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Strange FFT Behavior in MATLAB

```I think we can all agree that FFT { x*(n) } = X*(N-k), or the FFT of th
conj of x(n) is the conj of the reversed version of the FFT of x(n).

But in MATLAB if a= [1+2j 3+4j 5+6j 7+8j] then

fft(conj(a))          = [16+20j    -8+0j    -4-4j    0-8j]

and

conj(fliplr(fft(a))) = [-8+0j      -4-4j    0-8j   16+20j]

Any ideas?  Another engineer and I spent most of a day looking at th
model before finding the fundamental problem.

Cheers

/Patrick

```
 0
9/20/2007 12:01:01 PM
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```This is due to the fact that MATLAB index starts at 1 while when you say

FFT{x*(n)} = X*(N-k),

you are assuming that k = 0,...,N-1.

Note that when you do X(N-k) when k = 0, the first sample is X(N), whic
is the same as X(0).  However, if you use fliplr, this is no longer true.

HTH,

>I think we can all agree that FFT { x*(n) } = X*(N-k), or the FFT of the
>conj of x(n) is the conj of the reversed version of the FFT of x(n).
>
>But in MATLAB if a= [1+2j 3+4j 5+6j 7+8j] then
>
>fft(conj(a))          = [16+20j    -8+0j    -4-4j    0-8j]
>
>and
>
>conj(fliplr(fft(a))) = [-8+0j      -4-4j    0-8j   16+20j]
>
>Any ideas?  Another engineer and I spent most of a day looking at the
>model before finding the fundamental problem.
>
>
>Cheers
>
>/Patrick
>
>
>
>
```
 0
9/20/2007 2:04:32 PM
```Patrick wrote:
> I think we can all agree that FFT { x*(n) } = X*(N-k), or the FFT of the
> conj of x(n) is the conj of the reversed version of the FFT of x(n).

It's not exactly the reversed version. Your formula is correct, you
just have to look at it closer. Consider

X[k] := DFT{ x[n] }
X1[k] := DFT{ x*[n] }

So for k=0 you get

X1[0] = X*[N-0] = X*[N]. However, this index is beyond the limit
(indices go from k=0,1,...N-1), so X*[N] is in fact X*[0] (all indices
are modulo N). If you write it out you get

(X1[0], X1[1], ... , X1[N-1]) = (X*[0], X*[N-1], ...., X*[1]).

Presumably Matlab's 1-based indexing doesn't make this matter any
easier.

> But in MATLAB if a= [1+2j 3+4j 5+6j 7+8j] then
>
> fft(conj(a))          = [16+20j    -8+0j    -4-4j    0-8j]
>
> and
>
> conj(fliplr(fft(a))) = [-8+0j      -4-4j    0-8j   16+20j]

For that a, I get

X = fft(a)        = [16+20j    -8    -4-4j     -8j]
X1 = fft(conj(a)) = [16-20j     8j   -4+4j     -8 ]

which is just what the formula says.

> Any ideas?  Another engineer and I spent most of a day looking at the
> model before finding the fundamental problem.

No wonder. If I were to look all day at a model*, I would never find
any problems!

Regards,
Andor

*Favourite model to look at: Claudia.

```
 0
andor.bariska (1307)
9/20/2007 2:19:28 PM
```Ah ha!  X(N-k) isn't a "reversal" of X, it's a circular shift...  that wa
a silly mistake, thanks for the quick answers all.  If anyone is ever i
Dallas, TX beers are owed.

Cheers.

