David W. Cantrell wrote: ...> 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 CantrellNeat! Powerful! I like it. There's something Smith-charty about it. Thanks! 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 3, 20042004-03-03

Reply by ●March 10, 20042004-03-10

Bernhard Holzmayer <holzmayer.bernhard@deadspam.com> wrote in message news:<1810304.0co4jb6aFB@holzmayer.ifr.rt>...> Joenyim Kim wrote: > > > Can anybody tell me what is the z-transform of "sinc" function and > > what is its region of convergence? > > > > Thanks a lot, > > > > -Joenyim > > Does this help? > > sin(x) = x - (x^3)/3! + (x^5)/5! - .... (converges everywhere) > > so sinx/x evaluates to: > > 1 - (x^2)/3! + (x^4)/5! - ... > (still infinite converges everywhere, even at x=0) > > with BZT applied as f(z) = f(s) where s=(z-1)/(z+1) this gives: > > 1- ((z-1)/(z+1))^2/3! + ((z-1)/(z+1))^4/5! - .... > > I'm not sure if this converges at z=-1. > I guess, it should, but I'm not able to see it from the formula. > > > BernhardI published my FOUR-BOX algorithm at http://www.wehner.org/euler/ . To create sin(x) you subtract my box 4 from my box 2. The convergence is FACTORIAL. Sin(x)/(x) therefore has the same convergence as sin(x), as you multiply each term in the four-box structure by X and divide by the next number up. Eventually, that next number overtakes argument x, and the term converges. I call the standard McLaurin-Taylor formulae "unilateral" because there is a "bilateral" form, as described on that page. Charles Douglas Wehner

Reply by ●March 28, 20092009-03-28

I'm aware that this thread is really old, but there's a question still pending. It has been shown that the UNILATERAL Z transform exists for the sinc function, but what about the BILATERAL Z transform? By splitting the summation in two terms (onr for -inf..-1 and other for 1..+inf) one can easily show that one of them converges for |z|<1 and the other for |z|>1, so the Z transform wouldn't exist (ROCs don't overlap). But I don't think this is a valid procedure because i'm splitting the summation (that's only valid when each term converge individually). Any ideas on how to overcome this issue?>Bernhard Holzmayer <holzmayer.bernhard@deadspam.com> wrote in messagenews:<1810304.0co4jb6aFB@holzmayer.ifr.rt>...>> Joenyim Kim wrote: >> >> > Can anybody tell me what is the z-transform of "sinc" function and >> > what is its region of convergence? >> > >> > Thanks a lot, >> > >> > -Joenyim >> >> Does this help? >> >> sin(x) = x - (x^3)/3! + (x^5)/5! - .... (converges everywhere) >> >> so sinx/x evaluates to: >> >> 1 - (x^2)/3! + (x^4)/5! - ... >> (still infinite converges everywhere, even at x=0) >> >> with BZT applied as f(z) = f(s) where s=(z-1)/(z+1) this gives: >> >> 1- ((z-1)/(z+1))^2/3! + ((z-1)/(z+1))^4/5! - .... >> >> I'm not sure if this converges at z=-1. >> I guess, it should, but I'm not able to see it from the formula. >> >> >> Bernhard > >I published my FOUR-BOX algorithm at http://www.wehner.org/euler/ . > >To create sin(x) you subtract my box 4 from my box 2. > >The convergence is FACTORIAL. > >Sin(x)/(x) therefore has the same convergence as sin(x), as you >multiply each term in the four-box structure by X and divide by the >next number up. Eventually, that next number overtakes argument x, and >the term converges. > >I call the standard McLaurin-Taylor formulae "unilateral" because >there is a "bilateral" form, as described on that page. > >Charles Douglas Wehner >

Reply by ●March 29, 20092009-03-29

Mario Azcueta wrote:> I'm aware that this thread is really old, but there's a question > still pending. It has been shown that the UNILATERAL Z transform > exists for the sinc function, but what about the BILATERAL Z > transform?Instead of misunderstandings, you should have quoted the pertinent info which is Prof. Israel's first post: Robert Israel wrote (with corrections): | Well, if f(n) = sin(an)/(an) for n <> 0, f(0) = 1, the | z transform is | | 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) | | converging for |z| > 1 if a is real and nonzero (actually | also for |z|=1 except for z = exp(ia) and exp(-ia)). If you read carefully, the ROCs of Y(z) and Y(1/z) do overlap on the unit circle. In fact, at z = exp(i*w) we have Y(z) + Y(1/z) - 1 = pi/a * rect(w, [-a,a]). Martin -- I am an Indian and looked upon by the whites as a foolish man; but it must be because I follow the advice of the white man. --Shunka Witko

Reply by ●March 30, 20092009-03-30

Martin Thanks for your time writing. I've read carefully mr Israel's post, but I think you didn't pay that much attention to mine. Look at Israel's definition of Z transform: Y(z) = 1 + sum_{n=1}^infinity sin(an)/(an) z^(-n) That's the unilateral Z transform (look at the limits of the sum). My question was about the BILATERAL Z transform. And let my correct you one more thing: it's impossible that the ROC includes the whole unit circle, since it's a well known result that the sinc function is not an L2 function (not absolutely summable). I hope I made myself clear this time.>Mario Azcueta wrote: > >> I'm aware that this thread is really old, but there's a question >> still pending. It has been shown that the UNILATERAL Z transform >> exists for the sinc function, but what about the BILATERAL Z >> transform? > >Instead of misunderstandings, you should have quoted the pertinent >info which is Prof. Israel's first post: > >Robert Israel wrote (with corrections): > >| Well, if f(n) = sin(an)/(an) for n <> 0, f(0) = 1, the >| z transform is >| >| 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) >| >| converging for |z| > 1 if a is real and nonzero (actually >| also for |z|=1 except for z = exp(ia) and exp(-ia)). > >If you read carefully, the ROCs of Y(z) and Y(1/z) do overlap on the >unit circle. In fact, at z = exp(i*w) we have > Y(z) + Y(1/z) - 1 = pi/a * rect(w, [-a,a]). > > >Martin > >-- >I am an Indian and looked upon by the whites as a foolish man; >but it must be because I follow the advice of the white man. >--Shunka Witko >

