# Relationship between z and Fourier transforms

Started by July 30, 2013
```On 8/1/13 1:25 PM, Tim Wescott wrote:
> On Thu, 01 Aug 2013 12:48:45 -0700, robert bristow-johnson wrote:
>
>> On 8/1/13 12:17 PM, Tim Wescott wrote:
>>>
>>> Your classification system works for me (if not for my newsreader,
>>> which seems to have mangled your nice table).
>>>
>>>
>> careful, Tim.  you might inadvertently have taken a side in "_That_
>> argument".  note the upper right corner of the table.
>
> The only side I take in _that_ argument is that both sides are being
> remarkably inflexible.

that's an interesting take, because, as best as i understand it, only
*i* (and others taking this position) are inflexible as well as the math
and that the other side is insisting on a flexibility that the math does
not provide for (except in the trivial case where the only property of
the DFT one makes use of is linearity).

>  The math just is, and if you're getting the right
> results you are, ipso facto, using it correctly.

fully agree with that.  and, unless there is no shifting nor convolution
(in time or frequency) going on (which means the only property or
theorem of the DFT one can make use of is Linearity) then the *only*
(and this "only" is me being inflexible) correct mathematics is to
either explicitly periodically extend the data:

x[n+N] = x[n]   or   X[k+N] = X[k]

or to implicitly periodically extend the data:

x[n mod N]    or   X[k mod N]

either way you're periodically extending the data, the math says so, and
it seems to me silly to deny it.  there is no way to get away from it.
"resistance is futile."

--

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

```
```On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson
<rbj@audioimagination.com> wrote:

>On 8/1/13 1:25 PM, Tim Wescott wrote:
>> On Thu, 01 Aug 2013 12:48:45 -0700, robert bristow-johnson wrote:
>>
>>> On 8/1/13 12:17 PM, Tim Wescott wrote:
>>>>
>>>> Your classification system works for me (if not for my newsreader,
>>>> which seems to have mangled your nice table).
>>>>
>>>>
>>> careful, Tim.  you might inadvertently have taken a side in "_That_
>>> argument".  note the upper right corner of the table.
>>
>> The only side I take in _that_ argument is that both sides are being
>> remarkably inflexible.
>
>that's an interesting take, because, as best as i understand it, only
>*i* (and others taking this position) are inflexible as well as the math
>and that the other side is insisting on a flexibility that the math does
>not provide for (except in the trivial case where the only property of
>the DFT one makes use of is linearity).
>
>>  The math just is, and if you're getting the right
>> results you are, ipso facto, using it correctly.
>
>fully agree with that.  and, unless there is no shifting nor convolution
>(in time or frequency) going on (which means the only property or
>theorem of the DFT one can make use of is Linearity) then the *only*
>(and this "only" is me being inflexible) correct mathematics is to
>either explicitly periodically extend the data:
>
>     x[n+N] = x[n]   or   X[k+N] = X[k]
>
>or to implicitly periodically extend the data:
>
>     x[n mod N]    or   X[k mod N]
>
>
>either way you're periodically extending the data, the math says so, and
>it seems to me silly to deny it.  there is no way to get away from it.
>  "resistance is futile."

Except for the numerous counter-examples that have been pointed out to

>--
>
>r b-j                  rbj@audioimagination.com
>
>"Imagination is more important than knowledge."
>
>

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com
```
```eric.jacobsen@ieee.org (Eric Jacobsen) writes:

> On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson
> <rbj@audioimagination.com> wrote:
>
>>On 8/1/13 1:25 PM, Tim Wescott wrote:
>>> On Thu, 01 Aug 2013 12:48:45 -0700, robert bristow-johnson wrote:
>>>
>>>> On 8/1/13 12:17 PM, Tim Wescott wrote:
>>>>>
>>>>> Your classification system works for me (if not for my newsreader,
>>>>> which seems to have mangled your nice table).
>>>>>
>>>>>
>>>> careful, Tim.  you might inadvertently have taken a side in "_That_
>>>> argument".  note the upper right corner of the table.
>>>
>>> The only side I take in _that_ argument is that both sides are being
>>> remarkably inflexible.
>>
>>that's an interesting take, because, as best as i understand it, only
>>*i* (and others taking this position) are inflexible as well as the math
>>and that the other side is insisting on a flexibility that the math does
>>not provide for (except in the trivial case where the only property of
>>the DFT one makes use of is linearity).
>>
>>>  The math just is, and if you're getting the right
>>> results you are, ipso facto, using it correctly.
