sinc() question

In article 3EF90221.7789F30@ieee.org, Randy Yates at yates@ieee.org wrote on
06/24/2003 21:59:

> robert bristow-johnson wrote:
>> 
>> In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote
>> on 06/24/2003 08:03:
>> 
>>> What's the difference in the impulse response between a filter which
>>> has the frequency-domain response
>>> 
>>> |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
>>> 
>>> and
>>> 
>>> |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
>>> 
>>> ???
>> 
>> i think there would be no difference since the inverse FT integral of the
>> difference function of the two would be zero.
>> 
> 
> So does it pass Fs/2 or not???

that's a slightly different question than what you asked before.  i think
Dirk is correct.  the inverse FT of *either* frequency response above is the
sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function is

           { 1     |f| < Fs/2
           {
    H(f) = { 1/2   |f| = Fs/2
           {
           { 0     |f| > Fs/2

i think it has to be for reasons that i forgot, but Clay and i worked it out
once why.  do you remember Clay or Dirk?

r b-j

"Randy Yates" <yates@ieee.org> wrote in message
news:3EF90221.7789F30@ieee.org...
> So does it pass Fs/2 or not???
> --

With any sort of symmetric window applied, it will pass Fs/2 with 1/2
amplitude.

You probably mean without a window -- try it and see.  You need to perform
an infinite sum to find out, but the sum does not converge, so your question
has no good answer.  Thankfully, there is no way to test it in the real
world either, so an answer is unnecessary.

Randy Yates <yates@ieee.org> wrote in message news:<3EF90221.7789F30@ieee.org>...
> robert bristow-johnson wrote:
> > 
> > In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote
> > on 06/24/2003 08:03:
> > 
> > > What's the difference in the impulse response between a filter which
> > > has the frequency-domain response
> > >
> > > |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
> > >
> > > and
> > >
> > > |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
> > >
> > > ???
> > 
> > i think there would be no difference since the inverse FT integral of the
> > difference function of the two would be zero.
> > 
> > r b-j
> 
> So does it pass Fs/2 or not???

I don't think it does. First, the contribution of a finite amplitude 
component with bandwidth zero -- as the component exactly at Fs/2 -- 
contributes with nothing to the integral. That's what Piergiorgio said. 

Clay and others showed, in the pdf note Clay refered to in the other 
thread, that the even part of a sinusoidal sampled exactly at Fs/2 can 
be reconstructed. They mentioned only in passing that there were 
problems with the sampling of such a sinusoidal. 

The exercise I posted shows why there is a problem with the sampling. 

If I should attempt to draw a conclusion from all this, I would say that 
the f=Fs/2 sould be excluded from a "proper" reconstruction operator, 
because the sampling operator breaks down on that frequency. Of course, 
there are purely formal reasons. As Matt said, this is of no practical 
consequence, so I think I'll leave the question there.

Rune

Robert,

The proof is in "The Fourier Integral and its Applications" by Papoulis, p
30-31.

Proof starts out by decomposing function with jump discontinuity into
continuous function plus step function with amplitude equal to the jump.
Goes on from there to the stated conclusion. Too much notation to repeat
here.

Dirk

"robert bristow-johnson" <rbj@surfglobal.net> wrote in message
news:BB1E944B.1A4F%rbj@surfglobal.net...
> In article 3EF90221.7789F30@ieee.org, Randy Yates at yates@ieee.org wrote
on
> 06/24/2003 21:59:
>
> > robert bristow-johnson wrote:
> >>
> >> In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org
wrote
> >> on 06/24/2003 08:03:
> >>
> >>> What's the difference in the impulse response between a filter which
> >>> has the frequency-domain response
> >>>
> >>> |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
> >>>
> >>> and
> >>>
> >>> |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
> >>>
> >>> ???
> >>
> >> i think there would be no difference since the inverse FT integral of
the
> >> difference function of the two would be zero.
> >>
> >
> > So does it pass Fs/2 or not???
>
> that's a slightly different question than what you asked before.  i think
> Dirk is correct.  the inverse FT of *either* frequency response above is
the
> sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function is
>
>            { 1     |f| < Fs/2
>            {
>     H(f) = { 1/2   |f| = Fs/2
>            {
>            { 0     |f| > Fs/2
>
> i think it has to be for reasons that i forgot, but Clay and i worked it
out
> once why.  do you remember Clay or Dirk?
>
> r b-j
>
>

