FFT result not symetrical?

Quick question, I'm testing one of my own FFT routines with an input as a
complex waveform, I'm not sure what kind of output i'm suppose to get, but
one of the most obvious features is that the real part of the result is not
symmetrical ... is it possible to get a FFT result with such a feature or
is my routine incorrect?  (This may seem like a really silly question!)

If I input some samples with just the real part set, it behaves as I would
expect, and if I perform a Inverse FFT I get the original complex samples
returned, with no errors.

The input is generated from something called a gabor waveform, the real
input is (16 samples);

-0.04321392,
-0.03453337,
0.12078786,
0.27080447,
0.00000000,
-0.59395005,
-0.58104729,
0.36435210,
1.00000000,
0.36435210,
-0.58104729,
-0.59395005,
0.00000000,
0.27080447,
0.12078786,
-0.03453337,

and the imag. part is;

0.00000000,
-0.08337094,
-0.12078786,
0.11217089,
0.45593813,
0.24602216,
-0.58104729,
-0.87962379,
0.00000000,
0.87962379,
0.58104729,
-0.24602216,
-0.45593813,
-0.11217089,
0.12078786,
0.08337094,

Any help / input greatly appreciated :)

moosedude wrote:
> Quick question, I'm testing one of my own FFT routines with an input as a
> complex waveform, I'm not sure what kind of output i'm suppose to get, but
> one of the most obvious features is that the real part of the result is not
> symmetrical ... is it possible to get a FFT result with such a feature or
> is my routine incorrect?  (This may seem like a really silly question!)
>
> If I input some samples with just the real part set, it behaves as I would
> expect, and if I perform a Inverse FFT I get the original complex samples
> returned, with no errors.

Yes, what you see is expected. The reason is that Euler's equations no
longer apply for complex-valued signals. As you know, real-valued sines

and cosines can be expressed as

2*i*sin(x) = exp(ix) - exp(-ix)
2*  cos(x) = exp(ix) + exp(-ix)

i.e. the complex exponentials come in conjugate symmetric pairs.
These relations are only valid for real-valued signals, and are the
reasons why the DFT of a real-valued signals is conjugate
symmetric. 

Rune

On Fri, 07 Jul 2006 09:34:11 -0500, "moosedude"
<phill_holland@hotmail.com> wrote:

>
  (snipped)
>
>Any help / input greatly appreciated :)

Hello moosehead,

  Rune Allnor is, of course, correct.

Maybe the following website will 
help you a little.  Who knows.

www.dspguru.com/info/tutor/QuadSignals.pdf

Good Luck,
[-Rick-]