FPGA Central

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

In article [email protected], Bob Cain at
[email protected] 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

robert bristow-johnson wrote:

> In article [email protected], Bob Cain at
> [email protected] 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

"robert bristow-johnson" <[email protected]> wrote in message
news:BC51BE50.88A4%[email protected]..
> In article [email protected], Bob Cain at
> [email protected] 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

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

In article <[email protected]>,
Bob Cain <[email protected]> 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.

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.

"Martin Eisenberg" <[email protected]> wrote in message
news:[email protected]..
> 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

In article [email protected], Bob Cain at
[email protected] wrote on 02/13/2004 00:51:

> robert bristow-johnson wrote:
>
>> In article [email protected], Bob Cain at
>> [email protected] 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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Fred Marshall wrote:

>
> "Martin Eisenberg" <[email protected]> wrote in
> message news:[email protected]..
>> 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...

What I meant is this: The quantities s and H(s) are distinct, so
I don't think you can superimpose some of their components -- any
more than you can superimpose the x and y axes of a real->real
graph -- without losing something. In fact, you didn't actually *use*
the direction in question as Re(s) in your post, only as Re(H(s)).
Consider the spirals about both the jw line and the -1+jw line. Now
that Re(s) has come into play, there's no way to avoid their
potential crossing because space of perception lacks another offset
possibility. In other words, anywhere we can put that "-1", something
already exists. Whether that's "throwing out a dimension" or
"superposition" seems moot.


Martin

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

Quidquid latine dictum sit, altum viditur.

"Martin Eisenberg" <[email protected]> wrote in message
news:[email protected]..
> Fred Marshall wrote:
> >
> > Well, I don't think I "threw it out", I rather thought that I'd
> > superimposed...
>
> What I meant is this: The quantities s and H(s) are distinct, so
> I don't think you can superimpose some of their components -- any
> more than you can superimpose the x and y axes of a real->real
> graph -- without losing something. In fact, you didn't actually *use*
> the direction in question as Re(s) in your post, only as Re(H(s)).
> Consider the spirals about both the jw line and the -1+jw line. Now
> that Re(s) has come into play, there's no way to avoid their
> potential crossing because space of perception lacks another offset
> possibility. In other words, anywhere we can put that "-1", something
> already exists. Whether that's "throwing out a dimension" or
> "superposition" seems moot.
>

Martin,

Hmmmm... Well, there are many perspectives that work. For one thing, I
could do the same thing without thinking "superposition" and just create a
new 3-D figure. Otherwise, I think of it as plotting two curves or data
points on the same set of axes - thus superposition - both of them are
there. So, the axes do double duty to display different (and maybe related)
things on the same "chart".

In this case, as you point out, we have "s=x+jw" and we have
"H(jw)=A(jw)e^jwp" ... something like that.
So, if we share the "w" axis for both, then we have to figure out how to
display H(jw). The choice I made was to show the real part aligned with the
"x" axis and the imaginary part in the new third dimension. This means that
zero phase shift vectors point to the right in "x" and that 90 degree phase
shift vectors point up out of the x/w plane, etc. Just a 3-D plot of two
things superimposed on that plot.

Your analogy of real>real plots (which is really the same thing) might be
demonstrated like this:
(it doesn't mix up the x and y axes at all)
There is one function of x and y - so we can plot it in an x/y plane.
There is another function of x and y and z - so we can plot it by adding a z
axis to the plane above and making a 3-D plot without changing any of the
aspects of the original x/y plot.
Of course, it's going to be that the function of x,y,z can lie on or cross
the x,y plane (or not).
That's the idea.
Tying them together where there are common elements (in this case, "w")
seems useful.

Fred

Fred Marshall wrote:

> Hmmmm... Well, there are many perspectives that work.

Indeed. After reading your post I think the glass
was half empty to me and half full to you.


Martin

--
The power of accurate observation is commonly
called cynicism by those who don't have it.
--George Bernard Shaw

Ronald H. Nicholson Jr. wrote:

>
> 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.

Wonderful visualization aid, Ronald. Many thanks. I'm
still left wondering what it all means, however. All I can
see is that what is left in the sum of the integrators after
a single revolution is a minimum, zero, when the zeros are
inside the circle but the signifigance of that remains
obscure to me in a physical sense.


Bob
--

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

A. Einstein

Fred Marshall wrote:


>
> 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.

I dunno, I just thunk it up. As should be obvious my
experience has not been with traditional filter design at
all or with feedback systems so my feeling for the complex
plane representation never gelled.

