conjugate symmetric & mirror image of the spectrum

 I have a small doubt:

regarding the spectrum of the signal,
if the sequence is real even symmetric seq, then it is enough if we
calculate spectrum for 0:N/2. since (N/2+1) to N-1 points are
redundnats data points.

My doubt is
outputs 0 and N/2 will be real and unique,
and outputs 1 to N/2-1 will be conjugate symmetric with outputs N-1 to
N/2+1.

whether mirror image is folded version of the original spectrum????
and whether mirror image is same as conjugate symmetric. for points(N/
2+1 to N-1)  ??????


thank you

On 11 Mar, 06:11, rramya <ramyarav...@yahoo.co.in> wrote:
> &#4294967295;I have a small doubt:
>
> regarding the spectrum of the signal,
> if the sequence is real even symmetric seq, then it is enough if we
> calculate spectrum for 0:N/2. since (N/2+1) to N-1 points are
> redundnats data points.
>
> My doubt is
> outputs 0 and N/2 will be real and unique,
> and outputs 1 to N/2-1 will be conjugate symmetric with outputs N-1 to
> N/2+1.

Seems correct so far.

> whether mirror image is folded version of the original spectrum????
> and whether mirror image is same as conjugate symmetric. for points(N/
> 2+1 to N-1) &#4294967295;??????

What do you mean by 'mirror image' and 'original spectrum'?

Rune

rramya wrote:
> 
>  I have a small doubt:
> 
> regarding the spectrum of the signal,
> if the sequence is real even symmetric seq, then it is enough if we
> calculate spectrum for 0:N/2. since (N/2+1) to N-1 points are
> redundnats data points.

Yes it should be obvious you can calculate the spectrum from half the
points in your sequence. 

If your sequence is even numbered of samples and looks like this:

  ....... d c b a a b c d ........

You should be able to calculate the zero phase spectrum as follows:

cos(a*pi*f)+cos(3*b*pi*f)+cos(5*c*pi*f)+cos(7*d*pi*f) .........

You would probably want to evaluate that function in the range of 0 to
pi or maybe -pi to pi. And you may want to divide the whole thing by N/2
to scale it such that f0 evaluates to the total sum of the sequence.


If you wanted to you could include the appropriate linear phase factor
and sample the function at N points to get the same result as the DFT.


> 
> My doubt is
> outputs 0 and N/2 will be real and unique,
> and outputs 1 to N/2-1 will be conjugate symmetric with outputs N-1 to
> N/2+1.


If you are talking about real even sequence as the input then the
spectrum is just symmetric. 

But as others have already pointed out that in order to put that sort of
time domain sequence thru a DFT and get the result you want you would
have to organize the sequence differently. 


> 
> whether mirror image is folded version of the original spectrum????
> and whether mirror image is same as conjugate symmetric. for points(N/
> 2+1 to N-1)  ??????

It is just symmetric - the symmetry of your sequence should produce no
imaginary terms.

-jim

jim wrote:

> If your sequence is even numbered of samples and looks like this:
> 
>   ....... d c b a a b c d ........
> 
> You should be able to calculate the zero phase spectrum as follows:
> 
> cos(a*pi*f)+cos(3*b*pi*f)+cos(5*c*pi*f)+cos(7*d*pi*f) .........

Ooops that is wrong

It should be 

a*cos(pi*f)+b*cos(3*pi*f)+c*cos(5*pi*f)+d*cos(7*pi*f) .........


-jim

On Tue, 10 Mar 2009 22:11:09 -0700, rramya wrote:

> I have a small doubt:
> 
> regarding the spectrum of the signal, if the sequence is real even
> symmetric seq, then it is enough if we calculate spectrum for 0:N/2.
> since (N/2+1) to N-1 points are redundnats data points.
> 
> My doubt is
> outputs 0 and N/2 will be real and unique, and outputs 1 to N/2-1 will
> be conjugate symmetric with outputs N-1 to N/2+1.
> 
> whether mirror image is folded version of the original spectrum???? and
> whether mirror image is same as conjugate symmetric. for points(N/ 2+1
> to N-1)  ??????
> 
> 
> thank you

The English of your last paragraph is a bit mangled, so I'm going to 
guess that this statement answers your question:

If your input signal is real and even (i.e. symmetric around zero, such 
that s(t) = s(-t)), then the spectrum will be real and even -- so you 
don't even have to worry about those imaginary parts.

If your input signal is real and odd (i.e. anti-symmetric around zero, 
such that s(t) = -s(t)), then the spectrum will be purely imaginary and 
off -- so you don't have to worry about those real parts.

If your input signal is just plain real, without necessarily being odd or 
even, then you can decompose it to an even part and an odd part (check 
this for yourself -- it's true).  Then from the above two statements, the 
spectrum will be even in it's real part, and odd in it's imaginary part 
-- in other words, the it'll have conjugate symmetry around zero.

