what is the z-transform of sinc function?

Started by February 28, 2004
```Can anybody tell me what is the z-transform of "sinc" function and what is
its region of convergence?

Thanks a lot,

-Joenyim

```
```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

```
```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

```
```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.
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

```
```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

```
```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!=0

Certainly.

> 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 is

Does n take only integer values here? x? What gives?

...

Jerry
--
Engineering is the art of making what you want from things you can get.
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;

```
```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

```
```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

```
```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 1Z2

Let'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
```
```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.

-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
>
>

```