Reply by robert bristow-johnson●February 28, 20142014-02-28
On 2/28/14 6:06 PM, Tim Wescott wrote:
> On Thu, 27 Feb 2014 14:14:29 -0500, robert bristow-johnson wrote:
>
>> On 2/27/14 3:27 AM, Markus Grunwald wrote:
>>> Hello to all who answered :)
>>>
>>> Thank you very much, your Answers were most helpful. The one that I
>>> understood easily was clay's "Think of sampling a cosine every multiple
>>> of 180 degrees." while the others helped to understand further.
>>>
>>> And you're right, I got my indexing wrong.
>>
>> and we understand why. it's not your fault. but i won't say whose
>> fault it is.
>
> It's all our faults. Clearly the indexing should start at 'a' and go
> from there.
but neither you nor i have put out a math modeling tool along with a
toolbox directed toward signal processing.
i don't see how "it's all our faults".
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by Tim Wescott●February 28, 20142014-02-28
On Thu, 27 Feb 2014 14:14:29 -0500, robert bristow-johnson wrote:
> On 2/27/14 3:27 AM, Markus Grunwald wrote:
>> Hello to all who answered :)
>>
>> Thank you very much, your Answers were most helpful. The one that I
>> understood easily was clay's "Think of sampling a cosine every multiple
>> of 180 degrees." while the others helped to understand further.
>>
>> And you're right, I got my indexing wrong.
>
> and we understand why. it's not your fault. but i won't say whose
> fault it is.
It's all our faults. Clearly the indexing should start at 'a' and go
from there.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by robert bristow-johnson●February 27, 20142014-02-27
On 2/27/14 3:27 AM, Markus Grunwald wrote:
> Hello to all who answered :)
>
> Thank you very much, your Answers were most helpful. The one that I
> understood easily was clay's "Think of sampling a cosine every multiple
> of 180 degrees." while the others helped to understand further.
>
> And you're right, I got my indexing wrong.
and we understand why. it's not your fault. but i won't say whose
fault it is.
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by robert bristow-johnson●February 27, 20142014-02-27
On 2/26/14 10:08 PM, Eric Jacobsen wrote:
> On Wed, 26 Feb 2014 16:07:15 -0600, Tim Wescott
> <tim@seemywebsite.really> wrote:
>
>> On Wed, 26 Feb 2014 21:08:41 +0000, glen herrmannsfeldt wrote:
>>
>>> Tim Wescott<tim@seemywebsite.really> wrote:
...
>>>> Keep in mind that a[0 ... N/2] = a[N-1 ... N/2]*, where a* denotes the
>>>> complex conjugate of a. This (together with knowing that Im(a[0]) ==
>>>> Im(a [N/2] == 0) reflects the fact that you only put in N non-zero
>>>> points (because the N imaginary samples were zero), so you only get out
>>>> N unique points.
>>>
>>> That says why a[N/2] is special, but the OP asked about a[N/2+1].
>>> (Where the first element is a[0].)
>>
>> We all missed that.
>>
>> Hey! OP! You meant a[N/2]. Really. a[N/2 + 1] is just the complex
>> conjugate of a[N/2 - 1]. Or your indexing is off. Or something.
>
> I didn't miss it, I added the caveat about "depending on your
> coefficient numbering". I didn't point it out because I thought the
> only candidate in the region for specialness is the symmetry point, so
> I figured it was a typo or a numbering issue.
it's because of MATLAB (or Octave or R or whatever fixed 1-origin
indexing in whatever tool). blame it on err, ... MATLAB (Cleve, i won't
name names.)
jadedly,
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by Markus Grunwald●February 27, 20142014-02-27
Hello to all who answered :)
Thank you very much, your Answers were most helpful. The one that I
understood easily was clay's "Think of sampling a cosine every multiple
of 180 degrees." while the others helped to understand further.
And you're right, I got my indexing wrong.
cu
/me
Reply by Eric Jacobsen●February 26, 20142014-02-26
On Wed, 26 Feb 2014 16:07:15 -0600, Tim Wescott
<tim@seemywebsite.really> wrote:
>On Wed, 26 Feb 2014 21:08:41 +0000, glen herrmannsfeldt wrote:
>
>> Tim Wescott <tim@seemywebsite.really> wrote:
>>> On Wed, 26 Feb 2014 08:46:53 +0000, Markus Grunwald wrote:
>>
>> (snip)
>>>> If a is the result of the FFT, there are two "interesting" elements of
>>>> a:
>>
>> (snip)
>>
>>>> But... It seems that the element at a[N/2+1] is somehow "special",
>>>> like a[0]. Its imaginary part seems to be zero in this scenario.
>>
>> (snip)
>>>> Am I wrong? Is a[N/2+1] a "special" value? How's it interpreted?
>>
>>> What Clay and Eric said about your specific question. It more or less
>>> falls out of the fact that the input is real and N is even.
>>
>>> Keep in mind that a[0 ... N/2] = a[N-1 ... N/2]*, where a* denotes the
>>> complex conjugate of a. This (together with knowing that Im(a[0]) ==
>>> Im(a [N/2] == 0) reflects the fact that you only put in N non-zero
>>> points (because the N imaginary samples were zero), so you only get out
>>> N unique points.
>>
>> That says why a[N/2] is special, but the OP asked about a[N/2+1].
>> (Where the first element is a[0].)
>
>We all missed that.
>
>Hey! OP! You meant a[N/2]. Really. a[N/2 + 1] is just the complex
>conjugate of a[N/2 - 1]. Or your indexing is off. Or something.
I didn't miss it, I added the caveat about "depending on your
coefficient numbering". I didn't point it out because I thought the
only candidate in the region for specialness is the symmetry point, so
I figured it was a typo or a numbering issue.
