# what is the z-transform of sinc function?

Started by February 28, 2004
```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.
&#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 <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>
>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.
>&#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;
>

```
```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
```
```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

```
```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

```
```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.

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
```
```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>
>>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.
&#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;

```
```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.
```
```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

```
```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.
```