remember this, guys?:

robert bristow-johnson wrote:
>  given a perfectly periodic sequence, x[n] such that
>
>     x[n+N] = x[n]      for all n.
>
> so if you know one period, say
>
>     x[n]      for   0 <= n < N
>
> then you have a complete definition for x[n] for all n.  correct?
>
> so, if we were to reconstruct out of those real samples a
> continuous-time periodic signal, x(t) (let's assume the sampling period
> to be 1, with no loss of generality) so that x(t-N) = x(t) for all t,
> how is the function defined in terms of the finite set of samples x[0],
> x[1], ... x[N-1] that fully define the periodic sequence and, these
> discrete sequence should be able to fully define the continuous-time
> signal it was sampled from (assuming appropriate bandlimiting, which we
> are pushing the envelope a little).  now, usually, we think of the
> ideal interpolation formula for x(t)  is:
>
>            +inf
>     x(t) = SUM{ x[n] * sinc(t - n) }
>           n=-inf
>
>            N-1     +inf
>          = SUM{    SUM{ x[n+m*N] * sinc(t - (n+m*N)) } }
>            n=0    m=-inf
>
> but since x[n+m*N] = x[n] for any integer m, then
>
>
>            N-1     +inf
>     x(t) = SUM{    SUM{ x[n] * sinc(t - (n+m*N)) } }
>            n=0    m=-inf
>
>            N-1         +inf
>          = SUM{ x[n] * SUM{ sinc(t - n - m*N) } }
>            n=0        m=-inf
>
>            N-1
>          = SUM{ x[n] * h(t-n) }
>            n=0
>
> now this interpolation function,
>
>           +inf
>     h(t) = SUM{ sinc(t - m*N) } }
>           m=-inf
>
> is periodic with period N, h(t+N) = h(t) for all t, as we expect.  so
> what does that periodic interpolation function look like?  i've
> pondered this a while and first posted this to comp.dsp in 2001:
>
>  http://groups.google.com/group/comp.dsp/msg/2cbabcc52d43399c?fwc=1
>
> thread:
> http://groups.google.com/group/comp.dsp/browse_thread/thread/44a63e8177ff7365/2cbabcc52d43399c?#2cbabcc52d43399c
>
> take a look at that.   Adrian Hey got into it enough to see the thing
> that was bothering me a little.  anyway, the result i get for a closed
> for periodic h(t) is
>
>                { sin(pi*t)/(N*sin(pi*t/N))       N odd
>         h(t) = {
>                { sin(pi*t)/(N*tan(pi*t/N))       N even
>
> from above you get show
>
>           +inf
>     h(t) = SUM{ sin(pi*(t - m*N))/(pi*(t - m*N)) } }
>           m=-inf
>
>                       +inf
>          = sin(pi*t) * SUM{ 1/(pi*(t - m*N)) } }                N even
>                       m=-inf
>
>                       +inf
>          = sin(pi*t) * SUM{ ((-1)^m)/(pi*(t - m*N)) } }         N odd
>                       m=-inf
>
>
> >  I'm not sure if there is a significant difference in the two reconstruction
> > formulas for N and N+1 as N approaches infinity.  Thoughts?
>
> i just don't want to make that assumption and equality holds exactly
> for the two cases, even and odd.  one case the formula looks just like
> the Dirichlet interpolation commonly done for the FFT and the other,
> it's slightly different, but it *is* different.  in the even case, the
> tails of those sinc() functions have bumps of opposite polarity so they
> cancel themselves out more than in the odd case where the tails, though
> descreasing in amplitude as 1/x, have their bumps of the same polarity,
> so they team up.  so i'm pretty sure the formula is correct (the
> 1/tan() reduces the amplitude more than the 1/sin() as one approaches
> midway between the two main pulses.  it also came up (indirectly) with
> tthese two mathematical identities:
>
>                    +inf
>         1/tan(t) = SUM{ 1/(t - m*pi) }
>                   m=-inf
>
>                    +inf
>         1/sin(t) = SUM{ ((-1)^m)/(t - m*pi) }
>                   m=-inf
>
> which i have not been able to prove directly.
>
> if someone has it, i would like some insight for why the Dirichlet
> formula applies exactly only for N odd, not exactly for N even,
> although it applies exactly for both N odd or N even in the case of
> interpolating DFT output to get the DTFT of the time-limited x[n].
>
> do you guys see what it is that is bothering me?  or proving more
> directly the two math identities above would be gratifying.  i have not
> been able to do that directly, as of yet.

