equally to sin(x) / x and sin(x) / x^2 but > as you later notice they do not work for the second expression. You got > the facts correct for the specific examples but did not communicate > the rule required to address further examples. This is a pretty standard > introductory calculus exercise and the text books all give the full > statement which is a bit more elaborate than your statement. The comment > was that you dropped the extra conditions to make your rule work for > all examples. > > Or perhaps you have a nonstandard definition of computable value for what > most would call an indeterminant form.It's easy enough to compute that sin(x)/x^2 goes to infinity as x goes to 0. If you don't like "infinity" used that way, I'll accommodate you by saying that sin(x)/x^2 becomes unbounded as x goes to zero. Either way, it's computable. And sin^2(x)/x is computable too. It goes to zero when x does. Laplace is nice. Heavyside is nifty. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●March 3, 20042004-03-03

Jerry Avins wrote:> > It's easy enough to compute that sin(x)/x^2 goes to infinity as x goes > to 0. If you don't like "infinity" used that way, I'll accommodate you > by saying that sin(x)/x^2 becomes unbounded as x goes to zero. Either > way, it's computable. And sin^2(x)/x is computable too. It goes to zero > when x does. Laplace is nice. Heavyside is nifty. >From somewhere in the dim recesses of my mind I seem to recall that sin(x)/x can be expressed as an infinite product ~ cos(x)*cos(x/2)*cos(x/4)*cos(x/8)......cos(x/2^n) as n goes to infinity. In this form it is not particularly easy to evaluate except at multiples of pi/2. At x=0 it will clearly be one and at any other multiple of pi/2 it will be zero. -jim -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Reply by ●March 3, 20042004-03-03

In article <404603fd$0$3099$61fed72c@news.rcn.com>, Jerry Avins <jya@ieee.org> wrote:>Subject: Re: what is the z-transform of sinc function? >From: Jerry Avins <jya@ieee.org> >Reply-To: jya@ieee.org >Organization: The Hectic Eclectic >Date: Wed, 03 Mar 2004 11:12:40 -0500 >Newsgroups: sci.math,comp.dsp > >Gordon Sande wrote: > >>>>>Given the fraction A/B, A and B each being differentiable expressions, >>>>>A/B had a computable value where I come from even when each is zero. >> >> >> >> The words as written apply equally to sin(x) / x and sin(x) / x^2 but >> as you later notice they do not work for the second expression. You got >> the facts correct for the specific examples but did not communicate >> the rule required to address further examples. This is a pretty standard >> introductory calculus exercise and the text books all give the full >> statement which is a bit more elaborate than your statement. The comment >> was that you dropped the extra conditions to make your rule work for >> all examples. >> >> Or perhaps you have a nonstandard definition of computable value for what >> most would call an indeterminant form. > >It's easy enough to compute that sin(x)/x^2 goes to infinity as x goes >to 0. If you don't like "infinity" used that way, I'll accommodate you >by saying that sin(x)/x^2 becomes unbounded as x goes to zero. Either >way, it's computable. And sin^2(x)/x is computable too. It goes to zero >when x does. Laplace is nice. Heavyside is nifty.Which "infinity" do you mean as the indeterminant form sin(x) / x^2 at 0 does not have the same limit from both sides so it would not seem to fit any plausible notion of removable singularity. Usually the notion of computable value would be expected to have a well determined sign for the value when the value is away from 0. If one writes sin(x) / x^2 as { sin(x) / x } * { 1 / x } ones sees that you are trying to give meaning to 1 / x at 0. In the neighbourhood of 0 the values 1 / x can be arbitarily large and of either sign. It is a standard example of when the notion of a value gets into trouble.>Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >

Reply by ●March 3, 20042004-03-03

robert bristow-johnson <rbj@surfglobal.net> wrote in message news:<BC6ADF86.9186%rbj@surfglobal.net>...> In article e4a0829b.0403022146.693768d@posting.google.com, ZZBunker at > zzbunker@netscape.net wrote on 03/03/2004 00:46: > > > robert bristow-johnson <rbj@surfglobal.net> wrote in message > > news:<BC66C424.8FD7%rbj@surfglobal.net>... > >> In article c1qqaq$cn1$1@mozo.cc.purdue.edu, Joenyim Kim at > >> jeonyimkim80@yahoo.com wrote on 02/28/2004 14:35: > >> > >>> Can anybody tell me what is the z-transform of "sinc" function and what is > >>> its region of convergence? > >> > ... > >> > >> now if you were to scale the input a little: > >> > >> x[n] = sinc(a*n) = sin(pi*a*n)/(pi*a*n) > >> > >> for some real, a, then that's a little harder and i don't know what the > >> answer is right off the bat. come to think of it, i don't even know what > >> the Laplace Transform of the sinc() function is since i do not know how to > >> analytically extend the rect() function to use complex arguments. > > > > It's simple to extend rect to complex arguments, > > you get the donut function. > > > > donut(z) = rect(|z|). > > > > You don't need Laplace transforms, > > since they're not required for non-temporal functions. > > but if Im(z) turns out to be zero, does donut(z) = rect(z)?That depends on how rect is defined. Some people define it as an odd function, some people define it as an even function, and some people define it as a non-function, but rather as a limit. Using the minimal information interpretation of Signal Processing, Im(z) is zero everywhere in the complex plane, just like I stated it was. Hence for mathematicians it can also be written in the laborous form: R^2->R: donut(x,y;a) = lim [ ( a*sin(pi*sqrt(x^2+y^2)) ) ]. a->infinity+ Mod { shifting and scaling }.> hmmm. maybe since rect(x) is even. > > but does donut(z/a) = Z{sinc(a*n)}? or something similar?Does it? I don't know. Does it have too? No. Since the scaling properties of the Z transform are a property of the Z-transform, not of sinc(z).> > r b-j

