Robert Israel wrote:> In article <4043c9c4$0$3079$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote:> 0/0 is undefined where I come from. > limit_{x -> 0} sin(x)/x = 1 but that's not the same as saying > sin(0)/0 = 1.It's precisely the same. Lim(sin(x)/x) is how to calculate sin(0)/0. {x->0} 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. With sin(x)/x^2, there's a different result, which shows simply that not all zeros are the same. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
what is the z-transform of sinc function?
Started by ●February 28, 2004
Reply by ●March 2, 20042004-03-02
Reply by ●March 2, 20042004-03-02
In article <4044efdc$0$3088$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: Tue, 02 Mar 2004 15:34:31 -0500 >Newsgroups: sci.math,comp.dsp > >Robert Israel wrote: > >> In article <4043c9c4$0$3079$61fed72c@news.rcn.com>, >> Jerry Avins <jya@ieee.org> wrote: > >> 0/0 is undefined where I come from. >> limit_{x -> 0} sin(x)/x = 1 but that's not the same as saying >> sin(0)/0 = 1. > >It's precisely the same. Lim(sin(x)/x) is how to calculate sin(0)/0. > {x->0}It is usually called a "removable singularity" when the limit is well defined. Common practice is to assume that all removable singularities have been removed.>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. > >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.>Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >
Reply by ●March 2, 20042004-03-02
g.sande@worldnet.att.net (Gordon Sande) wrote:> In article <4044efdc$0$3088$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote: > >Robert Israel wrote: > > > >> In article <4043c9c4$0$3079$61fed72c@news.rcn.com>, > >> Jerry Avins <jya@ieee.org> wrote: > > > >> 0/0 is undefined where I come from. > >> limit_{x -> 0} sin(x)/x = 1 but that's not the same as saying > >> sin(0)/0 = 1. > > > >It's precisely the same. Lim(sin(x)/x) is how to calculate sin(0)/0. > > {x->0} > > It is usually called a "removable singularity" when the limit is well > defined.Right. At least, that's what I would say. I also think that most other mathematicians would agree with me. But perhaps I'm wrong in thinking that. See below.> Common practice is to assume that all removable singularities have been > removed.I suppose it's common in some contexts and not in others. [Certainly, in a typical beginning calculus course, for example, if the instructor says "f(x) = sin(x)/x", the student would be expected to claim that x = 0 is not in the (implied) domain of the function.] Readers might find a comment made by a referee to John Hauser to be of interest. <http://www.jhauser.us/publications/HandlingFloatingPointExceptions.html> Referee: "I find the assertion that the function sin(x)/x has a singularity at x = 0 to be hairsplitting. I'd wager $5 that 90% of mathematicians would claim that sin(x)/x does not have a singularity at x = 0, since to them it would be the name of the function defined as: f(0) = 1, and f(x) = sin(x)/x elsewhere." Hauser: "In computer programs, functions are effectively defined by the expressions that evaluate them. If a sinc subroutine is coded as `sin(x)/x', the subroutine will fail for x = 0. Mathematicians may automatically eliminate removable singularities in expressions like sin(x)/x, but computers won't. I have replaced the word function by expression in an attempt to be more precise." Clearly, I disagree with the referee about what "90% of mathematicians would claim". But perhaps I'm wrong. David Cantrell
Reply by ●March 2, 20042004-03-02
In article <20040302171102.808$cA@newsreader.com>, David W. Cantrell <DWCantrell@sigmaxi.org> wrote:>Readers might find a comment made by a referee to John Hauser to be of >interest. ><http://www.jhauser.us/publications/HandlingFloatingPointExceptions.html>>Referee: "I find the assertion that the function sin(x)/x has a singularity > at x = 0 to be hairsplitting. I'd wager $5 that 90% of mathematicians > would claim that sin(x)/x does not have a singularity at x = 0, since to > them it would be the name of the function defined as: f(0) = 1, and > f(x) = sin(x)/x elsewhere.">Clearly, I disagree with the referee about what "90% of mathematicians >would claim". But perhaps I'm wrong.I haven't conducted a survey, and I don't think the referee has either, but I would love to take that wager. I hope that just about every mathematician who has taught a beginning complex analysis class would say (in that context at least) that there is a removable singularity at x = 0. I suppose context is the key: in certain contexts we do assume that removable singularities have been removed. That's not the same as saying that they don't exist. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2