well, i suspected that this was scooped by someone else, but i didn't
expect so recently:

T. Schanze, "Sinc interpolation of discrete periodic signals", IEEE
Transactions on Signal Processing, vol. 43, pp. 1502-1503, June 1995.

i don't get IEEE, but if someone who has free access to the pdf file
wants to email it to me, i would not complain. (screw IEEE.)

anyway there is a free pdf of some comments about this at:

http://iul.eng.fiu.edu/candocia/Publications/candocia_sp.pdf

and they get the same results as me.  dunno how the methods differ, if
they do.

just FYI,

r b-j

Andor wrote:
> robert bristow-johnson wrote:
>
> > Andor wrote:
> ..
> > > I just saw your mistake. It should be
> > >
> > >              N-1
> > >      x(t) = SUM{ x[n] * sin(pi*(t-n))/(N*tan(pi/N*(t-n))) }
> > >             n=0
> > >
> > > (notice how the frequency is divided by N as compared to your formula).
> >
> > i'm happy to have mistakes pointed out (and i confess that i don't
> > understand this as throroughly as i think i should) but i don't see how
> > that is different from either formulae i put in this thread or in the
> > 2001 thread.  for N even, that's the result i have, Andor.
>
> No, you posted:
>
> >             N-1
> >     x(t) = SUM{ x[n] * sin(N*pi*(t-n))/(N*tan(pi*(t-n))) }
> >            n=0
>
>
> >     x(t+N) = x(t)

you're right.  in the Nov 14 post i had the scaling of the argument
bumped up by a factor of N.  i didn't even realize that.  previous and
later posts, i think i had the scaling correct.


> I told you where to look for the difference.

i was looking at other posts.  more current in this thread and earlier
in 2001.  it was a "typo" that i hadn't even been aware of.

> It may be minor, but it
> was enough not to get your interpolation formula to work when I tried.

cause it was scaled wrong.  how did it work with the proper scaling?

...
> > there is clearly a difference.  the N odd case, it looks just like
> > Dirichlet.  for the N even case, there is a tan() function in the
> > denominator instead of a sin() function.  but what makes this curious
> > is that in application of the Dirichlet kernal to DFT bin case, it
> > sin() in the denominator for any integer N, even or odd.  that
> > difference in form is the curiousity i don't yet have my head wrapped
> > around.
>
> What do you mean by the "application of the Dirichlet kernal to DFT bin
> case, ..., N even or odd" ?

if you assume the DFT to be a discrete-frequency uniform sampling of
the output of the DTFT of the same input x[n] (except, for the DTFT,
this x[n] is *zero* extended, not periodically extended), the
continuous-frequency DTFT X(e^jw) can be related to the DFT output X[k]
with this relationship:

                  N-1
      X(e^(jw)) = SUM{ X[k] *
sin(pi*(N*w/(2*pi)-k)/(N*sin(pi/N*(N*w/(2*pi)-k))) }
                  k=0

("t" becomes N*w/(2*pi))  and it's a sin() in the denominator (not
tan()) whether N is even or odd.  that is at least a bit curious to me.

> What I was getting at is that if you take the kernel ce for even N and
> you interpolate a an odd number of discrete points with it, you get an
> anti-symmetric (about N) interpolation function. The same happens when
> you take the odd kernel for even number of data points.

i don't quite understand what you mean here and i dunno exactly how to
ask a clarifying question.

r b-j

robert bristow-johnson wrote:

> Andor wrote:
..
> > I just saw your mistake. It should be
> >
> >              N-1
> >      x(t) = SUM{ x[n] * sin(pi*(t-n))/(N*tan(pi/N*(t-n))) }
> >             n=0
> >
> > (notice how the frequency is divided by N as compared to your formula).
>
> i'm happy to have mistakes pointed out (and i confess that i don't
> understand this as throroughly as i think i should) but i don't see how
> that is different from either formulae i put in this thread or in the
> 2001 thread.  for N even, that's the result i have, Andor.

No, you posted:

>             N-1
>     x(t) = SUM{ x[n] * sin(N*pi*(t-n))/(N*tan(pi*(t-n))) }
>            n=0


>     x(t+N) = x(t)

I told you where to look for the difference. It may be minor, but it
was enough not to get your interpolation formula to work when I tried.

...

