Mirror Image of Magnitude Response

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<adSdna95XZ1SS6nd4p2dnA@centurytel.net>...
> "Adel" <tadam@pisem.net> wrote in message
> news:c8d8c374.0402190218.2dec89ed@posting.google.com...
> > Hello, Fred.
> >
> > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
>  news:<C7idnbLi_4KCyq_d4p2dnA@centurytel.net>...
> > >
> > > Ihtiaq,
> > >
> > > Just to make sure: I think you're saying that you have an antisymmetric
> > > impulse response that, if centered at t=0, results in a purely imaginary
> > > frequency response H.  Right?  Aligning the center at t=0 is the same as
> >
> > If such an impulse response is not centered at t=0 it anyway results
> > in a purely imaginary frequency response H, doesn't it?
> 
> ***No.  If such an impulse response is not centered at t=0 then it has an
> Even part and an Odd (antisymmetric) part.
> By superposition, the Even part transforms to a real, even function and the
> Odd part transforms to an imaginary, odd (antisymmetric) function.
> If it's centered at zero, the Even part goes away.

Hello, Fred.

Yes, I am wrong here,
I have meant to say that the __DFT__ views the time domain as
circular, with sample zero inherently connected to sample N-1. Any
signal that is symmetrical / asymmetrical around sample zero will also
be symmetrical / asymmetrical around sample N/2, and vice versa. When
using members of the Fourier Transform family that do not view the
time domain as periodic (such as the DTFT), the asimmetry must be
around sample zero to produce a zero real part of frequency response.
Original question of I.R.Khan was about DTFT of course. So you are
write.
I am sorry.

Adel.

"I. R. Khan" <ir_khan@hotmail.com> wrote in message
news:c14a6g$1e6r72$1@ID-198607.news.uni-berlin.de...
> Fred, you are absolutely right. 1- Abs[H] =  1 + j H is not true for every
> H. Later Andor also realized it as he mentioned it here:
> > > > >
> > > > > I just noticed it's not quite so easy. While 1 + j H is real only,
> it
> > > > > might be greater than 1, whereas 1 - Abs[H] can never be greater
> than
> > > > > 1. This approach only works if you know that sign(j H(w)) = const.
>
> If H is all positive then 1- Abs[H] =  1 + j H.
> For example, if H = [j/2, j, j/2]
> then 1-Abs[H] = [1/2, 0, 1/2]
> and 1 + j H = [1/2, 0, 1/2]

Ishtiaq,

OK.  If you limit H to being positive and imaginary then you have to say:

1-Abs[H} = 1 + jH; for H imaginary and >=j0

because Abs[H] = -jH; for H imaginary and >=j0 ... and jH=-Abs[H]

so  1 +jH = 1 - Abs[H]; H imaginary >=j0

()therwise, the example might just be a singular point where the
*unqualified* expression happens to hold.
So you have to qualify the identities to be correct and complete.
If that qualification was given, it was elsewhere.....

Fred

"Adel" <tadam@pisem.net> wrote in message
news:c8d8c374.0402200002.481de512@posting.google.com...
> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:<adSdna95XZ1SS6nd4p2dnA@centurytel.net>...
> > "Adel" <tadam@pisem.net> wrote in message
> > news:c8d8c374.0402190218.2dec89ed@posting.google.com...
> > > Hello, Fred.
> > >
> > > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
> >  news:<C7idnbLi_4KCyq_d4p2dnA@centurytel.net>...
> > > >
> > > > Ihtiaq,
> > > >
> > > > Just to make sure: I think you're saying that you have an
antisymmetric
> > > > impulse response that, if centered at t=0, results in a purely
imaginary
> > > > frequency response H.  Right?  Aligning the center at t=0 is the
same as
> > >
> > > If such an impulse response is not centered at t=0 it anyway results
> > > in a purely imaginary frequency response H, doesn't it?
> >
> > ***No.  If such an impulse response is not centered at t=0 then it has
an
> > Even part and an Odd (antisymmetric) part.
> > By superposition, the Even part transforms to a real, even function and
the
> > Odd part transforms to an imaginary, odd (antisymmetric) function.
> > If it's centered at zero, the Even part goes away.
>
> Hello, Fred.
>
> Yes, I am wrong here,
> I have meant to say that the __DFT__ views the time domain as
> circular, with sample zero inherently connected to sample N-1. Any
> signal that is symmetrical / asymmetrical around sample zero will also
> be symmetrical / asymmetrical around sample N/2, and vice versa. When
> using members of the Fourier Transform family that do not view the
> time domain as periodic (such as the DTFT), the asimmetry must be
> around sample zero to produce a zero real part of frequency response.
> Original question of I.R.Khan was about DTFT of course. So you are
> write.
> I am sorry.

Adel,

No need to be sorry - we're all just trying to learn something.

I guess I have to respond by saying that it doesn't matter if the frequency
function is sampled or not (and thus, whether the time function is periodic
or not).  The discussion about time shifting doesn't really change - except
the time shift eventually gets back to the same point if it's a circular /
periodic function.  So, the pictures could have been on a circle instead of
an infinite line.

Fred