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I understand that the magnitude of the frequency response of 
an FIR filter is the intersection of a tube going down 
through the unit circle with the curvaceous surface created 
by the placement of the zeros of its roots.

Is there also a simple visualization of the phase of the 
frequency response that's related to the zero placement?


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

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In article c...@enews2.newsguy.com, Bob Cain at
a...@arcanemethods.com wrote on 02/12/2004 23:00:

> 
> I understand that the magnitude of the frequency response of
> an FIR filter is the intersection of a tube going down
> through the unit circle with the curvaceous surface created
> by the placement of the zeros of its roots.

wow! that's a weird, errr interesting way to look at it!

> Is there also a simple visualization of the phase of the
> frequency response that's related to the zero placement?

for magnitude, i would multiply the distance that e^(jw) is to each zero
(screw the poles since they all be at the origin) and for phase, add up the
angle of the difference "vectors" that connect each zero to e^(jw).  that's
the only visualization i can do.

r b-j

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robert bristow-johnson wrote:

> In article c...@enews2.newsguy.com, Bob Cain at
> a...@arcanemethods.com wrote on 02/12/2004 23:00:
> 
> 
>>I understand that the magnitude of the frequency response of
>>an FIR filter is the intersection of a tube going down
>>through the unit circle with the curvaceous surface created
>>by the placement of the zeros of its roots.
> 
> 
> wow! that's a weird, errr interesting way to look at it!

Is it accurate, though, and equivalent to what you give for 
it below?  That equivalence isn't obvious to my visualizer.

> 
> 
>>Is there also a simple visualization of the phase of the
>>frequency response that's related to the zero placement?
> 
> 
> for magnitude, i would multiply the distance that e^(jw) is to each zero
> (screw the poles since they all be at the origin) and for phase, add up the
> angle of the difference "vectors" that connect each zero to e^(jw).  that's
> the only visualization i can do.

Do you mean for each point on the unit circle add up the 
angles of all the vectors joining it to the zeros?

I'm trying to visualize how reflecting a zero from outside 
to inside the unit circle causes the phase function to 
become closer to zero everywhere.  Actually I'm trying to 
understand just what it is that's minimized in minimum phase 
filters.  If that's not it, please enlighten me.


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

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"robert bristow-johnson" <r...@surfglobal.net> wrote in message
news:BC51BE50.88A4%r...@surfglobal.net...
> In article c...@enews2.newsguy.com, Bob Cain at
> a...@arcanemethods.com wrote on 02/12/2004 23:00:
>
> >
> > I understand that the magnitude of the frequency response of
> > an FIR filter is the intersection of a tube going down
> > through the unit circle with the curvaceous surface created
> > by the placement of the zeros of its roots.

Well, not exactly I think ... and the answer ties the phase question into
it:

Consider a family of frequency response vectors with orgin on each point of
the unit circle and perpendicular to the unit circle.
Each vector is then free to rotate about the unit circle.  What's necessary
is to define what the third dimension represents ... and I suppose we're
free, for the moment, to make a choice.

I just realized that this is easier for me to do in s-plane rather than
z-plane space so let me do that first....

Envision a family of frequency response vectors with orgin on each point of
the jw axis and perpendicular to the jw axis.
Each vector is then free to rotate about the jw axis.  What's necessary is
to define what the third dimension represents ... and I suppose we're free,
for the moment, to make a choice.

Let's say that the real part of the vector is in the direction of "w" (which
is perpendicular to the jw axis) and the imaginary part is in the new 3rd
dimension we've created.

The magnitude response is the magnitude of the vector.
The phase response is the phase of the vector.

Now, let the end points of all the vectors connect into a continuum.  What
you have is a sort of spiral of varying diameter (the magnitude).  For
example, a linear phase FIR filter will have constant angular change with
frequency, so the vector rotates at a constant rate with frequency (i.e.
along the jw axis).  But, the magnitude, and thus the diameter of the
spiral, is free to change.

If you cut the jw axis with a plane, that's equivalent to cutting the
z-plane unit circle with a tube.
Since we selected the 3rd dimension to be the imaginary part, then the
vectors project the imaginary part onto this plane.  This projection isn't
the magnitude - it's the imaginary part.

Eventually we'll want to re-map from jw to z and we'll find the same thing
except the magnitude of the vector has to be mapped to match the circular
nature of the z-plane unit circle "axis".  That's a good exercise for the
student..... :-)

Now, I can imagine there might be another mapping or assignment of
dimensions that could work more along the lines you've stated (BTW, where
did you get this?) but I'm hard pressed to see how a simple projection onto
a plane or a tube (respectively) can yield magnitude - because magnitude
isn't represented here anywhere.

Let's get back to the idea of a tube:

The magnitude response *specification* for a filter is a "tube" of varying
diameter that surrounds the jw axis or a distorted one that surrounds the
z-plane unit circle.  If the magnitude response specificaton is |1.0|, then
the tube has diameter of 1.0 I guess.  If the magnitude response
specification is < 0.000001 then the tube has a very small diameter.
Something like that.
If the phase specification is given then it will give the angle of the
vector.
Something concerns me about scaling in the z-plane here...... but I think it
works otherwise.

