Re: the spectrum of symmetric vector

On 3月7日, 上午12时16分, Andor <andor.bari...@xxxxxxxxxx> wrote:
On 6 Mrz., 15:22, HyeeWang <hyeew...@xxxxxxxxx> wrote:





On 3月5日, 下午5时53分, Andor <andor.bari...@xxxxxxxxx> wrote:

On 5 Mrz., 07:25, HyeeWang <hyeew...@xxxxxxxxx> wrote:

On Mar 4, 10:36 pm, Andor <andor.bari...@xxxxxxxxx> wrote:

On 4 Mrz., 03:36, HyeeWang <hyeew...@xxxxxxxxx> wrote:

Help. Where it go wrong?

Get A N length even symmetric vector ,Fourier transform it and result
a spectrum vector with the same length
N,which the imaginary part must be zero.
In contrast to that, A N length odd symmetric vector should result
that the real part of spectrum is zero.

But I test and perform it and attain a wrong result. Where it go
wrong?

Here is matlab script to illustrate it.

N = 64;
x = randn(N,1);
y = [x;flipud(x)];
z = [x;-flipud(x)];

You are making a very common mistake. Even and DFT-even is not the
same thing. If you want a DFT-even sequnce of length 128 you have 65
values that you can specify, the other 63 follow by symmetry. Check
out this thread:

http://groups.google.com/group/comp.dsp/browse_frm/thread/8fa3856859c...

Also, if you want a DFT-odd vector, remember that the middle value
(Nyquist value, only if you have length(z) even) has to be zero.

Regards,
Andor- Hide quoted text -

- Show quoted text -

Thank you.Andor

I refered the link and got a lot.

Maybe I did not catch your comments fully. But I do not agree with
you.

1. Here, we talk symmetry in time domain,not in frequecny domain. So
it is none of the businesss with DFT-even.

The domain doesn't matter if you talk about symmetry for the DFT.
Consider your example:

N = 64;
x = randn(N,1);
y = [x;flipud(x)];
z = [x;-flipud(x)];

Y =fft(y);
Z =fft(z);
[Y,Z]

When you look at Y or Z, they are not purely real and imaginary as you
expect. The reason is that you are not using DFT symmetry. A real, DFT-
even vector has a real DFT-even spectrum. Your Y isn't real. Try this:

N = 64;
x1 = randn(N+1,1);
x2 = flipud(x1(2:end-1));
x = [x1;x2]
X =fft(x)

Notice that for a vector of length 128 you may specify 65 values (x1)
and the other 63 values (x2) follow from the necessary symmetry. X
has the same symmetry like x.

Regards,
Andor- 隐藏被引用文字 -

- 显示引用的文字 -

Andor. Thank you.

The topic is about that even symmetrical sequence produce real
spectrum.
What you talk is about that DFT-even symmetrical sequence prodeuce
real spectrum.

Both above are correct.

Although what you said do not correspond with the topic,you also get
the equivalent result.

Excuse me?

You start off this thread by (correclty) noting that an even vector
does not have a real spectrum, and you ask why that is so. I reply by
saying that only DFT-even vectors have a real spectrum. Then you say
that wasn't the question, but you agree that a DFT-even vector has a
real spectrum and suddenly claim that even vectors also have real
spectrum (which, just two days ago, you noted not to be the case).
Finally, you say that these two claims are equivalent, although they
obviously are not.

Doesn't this strike you as a little bit odd, too? (pun intended :-).

Regards,
Andor- 隐藏被引用文字 -

- 显示引用的文字 -

hi,Andor,thank you.

From what you talked,at begining,you have mistaken me and mistaken
what the topic is.

--- but you agree that a DFT-even vector has a
real spectrum and suddenly claim that even vectors also have real
spectrum (which, just two days ago, you noted not to be the case).

I never deny "even vectors also have real spectrum ". It is a basic
property of DFT.
You can retrieve it from any DSP textbook.

I started this thread because I wanna confirm the property and seek
some help to account for my problem.

You can read it again for your convenience.


Thank Rune for giving me help in solving it...
.

Relevant Pages

  • Re: the spectrum of symmetric vector
    ... N,which the imaginary part must be zero. ... that the real part of spectrum is zero. ... The domain doesn't matter if you talk about symmetry for the DFT. ... The example which you taked is a real DFT-even spectrum result. ...
    (comp.dsp)
  • Re: the spectrum of symmetric vector
    ... N,which the imaginary part must be zero. ... A N length odd symmetric vector should result ... that the real part of spectrum is zero. ... Here, we talk symmetry in time domain,not in frequecny domain. ...
    (comp.dsp)
  • Re: the spectrum of symmetric vector
    ... N,which the imaginary part must be zero. ... A N length odd symmetric vector should result ... that the real part of spectrum is zero. ... The reason is that you are not using DFT symmetry. ...
    (comp.dsp)
  • Re: conjugate symmetric & mirror image of the spectrum
    ... then it is enough if we calculate spectrum for 0:N/2. ... If your input signal is real and even (i.e. symmetric around zero, ... If your input signal is real and odd (i.e. anti-symmetric around zero, ... the it'll have conjugate symmetry around zero. ...
    (comp.dsp)
  • Re: the spectrum of symmetric vector
    ... N,which the imaginary part must be zero. ... that the real part of spectrum is zero. ...    For the symmetry argument to work, ... point concept about symmetry. ...
    (comp.dsp)