Thread Subject: About FFT and Even Function

Hello everyone,

I have a doubt about FFT function on Matlab. I know when I have an even and real function its FFT give only real values. But, when I calculate, for example, the FFT of a cosine function result real and imaginary values. I think I making a mistake on my interpretation of this result. Anyone can help me please :)

Thanks and sorry for my english!!!

On Oct 31, 10:42 pm, "Oscar " <oscarodrig...@yahoo.es> wrote:
> Hello everyone,
>
> I have a doubt about FFT function on Matlab. I know when I have an even and real function its FFT give only real values. But, when I calculate, for example, the FFT of a cosine function result real and imaginary values. I think I making a mistake on my interpretation of this result. Anyone can help me please :)
>
> Thanks and sorry for my english!!!

There are often mistakes made confusing 'symmetry' with 'fft-
evenness'. For examples, compare the difference between Matlab's
'symmetric' and 'periodic' forms of window functions. Use the windows
as the real signals to be transformed.

A good reference is fred harris' paper on windows. Look at the first
column of page 173. The paper is available a number of places. One is:

http://www.utdallas.edu/~cpb021000/EE%204361/Great%20DSP%20Papers/Harris%20on%20Windows.pdf

Dale B. Dalrymple

"Oscar " <oscarodrigo3k@yahoo.es> wrote in message
news:hcj73a$b5$1@fred.mathworks.com...
> Hello everyone,
>
> I have a doubt about FFT function on Matlab. I know when I have an even
> and real function its FFT give only real values. But, when I calculate,
> for example, the FFT of a cosine function result real and imaginary
> values. I think I making a mistake on my interpretation of this result.
> Anyone can help me please :)
>
> Thanks and sorry for my english!!!

If the data sequence itself is real and even (i.e. x[n]=x[N-n]) where N is
the sample points, then the discrete fourier transform (which fft is an
implementation of) should result in all real values. i.e. g[k] for k=1..N
should be real values.

Is this the case with your sampled data?

--Nasser

On Oct 31, 11:11 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> ...
> If the data sequence itself is real and even (i.e. x[n]=x[N-n]) where N is
> ...
> --Nasser

Let's try that:

>> % Nassar's formula, N = 6, x[n] = x[N-n] for n=1:N-1
>> fft([1 2 3 2 1 0])'
ans =
   9.0000
  -2.0000 + 3.4641i
        0
   1.0000
        0
  -2.0000 - 3.4641i
>> % "FFT-even" version, N = 6, x[n] = x[N-n+1] for n=2:N
>> fft([0 1 2 3 2 1])'
ans =
     9
    -4
     0
    -1
     0
    -4
>>

Dale B. Dalrymple

"dbd" <dbd@ieee.org> wrote in message
news:a220a52b-a008-4b16-acc7-0656a03275b3@w37g2000prg.googlegroups.com...
> On Oct 31, 11:11 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
>> ...
>> If the data sequence itself is real and even (i.e. x[n]=x[N-n]) where N
>> is
>> ...
>> --Nasser
>
> Let's try that:
>

>>> % Nassar's formula, N = 6, x[n] = x[N-n] for n=1:N-1
>>> fft([1 2 3 2 1 0])'
> ans =
> 9.0000
> -2.0000 + 3.4641i
> 0
> 1.0000
> 0
> -2.0000 - 3.4641i

Ofcourse, I said above n=1:N-1 which is clearly wrong (the number of points
is N now N-1 points), I meant to write n=0:N-1 to make it N points. I made a
typo starting it from 1 not 0.

May be now you can try my formula? ;) but watch for the index.

--Nasser

dbd <dbd@ieee.org> wrote in message <a220a52b-a008-4b16-acc7-0656a03275b3@w37g2000prg.googlegroups.com>...
> On Oct 31, 11:11 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> > ...
> > If the data sequence itself is real and even (i.e. x[n]=x[N-n]) where N is
> > ...
> > --Nasser
>
> Let's try that:
>
> >> % Nassar's formula, N = 6, x[n] = x[N-n] for n=1:N-1
> >> fft([1 2 3 2 1 0])'
> ans =
> 9.0000
> -2.0000 + 3.4641i
> 0
> 1.0000
> 0
> -2.0000 - 3.4641i
> >> % "FFT-even" version, N = 6, x[n] = x[N-n+1] for n=2:N
> >> fft([0 1 2 3 2 1])'
> ans =
> 9
> -4
> 0
> -1
> 0
> -4
> >>
>
> Dale B. Dalrymple


ohhh!! Thank you for your help!!!!, :)