>>
>>fully agree with that.  and, unless there is no shifting nor convolution
>>(in time or frequency) going on (which means the only property or
>>theorem of the DFT one can make use of is Linearity) then the *only*
>>(and this "only" is me being inflexible) correct mathematics is to
>>either explicitly periodically extend the data:
>>
>>     x[n+N] = x[n]   or   X[k+N] = X[k]
>>
>>or to implicitly periodically extend the data:
>>
>>     x[n mod N]    or   X[k mod N]
>>
>>
>>either way you're periodically extending the data, the math says so, and
>>it seems to me silly to deny it.  there is no way to get away from it.
>>  "resistance is futile."
>
> Except for the numerous counter-examples that have been pointed out to

A thought, perhaps erroneous, just occurred to me. Shouldn't we be able
to distinguish between the interpretation as a periodic extension versus
that as a windowed input as follows: An implicit periodic extension
would have no windowing involved and there would be no "smearing"; A
windowed input would have smearing.

Without cranking through a few examples, my gut tells me it's the
former.
--
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com
```
```On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson wrote:

> On 8/1/13 1:25 PM, Tim Wescott wrote:
>> On Thu, 01 Aug 2013 12:48:45 -0700, robert bristow-johnson wrote:
>>
>>> On 8/1/13 12:17 PM, Tim Wescott wrote:
>>>>
>>>> Your classification system works for me (if not for my newsreader,
>>>> which seems to have mangled your nice table).
>>>>
>>>>
>>> careful, Tim.  you might inadvertently have taken a side in "_That_
>>> argument".  note the upper right corner of the table.
>>
>> The only side I take in _that_ argument is that both sides are being
>> remarkably inflexible.
>
> that's an interesting take, because, as best as i understand it, only
> *i* (and others taking this position) are inflexible as well as the math
> and that the other side is insisting on a flexibility that the math does
> not provide for (except in the trivial case where the only property of
> the DFT one makes use of is linearity).
>
>>  The math just is, and if you're getting the right
>> results you are, ipso facto, using it correctly.
>
> fully agree with that.  and, unless there is no shifting nor convolution
> (in time or frequency) going on (which means the only property or
> theorem of the DFT one can make use of is Linearity) then the *only*
> (and this "only" is me being inflexible) correct mathematics is to
> either explicitly periodically extend the data:
>
>      x[n+N] = x[n]   or   X[k+N] = X[k]
>
> or to implicitly periodically extend the data:
>
>      x[n mod N]    or   X[k mod N]
>
>
> either way you're periodically extending the data, the math says so, and
> it seems to me silly to deny it.  there is no way to get away from it.
>   "resistance is futile."

It works either if you periodically extend the data, or if you say that
the data set is finite.

A pair of examples pertaining to the Fourier series comes from electrical
engineering, one involving periodic phenomena, and the other involving
bounded phenomena.

The periodic example is the use of the Fourier series to find the
harmonics of a periodic waveform.  This is your "periodic extension" case.

The aperiodic example (and I'm rustier at this one, but I remember it
being done) is in the use of the Fourier series to find the potential
variation within a square box that's bounded by conductors.  In this case
there is no periodicity -- there could be anything outside the box, the
problem stops as soon as you hit that metal wall with its imposed voltage.

The math fits in either case.  So, either view is equally valid.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

```
```On Thu, 01 Aug 2013 17:15:27 -0400, Randy Yates
<yates@digitalsignallabs.com> wrote:

>eric.jacobsen@ieee.org (Eric Jacobsen) writes:
>
>> On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson
>> <rbj@audioimagination.com> wrote:
>>
>>>On 8/1/13 1:25 PM, Tim Wescott wrote:
>>>> On Thu, 01 Aug 2013 12:48:45 -0700, robert bristow-johnson wrote:
>>>>
>>>>> On 8/1/13 12:17 PM, Tim Wescott wrote:
>>>>>>
>>>>>> Your classification system works for me (if not for my newsreader,
>>>>>> which seems to have mangled your nice table).
>>>>>>
>>>>>>
>>>>> careful, Tim.  you might inadvertently have taken a side in "_That_
>>>>> argument".  note the upper right corner of the table.
>>>>
>>>> The only side I take in _that_ argument is that both sides are being
>>>> remarkably inflexible.
>>>
>>>that's an interesting take, because, as best as i understand it, only
>>>*i* (and others taking this position) are inflexible as well as the math
>>>and that the other side is insisting on a flexibility that the math does
>>>not provide for (except in the trivial case where the only property of
>>>the DFT one makes use of is linearity).
>>>
>>>>  The math just is, and if you're getting the right
>>>> results you are, ipso facto, using it correctly.