Actually, I've always thought, for some reason I can't
remember, perhaps an analogy I was given that I totally
misunderstood, that if you plotted the magnitude of the
transfer function (I failed to state the critical word
"magnitude" in my first post) as a surface above the plane
with the poles and zeros deforming it appropriately that the
magnitude function we usually think about and plot linearly
in a frequency response plot was the intersection of that
surface with a cylinder perpindicular to the complex plane
passing through the unit circle. In the s plane it would be
the intersection with a plane perpindicular to it passing
through the imaginary axis.

From what I'm reading in the thread I'm still not sure
whether that is right or dead wrong. :-)

Now I am trying to understand how the phase function varies
as you traverse that circle. I do understand Ronald's
explanation and see that (since I'm only thinking now about
FIR's) the placement of the zeros inside or outside
determine whether the integral of the total of the
contributions of all the zeros can only be a multiple of
2*pi if they are all inside but I'm still not making the
connection as to why that would cause the energy in an
impulse response to all bunch up at the start when it's
minimum phase and what is really being minimized if I look
at just the usual (unwraped) phase plot (assuming it even
shows up there.)


Bob
--

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

A. Einstein

"Ronald H. Nicholson Jr." <[email protected]> wrote in message
news:c0jn3f$6dn$[email protected]..
> In article <[email protected]>,
>
> The location of both the poles and the zeros is important.

The way I learned to visualize this was to imagine that the s-plane was
covered with a thin rubber sheet.
Then, imagine that a pencil were stood on end, under the sheet, like a tent
pole at each pole location.
Then, imagine that a thumbtack is set to hold the sheet down to the plane
wherever there's a zero.
It's helped now and then!

Fred

robert bristow-johnson wrote:


>
>>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

Ok. I do basically understand the mapping between them.

>> 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.

Understood. I'm only thinking about all-zero systems right
now anyway.

> 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.

Doh! Pretty simple, huh? <Bob blushes>

>
> does that do it, Bob?

At last. Thanks for the patience.


Bob
--

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

A. Einstein

Bob Cain wrote:

...

> Now I am trying to understand how the phase function varies as you
> traverse that circle. I do understand Ronald's explanation and see that
> (since I'm only thinking now about FIR's) the placement of the zeros
> inside or outside determine whether the integral of the total of the
> contributions of all the zeros can only be a multiple of 2*pi if they
> are all inside but I'm still not making the connection as to why that
> would cause the energy in an impulse response to all bunch up at the
> start when it's minimum phase and what is really being minimized if I
> look at just the usual (unwraped) phase plot (assuming it even shows up
> there.)

I'm out in a limb here, waving my arms rather than doing math, but here
goes <deep breath>:

Adding phase shift implies adding or subtracting time. In real life,
subtracting isn't possible, so extra phase shift means extra delay.
I seems that simple to me.

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

"Bob Cain" <[email protected]> wrote in message
news:[email protected]..
> Fred Marshall wrote:
>
>
> >
> > 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.
>
...............
>that if you plotted the magnitude of the
> transfer function (I failed to state the critical word
> "magnitude" in my first post) as a surface above the plane
> with the poles and zeros deforming it appropriately that the
> magnitude function we usually think about and plot linearly
> in a frequency response plot was the intersection of that
> surface with a cylinder perpindicular to the complex plane
> passing through the unit circle. In the s plane it would be
> the intersection with a plane perpendicular to it passing
> through the imaginary axis.

Bob,

Oh, well then..... you already *have* the magnitude surface for all values
of s or z.
So, sure, a perpendicular plane or cylinder will simply be cut by the
magnitude surface at those values on the jw axis in s or on the unit circle
in z.

Fred

In article <40321b52$0$3084$[email protected]>,
Jerry Avins <[email protected]> wrote:
>Bob Cain wrote:
> ...
>> Now I am trying to understand how the phase function varies as you
>> traverse that circle. I do understand Ronald's explanation and see that
>> (since I'm only thinking now about FIR's) the placement of the zeros
>> inside or outside determine whether the integral of the total of the
>> contributions of all the zeros can only be a multiple of 2*pi if they
>> are all inside but I'm still not making the connection as to why that
>> would cause the energy in an impulse response to all bunch up at the
>> start when it's minimum phase and what is really being minimized if I
>> look at just the usual (unwraped) phase plot (assuming it even shows up
>> there.)
>
>I'm out in a limb here, waving my arms rather than doing math, but here
>goes <deep breath>:
>
>Adding phase shift implies adding or subtracting time. In real life,
>subtracting isn't possible, so extra phase shift means extra delay.