So, an all-pass filter specification, with no phase response spec, is a tube
with constant diameter of 1.0 in the s-plane.  The actual filter response
approximates this perfect tube with a wiggly-diameter spiral that can change
direction of rotation.

Oh...
Maybe you were thinking that the real part of the frequency response could
be projected on that unit circle tube.  It could.
Maybe you were thinking that a *symmetric* FIR filter has only real
response - which it does.
And, maybe you were thinking that a symmetric FIR filter with no zeros, or
only even ordered multiple zeros, on the unit circle will have only a
positive real response - which it can have with a usually trivial constraint
that the sum of the coefficients is positive rather than negative (a simple
sign change on all of them).
In that case, then plotting the real part onto the new 3rd dimension will
project the magnitude onto the tube.
If there are odd-ordered zeros on the unit circle then you will have to take
the magnitude of the negative values - so the more general statement doesn't
work does it?

So, I backed into it.... and it only applies to the more narrowly defined
case:
Symmetric, real, FIR with no odd-ordered zeros on the unit circle.
The placement of poles or other zeros doesn't matter - as long as it's a
symmetric FIR filter.  They are constrained by that structure of course so
that the poles are at the orgin and the zeros are in complex conjugate pairs
and those pairs are in reciprocal pairs to form quads.

I think that's all correct.....

Fred

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Fred Marshall wrote:

   ...

> I just realized that this is easier for me to do in s-plane rather than
> z-plane space so let me do that first....

Yes. See "Spirule"

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

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In article <c...@enews1.newsguy.com>,
Bob Cain  <a...@arcanemethods.com> wrote:
>>>I understand that the magnitude of the frequency response of
>>>an FIR filter is the intersection of a tube going down
>>>through the unit circle with the curvaceous surface created
>>>by the placement of the zeros of its roots.
...
>>>Is there also a simple visualization of the phase of the
>>>frequency response that's related to the zero placement?
...
>I'm trying to visualize how reflecting a zero from outside 
>to inside the unit circle causes the phase function to 
>become closer to zero everywhere.  Actually I'm trying to 
>understand just what it is that's minimized in minimum phase 
>filters.  If that's not it, please enlighten me.

The location of both the poles and the zeros is important.  Imagine a
crank handle at the location of every pole or zero on the complex
plane. These crank handles are attached to rotation counters (funny
counters that count by 2*pi instead of just once per revolution).
The counters count up for zeros, but count down for poles.  Every crank
handle is automated to aim at where you are standing on the a unit circle
(perhaps operated by little demons who are tired of playing with all
those doors over in the thermodynamics lab).

First consider a single pole or zero.  If it's inside the unit circle then
the counter will count up or down an entire 2*pi every time you complete
a walk around the entire circle.  If the pole or zero is outside the unit
circle (say far away) then the handle may only wiggle back and forth,
but never move in a complete revolution, no matter how many times you
walks around in unit circles.

Consider not just one pole or zero, but the sum of several.  Say you
are plotting the sum of all the rotation counters on the cylinder
wall as you take a walk around the unit circle.  Unless you have
a handy time machine, all the poles will be inside the unit circle.
So for every trip around, the sum will go down by 2*pi*number_of_poles.
If all the zeros are also inside the unit circle the the sum will go up
by 2*pi*number_of_zeros.  If the the number of poles is the same as the
number of zeros, then the sum will be zero after a round trip, which is
usually considered a fairly small value.

But if some of the zeros are outside the unit circle, and you have the
same number of zeros as poles (why?), then the sum of the positive and
negative counters will not cancel after one round trip.  The sum
will continue to diverge as you walk around and around the unit circle,
and you will eventually need a ladder to keep plotting points.  For
some reason, this ever-rising curve is unlikely to be called "minimum".

An interesting question might be whether, if you actually came across
a working handy-dandy (z^+1) time machine, and could move some of the
poles outside the unit circle, would a minimum phase design also require
moving an equal number of zeros outside?

Whether or not these demons standing on a complex plane and operating
funny crank handles with counters attached has anything to do with
frequency response, and if so under what conditions, is left as an
exercise for the student.


IMHO. YMMV.
-- 
Ron Nicholson   rhn AT nicholson DOT com   http://www.nicholson.com/rhn/ 
#include <canonical.disclaimer>        // only my own opinions, etc.

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Ronald H. Nicholson Jr. wrote:

   ...

> Whether or not these demons standing on a complex plane and operating
> funny crank handles with counters attached has anything to do with
> frequency response, and if so under what conditions, is left as an
> exercise for the student.
> 
> 
> IMHO. YMMV.

Nifty!
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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Bob Cain wrote:

> I understand that the magnitude of the frequency response of 
> an FIR filter is the intersection of a tube going down 
> through the unit circle with the curvaceous surface created 
> by the placement of the zeros of its roots.