>>>
>>>fully agree with that.  and, unless there is no shifting nor convolution
>>>(in time or frequency) going on (which means the only property or
>>>theorem of the DFT one can make use of is Linearity) then the *only*
>>>(and this "only" is me being inflexible) correct mathematics is to
>>>either explicitly periodically extend the data:
>>>
>>>     x[n+N] = x[n]   or   X[k+N] = X[k]
>>>
>>>or to implicitly periodically extend the data:
>>>
>>>     x[n mod N]    or   X[k mod N]
>>>
>>>
>>>either way you're periodically extending the data, the math says so, and
>>>it seems to me silly to deny it.  there is no way to get away from it.
>>>  "resistance is futile."
>>
>> Except for the numerous counter-examples that have been pointed out to
>
>A thought, perhaps erroneous, just occurred to me. Shouldn't we be able
>to distinguish between the interpretation as a periodic extension versus
>that as a windowed input as follows: An implicit periodic extension
>would have no windowing involved and there would be no "smearing"; A
>windowed input would have smearing.

What would be examples of the differences?   It's really just an
interpretational issue, as there is no difference in the input or
output of a DFT regardless of the point of view one takes.

>Without cranking through a few examples, my gut tells me it's the
>former.
>--
>Randy Yates
>Digital Signal Labs
>http://www.digitalsignallabs.com

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com
```
```On 8/1/13 2:06 PM, Eric Jacobsen wrote:
> On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson
> <rbj@audioimagination.com>  wrote:
>
>> On 8/1/13 1:25 PM, Tim Wescott wrote:
...
>>>   The math just is, and if you're getting the right
>>> results you are, ipso facto, using it correctly.
>>
>> fully agree with that.  and, unless there is no shifting nor convolution
>> (in time or frequency) going on (which means the only property or
>> theorem of the DFT one can make use of is Linearity) then the *only*
>> (and this "only" is me being inflexible) correct mathematics is to
>> either explicitly periodically extend the data:
>>
>>      x[n+N] = x[n]   or   X[k+N] = X[k]
>>
>> or to implicitly periodically extend the data:
>>
>>      x[n mod N]    or   X[k mod N]
>>
>>
>> either way you're periodically extending the data, the math says so, and
>> it seems to me silly to deny it.  there is no way to get away from it.
>>   "resistance is futile."
>
> Except for the numerous counter-examples that have been pointed out to

you have numerous counter-examples which involve shifting the data or
convolving the data (or their counterparts in the other domain which
means multiplying by something that is not constant) where you can
ignore both

x[n+N] = x[n]

or

x[n mod N]

and the X[k] counterparts?

would you mind repeating just one of those counter examples?  and in
these numerous counter examples you are able to show mathematical
correctness while ignoring x[n+N] = x[n] and x[n mod N] (or the same
thing in X[k])?

zero-padding is a questionable counter-example because if you claim that
the zero shifted in isn't the same zero that was shifted out the other
end, then i would dispute that.

--

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

```
```robert bristow-johnson <rbj@audioimagination.com> wrote:
> On 8/1/13 1:25 PM, Tim Wescott wrote:

(snip)
>> The only side I take in _that_ argument is that both sides are being
>> remarkably inflexible.

> that's an interesting take, because, as best as i understand it, only
> *i* (and others taking this position) are inflexible as well as
> the math and that the other side is insisting on a flexibility
> that the math does not provide for (except in the trivial case
> where the only property of the DFT one makes use of is linearity).

>>  The math just is, and if you're getting the right
>> results you are, ipso facto, using it correctly.

> fully agree with that.  and, unless there is no shifting nor convolution
> (in time or frequency) going on (which means the only property or
> theorem of the DFT one can make use of is Linearity) then the *only*
> (and this "only" is me being inflexible) correct mathematics is to
> either explicitly periodically extend the data:

I am not sure by now which side who is on.

Consider a least squares fit of some data to an even polynomical.
(All powers even, so f(x)=f(-x).

If one fits some data to such polynomial, one should not be
surprised to find the result an even function.

One might say that, given a basis set of even functions, that
the fit "assumes" the data is even.

But otherwise, the Fourier transform is a specific case of a
series solution to a linear differential equation.

In the case of a boundary value problem, one chooses the basis set
to satisfy the boundary conditions, usually not the other way around.

Now, there are a number of cases for real problems where one assumes
periodicity when the actual problem might not have that symmetry.

For one, it is commonly done in solid-state physics when considering
the properties of a crystal, ignoring the surface states. In a
crystal with 1e24 atoms, only a small fraction are on the surface.