Reply by ●March 2, 20042004-03-02
In article 20040302171102.808$cA@newsreader.com, David W. Cantrell at DWCantrell@sigmaxi.org wrote on 03/02/2004 17:11:> I suppose it's common in some contexts and not in others. [Certainly, in a > typical beginning calculus course, for example, if the instructor says > "f(x) = sin(x)/x", the student would be expected to claim that x = 0 is not > in the (implied) domain of the function.] > > Readers might find a comment made by a referee to John Hauser to be of > interest. > <http://www.jhauser.us/publications/HandlingFloatingPointExceptions.html> > > Referee: "I find the assertion that the function sin(x)/x has a singularity > at x = 0 to be hairsplitting. I'd wager $5 that 90% of mathematicians > would claim that sin(x)/x does not have a singularity at x = 0, since to > them it would be the name of the function defined as: f(0) = 1, and > f(x) = sin(x)/x elsewhere." > > Hauser: "In computer programs, functions are effectively defined by the > expressions that evaluate them. If a sinc subroutine is coded as > `sin(x)/x', the subroutine will fail for x = 0. Mathematicians may > automatically eliminate removable singularities in expressions like > sin(x)/x, but computers won't. > > I have replaced the word function by expression in an attempt to be more > precise." > > Clearly, I disagree with the referee about what "90% of mathematicians > would claim". But perhaps I'm wrong.would you say it's 0% or 100%. from what they taught me, sin(x)/x clearly has a "singularity" at 0 but it is a "removable singularity". sin(x)/x and *all* its derivatives exist everywhere and it is analytic everywhere. in fact, a decent McLaurin series can be determined for it. r b-j
Reply by ●March 2, 20042004-03-02
robert bristow-johnson <rbj@surfglobal.net> wrote:> In article 20040302171102.808$cA@newsreader.com, David W. Cantrell at > DWCantrell@sigmaxi.org wrote on 03/02/2004 17:11: > > > I suppose it's common in some contexts and not in others. [Certainly, > > in a typical beginning calculus course, for example, if the instructor > > says "f(x) = sin(x)/x", the student would be expected to claim that x = > > 0 is not in the (implied) domain of the function.] > > > > Readers might find a comment made by a referee to John Hauser to be of > > interest. > > <http://www.jhauser.us/publications/HandlingFloatingPointExceptions.htm > > l> > > > > Referee: "I find the assertion that the function sin(x)/x has a > > singularity > > at x = 0 to be hairsplitting. I'd wager $5 that 90% of mathematicians > > would claim that sin(x)/x does not have a singularity at x = 0, since > > to them it would be the name of the function defined as: f(0) = 1, and > > f(x) = sin(x)/x elsewhere." > > > > Hauser: "In computer programs, functions are effectively defined by the > > expressions that evaluate them. If a sinc subroutine is coded as > > `sin(x)/x', the subroutine will fail for x = 0. Mathematicians may > > automatically eliminate removable singularities in expressions like > > sin(x)/x, but computers won't. > > > > I have replaced the word function by expression in an attempt to be > > more precise." > > > > Clearly, I disagree with the referee about what "90% of mathematicians > > would claim". But perhaps I'm wrong. > > would you say it's 0% or 100%.Neither extreme. I suspect that more than 50% (but less than 100%) of mathematicians would say that f(x) = sin(x)/x _does_ have a singularity (albeit a removable one) at x = 0.> from what they taught me, sin(x)/x clearly has a "singularity" at 0 but > it is a "removable singularity".Fine!> sin(x)/x and *all* its derivatives exist everywhere and it is analytic > everywhere.But above, you're really talking about sinc(x), rather than about a function having a singularity.> in fact, a decent McLaurin series can be determined for it.Of course. Just take the Maclaurin series for sin(x) and divide it, term-by-term, by x. David
Reply by ●March 2, 20042004-03-02
