Real and Complex FFT responses

Rune Allnor wrote:

> stenasc@yahoo.com (Bob) wrote in message news:<20540d3a.0308190001.7afc6596@posting.google.com>...
> > Hi Folks
> >
> > Just a quick question to verify if I'm correct.
> >
> > In a real only FFT, I get a mirror image of the peaks in the output
> > symmetrical
> > about the half way number of points of the FFT. I know this is
> > correct.
> >
> > However with the complex FFT, I only get the peaks(on one side
> > only)...no mirror image. They are at the correct position. Can someone
> > tell me if this is correct or not, and explain how this works?
> >
> > As usual, thanks in advance for all replies.
> >
> > Bob
>
> What you see is correct. To see why, check out the chapter on the
> properties of the Fourier transform that is found in almost all basic
> texts on DSP or Fourier analysis. There you will find an example
> that amounts to
>
>    F(-w) = F'(w)   for f(t) real
>
> where F(w) is the Fourier transform of f(t) and F' is the complex
> conjugate of F.
>
> So the "mirror" appears because of the "real-valuedness" of f(t)
> and is not a property of the Fourier tranform as such. Complex-valued
> data give no mirror images.
>
> Rune

Sounds right since the Fourier TF of a complex frequency exp(I.w0.t) = delta(w-w0) ie a single impulse

 - not two. If you take the FT of cos(wot) you will get two impulses however.

Tom

"Bernhard Holzmayer" <holzmayer.bernhard@deadspam.com> a &#4294967295;crit dans le
message news: 2296981.FC1uc1BQzq@holzmayer.ifr.rt...
> Bernhard Holzmayer wrote:
>
> > Bob wrote:
> >
> >> Hi Folks
> >>
> >> Just a quick question to verify if I'm correct.
> >>
> >> In a real only FFT, I get a mirror image of the peaks in the
> >> output symmetrical
> >> about the half way number of points of the FFT. I know this is
> >> correct.
> >>
> >> However with the complex FFT, I only get the peaks(on one side
> >> only)...no mirror image. They are at the correct position. Can
> >> someone tell me if this is correct or not, and explain how this
> >> works?
> >>
> >> As usual, thanks in advance for all replies.
> >>
> >> Bob
> >
> > A very impressive demonstration (at least for me) is this:
> > Given signals
> > m1(t) = cos(t) + j*sin(t) and
> > m2(t) = cos(t) - j*sin(t)  (conjugate of m1)
> > FFT(m1) gives you exactly one peak in the frequency domain,
> > FFT(m2) gives another peak.
> > FFT(m1+m2) gives both peaks (superposition, clear)
> > Superposition in the time domain gives:
> > m1(t)+m2(t) = 2cos(t), because the imaginary parts compensate.
> >
> > So you end up with the real function giving two peaks in the
> > frequency domain.
> > If you have access to Matlab, try it:
> >
> > clear i;
> > t=[ 0; 0.1; 100];
> > m1=cos(t)+i*sin(t)
> > m2=cos(t)-i*sin(t)
> > plot(t,fft(m1));
> > plot(t,fft(m2));
> > plot(t,fft(m1+m2));
> >
> > Bernhard
> >
> Sorry, I sent too fast :-)
> It's the same thing for real and imaginary part of the FFT.
> If you do the above, you get the result only by evaluation of the
> real part of the FFT.
>  clear i;
>  t=[ 0; 0.1; 100];

for me , I have to type :
t=[ 0: 0.1: 100];


>  m1=cos(t)+i*sin(t)
>  m2=cos(t)-i*sin(t)
>  plot(t,imag(fft(m1)));
>  plot(t,imag(fft(m2)));
>  plot(t,imag(fft(m1-m2)));
>
> The other sign in the last line comes from this:
> m1-m2 = cos(t)-cos(t)+j*sin(t)-(-j*sin(t)) = 2jsin(t)
>
> Bernhard
>
> --
> before sending to the above email-address:
> replace deadspam.com by foerstergroup.de

Lotfi wrote:

> for me , I have to type :
> t=[ 0: 0.1: 100];

Thanks for the correction. Sorry, I typed it manually - 
Matlab is on the other PC.. not here.
So you get this:

clear i;
t=[ 0: 0.1: 100];

% for real part:

m1=cos(t)+i*sin(t)
m2=cos(t)-i*sin(t)
plot(t,fft(m1));
plot(t,fft(m2));
plot(t,fft(m1+m2));

% and, for imag part:

 m1=cos(t)+i*sin(t)
 m2=cos(t)-i*sin(t)
 plot(t,imag(fft(m1)));
 plot(t,imag(fft(m2)));
 plot(t,imag(fft(m1-m2)));


Bernhard

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On Thu, 11 Sep 2003 12:11:57 +0200, Bernhard Holzmayer
<holzmayer.bernhard@deadspam.com> wrote:

   (snipped)
>
>clear i;
>t=[ 0: 0.1: 100];
>
>% for real part:
>
>m1=cos(t)+i*sin(t)
>m2=cos(t)-i*sin(t)
>plot(t,fft(m1));
>plot(t,fft(m2));
>plot(t,fft(m1+m2));
>
>% and, for imag part:
>
> m1=cos(t)+i*sin(t)
> m2=cos(t)-i*sin(t)
> plot(t,imag(fft(m1)));
> plot(t,imag(fft(m2)));
> plot(t,imag(fft(m1-m2)));
>
>
>Bernhard

Hi Bernhard,

    For anyone new to signal processing, 
it's *VERY *VERY* useful to experiment with 
code like yours.  It helps give physical 
meaning to Euler's equations.  It helps 
make all of these 'real' and 'imaginary' 
signals "real".   :-)

[-Rick-]
 

Rick Lyons wrote:

> On Thu, 11 Sep 2003 12:11:57 +0200, Bernhard Holzmayer
> <holzmayer.bernhard@deadspam.com> wrote:
> 
>    (snipped)
>>
>>clear i;
>>t=[ 0: 0.1: 100];
>>
>>% for real part:
>>
>>m1=cos(t)+i*sin(t)
>>m2=cos(t)-i*sin(t)
>>plot(t,fft(m1));
>>plot(t,fft(m2));
>>plot(t,fft(m1+m2));
>>
>>% and, for imag part:
>>
>> m1=cos(t)+i*sin(t)
>> m2=cos(t)-i*sin(t)
>> plot(t,imag(fft(m1)));
>> plot(t,imag(fft(m2)));
>> plot(t,imag(fft(m1-m2)));
>>
>>
>>Bernhard
> 
> Hi Bernhard,
> 
>     For anyone new to signal processing,
> it's *VERY *VERY* useful to experiment with
> code like yours.  It helps give physical
> meaning to Euler's equations.  It helps
> make all of these 'real' and 'imaginary'
> signals "real".   :-)
> 
> [-Rick-]

Thanks, I appreciate your opinion.

Bernhard
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