robert bristow-johnson <rbj@surfglobal.net> wrote in message news:<BB1E944B.1A4F%rbj@surfglobal.net>...
> In article 3EF90221.7789F30@ieee.org, Randy Yates at yates@ieee.org wrote on
> 06/24/2003 21:59:
> 
> > robert bristow-johnson wrote:
> >> 
> >> In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote
> >> on 06/24/2003 08:03:
> >> 
> >>> What's the difference in the impulse response between a filter which
> >>> has the frequency-domain response
> >>> 
> >>> |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
> >>> 
> >>> and
> >>> 
> >>> |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
> >>> 
> >>> ???
> >> 
> >> i think there would be no difference since the inverse FT integral of the
> >> difference function of the two would be zero.
> >> 
> > 
> > So does it pass Fs/2 or not???
> 
> that's a slightly different question than what you asked before.  

So it is.

> i think
> Dirk is correct.  the inverse FT of *either* frequency response above is the
> sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function is
> 
>            { 1     |f| < Fs/2
>            {
>     H(f) = { 1/2   |f| = Fs/2
>            {
>            { 0     |f| > Fs/2
> 
> i think it has to be for reasons that i forgot, but Clay and i worked it out
> once why.  

Then the inverse FT lies? I am asking it two different questions that
have should have two different behaviors and I get one response? 

Of course this thread is totally academic, but I am curious. Thanks
for entertaining the idea, Robert et alius.

--Randy

In article 567ce618.0306250841.698282f8@posting.google.com, Randy Yates at
yates@ieee.org wrote on 06/25/2003 12:41:

> robert bristow-johnson <rbj@surfglobal.net> wrote in message
> news:<BB1E944B.1A4F%rbj@surfglobal.net>...
>> In article 3EF90221.7789F30@ieee.org, Randy Yates at yates@ieee.org wrote on
>> 06/24/2003 21:59:
>> 
>>> robert bristow-johnson wrote:
>>>> 
>>>> In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote
>>>> on 06/24/2003 08:03:
>>>> 
>>>>> What's the difference in the impulse response between a filter which
>>>>> has the frequency-domain response
>>>>> 
>>>>> |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
>>>>> 
>>>>> and
>>>>> 
>>>>> |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
>>>>> 
>>>>> ???
>>>> 
>>>> i think there would be no difference since the inverse FT integral of the
>>>> difference function of the two would be zero.
>>>> 
>>> 
>>> So does it pass Fs/2 or not???
>> 
>> that's a slightly different question than what you asked before.
> 
> So it is.
> 
>> i think
>> Dirk is correct.  the inverse FT of *either* frequency response above is the
>> sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function is
>> 
>>        { 1     |f| < Fs/2
>>        {
>> H(f) = { 1/2   |f| = Fs/2
>>        {
>>        { 0     |f| > Fs/2
>> 
>> i think it has to be for reasons that i forgot, but Clay and i worked it out
>> once why.  
> 
> Then the inverse FT lies?

only on a set of "f" with measure zero.

> I am asking it two different questions that
> have should have two different behaviors

only on a set of "f" with measure zero.

> and I get one response?

yowsem.