-- 
http://www.wescottdesign.com

On Mar 11, 7:45&#4294967295;pm, Tim Wescott <t...@seemywebsite.com> wrote:
> On Tue, 10 Mar 2009 22:11:09 -0700, rramya wrote:
> > I have a small doubt:
>
> > regarding the spectrum of the signal, if the sequence is real even
> > symmetric seq, then it is enough if we calculate spectrum for 0:N/2.
> > since (N/2+1) to N-1 points are redundnats data points.
>
> > My doubt is
> > outputs 0 and N/2 will be real and unique, and outputs 1 to N/2-1 will
> > be conjugate symmetric with outputs N-1 to N/2+1.
>
> > whether mirror image is folded version of the original spectrum???? and
> > whether mirror image is same as conjugate symmetric. for points(N/ 2+1
> > to N-1) &#4294967295;??????
>
> > thank you
>
> The English of your last paragraph is a bit mangled, so I'm going to
> guess that this statement answers your question:
>
> If your input signal is real and even (i.e. symmetric around zero, such
> that s(t) = s(-t)), then the spectrum will be real and even -- so you
> don't even have to worry about those imaginary parts.
>
> If your input signal is real and odd (i.e. anti-symmetric around zero,
> such that s(t) = -s(t)), then the spectrum will be purely imaginary and
> off -- so you don't have to worry about those real parts.
>
> If your input signal is just plain real, without necessarily being odd or
> even, then you can decompose it to an even part and an odd part (check
> this for yourself -- it's true). &#4294967295;Then from the above two statements, the
> spectrum will be even in it's real part, and odd in it's imaginary part
> -- in other words, the it'll have conjugate symmetry around zero.
>
> --http://www.wescottdesign.com- Hide quoted text -
>
> - Show quoted text -

Actually , I studied ....
real DFT &#4294967295;assumes&#4294967295; a mirrored set of negative
frequencies due to the fact that the real DFT only ever transforms
real time domain signals and never complex ones (thus producing
mirrored negative frequencies).

then only doubt came into my mind is "whether mirrored negative freq
(spectrum)is exactly same as original spectrum."


 if we take sine wave when we take Fourier series, its Fourier series
coefficents we get the conjugate symmetric i.e,
                                                A       =  A *
                                                  -k            k

Whether mirrored negative freq coefficients is same as    A*
 
k
or its fourier coefficients is A
                                         k


Thank you.

Tim Wescott <tim@seemywebsite.com> wrote:
 
> If your input signal is real and even (i.e. symmetric around zero, such 
> that s(t) = s(-t)), then the spectrum will be real and even -- so you 
> don't even have to worry about those imaginary parts.
 
> If your input signal is real and odd (i.e. anti-symmetric around zero, 
> such that s(t) = -s(t)), then the spectrum will be purely imaginary and 
> off -- so you don't have to worry about those real parts.

I thought the OP was asking about the N/2 term.

That is, that the N point transform has (N/2)+1 cos terms
and (N/2)-1 sine terms.  

-- glen

On Mar 11, 11:41&#4294967295;pm, glen herrmannsfeldt <g...@ugcs.caltech.edu>
wrote:
> Tim Wescott <t...@seemywebsite.com> wrote:
> > If your input signal is real and even (i.e. symmetric around zero, such
> > that s(t) = s(-t)), then the spectrum will be real and even -- so you
> > don't even have to worry about those imaginary parts.
> > If your input signal is real and odd (i.e. anti-symmetric around zero,
> > such that s(t) = -s(t)), then the spectrum will be purely imaginary and
> > off -- so you don't have to worry about those real parts.
>
> I thought the OP was asking about the N/2 term.
>
> That is, that the N point transform has (N/2)+1 cos terms
> and (N/2)-1 sine terms. &#4294967295;
>
> -- glen

actually i cant trace out THE WORD "MIRRORED NEGATIVE SET OF FREQ
" (FROM N/1+1 TO N-1)

THANKYOU

On Mar 12, 7:11&#4294967295;am, rramya <ramyarav...@yahoo.co.in> wrote:
> On Mar 11, 11:41&#4294967295;pm, glen herrmannsfeldt <g...@ugcs.caltech.edu>
> wrote:
>
> > Tim Wescott <t...@seemywebsite.com> wrote:
> > > If your input signal is real and even (i.e. symmetric around zero, such
> > > that s(t) = s(-t)), then the spectrum will be real and even -- so you
> > > don't even have to worry about those imaginary parts.
> > > If your input signal is real and odd (i.e. anti-symmetric around zero,
> > > such that s(t) = -s(t)), then the spectrum will be purely imaginary and
> > > off -- so you don't have to worry about those real parts.
>
> > I thought the OP was asking about the N/2 term.
>
> > That is, that the N point transform has (N/2)+1 cos terms
> > and (N/2)-1 sine terms. &#4294967295;
>
> > -- glen
>
> actually i cant trace out THE WORD "MIRRORED NEGATIVE SET OF FREQ
> " (FROM N/1+1 TO N-1)
>
> THANKYOU

but still i can't get the meaning of " mirorred set of negative freq"
What do you infer from the above statement.