Gordon Sande wrote:> In article <4044efdc$0$3088$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: Tue, 02 Mar 2004 15:34:31 -0500 >>Newsgroups: sci.math,comp.dsp >> >>Robert Israel wrote: >> >> >>>In article <4043c9c4$0$3079$61fed72c@news.rcn.com>, >>>Jerry Avins <jya@ieee.org> wrote: >> >>>0/0 is undefined where I come from. >>>limit_{x -> 0} sin(x)/x = 1 but that's not the same as saying >>>sin(0)/0 = 1. >> >>It's precisely the same. Lim(sin(x)/x) is how to calculate sin(0)/0. >> {x->0} > > > It is usually called a "removable singularity" when the limit is well > defined. Common practice is to assume that all removable singularities > have been removed. > > >>Given the fraction A/B, A and B each being differentiable expressions, >>A/B had a computable value where I come from even when each is zero. >> >>With sin(x)/x^2, there's a different result, which shows simply that not >>all zeros are the same. > > > You just provided a counterexample to your claim so the conditions are > a bit stronger than your statement.Gordon, I'm surprised! I didn't claim that 0/0 was known without separate specification, but only that sin(x)/x, x being zero, is. The claim is no stronger than the what can be supported. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 3, 20042004-03-03
robert bristow-johnson <rbj@surfglobal.net> wrote in message news:<BC66C424.8FD7%rbj@surfglobal.net>...> In article c1qqaq$cn1$1@mozo.cc.purdue.edu, Joenyim Kim at > jeonyimkim80@yahoo.com wrote on 02/28/2004 14:35: > > > Can anybody tell me what is the z-transform of "sinc" function and what is > > its region of convergence? > > i thought originally that it's a homework problem, but i wonder if it is > since i have never thought of using the sinc() function in the discrete-time > world. without scaling the independent variable, then, for discrete-time, > n, > > { 1 n=0 > x[n] = sinc(n) = sin(pi*n)/(pi*n) = { > { 0 n!=0 > > so the Z transform of it is the same as for the discrete impulse function > (sometimes called the Kronecker Delta) > > { 1 n=0 > d[n] = { > { 0 n!=0 > > and that is > > Z{d[n]} = 1 for all z > > > now if you were to scale the input a little: > > x[n] = sinc(a*n) = sin(pi*a*n)/(pi*a*n) > > for some real, a, then that's a little harder and i don't know what the > answer is right off the bat. come to think of it, i don't even know what > the Laplace Transform of the sinc() function is since i do not know how to > analytically extend the rect() function to use complex arguments.It's simple to extend rect to complex arguments, you get the donut function. donut(z) = rect(|z|). You don't need Laplace transforms, since they're not required for non-temporal functions.
Reply by ●March 3, 20042004-03-03
In article e4a0829b.0403022146.693768d@posting.google.com, ZZBunker at zzbunker@netscape.net wrote on 03/03/2004 00:46:> robert bristow-johnson <rbj@surfglobal.net> wrote in message > news:<BC66C424.8FD7%rbj@surfglobal.net>... >> In article c1qqaq$cn1$1@mozo.cc.purdue.edu, Joenyim Kim at >> jeonyimkim80@yahoo.com wrote on 02/28/2004 14:35: >> >>> Can anybody tell me what is the z-transform of "sinc" function and what is >>> its region of convergence? >>...>> >> now if you were to scale the input a little: >> >> x[n] = sinc(a*n) = sin(pi*a*n)/(pi*a*n) >> >> for some real, a, then that's a little harder and i don't know what the >> answer is right off the bat. come to think of it, i don't even know what >> the Laplace Transform of the sinc() function is since i do not know how to >> analytically extend the rect() function to use complex arguments. > > It's simple to extend rect to complex arguments, > you get the donut function. > > donut(z) = rect(|z|). > > You don't need Laplace transforms, > since they're not required for non-temporal functions.but if Im(z) turns out to be zero, does donut(z) = rect(z)? hmmm. maybe since rect(x) is even. but does donut(z/a) = Z{sinc(a*n)}? or something similar? r b-j
Reply by ●March 3, 20042004-03-03
israel@math.ubc.ca (Robert Israel) wrote in message news:<c21bl7$e3e$1@nntp.itservices.ubc.ca>...> In article <BC692709.90C5%rbj@surfglobal.net>, > robert bristow-johnson <rbj@surfglobal.net> wrote: > >In article c20d3h$97$1@nntp.itservices.ubc.ca, Robert Israel at > >israel@math.ubc.ca wrote on 03/01/2004 17:22: > > >> Y(z) = 1 + sum_{n=1}^infinity sin(an)/(an) z^(-n) > >> = 1 + (2ia)^(-1) sum_{n=1}^infinity ((exp(ia)/z)^n - (exp(-ia)/z)^n)/n > >> = 1 + (-ln(1 - exp(ia)/z) + ln(1 - exp(-ia)/z))/(2 i a) > > Let's write this as Y(a,z). > > >so the SUM{ c^n / n } (for some c) is a known summation form? i just didn't > >know that offhand. > > Yes, it's the Maclaurin series of -ln(1-c).Well, no it's not. Since if the Maclaurin series of ln(x) then it's only as well-known as ln(x) is known. Which is only up to about how well sin(x) is known. Which is the reason that people who know what they're doing write: 1) sinc(x) == sin(pi*x)/(pi*x) and not as 2) sinc(x) == sin(x)/x which is what Maclaurin series reference manuals do.