There are also a fair number of problems that naturally decay to zero
toward the ends, and so assuming periodicity doesn't matter.

-- glen

```
```On Thursday, August 1, 2013 12:48:45 PM UTC-7, robert bristow-johnson wrote:
> On 8/1/13 12:17 PM, Tim Wescott wrote:
> >
> > Your classification system works for me (if not for my newsreader, which
> > seems to have mangled your nice table).
> >
> careful, Tim.  you might inadvertently have taken a side in "_That_
> argument".  note the upper right corner of the table.
>
> r b-j

That you chose to add a notation in a manner that you expected to be hard to notice provides more illumination about your nature than the notation tells us about the nature of transforms.

The existence of the DFS in the chart already indicates that the assumptions of periodicity and of period N already apply to signals at that point in the diagram.

A better graphic presentation of some of the transform relationships is available at:
http://www-mmsp.ece.mcgill.ca/Documents/Reports/2011/KabalR2011c.pdf
Frequency Domain Representations
of Sampled and Wrapped Signals
Peter Kabal

Dale B. Dalrymple

```
```Tim Wescott <tim@seemywebsite.really> wrote:
> On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson wrote:

(snip)

>> either way you're periodically extending the data, the math says so, and
>> it seems to me silly to deny it.  there is no way to get away from it.
>>   "resistance is futile."

> It works either if you periodically extend the data, or if you say that
> the data set is finite.

> A pair of examples pertaining to the Fourier series comes from electrical
> engineering, one involving periodic phenomena, and the other involving
> bounded phenomena.

> The periodic example is the use of the Fourier series to find the
> harmonics of a periodic waveform.  This is your "periodic
> extension" case.

> The aperiodic example (and I'm rustier at this one, but I remember it
> being done) is in the use of the Fourier series to find the potential
> variation within a square box that's bounded by conductors.  In this case
> there is no periodicity -- there could be anything outside the box, the
> problem stops as soon as you hit that metal wall with its imposed voltage.

OK, but consider a related problem, either in 1D (signal on a coax
cable) or 3D (EM wave in a metal box). When the wave hits the boundary,
there is a reflection. The solution looks the same as if the original
one went through the boundary, and another comes from the outside
into the problem at the boundary.

Or, consider when you look at the reflection of an object in a
mirror, it "looks" the same as it would if there was no mirror,
and an actual object on the other side.

> The math fits in either case.  So, either view is equally valid.

-- glen
```
```On Thu, 01 Aug 2013 14:43:37 -0700, robert bristow-johnson
<rbj@audioimagination.com> wrote:

>On 8/1/13 2:06 PM, Eric Jacobsen wrote:
>> On Thu, 01 Aug 2013 13:42:19 -0700, robert bristow-johnson
>> <rbj@audioimagination.com>  wrote:
>>
>>> On 8/1/13 1:25 PM, Tim Wescott wrote:
>...
>>>>   The math just is, and if you're getting the right
>>>> results you are, ipso facto, using it correctly.
>>>
>>> fully agree with that.  and, unless there is no shifting nor convolution
>>> (in time or frequency) going on (which means the only property or
>>> theorem of the DFT one can make use of is Linearity) then the *only*
>>> (and this "only" is me being inflexible) correct mathematics is to
>>> either explicitly periodically extend the data:
>>>
>>>      x[n+N] = x[n]   or   X[k+N] = X[k]
>>>
>>> or to implicitly periodically extend the data:
>>>
>>>      x[n mod N]    or   X[k mod N]
>>>
>>>
>>> either way you're periodically extending the data, the math says so, and
>>> it seems to me silly to deny it.  there is no way to get away from it.
>>>   "resistance is futile."
>>
>> Except for the numerous counter-examples that have been pointed out to
>
>you have numerous counter-examples which involve shifting the data or
>convolving the data (or their counterparts in the other domain which
>means multiplying by something that is not constant) where you can
>ignore both
>
>       x[n+N] = x[n]
>
>or
>
>       x[n mod N]
>
>and the X[k] counterparts?
>
>would you mind repeating just one of those counter examples?  and in
>these numerous counter examples you are able to show mathematical
>correctness while ignoring x[n+N] = x[n] and x[n mod N] (or the same
>thing in X[k])?

It's been said many times already.   I've long grown tired of
repeating them over again to you.

>zero-padding is a questionable counter-example because if you claim that
>the zero shifted in isn't the same zero that was shifted out the other
>end, then i would dispute that.

You dispute a lot of things.

>
>--
>
>r b-j                  rbj@audioimagination.com
>
>"Imagination is more important than knowledge."
>
>

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com
```