With the surface defined by the modulus of the complex value assigned 
to each point in the z-plane, yes. Since few are able to imagine 
objects of the four spatial dimensions the actual situation calls 
for, something's got to give. Fred threw out the real direction of 
the s-plane to make room for his spirals; you kept that over the 
response's phase component. Your choice is in line with the answer to 
your question -- with practice in using the sum-of-angles rule Robert 
stated, one may do reasonably well. But IMHO it's not going to get as 
intuitive as for magnitude. Incidentally, note that Robert's rules 
simply break down the complex product that is the factored form of 
the FIR transfer function.

I did a little experiment just to convince myself that you might want 
to carry further. I plotted magnitude and phase response of a few 
filters with one through three spaced, conjugate zero pairs. The 
magnitudes are rather well-behaved, as you know -- just seeing some 
plots quickly enables one to predict what moving the zeros around 
will do. Now, each zero introduces a phase jump (of pi, at its 
frequency, when the radius is unity), but whether it will be up or 
down and exactly how the pieces look is hardly deducible from my few 
examples. Still, further experimentation could lead to more of a gut 
feel there.


Martin

-- 
   Please help refine my English usage!
    -= Send your critique by email. =-

Quidquid latine dictum sit, altum viditur.

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"Martin Eisenberg" <m...@PAMudo.edu> wrote in message
news:1...@ostenberg.wh.uni-dortmund.de...
> Bob Cain wrote:
>
> > I understand that the magnitude of the frequency response of
> > an FIR filter is the intersection of a tube going down
> > through the unit circle with the curvaceous surface created
> > by the placement of the zeros of its roots.
>
> With the surface defined by the modulus of the complex value assigned
> to each point in the z-plane, yes. Since few are able to imagine
> objects of the four spatial dimensions the actual situation calls
> for, something's got to give. Fred threw out the real direction of
> the s-plane to make room for his spirals;

Well, I don't think I "threw it out", I rather thought that I'd
superimposed...

Fred

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In article c...@enews1.newsguy.com, Bob Cain at
a...@arcanemethods.com wrote on 02/13/2004 00:51:

> robert bristow-johnson wrote:
> 
>> In article c...@enews2.newsguy.com, Bob Cain at
>> a...@arcanemethods.com wrote on 02/12/2004 23:00:
>> 
>> 
>>> I understand that the magnitude of the frequency response of
>>> an FIR filter is the intersection of a tube going down
>>> through the unit circle with the curvaceous surface created
>>> by the placement of the zeros of its roots.
>> 
>> 
>> wow! that's a weird, errr interesting way to look at it!
> 
> Is it accurate, though, and equivalent to what you give for
> it below?  That equivalence isn't obvious to my visualizer.
> 
>> 
>> 
>>> Is there also a simple visualization of the phase of the
>>> frequency response that's related to the zero placement?
>> 
>> 
>> for magnitude, i would multiply the distance that e^(jw) is to each zero
>> (screw the poles since they all be at the origin) and for phase, add up the
>> angle of the difference "vectors" that connect each zero to e^(jw).  that's
>> the only visualization i can do.
> 
> Do you mean for each point on the unit circle add up the
> angles of all the vectors joining it to the zeros?

yes and you do the same for the poles but subtract all of those angles.


> I'm trying to visualize how reflecting a zero from outside
> to inside the unit circle causes the phase function to
> become closer to zero everywhere.

it's a lot easier visualizing it on the s-plane for continuous-time LTI
systems ( arg(jw-q) < arg(jw+q)  for q<0 and real ) but i think it's true
for the z-plane, too.  i can think of a round-about justification of that by
using the previous result for the s-plane and the BiLinear Transform and
warped frequency mapping of the BLT.  do you want me to go through the hand
waving?

>  Actually I'm trying to
> understand just what it is that's minimized in minimum phase
> filters.  If that's not it, please enlighten me.

oh, it is it.  this is much easier to see for the s-plane.  given a
particular magnitude response that can be attained with poles and zeros in
the s-plane, the only "constellations" of poles and zeros that will give you
*that* particular magnitude response (let's not consider different constant
gain factors) is the set of constellations that have the poles and zeros in
some position or in reflections about the jw axis.  of course poles must be
in the left-half plane so they can't be reflected to the other side of the
jw axis.  but zeros can.  but if they are reflected from the left-hand plane
to the right-hand plane, you can see that their angle will always be greater
than 90 degrees whereas they were always less than 90 degrees before.  so to
get the smallest set of angles on those zeros, you gotta have them all in
the left-half plane.

now if you consider the BiLinear Transform and the frequency warping
property of it, for some continuous-time filter with poles and zeros in the
s-plane, there is an equal order discrete-time filter with the same number
of poles and zeros in the z-plane.  (if the original number of zeros was
less than the number of poles, then those "missing" zeros get mapped to z=-1
in the z-plane.  think of them as zeros out at s=-inf.)  now the frequency
warping property says that the discrete-time filter will have exactly the
same frequency response (that is the same magnitude and phase) BUT AT
DIFFERENT FREQUENCIES defined by that arctan() relationship.  so that means
if, at some frequency, reflecting some s-plane zero to the right-half plane
increases the phase angle, then, at some other frequency, reflecting the
corresponding z-plane zero to outside the unit circle will do the same and
increase the phase angle.

does that do it, Bob?

r b-j





















































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