> Of course this thread is totally academic, but I am curious. Thanks
> for entertaining the idea, Robert et alius.


consider this function, Randy:

           { 1      t=0
    x(t) = {
           { 0      t<>0    (alternate notation t!=0)

this is not the Dirac impulse function and, for the purpose of amplification
(both literally and physically), that "1" could be any finite number.

so what's X(f), the FT of x(t)?

r b-j

"Randy Yates" <yates@ieee.org> wrote in message
news:567ce618.0306250841.698282f8@posting.google.com...
> robert bristow-johnson <rbj@surfglobal.net> wrote in message
news:<BB1E944B.1A4F%rbj@surfglobal.net>...

> > i think
> > Dirk is correct.  the inverse FT of *either* frequency response above is
the
> > sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function
is
> >
> >            { 1     |f| < Fs/2
> >            {
> >     H(f) = { 1/2   |f| = Fs/2
> >            {
> >            { 0     |f| > Fs/2
> >
> > i think it has to be for reasons that i forgot, but Clay and i worked it
out
> > once why.
>
> Then the inverse FT lies? I am asking it two different questions that
> have should have two different behaviors and I get one response?
>
> Of course this thread is totally academic, but I am curious. Thanks
> for entertaining the idea, Robert et alius.


Hello Randy,

The above problem is known (with the particular integral of sin(x) over x)
as the Dirichlet's discontinuous factor. There is a theorem for Fourier
transforms that says at a point of discontinuity that the transform will be
the average of the left and right hand limits. A proof may be found in
appendix 1 of R. V. Churchill's "Fourier Series and Boundary Value
Problems", 2nd Ed. 1963.

Since your two questions differ by only a single point of discontinuity,
application of Dirichlet's theorem, makes them have the same answer. But as
others have stated, sampling at this magic rate has problems with phase
sensitivity. If sampling is done at the extrema, then all of the info is
gleaned. However, if sampling at the zero crossings, no info is gained. I
know your mentioned this as an academic problem, so I that the academic
resolution is the Dirichlet theorem.

I hope this helps.

Clay








>
> --Randy

"Clay S. Turner" wrote:
> 
> "Randy Yates" <yates@ieee.org> wrote in message
> news:567ce618.0306250841.698282f8@posting.google.com...
> > robert bristow-johnson <rbj@surfglobal.net> wrote in message
> news:<BB1E944B.1A4F%rbj@surfglobal.net>...
> 
> > > i think
> > > Dirk is correct.  the inverse FT of *either* frequency response above is
> the
> > > sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function
> is
> > >
> > >            { 1     |f| < Fs/2
> > >            {
> > >     H(f) = { 1/2   |f| = Fs/2
> > >            {
> > >            { 0     |f| > Fs/2
> > >
> > > i think it has to be for reasons that i forgot, but Clay and i worked it
> out
> > > once why.
> >
> > Then the inverse FT lies? I am asking it two different questions that
> > have should have two different behaviors and I get one response?
> >
> > Of course this thread is totally academic, but I am curious. Thanks
> > for entertaining the idea, Robert et alius.
> 
> Hello Randy,
> 
> The above problem is known (with the particular integral of sin(x) over x)
> as the Dirichlet's discontinuous factor. There is a theorem for Fourier
> transforms that says at a point of discontinuity that the transform will be
> the average of the left and right hand limits. A proof may be found in
> appendix 1 of R. V. Churchill's "Fourier Series and Boundary Value
> Problems", 2nd Ed. 1963.
> 
> Since your two questions differ by only a single point of discontinuity,
> application of Dirichlet's theorem, makes them have the same answer. But as
> others have stated, sampling at this magic rate has problems with phase
> sensitivity. If sampling is done at the extrema, then all of the info is
> gleaned. However, if sampling at the zero crossings, no info is gained. I
> know your mentioned this as an academic problem, so I that the academic
> resolution is the Dirichlet theorem.
> 
> I hope this helps.

Oh, yeah - Dirk mentioned the solution was in the Papoulis text but that
book is out-of-print and do you know what Books-a-million want for a used
copy? On the other hand, I have the Churchill text and can look it up at
work tomorrow. 