> > That's not too hard. Just insert the DFT definition for X[k], then
> > factor out the whole kasunkle and you get (for N even)
> >
> > f(t) = 1/N sum_{n=0}^{N-1} ce_{N,n}(t) x[n],
> >
> > where  ce_{N,n}(t) is the periodic interpolation kernel for even N,
> > given by
> >
> > ce_{N,n}(t)  = 1  - cos(pi n) cos(pi t) + 2 sum_{k=1}^{N/2} cos(2 pi n
> > k/N) cos(2 pi k / N t) + sin(2 pi n k / N) sin(2 pi k / N t)
> > = 1  - cos(pi n) cos(pi t) + 2 sum_{k=1}^{N/2} cos(2 pi k/N (n-t))
> > = D_{N/2}(2 pi (n-t) / N) - cos(pi (n-t),
> >
> > where D_M(x) is the Dirichlet kernel, ie.
> >
> > D_M(x) = sin( (M+1/2) x) / sin(x/2).
> >
> > > in that reference to 2001, i think i *did*, and the
> > > result was not surprizing for N odd, but a little surprizing for N
> > > even.  i used superposition and defined x[n] to be 1 for n = 0 (or an
> > > integer multiple of N) and zero for all other n.
> >
> > For N odd I get the interpolation kernel
> >
> > co_{N,n(t) = D_{(N-1)/2}(2 pi (n-t) / N).
> >
> > This is just a consequence, as you said, of the Nyquist term, which is
> > added in for the even case. By the way, both kernels, ce and co, result
> > in a pi-bandlimited, periodic interpolation of the data, if applied
> > naively (ie. without regard to the parity of N). Do you spot the
> > difference of the two interpolation functions (using the even and the
> > odd kernel)?
>
> there is clearly a difference.  the N odd case, it looks just like
> Dirichlet.  for the N even case, there is a tan() function in the
> denominator instead of a sin() function.  but what makes this curious
> is that in application of the Dirichlet kernal to DFT bin case, it
> sin() in the denominator for any integer N, even or odd.  that
> difference in form is the curiousity i don't yet have my head wrapped
> around.

What do you mean by the "application of the Dirichlet kernal to DFT bin
case, ..., N even or odd" ?

What I was getting at is that if you take the kernel ce for even N and
you interpolate a an odd number of discrete points with it, you get an
anti-symmetric (about N) interpolation function. The same happens when
you take the odd kernel for even number of data points.

Regards,
Andor

robert bristow-johnson wrote:
> Jerry Avins wrote:
>> Of course. What's more, if it the signal isn't strictly bandlimited and
>> sampled for at least k/(2Fs-Fmax) (k~=1) it can be only approximately
>> reconstructed. The exercise is academic, but interesting nonetheless.
> 
> i don't think it's ***purely*** academic (oh, maybe it is).  but this
> bandlimited reconstruction of periodic discrete-time data is a
> well-defined problem:
> 
> assuming your discrete-time periodic function is:
> 
>                { 1   for n = m*N  where m = any integer
>         x[n] = {
>                { 0   otherwise
> 
> (there is no loss of generality if you understand this animal to be
> linear and time-invariant.)
> 
> assuming the sampling period to be 1, the straight-forward
> reconstruction formula is
> 
>                +inf
>         x(t) = SUM{ sinc(t - m*N) }
>               m=-inf
> 
> it's clear that when t = n = integer, that x(t) = x[n].  we can
> manipulate it to be:
> 
>                            +inf
>         x(t) = sin(pi*t) * SUM{ ((-1)^(m*N))/(pi*(t - m*N)) }
>                           m=-inf
> 
> from that you can see that the (-1)^(m*N) factor toggles with m for odd
> N and does not toggle for even N.
> 
> but from knowledge that the DFT of that periodic sequence is X[k] = 1
> for all k, we can get for N odd:
> 
>         x(t) = sin(pi*t)/(N*sin(pi*t/N))       N odd
> 
> but for even N you get, if you split the X[N/2] into two, one for
> postive and the other for negative frequency,
> 
>         x(t) = sin(pi*t)/(N*tan(pi*t/N))       N even

You already pointed that out. I find it interesting to the point of 
being astounding; I certainly wouldn't have guessed it. Nevertheless, 
any energy in the last bin violates the Fs/2 > Fmax requirement for 
exact reconstruction, which is what I meant when I called the exercise 
academic.

Look: I can derive a formula for the flexure of a beam that's accurate 
enough so measurement can't disprove it. (I claim no credit; it's in the 
textbooks.) But it must be wrong because somewhere along the development 
is the implicit assumption that the beam doesn't flex. (Specifically, 
that cross sections of the beam remain parallel through the analysis.)