On Wed, 26 Feb 2014 08:46:53 +0000, Markus Grunwald wrote:
> Hello,
>
> while working daily with FFTs, I seem to have forgotten some of my
> theory :(
>
> Given:
> - the "usual" FFT algorithm that turns N complex samples in time domain
> to N complex samples in frequency domain.
>
> - a real valued input signal (= some sampled data, sampling rate "fa").
>
> If a is the result of the FFT, there are two "interesting" elements of
> a:
>
> a[0] : This is the DC part of the signal. That much is clear. Its
> imaginary part is zero in our scenario.
>
> a[1…N/2] : correspond to the multiples of Δf=fa/N. That's clear, too.
>
> But... It seems that the element at a[N/2+1] is somehow "special", like
> a[0]. Its imaginary part seems to be zero in this scenario. I don't
> know,
> but it seems it's not the first of the negative frequencies that make up
> the rest of the spectrum.
>
> Am I wrong? Is a[N/2+1] a "special" value? How's it interpreted?
Your indexing is off. See my response to Glen Herrmannsfeldt's response
to my response. (Speaking of indexing).
If a's indexing starts at 0, then a[N/2] is special, with a zero
imaginary part, and what Eric and Clay said holds.
If a's indexing starts at 0, then a[N/2 + 1] is only special in that with
a real-valued input it equals the complex conjugate of a[N/2 - 1] -- but
that holds for any pair a[N/2 + n] vs. a[N/2 - n], 0 <= n <= N/2.
If a's indexing starts at 1 (i.e., a[1] = DC), then a[N/2 + 1] is
special, and what Eric and Clay said holds with appropriate adjustment of
indexes.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by Tim Wescott●February 26, 20142014-02-26
On Wed, 26 Feb 2014 21:08:41 +0000, glen herrmannsfeldt wrote:
> Tim Wescott <tim@seemywebsite.really> wrote:
>> On Wed, 26 Feb 2014 08:46:53 +0000, Markus Grunwald wrote:
>
> (snip)
>>> If a is the result of the FFT, there are two "interesting" elements of
>>> a:
>
> (snip)
>
>>> But... It seems that the element at a[N/2+1] is somehow "special",
>>> like a[0]. Its imaginary part seems to be zero in this scenario.
>
> (snip)
>>> Am I wrong? Is a[N/2+1] a "special" value? How's it interpreted?
>
>> What Clay and Eric said about your specific question. It more or less
>> falls out of the fact that the input is real and N is even.
>
>> Keep in mind that a[0 ... N/2] = a[N-1 ... N/2]*, where a* denotes the
>> complex conjugate of a. This (together with knowing that Im(a[0]) ==
>> Im(a [N/2] == 0) reflects the fact that you only put in N non-zero
>> points (because the N imaginary samples were zero), so you only get out
>> N unique points.
>
> That says why a[N/2] is special, but the OP asked about a[N/2+1].
> (Where the first element is a[0].)
We all missed that.
Hey! OP! You meant a[N/2]. Really. a[N/2 + 1] is just the complex
conjugate of a[N/2 - 1]. Or your indexing is off. Or something.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by glen herrmannsfeldt●February 26, 20142014-02-26
Tim Wescott <tim@seemywebsite.really> wrote:
> On Wed, 26 Feb 2014 08:46:53 +0000, Markus Grunwald wrote:
(snip)
>> If a is the result of the FFT, there are two "interesting" elements of
>> a:
(snip)
>> But... It seems that the element at a[N/2+1] is somehow "special", like
>> a[0]. Its imaginary part seems to be zero in this scenario.
(snip)
>> Am I wrong? Is a[N/2+1] a "special" value? How's it interpreted?
> What Clay and Eric said about your specific question. It more or less
> falls out of the fact that the input is real and N is even.
> Keep in mind that a[0 ... N/2] = a[N-1 ... N/2]*, where a* denotes the
> complex conjugate of a. This (together with knowing that Im(a[0]) == Im(a
> [N/2] == 0) reflects the fact that you only put in N non-zero points
> (because the N imaginary samples were zero), so you only get out N unique
> points.
That says why a[N/2] is special, but the OP asked about a[N/2+1].
(Where the first element is a[0].)
-- glen
Reply by Tim Wescott●February 26, 20142014-02-26
On Wed, 26 Feb 2014 08:46:53 +0000, Markus Grunwald wrote:
> Hello,
>
> while working daily with FFTs, I seem to have forgotten some of my
> theory :(
>
> Given:
> - the "usual" FFT algorithm that turns N complex samples in time domain
> to N complex samples in frequency domain.
>
> - a real valued input signal (= some sampled data, sampling rate "fa").
>
> If a is the result of the FFT, there are two "interesting" elements of
> a:
>
> a[0] : This is the DC part of the signal. That much is clear. Its
> imaginary part is zero in our scenario.
>
> a[1…N/2] : correspond to the multiples of Δf=fa/N. That's clear, too.
>
> But... It seems that the element at a[N/2+1] is somehow "special", like
> a[0]. Its imaginary part seems to be zero in this scenario. I don't
> know,
> but it seems it's not the first of the negative frequencies that make up
> the rest of the spectrum.
>
> Am I wrong? Is a[N/2+1] a "special" value? How's it interpreted?
What Clay and Eric said about your specific question. It more or less
falls out of the fact that the input is real and N is even.
Keep in mind that a[0 ... N/2] = a[N-1 ... N/2]*, where a* denotes the
complex conjugate of a. This (together with knowing that Im(a[0]) == Im(a
[N/2] == 0) reflects the fact that you only put in N non-zero points
(because the N imaginary samples were zero), so you only get out N unique
points.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com