Yes, I was aware of this theorem, but thanks for refreshing my memory
and having the presence of mind to see it applies here. 

Yes, these border-line problems with Fs/2 frequencies were the point. If you
could ensure they wouldn't come through via an analytical (as opposed to a
practical, thank you very much Timmermans) solution, then the problems 
could be bypassed. 

Thanks for your input, Clay. 
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

robert bristow-johnson wrote:
> 
> In article 567ce618.0306250841.698282f8@posting.google.com, Randy Yates at
> yates@ieee.org wrote on 06/25/2003 12:41:
> 
> > robert bristow-johnson <rbj@surfglobal.net> wrote in message
> > news:<BB1E944B.1A4F%rbj@surfglobal.net>...
> >> In article 3EF90221.7789F30@ieee.org, Randy Yates at yates@ieee.org wrote on
> >> 06/24/2003 21:59:
> >>
> >>> robert bristow-johnson wrote:
> >>>>
> >>>> In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org wrote
> >>>> on 06/24/2003 08:03:
> >>>>
> >>>>> What's the difference in the impulse response between a filter which
> >>>>> has the frequency-domain response
> >>>>>
> >>>>> |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
> >>>>>
> >>>>> and
> >>>>>
> >>>>> |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
> >>>>>
> >>>>> ???
> >>>>
> >>>> i think there would be no difference since the inverse FT integral of the
> >>>> difference function of the two would be zero.
> >>>>
> >>>
> >>> So does it pass Fs/2 or not???
> >>
> >> that's a slightly different question than what you asked before.
> >
> > So it is.
> >
> >> i think
> >> Dirk is correct.  the inverse FT of *either* frequency response above is the
> >> sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function is
> >>
> >>        { 1     |f| < Fs/2
> >>        {
> >> H(f) = { 1/2   |f| = Fs/2
> >>        {
> >>        { 0     |f| > Fs/2
> >>
> >> i think it has to be for reasons that i forgot, but Clay and i worked it out
> >> once why.
> >
> > Then the inverse FT lies?
> 
> only on a set of "f" with measure zero.
> 
> > I am asking it two different questions that
> > have should have two different behaviors
> 
> only on a set of "f" with measure zero.

Is that a liquid or solid measure? English or metric?

What in the confounded transform discontinuity is a "set of 'f'
with measure zero"??? Yeah, I know - you're gonna say it has
to do with functional analysis and when I grow up to be a big
mathematician I'll understand. 

> > and I get one response?
> 
> yowsem.

Okay fine.

> > Of course this thread is totally academic, but I am curious. Thanks
> > for entertaining the idea, Robert et alius.
> 
> consider this function, Randy:
> 
>            { 1      t=0
>     x(t) = {
>            { 0      t<>0    (alternate notation t!=0)
> 
> this is not the Dirac impulse function

Right, it sure ain't.

> and, for the purpose of amplification
> (both literally and physically), that "1" could be any finite number.

Even 0? No, I get your point.

> so what's X(f), the FT of x(t)?

Yeah, I see your point. That still leaves me high and dry, though. 
There is no function that passes all frequencies between -Fs/2 and
+Fs/2 but not the endpoints? This seems like such a simple requirement,
and you're telling me it can't be done? 
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side 
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO   
http://home.earthlink.net/~yatescr

In article 3EFA6355.47180E19@ieee.org, Randy Yates at yates@ieee.org wrote
on 06/25/2003 23:06:

> robert bristow-johnson wrote:
>> 
>> In article 567ce618.0306250841.698282f8@posting.google.com, Randy Yates at
>> yates@ieee.org wrote on 06/25/2003 12:41:
>> 
>>> robert bristow-johnson <rbj@surfglobal.net> wrote in message
>>> news:<BB1E944B.1A4F%rbj@surfglobal.net>...
>>>> In article 3EF90221.7789F30@ieee.org, Randy Yates at yates@ieee.org wrote
>>>> on
>>>> 06/24/2003 21:59:
>>>> 
>>>>> robert bristow-johnson wrote:
>>>>>> 
>>>>>> In article 3EF83ECD.72947E50@ieee.org, Randy Yates at yates@ieee.org
>>>>>> wrote
>>>>>> on 06/24/2003 08:03:
>>>>>> 
>>>>>>> What's the difference in the impulse response between a filter which
>>>>>>> has the frequency-domain response
>>>>>>> 
>>>>>>> |H(w)| = 1, |f| < Fs/2, 0 otherwise, angle(H(w)) = 0
>>>>>>> 
>>>>>>> and
>>>>>>> 
>>>>>>> |H(w)| = 1, |f| <= Fs/2, 0 otherwise, angle(H(w)) = 0
>>>>>>> 
>>>>>>> ???
>>>>>> 
>>>>>> i think there would be no difference since the inverse FT integral of the
>>>>>> difference function of the two would be zero.
>>>>>> 
>>>>> 
>>>>> So does it pass Fs/2 or not???
>>>> 
>>>> that's a slightly different question than what you asked before.
>>> 
>>> So it is.
>>> 
>>>> i think
>>>> Dirk is correct.  the inverse FT of *either* frequency response above is
>>>> the
>>>> sinc(Fs*t) (i think scaled by Fs) and the FT of the sinc(Fs*t) function is
>>>> 
>>>>        { 1     |f| < Fs/2
>>>>        {
>>>> H(f) = { 1/2   |f| = Fs/2
>>>>        {
>>>>        { 0     |f| > Fs/2
>>>> 
>>>> i think it has to be for reasons that i forgot, but Clay and i worked it
>>>> out once why.
>>> 
>>> Then the inverse FT lies?
>> 
>> only on a set of "f" with measure zero.
>> 
>>> I am asking it two different questions that
>>> have should have two different behaviors
>> 
>> only on a set of "f" with measure zero.
> 
> Is that a liquid or solid measure? English or metric?
> 
> What in the confounded transform discontinuity is a "set of 'f'
> with measure zero"??? Yeah, I know - you're gonna say it has
> to do with functional analysis and when I grow up to be a big
> mathematician I'll understand.

no "f" is the f in "H(f)".  that is frequency.  in Real Analysis (gives
meaning to the truism: "you can't spell 'analysis' without 'anal'") there is
a concept of measure theory.  it has a little to do with countably infinite
and uncountably infinite numbers.

the measure of the set of reals from 0 to 1 is 1.  from -1 to 1 is 2.  from
-32 to +64 is 96.  sound simple?  the measure of the set of rationals in the
same segments is zero.

the measure of a set of a single point on the number line is zero.

>>> and I get one response?
>> 
>> yowsem.
> 
> Okay fine.
> 
>>> Of course this thread is totally academic, but I am curious. Thanks
>>> for entertaining the idea, Robert et alius.
>> 
>> consider this function, Randy:
>> 
>>        { 1      t=0
>> x(t) = {
>>        { 0      t<>0    (alternate notation t!=0)
>> 
>> this is not the Dirac impulse function
> 
> Right, it sure ain't.
> 
>> and, for the purpose of amplification
>> (both literally and physically), that "1" could be any finite number.
> 
> Even 0?

sure.  but it won't say much.

> No, I get your point.
> 
>> so what's X(f), the FT of x(t)?
> 
> Yeah, I see your point. That still leaves me high and dry, though.
> There is no function that passes all frequencies between -Fs/2 and
> +Fs/2 but not the endpoints?

well, the sinc() does 1/2 of the endpoints.

> This seems like such a simple requirement,
> and you're telling me it can't be done?

discontinuity is a problematic thing.  i think when there is a discontinuity
in a "continuous" (as far as definition) function, the best we can settle
for in a real physical system is the average of the two sides of the
discontinuity.  i dunno.

r b-j