Can anybody tell me what is the z-transform of "sinc" function and what is its region of convergence? Thanks a lot, -Joenyim
what is the z-transform of sinc function?
Started by ●February 28, 2004
Reply by ●February 28, 20042004-02-28
In article c1qqaq$cn1$1@mozo.cc.purdue.edu, Joenyim Kim at jeonyimkim80@yahoo.com wrote on 02/28/2004 14:35:> Can anybody tell me what is the z-transform of "sinc" function and what is > its region of convergence?i thought originally that it's a homework problem, but i wonder if it is since i have never thought of using the sinc() function in the discrete-time world. without scaling the independent variable, then, for discrete-time, n, { 1 n=0 x[n] = sinc(n) = sin(pi*n)/(pi*n) = { { 0 n!=0 so the Z transform of it is the same as for the discrete impulse function (sometimes called the Kronecker Delta) { 1 n=0 d[n] = { { 0 n!=0 and that is Z{d[n]} = 1 for all z now if you were to scale the input a little: x[n] = sinc(a*n) = sin(pi*a*n)/(pi*a*n) for some real, a, then that's a little harder and i don't know what the answer is right off the bat. come to think of it, i don't even know what the Laplace Transform of the sinc() function is since i do not know how to analytically extend the rect() function to use complex arguments.> Thanks a lot,FWIW r b-j
Reply by ●March 1, 20042004-03-01
Joenyim Kim wrote:> Can anybody tell me what is the z-transform of "sinc" function and > what is its region of convergence? > > Thanks a lot, > > -JoenyimDoes this help? sin(x) = x - (x^3)/3! + (x^5)/5! - .... (converges everywhere) so sinx/x evaluates to: 1 - (x^2)/3! + (x^4)/5! - ... (still infinite converges everywhere, even at x=0) with BZT applied as f(z) = f(s) where s=(z-1)/(z+1) this gives: 1- ((z-1)/(z+1))^2/3! + ((z-1)/(z+1))^4/5! - .... I'm not sure if this converges at z=-1. I guess, it should, but I'm not able to see it from the formula. Bernhard
Reply by ●March 1, 20042004-03-01
In article <BC66C424.8FD7%rbj@surfglobal.net>, robert bristow-johnson <rbj@surfglobal.net> wrote:>In article c1qqaq$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!=0The more usual definition, I think, is sinc(x) = sin(x)/x for x <> 0. Well, if f(n) = sin(ax)/(ax) for x <> 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)). 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 1, 20042004-03-01
In article c20d3h$97$1@nntp.itservices.ubc.ca, Robert Israel at israel@math.ubc.ca wrote on 03/01/2004 17:22:> In article <BC66C424.8FD7%rbj@surfglobal.net>, > robert bristow-johnson <rbj@surfglobal.net> wrote: >> In article c1qqaq$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 > > > The more usual definition, I think, is sinc(x) = sin(x)/x for x <> 0.perhaps in mathematics circles. but in electrical engineering and in signal processing circles, the pi is in there, almost without exception. no big deal.> Well, if f(n) = sin(ax)/(ax) for x <> 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)so the SUM{ c^n / n } (for some c) is a known summation form? i just didn't know that offhand.> converging for |z| > 1 if a is real and nonzero (actually also for |z|=1 > except for z= exp(ia) and exp(-ia)).it better work for |z| = 1 because that would be the Fourier Transform for sinc() which is well known. does this result look like a rectangular function (of omega) for z = exp(j*omega)? (my "j" is your "i".) r b-j
Reply by ●March 1, 20042004-03-01
Robert Israel wrote:> In article <BC66C424.8FD7%rbj@surfglobal.net>, > robert bristow-johnson <rbj@surfglobal.net> wrote: >...>>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!=0Certainly.> The more usual definition, I think, is sinc(x) = sin(x)/x for x <> 0.No. sinc(x) = sin(x)/x for all x, including zero. Evaluation may not be trivial, but it's easy.> Well, if f(n) = sin(ax)/(ax) for x <> 0, f(0) = 1, the z transform isDoes n take only integer values here? x? What gives? ... Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 2, 20042004-03-02