Reply by ●March 3, 20042004-03-03

Gordon Sande wrote:> Which "infinity" do you mean as the indeterminant form sin(x) / x^2 at 0 > does not have the same limit from both sides so it would not seem to fit > any plausible notion of removable singularity. Usually the notion of > computable value would be expected to have a well determined sign for > the value when the value is away from 0. > > If one writes sin(x) / x^2 as { sin(x) / x } * { 1 / x } ones sees that > you are trying to give meaning to 1 / x at 0. In the neighbourhood of > 0 the values 1 / x can be arbitarily large and of either sign. It is > a standard example of when the notion of a value gets into trouble.Does infinity need to have a sign? I like to think of tan(x) and 1/x as "wrapping around" and coming out the other side. Just as 0 and j0 map to the same place, why can't their reciprocals be the same also? In polar coordinates the angles of zero and 1/0 don't matter. The really screwy case is exp(1/x). That keeps me awake nights. Functions of two kinds need to be dealt with: "blows up" and "doesn't blow up". Purists might want to add another: "blows up real fast". What's the point of asking about the sign of sin(x)/x^2 @ x=0? It blew up. It blows up all by itself, but to it blow it down takes a lot of huffing and puffing. I can only pretend to Philistinism so long, then I have to smile behind my fan. In the real world, a supposed singularity is a sign of an overly simple model. Simplicity is good, so we adapt. We need to be careful with the fine distinctions when exploring new territory*. Once the terrain is familiar, we can relax. Relax! You have it charted! Thanks. Jerry __________________ * Funny things happen with zeros. I once heard a lecturer claim (and "prove"!) that an airplane wing won't produce lift in an ideal (i.e., lossless) fluid. The argument boiled down to something like "producing lift necessarily creates drag ...". Once I juxtaposed the pieces of the argument to expose the fallacy, he claimed (using many more words) that an action didn't _necessarily_ produce a reaction. -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●March 3, 20042004-03-03

jim <"N0sp"@m.sjedging@mwt.net> writes:> From somewhere in the dim recesses of my mind I seem to recall that sin(x)/x can > be expressed as an infinite product ~ > cos(x)*cos(x/2)*cos(x/4)*cos(x/8)......cos(x/2^n) as n goes to infinity. In this > form it is not particularly easy to evaluate except at multiples of pi/2. At x=0 > it will clearly be one and at any other multiple of pi/2 it will be zero.Close. Omit the first term. sin(x)/x = cos(x/2) * cos(x/4) * ... * cos(x/2^n) * ... Scott -- Scott Hemphill hemphill@alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear

Reply by ●March 3, 20042004-03-03

Jerry Avins <jya@ieee.org> wrote:> Gordon Sande wrote: > > > Which "infinity" do you mean as the indeterminant form sin(x) / x^2 at > > 0 does not have the same limit from both sides so it would not seem to > > fit any plausible notion of removable singularity.Maybe, as Jerry indicates below, he was thinking of the unsigned infinity of R*, the one-point compactification of the reals. If R* is taken to be the codomain of sin(x) / x^2, then at x = 0 we do have the same limit from both sides, and so the singularity is removable.> > Usually the notion > > of computable value would be expected to have a well determined sign > > for the value when the value is away from 0.But that infinity and 0 are both unusual, being unsigned.> > If one writes sin(x) / x^2 as { sin(x) / x } * { 1 / x } ones sees that > > you are trying to give meaning to 1 / x at 0. In the neighbourhood of > > 0 the values 1 / x can be arbitarily large and of either sign. It is > > a standard example of when the notion of a value gets into trouble.In R*, 1/0 is unsigned infinity.> Does infinity need to have a sign? I like to think of tan(x) and 1/x as > "wrapping around" and coming out the other side.Right. You might like the diagram of R* shown at <http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html>. If the codomain of tan(x) is taken to be R*, then it is continuous on R, and its graph is simply a helix (on the surface of a cylinder, with the y-axis wrapped around as shown at the link above). For f(x) = 1/x, if we take both domain and range to be R*, then we can visualize its graph nicely on the surface of a torus. (Wrap the y-axis around to get a cylinder, then wrap the x-axis around, giving a torus.) The graph forms a simple closed curve. David Cantrell

Reply by ●March 3, 20042004-03-03

Scott Hemphill wrote:> > jim <"N0sp"@m.sjedging@mwt.net> writes: > > > From somewhere in the dim recesses of my mind I seem to recall that sin(x)/x can > > be expressed as an infinite product ~ > > cos(x)*cos(x/2)*cos(x/4)*cos(x/8)......cos(x/2^n) as n goes to infinity. In this > > form it is not particularly easy to evaluate except at multiples of pi/2. At x=0 > > it will clearly be one and at any other multiple of pi/2 it will be zero. > > Close. Omit the first term. > > sin(x)/x = cos(x/2) * cos(x/4) * ... * cos(x/2^n) * ... >Ah yes, of course, what I had was sin(2x)/2x. -jim -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =-----