Mirror Image of Magnitude Response

I have a real impulse response h of an FIR filter, and its frequency
response H is pure imaginary (if I rotate h to left by half of its length).
I want the impulse response of a filter whose magnitude response is
1-Abs[H]. Is it possible to do this with some trick on h?

Regards,
Ishtiaq.

I. R. Khan wrote:

 > I have a real impulse response h of an FIR filter, and its frequency
 > response H is pure imaginary (if I rotate h to left by half of its 
length).
 > I want the impulse response of a filter whose magnitude response is
 > 1-Abs[H]. Is it possible to do this with some trick on h?
 >
 > Regards,
 > Ishtiaq.
 >

I don't offhand know. Rotation is measured by angle. How should I
interpret rotating by a length?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

"I. R. Khan" <ir_khan@hotmail.com> wrote in message
news:c0sou8$1atqq3$1@ID-198607.news.uni-berlin.de...
> I have a real impulse response h of an FIR filter, and its frequency
> response H is pure imaginary (if I rotate h to left by half of its
length).
> I want the impulse response of a filter whose magnitude response is
> 1-Abs[H]. Is it possible to do this with some trick on h?

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
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
.. the imaginary part of H is antisymmetric (around f=0).
.. Therefore,the imaginary part has to be zero at f=0, f=fs/2,f=fs

So far, these are just a truisms.

Now you want a filter whose magnitude response is 1-Abs[H] where H is the
one mentioned above.
Well, Abs[H] is a real, even function isn't it?  It also has zeros at f=0
and f=fs.
So, 1-Abs[H] is also real and even because 1 is real and even and real even
+/- another real even = a new real even, right?
Only now, instead of there being zeros at f=0 and f=fs, the value will be
1.0.  All the other values are arbitrary I believe.

So, you can figure it out using an FFT/IFFT process easily enough.
However, ABS in frequency has no dual in time as far as I know.
So, the answer to your question is: In general, No.

Fred

Thanks Fred.

> 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
> .. the imaginary part of H is antisymmetric (around f=0).
> .. Therefore,the imaginary part has to be zero at f=0, f=fs/2,f=fs

Yes, yes, yes and yes. God! how did you know?

 For example h = {1/(2pi), 0, -1/(2pi)}, and H is pure imaginary if centered
at t=0.
I want to shift H up by 1, and can do that by replacing zero in h by j
(=sqrt(-1)). I was wondering if it is possible to do it keeping the
coefficients all real.

This leads to another question. Is the impulse response h = {1/(2pi),
j, -1/(2pi)} of some importance to real inputs? I seems it will change them
to complex.

Infact h = {1/(2pi), j, -1/(2pi)} is a differentiator highly accurate at
fs/2, but I want a similar differentiator with real coefficients.

Regards,
Ishtiaq.

> I don't offhand know. Rotation is measured by angle. How should I
> interpret rotating by a length?
>
> Jerry

Jerry, I was using Mathematica that time and I just translated the follwoing
command:

h = RotateLeft[h, Length[h]/2];

Regards,
Ishtaiq.

I. R. Khan wrote:

>>I don't offhand know. Rotation is measured by angle. How should I
>>interpret rotating by a length?
>>
>>Jerry
> 
> 
> Jerry, I was using Mathematica that time and I just translated the follwoing
> command:
> 
> h = RotateLeft[h, Length[h]/2];
> 
> Regards,
> Ishtaiq.

I suppose that is a circumferential length on a number circle?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

"Jerry Avins" <jya@ieee.org> wrote in message
news:4032e7cc$0$3072$61fed72c@news.rcn.com...
> I. R. Khan wrote:
>
> >>I don't offhand know. Rotation is measured by angle. How should I
> >>interpret rotating by a length?
> >>
> >>Jerry
> >
> >
> > Jerry, I was using Mathematica that time and I just translated the
follwoing
> > command:
> >
> > h = RotateLeft[h, Length[h]/2];
> >
> > Regards,
> > Ishtaiq.
>
> I suppose that is a circumferential length on a number circle?

If h={a,b,c,d,e,f}
then
Length[h] = 6
and
RotateLeft[h,3] = {d,e,f,a,b,c}

Regards,
Ishtiaq.

Fred Marshall wrote:
> "I. R. Khan" wrote:
> > I have a real impulse response h of an FIR filter, and its frequency
> > response H is pure imaginary (if I rotate h to left by half of its
>  length).
> > I want the impulse response of a filter whose magnitude response is
> > 1-Abs[H]. Is it possible to do this with some trick on h?
 
...

> So, you can figure it out using an FFT/IFFT process easily enough.
> However, ABS in frequency has no dual in time as far as I know.
> So, the answer to your question is: In general, No.

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?

Regards,
Andor

"I. R. Khan" wrote:
> 
> Thanks Fred.
> 
> > 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
> > .. the imaginary part of H is antisymmetric (around f=0).
> > .. Therefore,the imaginary part has to be zero at f=0, f=fs/2,f=fs
> 
> Yes, yes, yes and yes. God! how did you know?
> 
>  For example h = {1/(2pi), 0, -1/(2pi)}, and H is pure imaginary if centered
> at t=0.
> I want to shift H up by 1, and can do that by replacing zero in h by j
> (=sqrt(-1)). I was wondering if it is possible to do it keeping the
> coefficients all real.
> 
> This leads to another question. Is the impulse response h = {1/(2pi),
> j, -1/(2pi)} of some importance to real inputs? I seems it will change them
> to complex.
> 
> Infact h = {1/(2pi), j, -1/(2pi)} is a differentiator highly accurate at
> fs/2, but I want a similar differentiator with real coefficients.
>

The center tap must be 0 for the filter to be real and anti-symmetric if there are
an odd number of taps.

-jim


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"Andor" <an2or@mailcircuit.com> wrote in message
news:ce45f9ed.0402180215.7292a111@posting.google.com...
> Fred Marshall wrote:
> > "I. R. Khan" wrote:
> > > I have a real impulse response h of an FIR filter, and its frequency
> > > response H is pure imaginary (if I rotate h to left by half of its
> >  length).
> > > I want the impulse response of a filter whose magnitude response is
> > > 1-Abs[H]. Is it possible to do this with some trick on h?
>
> ...
>
> > So, you can figure it out using an FFT/IFFT process easily enough.
> > However, ABS in frequency has no dual in time as far as I know.
> > So, the answer to your question is: In general, No.
>
> 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?

 Abs doesn't carry the "j" to the output.  Abs yields a real number.
So, 1-Abs[anything] is a real number.

Fred