Engineering is like that. It doesn't bother me, but it probably drives 
my mathematically minded colleagues crazy.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins wrote:
>
> Of course. What's more, if it the signal isn't strictly bandlimited and
> sampled for at least k/(2Fs-Fmax) (k~=1) it can be only approximately
> reconstructed. The exercise is academic, but interesting nonetheless.

i don't think it's ***purely*** academic (oh, maybe it is).  but this
bandlimited reconstruction of periodic discrete-time data is a
well-defined problem:

assuming your discrete-time periodic function is:

               { 1   for n = m*N  where m = any integer
        x[n] = {
               { 0   otherwise

(there is no loss of generality if you understand this animal to be
linear and time-invariant.)

assuming the sampling period to be 1, the straight-forward
reconstruction formula is

               +inf
        x(t) = SUM{ sinc(t - m*N) }
              m=-inf

it's clear that when t = n = integer, that x(t) = x[n].  we can
manipulate it to be:

                           +inf
        x(t) = sin(pi*t) * SUM{ ((-1)^(m*N))/(pi*(t - m*N)) }
                          m=-inf

from that you can see that the (-1)^(m*N) factor toggles with m for odd
N and does not toggle for even N.

but from knowledge that the DFT of that periodic sequence is X[k] = 1
for all k, we can get for N odd:

        x(t) = sin(pi*t)/(N*sin(pi*t/N))       N odd

but for even N you get, if you split the X[N/2] into two, one for
postive and the other for negative frequency,

        x(t) = sin(pi*t)/(N*tan(pi*t/N))       N even

Ron N. wrote:
> robert bristow-johnson wrote:
>> Ron N. wrote:
>>
>>> If the signal is strongly bandlimited, then there is no content in the N/2 bin.
>> let's say we don't know how strongly bandlimited it was originally,
>> except there was nothing above Nyquist.  maybe *some* energy at
>> Nyquist.
>>
>>>  If the signal is weakly bandlimited, then N has to approach infinity for
>>> the N/2 bin to determine the content at that frequency.
>> i don't really get that at all.  N is fixed.
> 
> If there is any content in the N/2 bin, is the energy at the exact
> Nyquist frequency determinate?  Wouldn't any amplitude of
> sinusoid at the Nyquist frequency produce alternating sign
> samples of any equal or lower amplitude at the sample rate
> for some phase?

Of course. What's more, if it the signal isn't strictly bandlimited and 
sampled for at least k/(2Fs-Fmax) (k~=1) it can be only approximately 
reconstructed. The exercise is academic, but interesting nonetheless.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

robert bristow-johnson wrote:
> Ron N. wrote:
>
> > If the signal is strongly bandlimited, then there is no content in the N/2 bin.
>
> let's say we don't know how strongly bandlimited it was originally,
> except there was nothing above Nyquist.  maybe *some* energy at
> Nyquist.
>
> >  If the signal is weakly bandlimited, then N has to approach infinity for
> > the N/2 bin to determine the content at that frequency.
>
> i don't really get that at all.  N is fixed.

If there is any content in the N/2 bin, is the energy at the exact
Nyquist frequency determinate?  Wouldn't any amplitude of
sinusoid at the Nyquist frequency produce alternating sign
samples of any equal or lower amplitude at the sample rate
for some phase?


IMHO. YMMV.
-- 
rhn A.T nicholson d.0.t C-o-M

Andor wrote:
> robert bristow-johnson wrote:
>
> > Andor wrote:
> > >
> > > are you claiming that x(n) = x[n]?
> >
> > yes, for n = integer.
> >
> > > I tried your formula in a maths
> > > program, and it seems that x(t) is periodic with period 1, and does not
> > > at all interpolate x[n] (at any scale).
> >
> > it should be periodic with period N.
>
> I just saw your mistake. It should be
>
>              N-1
>      x(t) = SUM{ x[n] * sin(pi*(t-n))/(N*tan(pi/N*(t-n))) }
>             n=0
>
> (notice how the frequency is divided by N as compared to your formula).

i'm happy to have mistakes pointed out (and i confess that i don't
understand this as throroughly as i think i should) but i don't see how
that is different from either formulae i put in this thread or in the
2001 thread.  for N even, that's the result i have, Andor.

