Reply by Fred Marshall●February 20, 20042004-02-20
"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
Reply by Fred Marshall●February 20, 20042004-02-20
"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
Reply by Adel●February 20, 20042004-02-20
"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.
Reply by Andor●February 20, 20042004-02-20
I. R. Khan wrote:
> > 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.
>
> Between 0 and fs/2, H is all -ive, but for fs/2 to fs, it is all positive.
> Yes you are right,
>
> > BTW: If you still want to look at the Hilbert Transform, you can do
> > that in Matlab, for example.
>
>
> For h = [-1 0 1]
> Matlab gives hilbert(h) = [-1 j 1];
> seems that Hilbert Transform of a real sequence is not real.
The name of this function is misleading. hilbert(h) uses the Hilbert
Transform to calculate the appropriate imaginary part to make an
analytic vector with real part h.
So imag(hilbert(h)) is the actual Hilbert Transform of h. This is just
a wide-band 90� phase shift (= arg(j) ). In your case, [0 1 0] is the
Hilbert Transform of [-1 0 1].
Regards,
Andor
Reply by I. R. Khan●February 20, 20042004-02-20
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]
Regards,
Ishtiaq.
Reply by Fred Marshall●February 20, 20042004-02-20
"I. R. Khan" <ir_khan@hotmail.com> wrote in message
news:c13uk2$1debqf$1@ID-198607.news.uni-berlin.de...
> > >
> > > 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.
> > >
> >
> > I'm confused. It appears you're using an identity that's a new one:
> >
> > H complex
> > Abs[H] complex ??????
> >
> > From this follows:
> >
> > H real
> > Abs[H] real .... OK
> >
> > H imaginary
> > Abs[H] imaginary ?????
> >
> > Here are simple counter examples:
> >
> > H=1
> > Abs[H]=1
> >
> > H=j
> > Abs[H]=1
> >
> > H=1+j
> > Abs[H]=sqrt[2] a real scalar quantity.
>
> Fred, I don't understand where you are confused. H is imaginary so jH is
> real, and so is Abs[H]. But jH = Abs[H] iff H has all -ive.
>
Ishtiaq,
I wasn't so much confused as I was concerned by what Andor said:
>But if H is purely imaginary, ie. H[w] = j Hr[w], where Hr is real,
>then
>1- Abs[H] = 1 + j H
I believe this is incorrect. Perhaps it's notation. Let's see:
H is purely imaginary - let's call it H=jP where P is real. I just used "P"
instead of "Hr" to avoid confusion.
jH is then purely real and has the opposite sign of P, i.e. jH = (j^2)*P
= -P
1 - Abs[H] = 1 - Abs[jP] = 1 - Abs[P] ... the latter because sgn(P) goes
away with Abs and so does j.
at the same time:
1 + jH = 1 - P which is *not* the same as 1 - Abs[P] = 1 - Abs[H]
I think the issue here is not so much whether jH is real, it's that Abs
takes away all the negative values.
If I've gone wrong, where is it?
Fred
Reply by I. R. Khan●February 19, 20042004-02-19
> 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.
Between 0 and fs/2, H is all -ive, but for fs/2 to fs, it is all positive.
Yes you are right,
> BTW: If you still want to look at the Hilbert Transform, you can do
> that in Matlab, for example.
For h = [-1 0 1]
Matlab gives hilbert(h) = [-1 j 1];
seems that Hilbert Transform of a real sequence is not real.
Regards,
Ishtiaq.
Reply by I. R. Khan●February 19, 20042004-02-19
> >
> > 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.
> >
>
> I'm confused. It appears you're using an identity that's a new one:
>
> H complex
> Abs[H] complex ??????
>
> From this follows:
>
> H real
> Abs[H] real .... OK
>
> H imaginary
> Abs[H] imaginary ?????
>
> Here are simple counter examples:
>
> H=1
> Abs[H]=1
>
> H=j
> Abs[H]=1
>
> H=1+j
> Abs[H]=sqrt[2] a real scalar quantity.
Fred, I don't understand where you are confused. H is imaginary so jH is
real, and so is Abs[H]. But jH = Abs[H] iff H has all -ive.
Ishtiaq.
Reply by Fred Marshall●February 19, 20042004-02-19
"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.
>
> > rotating or shifting the impulse response by half its length (length
> > measured either in time or in samples = time).
> >
> > Assuming that H is periodic at fs, I believe you will also find that
> > .. H is zero at f=0 and f=fs
> > .. the real part of H is symmetric
>
> .. and also real part of H is zero I think
Adel,
No. As above. Shifting in time introduces an even or odd part like this:
Take any arbitrary f(t) that is not either all even or all odd. You can
split it into even and odd parts like this:
fe(t)= [f(t)+f(-t)]/2
fo(t)= [f(t)-f(-t)]/2
|
|
1.0 +---------+---------+
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
-----------------------+---------+---------+---------------------
-T T
time -> 0
Even function of time f(t) ^
|
|
1.0 +-------------------+
| |
| |
| |
| |
| |
| |
| |
| |
| |
| |
---------------------------------+-------------------+-----------
time -> 0 2T
Same function of time, shifted is no longer even f(t-T) ^
|
|
0.5+-------------------+-------------------+
| | |
| | |
| | |
| | |
| | |
-------------+-------------------+-------------------+-----------
-2T
time -> 0 2T
The even part of f(t-T) ^
|
|
+-------------------+0.5
| |
| |
| |
| |
| |
--------------+------------------0-------------------+-----------
-2T| |
|time -> | 2T
| |
| |
| |
-0.5+------------------+
The odd part of f(t-T)
The sum of the even part and odd part yield f(t-T) ^
So, the real and imaginary parts of a Fourier Transform do change if a real
time signal is shifted relative to where you decide to define t=0. Only if
you shift t=0 along with it does it stay the same - in which case you have a
"floating" definition of your time frame. The only exception one might note
is a periodic (thus infinite) waveform where you may or may not deal with
ambiguities of time shifts in increments of 2*pi radians.
Fred
Reply by Fred Marshall●February 19, 20042004-02-19
"Andor" <an2or@mailcircuit.com> wrote in message
news:ce45f9ed.0402190352.3d5aa74d@posting.google.com...
> I. R. Khan wrote:
> > > But if H is purely imaginary, ie. H[w] = j Hr[w], where Hr is real,
> > > then
> > >
> > > 1- Abs[H] = 1 + j H
> > >
> > > That would mean you get the coefficients of the new filter by taking
> > > the Hilbert-Transform of the original filter h and add 1.0 to the
> > > middle coefficient of the resulting filter (make sure the length of
> > > this new filter is odd).
> > >
> > > What do you think?
> >
> > Thanks Andor. I think this should work. You are right. I need 1+jH,
which
> > will be real because H is imaginary. If you can kindly give me some
> > information on how to get the Hilbert-Transform of the original filter
h, I
> > shall be thankful. For example, what will be the Hilbert-Transform of
> > {-1/6, -1/2, 0, 1/2, 1/6}.
>
> 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.
>
I'm confused. It appears you're using an identity that's a new one:
H complex
Abs[H] complex ??????
From this follows:
H real
Abs[H] real .... OK
H imaginary
Abs[H] imaginary ?????
Here are simple counter examples:
H=1
Abs[H]=1
H=j
Abs[H]=1
H=1+j
Abs[H]=sqrt[2] a real scalar quantity.