On Nov 1, 8:52 am, "Nasser M. Abbasi" <n...@12000.org> wrote:
> "dbd" <d...@ieee.org> wrote in message
>
> news:a220a52b-a008-4b16-acc7-0656a03275b3@w37g2000prg.googlegroups.com...
>
>
>
> > On Oct 31, 11:11 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> >> ...
> >> If the data sequence itself is real and even (i.e. x[n]=x[N-n]) where N
> >> is
> >> ...
> >> --Nasser
>
> > Let's try that:
>
> >>> % Nassar's formula, N = 6, x[n] = x[N-n] for n=1:N-1
> >>> fft([1 2 3 2 1 0])'
> > ans =
> > 9.0000
> > -2.0000 + 3.4641i
> > 0
> > 1.0000
> > 0
> > -2.0000 - 3.4641i
>
> Ofcourse, I said above n=1:N-1 which is clearly wrong (the number of points
> is N now N-1 points), I meant to write n=0:N-1 to make it N points. I made a
> typo starting it from 1 not 0.
>
> May be now you can try my formula? ;) but watch for the index.

No, Matlab indices start at 1. I applied your formula to 1:N-1 because
it doesn't work for N, but I did fft a set of N values. (These are
samples of a triangular waveform.)

Anyway, did you mean something like this?

>> fft([0 1 2 3 2 1 0])'
ans =
   9.0000
  -4.5489 + 2.1906i
   0.1920 - 0.2408i
  -0.1431 + 0.6270i
  -0.1431 - 0.6270i
   0.1920 + 0.2408i
  -4.5489 - 2.1906i
>>

These are not real. The point is that symmetry is not the same as "fft-
evenness".

>
> --Nasser

Dale B. Dalrymple

"dbd" <dbd@ieee.org> wrote in message
news:e605aa13-a63b-4032-8b9d-039cdd5efa7a@b36g2000prf.googlegroups.com...
>

> No, Matlab indices start at 1.

Thanks for letting me know that, I'll have to remember that.

>I applied your formula to 1:N-1 because
> it doesn't work for N, but I did fft a set of N values. (These are
> samples of a triangular waveform.)
>
> Anyway, did you mean something like this?
>
>>> fft([0 1 2 3 2 1 0])'
> ans =
> 9.0000
> -4.5489 + 2.1906i
> 0.1920 - 0.2408i
> -0.1431 + 0.6270i
> -0.1431 - 0.6270i
> 0.1920 + 0.2408i
> -4.5489 - 2.1906i
>>>
>


No, that is not what I mean. In the above, you have N=7, but "my formula"
said

x(n)=x(N-n) for n=0..N-1 which means x(7) is not valid part of the
sequence since 7 is more than 7-1 which is 6. So the sequence should have
been

x(0)=x(7) (n=0 case)
x(1)=x(6) (n=1 case)
x(2)=x(5) (n=2 case)
x(3)=x(4) (n=3 case)

So, assume, using your numbers, that x(0)=0, x(1)=1, x(2)=2, x(3)=3 one
gets

x(0)=0
x(1)=1
x(2)=2
x(3)=3
x(4)=3
x(5)=2
x(6)=1 and you STOP here, since n goes only up to N-1 whichis 6.

So now the x sequence is [0,1,2,3,3,2,1]

EDU>> fft([0 1 2 3 3 2 1])

ans =

   12.0000 -5.0489 -0.3080 -0.6431 -0.6431 -0.3080 -5.0489


So, I guess Nasser's formula works after all? ;)

--Nasser

On Nov 1, 11:19 am, "Nasser M. Abbasi" <n...@12000.org> wrote:
...
> ...
> No, that is not what I mean. In the above, you have N=7, but "my formula"
> said
>
> x(n)=x(N-n) for n=0..N-1 which means x(7) is not valid part of the

No. Your first post did not say n=0...N-1. In fact, you chose 1 to N
for the fft indices.

> sequence since 7 is more than 7-1 which is 6. So the sequence should have
> been
>
> x(0)=x(7) (n=0 case)
> x(1)=x(6) (n=1 case)
> x(2)=x(5) (n=2 case)
> x(3)=x(4) (n=3 case)
>
> So, assume, using your numbers, that x(0)=0, x(1)=1, x(2)=2, x(3)=3 one
> gets
>
> x(0)=0
> x(1)=1
> x(2)=2
> x(3)=3
> x(4)=3
> x(5)=2
> x(6)=1 and you STOP here, since n goes only up to N-1 whichis 6.
>
> So now the x sequence is [0,1,2,3,3,2,1]
>
> EDU>> fft([0 1 2 3 3 2 1])
>
> ans =
>
> 12.0000 -5.0489 -0.3080 -0.6431 -0.6431 -0.3080 -5.0489
>
> So, I guess Nasser's formula works after all? ;)

No Nasser, this is a Matlab group. There is no x(0). That is broken
here. :(

You gave 1 to N as the indices of the fft in your original post. That
was the thing you got right at first. :)

Your follow-up posts are different from the content of your original
post. This demonstrates that "There are often mistakes made" as I
originally posted to Oscar. Thank you for the extended
demonstration. :)

>
> --Nasser

Dale B. Dalrymple