> > > The (unique) pi-bandlimited interpolation function f(t) of the
> > > uniformly spaced discrete N-periodic data x[n] is given (for n even) by
> > >
> > > f(t) = 1/N ( a_0 + 2 sum_{k=1}^{N/2-1} ( a_k cos(2 pi k / N t)
> > >         - b_k sin(2 pi k / N t) ) + a_{N/2} cos(pi t) ),
> > >
> > > where a_k = Re[ X[k] ] and b_k = Im[ X[k] ], and X[k] = DFT[ x[n] ].
> > > This results in f(n) = x[n], and f(t + N) = f(t) for all t. For odd N,
> > > you can scratch the Nyquist cosine term and sum to (N-1)/2.
> >
> > well, if you have an arbitrary set of N values of x[n] for 0 <= n < N
> > so that x[n+N] = x[n] for all n, when N is even, there is no guarantee
> > that there is no energy in the X[N/2] bin.  what do you do with that
> > component?  i think split it in half in the only reasonable thing.
>
> Don't know what you mean.

i'm saying that, in general for the problem of bandlimited
reconstruction of a periodic discrete-time function (or "sequence")
that, for N even, we cannot count on there being no energy in the
Nyquist bin, X[N/2].  indeed for the "special case" of

           { 1    for  n = m*N     m = integer
    x[n] = {
           { 0    otherwise

all X[k] = 1, including X[N/2].  so then, when creating a real
continuous-time function (assume the sampling time T = 1 so that x(n) =
x[n] for integer n), how do we construct that real c-t function?  we
know X[0]/N is the DC component, and for 1 <= k <= N/2 - 1, we know
that X[N-k] are the negative frequency components, X[k] are positive
frequency, and (if x[n] is real, then these are complex conj) they
combine to real sinusoidal components that we can add up.  what do we
do with X[N/2] (which does not exist for N odd)?  i think the only
reasonable thing to do is split it in half and put half of the
amplitude at a frequency of -(N/2)/N and the other at +(N/2)/N.  what
other reasonable thing can we do with that component?

> >
> > so, you have your a_k and b_k defined in terms of X[k] which are
> > defined in terms of x[n]. can you come up with an expression for f(t)
> > in terms of x[n]?
>
> That's not too hard. Just insert the DFT definition for X[k], then
> factor out the whole kasunkle and you get (for N even)
>
> f(t) = 1/N sum_{n=0}^{N-1} ce_{N,n}(t) x[n],
>
> where  ce_{N,n}(t) is the periodic interpolation kernel for even N,
> given by
>
> ce_{N,n}(t)  = 1  - cos(pi n) cos(pi t) + 2 sum_{k=1}^{N/2} cos(2 pi n
> k/N) cos(2 pi k / N t) + sin(2 pi n k / N) sin(2 pi k / N t)
> = 1  - cos(pi n) cos(pi t) + 2 sum_{k=1}^{N/2} cos(2 pi k/N (n-t))
> = D_{N/2}(2 pi (n-t) / N) - cos(pi (n-t),
>
> where D_M(x) is the Dirichlet kernel, ie.
>
> D_M(x) = sin( (M+1/2) x) / sin(x/2).
>
> > in that reference to 2001, i think i *did*, and the
> > result was not surprizing for N odd, but a little surprizing for N
> > even.  i used superposition and defined x[n] to be 1 for n = 0 (or an
> > integer multiple of N) and zero for all other n.
>
> For N odd I get the interpolation kernel
>
> co_{N,n(t) = D_{(N-1)/2}(2 pi (n-t) / N).
>
> This is just a consequence, as you said, of the Nyquist term, which is
> added in for the even case. By the way, both kernels, ce and co, result
> in a pi-bandlimited, periodic interpolation of the data, if applied
> naively (ie. without regard to the parity of N). Do you spot the
> difference of the two interpolation functions (using the even and the
> odd kernel)?

there is clearly a difference.  the N odd case, it looks just like
Dirichlet.  for the N even case, there is a tan() function in the
denominator instead of a sin() function.  but what makes this curious
is that in application of the Dirichlet kernal to DFT bin case, it
sin() in the denominator for any integer N, even or odd.  that
difference in form is the curiousity i don't yet have my head wrapped
around.

r b-j

robert bristow-johnson wrote:

> Andor wrote:
> >
> > are you claiming that x(n) = x[n]?
>
> yes, for n = integer.
>
> > I tried your formula in a maths
> > program, and it seems that x(t) is periodic with period 1, and does not
> > at all interpolate x[n] (at any scale).
>
> it should be periodic with period N.