/ptj

>Patrick wrote:
>> I think we can all agree that FFT { x*(n) } = X*(N-k), or the FFT o
the
>> conj of x(n) is the conj of the reversed version of the FFT of x(n).
>
>It's not exactly the reversed version. Your formula is correct, you
>just have to look at it closer. Consider
>
>X[k] := DFT{ x[n] }
>X1[k] := DFT{ x*[n] }
>
>So for k=0 you get
>
>X1[0] = X*[N-0] = X*[N]. However, this index is beyond the limit
>(indices go from k=0,1,...N-1), so X*[N] is in fact X*[0] (all indices
>are modulo N). If you write it out you get
>
>(X1[0], X1[1], ... , X1[N-1]) = (X*[0], X*[N-1], ...., X*[1]).
>
>Presumably Matlab's 1-based indexing doesn't make this matter any
>easier.
>
>> But in MATLAB if a= [1+2j 3+4j 5+6j 7+8j] then
>>
>> fft(conj(a))          = [16+20j    -8+0j    -4-4j    0-8j]
>>
>> and
>>
>> conj(fliplr(fft(a))) = [-8+0j      -4-4j    0-8j   16+20j]
>
>For that a, I get
>
>X = fft(a)        = [16+20j    -8    -4-4j     -8j]
>X1 = fft(conj(a)) = [16-20j     8j   -4+4j     -8 ]
>
>which is just what the formula says.
>
>> Any ideas?  Another engineer and I spent most of a day looking at the
>> model before finding the fundamental problem.
>
>No wonder. If I were to look all day at a model*, I would never find
>any problems!
>
>Regards,
>Andor
>
>*Favourite model to look at: Claudia.
>
>
```
 0
9/20/2007 4:14:14 PM
```On Sep 20, 8:01 am, "patrickjennings" <patrick.t.jenni...@gmail.com>
wrote:
> I think we can all agree that FFT { x*(n) } = X*(N-k),

actually the folks at The Math Works do not agree.  as fatnbafan said,
MATLAB is hard-wired or hard-coded so that the indices of all arrays
begin with 1, not 0 as it should for the DFT or FFT.

so in MATLAB, if N=length(x); y = conj(x);  X = fft(x);  Y = =
fft(y);  then

Y(k+1) = conj( X(mod(N-k+1, N)) );         % for  0 <= k < N

or stated so elegantly that it's amazing we all don't just sing the
praises of MATLAB,

Y(k) = conj( X(mod(N-k+2, N) );            % for  1 <= k <= N

gee, isn't that elegant?

r b-j

> or the FFT of the
> conj of x(n) is the conj of the reversed version of the FFT of x(n).
>
> But in MATLAB if a= [1+2j 3+4j 5+6j 7+8j] then
>
> fft(conj(a))          = [16+20j    -8+0j    -4-4j    0-8j]
>
> and
>
> conj(fliplr(fft(a))) = [-8+0j      -4-4j    0-8j   16+20j]
>
> Any ideas?  Another engineer and I spent most of a day looking at the
> model before finding the fundamental problem.
>
> Cheers
>
> /Patrick

```
 0
rbj (4086)
9/20/2007 4:56:05 PM
```robert bristow-johnson <rbj@audioimagination.com> wrote in

> actually the folks at The Math Works do not agree.  as fatnbafan said,
> MATLAB is hard-wired or hard-coded so that the indices of all arrays
> begin with 1, not 0 as it should for the DFT or FFT.

It's an index-- just a conceptual place holder, not a definition. X(1),
in a ones indexed language, merely points to the first memory location in
array X.

Most likely the ones indexing in Matlab has its roots in Fortran.

Aside from having to remember what language you're working in and getting
bitten every now and again when moving back and forth between languages,
it really isn't that problematic.

> Y(k) = conj( X(mod(N-k+2, N) );            % for  1 <= k <= N
>
> gee, isn't that elegant?

c implemenation of fftw, compiling it, linking to it, then figuring out
which of the many arcane fftw routines you need to be calling, and then
figuring out what arcane pattern your returned arrays are in, maybe
creating complex variable types from two doubles, and then first doing
your elegant zero-indexed circular shift, its a veritable tango.  If
you're talking about doing the same in some zero-indexed matlab clone,
probably not as elegant.

--
Scott
```
 0
namdiesttocs (1203)
9/20/2007 5:10:51 PM
```<snip, ML's fft behaviour...

I don't buy all this <expletive> stuff.  If you don't like
the wrapper TMW have put around the core FFT/FFTW call,
preferred way - don't blame TMW for not sharing your own
choice.

Phew, glad that's off my chest.  Now all we need is support
for zero-based indexing ;-)
```
 0
9/20/2007 5:25:47 PM
```When analyzing (real) time series data, the complex
frequency spectrum is symmetric in the real part and
anti-symmetric in the imaginary part.