if we take the Fourier series of cosine waveform (even symmetry)
Then Fourier series coefficients A = A
                                               -k     k

whether we can say it as a mirrored set of negative freq. ( whose
Fourier coefficient values is same as that of positive k fourier
coefficients) especially when the signal is even signal. ???????



if we take fourier series of sinewaveform (odd symmetry)
Then Fourier series coefficients A = A*
                                               -k     k
here only imaginary part of the A is conjugated.
                                               k

Whether we can say it as a conjugate symmetric set of negative
frequecies (whose fourier coefficient values is complex conjugate of
positive k fourier coefficients) especially when signal is
odd. ???????

As we all know, digital dreq lies bet -pi to pi or (0,2*pi).
if signal is real & even , it is enough if we calculate o to pi
then from pi to 2*pi or (0 to -pi), the spectrum will be folded with
respect to pi (or with respect to 0)
to extend up to 2*pi ( - pi).

My question is :
whether mirored set of negative freq is same as positive freq (range
of positive freq= 0 to pi,range of negative freq -pi to 0 )which when
folded w.r.to zero freq???????? ( if incase signal is even)


Thank you.

rramya wrote:
> 
> On Mar 12, 7:11 am, rramya <ramyarav...@yahoo.co.in> wrote:
> > On Mar 11, 11:41 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu>
> > wrote:
> >
> > > Tim Wescott <t...@seemywebsite.com> wrote:
> > > > If your input signal is real and even (i.e. symmetric around zero, such
> > > > that s(t) = s(-t)), then the spectrum will be real and even -- so you
> > > > don't even have to worry about those imaginary parts.
> > > > If your input signal is real and odd (i.e. anti-symmetric around zero,
> > > > such that s(t) = -s(t)), then the spectrum will be purely imaginary and
> > > > off -- so you don't have to worry about those real parts.
> >
> > > I thought the OP was asking about the N/2 term.
> >
> > > That is, that the N point transform has (N/2)+1 cos terms
> > > and (N/2)-1 sine terms.
> >
> > > -- glen
> >
> > actually i cant trace out THE WORD "MIRRORED NEGATIVE SET OF FREQ
> > " (FROM N/1+1 TO N-1)
> >
> > THANKYOU
> 
> but still i can't get the meaning of " mirorred set of negative freq"
> What do you infer from the above statement.
> 
> if we take the Fourier series of cosine waveform (even symmetry)
> Then Fourier series coefficients A = A
>                                                -k     k
> 
> whether we can say it as a mirrored set of negative freq. ( whose
> Fourier coefficient values is same as that of positive k fourier
> coefficients) especially when the signal is even signal. ???????
> 
> if we take fourier series of sinewaveform (odd symmetry)
> Then Fourier series coefficients A = A*
>                                                -k     k
> here only imaginary part of the A is conjugated.
>                                                k
> 
> Whether we can say it as a conjugate symmetric set of negative
> frequecies (whose fourier coefficient values is complex conjugate of
> positive k fourier coefficients) especially when signal is
> odd. ???????
> 
> As we all know, digital dreq lies bet -pi to pi or (0,2*pi).


What you are asking is not easy to understand.... 

As far as  can tell the root of the confusion is a lack of any reference
frames. 

Use fixed width fonts. This would give your ascii symbolism a meaningful
frame of reference.

You talk about cosines and sines and their symmetry. These are
continuous functions. If you digitize a sinusoid then you need to
establish a frame of reference to avoid confusion. You need to call some
point in the sequence zero.

Here are 4 digital sequences where length N=4. If you think that one of
them represents a cosine of frequency fs/4 which one is it?

 1  1  0  0
-1  1  1 -1
 1  0 -1  0

The fact is any one of those could be a sampled cosine wave. It all just
depends where you think the point in the sequence is that  you refer to
as 0. Frankly I don't know which one of these (if any) you would pick to
be a cosine. 

The same is true when you speak of symmetry. Let's assume that the
sequence repeats over and over endlessly, when you talk about symmetry I
don't know where you consider the point of reference to be for that
symmetry. 

-jim








> if signal is real & even , it is enough if we calculate o to pi
> then from pi to 2*pi or (0 to -pi), the spectrum will be folded with
> respect to pi (or with respect to 0)
> to extend up to 2*pi ( - pi).
> 
> My question is :
> whether mirored set of negative freq is same as positive freq (range
> of positive freq= 0 to pi,range of negative freq -pi to 0 )which when
> folded w.r.to zero freq???????? ( if incase signal is even)
> 
> Thank you.