I just saw your mistake. It should be

             N-1
     x(t) = SUM{ x[n] * sin(pi*(t-n))/(N*tan(pi/N*(t-n))) }
            n=0

(notice how the frequency is divided by N as compared to your formula).

>
> > The (unique) pi-bandlimited interpolation function f(t) of the
> > uniformly spaced discrete N-periodic data x[n] is given (for n even) by
> >
> > f(t) = 1/N ( a_0 + 2 sum_{k=1}^{N/2-1} ( a_k cos(2 pi k / N t)
> >         - b_k sin(2 pi k / N t) ) + a_{N/2} cos(pi t) ),
> >
> > where a_k = Re[ X[k] ] and b_k = Im[ X[k] ], and X[k] = DFT[ x[n] ].
> > This results in f(n) = x[n], and f(t + N) = f(t) for all t. For odd N,
> > you can scratch the Nyquist cosine term and sum to (N-1)/2.
>
> well, if you have an arbitrary set of N values of x[n] for 0 <= n < N
> so that
> x[n+N] = x[n] for all n, when N is even, there is no guarantee that
> there is no energy
> in the X[N/2] bin.  what do you do with that component?  i think split
> it in half in the
> only reasonable thing.

Don't know what you mean.

>
> so, you have your a_k and b_k defined in terms of X[k] which are
> defined in terms of x[n]. can you come up with an expression for f(t)
> in terms of x[n]?

That's not too hard. Just insert the DFT definition for X[k], then
factor out the whole kasunkle and you get (for N even)

f(t) = 1/N sum_{n=0}^{N-1} ce_{N,n}(t) x[n],

where  ce_{N,n}(t) is the periodic interpolation kernel for even N,
given by

ce_{N,n}(t)  = 1  - cos(pi n) cos(pi t) + 2 sum_{k=1}^{N/2} cos(2 pi n
k/N) cos(2 pi k / N t) + sin(2 pi n k / N) sin(2 pi k / N t)
= 1  - cos(pi n) cos(pi t) + 2 sum_{k=1}^{N/2} cos(2 pi k/N (n-t))
= D_{N/2}(2 pi (n-t) / N) - cos(pi (n-t),

where D_M(x) is the Dirichlet kernel, ie.

D_M(x) = sin( (M+1/2) x) / sin(x/2).

> in that reference to 2001, i think i *did*, and the
> result was not surprizing for N odd, but a little surprizing for N
> even.  i used superposition and defined x[n] to be 1 for n = 0 (or an
> integer multiple of N) and zero for all other n.

For N odd I get the interpolation kernel

co_{N,n(t) = D_{(N-1)/2}(2 pi (n-t) / N).

This is just a consequence, as you said, of the Nyquist term, which is
added in for the even case. By the way, both kernels, ce and co, result
in a pi-bandlimited, periodic interpolation of the data, if applied
naively (ie. without regard to the parity of N). Do you spot the
difference of the two interpolation functions (using the even and the
odd kernel)?

Regards,
Andor

Andor wrote:
>
> are you claiming that x(n) = x[n]?

yes, for n = integer.

> I tried your formula in a maths
> program, and it seems that x(t) is periodic with period 1, and does not
> at all interpolate x[n] (at any scale).

it should be periodic with period N.

> The (unique) pi-bandlimited interpolation function f(t) of the
> uniformly spaced discrete N-periodic data x[n] is given (for n even) by
>
> f(t) = 1/N ( a_0 + 2 sum_{k=1}^{N/2-1} ( a_k cos(2 pi k / N t)
>         - b_k sin(2 pi k / N t) ) + a_{N/2} cos(pi t) ),
>
> where a_k = Re[ X[k] ] and b_k = Im[ X[k] ], and X[k] = DFT[ x[n] ].
> This results in f(n) = x[n], and f(t + N) = f(t) for all t. For odd N,
> you can scratch the Nyquist cosine term and sum to (N-1)/2.

well, if you have an arbitrary set of N values of x[n] for 0 <= n < N
so that
x[n+N] = x[n] for all n, when N is even, there is no guarantee that
there is no energy
in the X[N/2] bin.  what do you do with that component?  i think split
it in half in the
only reasonable thing.

so, you have your a_k and b_k defined in terms of X[k] which are
defined in terms of x[n]. can you come up with an expression for f(t)
in terms of x[n]?  in that reference to 2001, i think i *did*, and the
result was not surprizing for N odd, but a little surprizing for N
even.  i used superposition and defined x[n] to be 1 for n = 0 (or an
integer multiple of N) and zero for all other n.

r b-j