Way back in Matlab 3, when memory and performance were at a
premium, I formulated MEX files that used this symmetry to
reduce the data size.

It is possible to arrange a real time series into a complex
series of half the length and post process the fft output to
get the spectrum. Or the other way around - arranging half
the spectrum into a real series. For lengths of a power of
2, Matlab 3 used a fast Cooley Tukey algorithm which was
most efficient going from real to complex.

For even lengths that are not a power of 2, Matlab 3 only
used a complex FFT. In those cases, I took the real series
(either time series or half the spectrum), and arranged it
as a complex series of half the length. Again, I could take
half a frequency spectrum and rearrange it as a real series
then rearrange it again as a complex series of half the length.

It sounds more complicated than it was and it ran several
times faster.
```
 0
ajrobb (28)
8/16/2008 3:40:20 AM

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Hi Paul, Thanks for your answer to my problem "Subject: Strange behavior in run-time vs. dbx behavior when using shmget". I don't understand how your hint will help me. Maybe i was misunderstood. I'm writing to the 0x60000000 segment only once in the beginning of my application, so the code there isn't supposed to be changed, and in this segment there are no calls to other procedures in this segment. So only a code from segment 0x10000000 will branch to this segment. Maybe you mean that before writing to this segment i should flush the buffer? Thanks Israel israelaix@hotmail.com (Israel Waldman) writes: > Hi Paul, If you wanted to ask me personally, it is probably would have been better to just send a private e-mail. However, having decided to use the usenet, you should have really continued the thread you have already started, not start a new one. [You do know there is a difference, don't you?] > Maybe you mean that before writing to this segment i should flush the > buffer? What I meant is that this sequence doesn't work on PowerPC: void *ip = ... // (1) some operation that returns a pointer to // writable and executable page, e.g. 0x60000000 // (2) code to construct executable instructions at address ip // (3) [missing steps] void (*func)(void) = (void (*)())ip; (*ip)(); // call the dynamically-generated code @ 0x600... The missing steps in (3) are flushing D-cache and invalidating I-cache. You ma...

Why strange IF...ELSE behavior
Hi all, I'm getting a strange result with the following IF statement: \$bar = (\$foo == 'last') ? true : false; In my script \$foo normaly has a integer value greater than 0, but can have last as value. But whatever value of \$foo \$bar always becomes true. When I use the regular if {....}else{....} I get the same result, always \$bar becomes true. I've played around with a couple things, like turning the statement around (\$foo != 'last') but nothing seems to give the desired result. Can anyone tell me what I'm doing wrong or how to fix this. Thanx, J-P J-P �crivit: > Hi all, > > I'm getting a strange result with the following IF statement: > > \$bar = (\$foo == 'last') ? true : false; > > In my script \$foo normaly has a integer value greater than 0, but can > have last as value. But whatever value of \$foo \$bar always becomes > true. It's due to the silent type casting of php you should use === instead of == -- P'tit Marcel > > It's due to the silent type casting of php > you should use === instead of == > Thanx that did it, never to late to learn something. ...

Strange findfirst behavior
I have a combo box that finds a record on the form, and for the most part it works fine. But, on occasion it brings me to the record just before the one I am looking for. If I compact and repair the database it works correctly again. Sometimes I have to compact and repair twice. Any help would be appreciated It sounds as if something is corrupting the index on the field you are doing the search on. Please see this link for information on avoiding corruption. http://www.allenbrowne.com/ser-25.html -- Wayne Morgan MS Access MVP "Shyguy" <shyguy@aol.com> wrote in message news:j88cq1lf21ndsq90b49t5e1oqlll08ovr0@4ax.com... >I have a combo box that finds a record on the form, and for the most > part it works fine. But, on occasion it brings me to the record just > before the one I am looking for. If I compact and repair the database > it works correctly again. Sometimes I have to compact and repair > twice. > > Any help would be appreciated > Thanks for the link. I'll have to go through it later. ;-) On Mon, 19 Dec 2005 04:45:03 GMT, "Wayne Morgan" <comprev_gothroughthenewsgroup@hotmail.com> wrote: >It sounds as if something is corrupting the index on the field you are doing >the search on. Please see this link for information on avoiding corruption. > >http://www.allenbrowne.com/ser-25.html ...

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