In article <4043c9c4$0$3079$61fed72c@news.rcn.com>, Jerry Avins <jya@ieee.org> wrote:>Robert Israel wrote:>> The more usual definition, I think, is sinc(x) = sin(x)/x for x <> 0.>No. sinc(x) = sin(x)/x for all x, including zero. Evaluation may not be >trivial, but it's easy.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.>> Well, if f(n) = sin(ax)/(ax) for x <> 0, f(0) = 1, the z transform isSorry, of course I meant f(n) = sin(an)/(an) for n <> 0.>Does n take only integer values here? x? What gives?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 <BC692709.90C5%rbj@surfglobal.net>, robert bristow-johnson <rbj@surfglobal.net> wrote:>In article c20d3h$97$1@nntp.itservices.ubc.ca, Robert Israel at >israel@math.ubc.ca wrote on 03/01/2004 17:22:>> Y(z) = 1 + sum_{n=1}^infinity sin(an)/(an) z^(-n) >> = 1 + (2ia)^(-1) sum_{n=1}^infinity ((exp(ia)/z)^n - (exp(-ia)/z)^n)/n >> = 1 + (-ln(1 - exp(ia)/z) + ln(1 - exp(-ia)/z))/(2 i a)Let's write this as Y(a,z).>so the SUM{ c^n / n } (for some c) is a known summation form? i just didn't >know that offhand.Yes, it's the Maclaurin series of -ln(1-c).>> converging for |z| > 1 if a is real and nonzero (actually also for |z|=1 >> except for z= exp(ia) and exp(-ia)).>it better work for |z| = 1 because that would be the Fourier Transform for >sinc() which is well known. does this result look like a rectangular >function (of omega) for z = exp(j*omega)? (my "j" is your "i".)It's a bit tricky for z=exp(i w). Note that in the real part of the sum we have sum_{n=1}^infinity sin(an) cos(wn)/n = 1/2 sum_{n=1}^infinity sin((a+w)n)/n + 1/2 sum_{n=1}^infinity sin((a-w)n)/n which is indeed a "rectangular wave" in w, with jumps at w=(+/-) a + 2 pi m. But there's also an imaginary part, which goes to +/- infinity at w=(+/-) a. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2
Reply by ●March 2, 20042004-03-02
israel@math.ubc.ca (Robert Israel) wrote in message news:<c218bd$cob$1@nntp.itservices.ubc.ca>...> In article <4043c9c4$0$3079$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote: > >Robert Israel wrote: > > >> The more usual definition, I think, is sinc(x) = sin(x)/x for x <> 0. > > >No. sinc(x) = sin(x)/x for all x, including zero. Evaluation may not be > >trivial, but it's easy. > > 0/0 is undefined where I come from. > limit_{x -> 0} sin(x)/x = 1 but that's not the same as saying > sin(0)/0 = 1. > > > >> Well, if f(n) = sin(ax)/(ax) for x <> 0, f(0) = 1, the z transform is > > Sorry, of course I meant f(n) = sin(an)/(an) for n <> 0. > > >Does n take only integer values here? x? What gives? > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2Let's take RBJ's general structure where a is complex defined as a=ar+j*ai Let me take Israel general form, f(n)=sin(an)/(an) for n<>0, assume, a involves both "pi" and scale factor. After a little mathematics we arrieve, n= + inf f(z)=1/A * SUM r^(-n)/n*((exp^(j*Br*n)*exp^(-Bi*n)) - (exp^(j*Cr*n)*exp^(Bi*n))) n = -inf where, ar-w=Br, ai=Bi, Cr=ar+w and z is written in polar form as z=r*exp(jw) A=1/(2*j(ar+j*ai)) is constant It looks to me that the series converges both at -inf and +inf and question comes whether to take r>1 when n>0 and r<1 when n<0 ? Probably someone can explain whether the claim is right or undefined or wrong? Santosh
Reply by ●March 2, 20042004-03-02
As far as I know, it's just a rectangle (i.e. the "ideal" lowpass filter). From what I understand, that's why sinc functions are used in the time-domain for signal reconstruction. HTH, and if it's not correct, please advise. -Mike "Joenyim Kim" <jeonyimkim80@yahoo.com> wrote in message news:c1qqaq$cn1$1@mozo.cc.purdue.edu...> Can anybody tell me what is the z-transform of "sinc" function and what is > its region of convergence? > > Thanks a lot, > > -Joenyim > >