Reply by ●March 30, 20092009-03-30

Mario Azcueta wrote:> Look at Israel's definition of Z transform: > > Y(z) = 1 + sum_{n=1}^infinity sin(an)/(an) z^(-n) > > That's the unilateral Z transform (look at the limits of the > sum). My question was about the BILATERAL Z transform.Which is, formally, Y(z) + Y(1/z) - 1 as I wrote before.> And let my correct you one more thing: it's impossible that the > ROC includes the whole unit circleI didn't say that. It is the case that the intersections of the two ROCs with the unit circle coincide. And that set differs from the whole circle in those points where the rectangle function is undefined too, namely at the jumps, like Robert Israel had remarked.>>Robert Israel wrote (with corrections): >> >>| Well, if f(n) = sin(an)/(an) for n <> 0, f(0) = 1, the >>| z transform is >>| >>| 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) >>| >>| converging for |z| > 1 if a is real and nonzero (actually >>| also for |z|=1 except for z = exp(ia) and exp(-ia)). >> >>If you read carefully, the ROCs of Y(z) and Y(1/z) do overlap on >>the unit circle. In fact, at z = exp(i*w) we have >> Y(z) + Y(1/z) - 1 = pi/a * rect(w, [-a,a]).Martin -- The man who does not read good books has no advantage over the man who cannot read them. --Mark Twain

Reply by ●March 30, 20092009-03-30

I don't see why you say that the ROCs of Y(z) and Y(1/z) do overlap on the unit circle (you haven't provided a proof either). To make my point stronger, let me quote Oppenheim, Schaffer & Buck's book "Discrete time signal processing", Sec. 3.1: ............. Uniform convergence of the z-transform requires absolute summability of the exponentially weighted sequence [...] Neither of the sequences x1(n)=sin(wn)/(pi.n) x2(n)=cos(wn) multiplied by r^(-n) would be absolutly summable for any value of r. Thus, these sequences do not have a z-transform that converges absolutely for any z. ·············· Stronger argumentation can be made on the basis of analytic functions, which can be found in the same section of that book. If I had read this before, i wouldn't had revived this thread. I think this also answers my original question :)>Mario Azcueta wrote: > >> Look at Israel's definition of Z transform: >> >> Y(z) = 1 + sum_{n=1}^infinity sin(an)/(an) z^(-n) >> >> That's the unilateral Z transform (look at the limits of the >> sum). My question was about the BILATERAL Z transform. > >Which is, formally, Y(z) + Y(1/z) - 1 as I wrote before. > >> And let my correct you one more thing: it's impossible that the >> ROC includes the whole unit circle > >I didn't say that. It is the case that the intersections of the two >ROCs with the unit circle coincide. And that set differs from the >whole circle in those points where the rectangle function is >undefined too, namely at the jumps, like Robert Israel had remarked. > >>>Robert Israel wrote (with corrections): >>> >>>| Well, if f(n) = sin(an)/(an) for n <> 0, f(0) = 1, the >>>| z transform is >>>| >>>| 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) >>>| >>>| converging for |z| > 1 if a is real and nonzero (actually >>>| also for |z|=1 except for z = exp(ia) and exp(-ia)). >>> >>>If you read carefully, the ROCs of Y(z) and Y(1/z) do overlap on >>>the unit circle. In fact, at z = exp(i*w) we have >>> Y(z) + Y(1/z) - 1 = pi/a * rect(w, [-a,a]). > > >Martin > >-- >The man who does not read good books has no >advantage over the man who cannot read them. >--Mark Twain >

Reply by ●March 31, 20092009-03-31

Martin, after thinking a little I finally see that they should overlap, but still, that wouldn't be consistent with what the book says.. there must be something I'm missing here.

Reply by ●March 31, 20092009-03-31

Mario Azcueta wrote:> Martin, after thinking a little I finally see that they should > overlap, but still, that wouldn't be consistent with what the > book says.. there must be something I'm missing here.The book quote was:>> Uniform convergence of the z-transform requires absolute >> summability of the exponentially weighted sequence [...] >> Neither of the sequences >> >> x1(n)=sin(wn)/(pi.n) >> x2(n)=cos(wn) >> >> multiplied by r^(-n) would be absolutly summable for any value >> of r. Thus, these sequences do not have a z-transform that >> converges absolutely for any z.The incongruence is that in the present case we know a priori that the convergent, if it indeed exists, is discontinuous on the unit circle. But the partial sums of the sinc z-transform are all continuous functions. Hence no more than pointwise convergence could be expected in the first place. Perhaps O&S prefer to exclude such corner cases from discussion? Martin -- A trick that is used more than once is called a method. --Ron Getoor paraphrasing George Polya and Gabor Szego