Here's another "hand-wavey" one:

Consider the unit delay. That's a zero at the origin. It's phase
response will spiral by 2*pi per trip around the unit cylindar.
The phase response of an N unit delay line will spiral by 2*pi*N.
Wave hands:
The more any arbitrary frequency response in the z plane "resembles"
one of these spirals, the more the time response will "resemble" a
delay line of N taps. A minimum phase filter is the one closest to
"resembling" a zero-delay delay line.


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

Ronald H. Nicholson Jr. wrote:

> In article <40321b52$0$3084$[email protected]>,
> Jerry Avins <[email protected]> wrote:
>
>>Bob Cain wrote:
>> ...
>>
>>> ... I do understand Ronald's explanation and see that
>>>(since I'm only thinking now about FIR's) the placement of the zeros
>>>inside or outside determine whether the integral of the total of the
>>>contributions of all the zeros can only be a multiple of 2*pi if they
>>>are all inside but I'm still not making the connection as to why that
>>>would cause the energy in an impulse response to all bunch up at the
>>>start when it's minimum phase and what is really being minimized if I
>>>look at just the usual (unwraped) phase plot (assuming it even shows up
>>>there.)

...

>>Adding phase shift implies adding or subtracting time. In real life,
>>subtracting isn't possible, so extra phase shift means extra delay.
>
>
> Here's another "hand-wavey" one:
>
> Consider the unit delay. That's a zero at the origin. It's phase
> response will spiral by 2*pi per trip around the unit cylindar.
> The phase response of an N unit delay line will spiral by 2*pi*N.
> Wave hands:
> The more any arbitrary frequency response in the z plane "resembles"
> one of these spirals, the more the time response will "resemble" a
> delay line of N taps. A minimum phase filter is the one closest to
> "resembling" a zero-delay delay line.

It just goes to show: when an idea is sound, it hangs together from any
angle.

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

"Bob Cain" <[email protected]> wrote in message
news:[email protected]..
> Actually, I've always thought [...]
> From what I'm reading in the thread I'm still not sure
> whether that is right or dead wrong. :-)

Yes, what you say works fine if you measure log amplitude, and the
deformations sum.

> Now I am trying to understand how the phase function varies
> as you traverse that circle. [...] I'm still not making the
> connection as to why that would cause the energy in an
> impulse response to all bunch up at the start when it's
> minimum phase [...]

The easiest explantion is in tems of impulse responses. The impulse
response for the filter is the convolution of the impulse responses for all
the poles and zeros. Swapping any zero outside the unit circle for the
matching one inside the unit circle will pack more of the impulse response
near t=0, because the impulse response for that zero will be packed more
towards t=0, and that is easy to see, because H(z) for a single-zero FIR is
zero at z=-h[1]/h[0].

>and what is really being minimized if I look
> at just the usual (unwraped) phase plot (assuming it even
> shows up there.)

Now that's an interesting question that I would like the answer to as well.
In particular, knowing that a filter is minimum phase for a particlar
magnitude response doesn't convince me that there isn't a *nearly identical*
magnitude response with a minimum phase filter that is *much* more
concentrated towards t=0. I would be very happy to have a conclusive
demonstration of that.

Matt Timmermans wrote:

> "Bob Cain" <[email protected]> wrote in message
> news:[email protected]..
>
>>Actually, I've always thought [...]
>> From what I'm reading in the thread I'm still not sure
>>whether that is right or dead wrong. :-)
>
>
> Yes, what you say works fine if you measure log amplitude, and the
> deformations sum.
>
>
>>Now I am trying to understand how the phase function varies
>>as you traverse that circle. [...] I'm still not making the
>>connection as to why that would cause the energy in an
>>impulse response to all bunch up at the start when it's
>>minimum phase [...]
>
>
> The easiest explantion is in tems of impulse responses. The impulse
> response for the filter is the convolution of the impulse responses for all
> the poles and zeros. Swapping any zero outside the unit circle for the
> matching one inside the unit circle will pack more of the impulse response
> near t=0, because the impulse response for that zero will be packed more
> towards t=0, and that is easy to see, because H(z) for a single-zero FIR is
> zero at z=-h[1]/h[0].
>
>
>>and what is really being minimized if I look
>>at just the usual (unwraped) phase plot (assuming it even
>>shows up there.)
>
>
> Now that's an interesting question that I would like the answer to as well.
> In particular, knowing that a filter is minimum phase for a particlar
> magnitude response doesn't convince me that there isn't a *nearly identical*
> magnitude response with a minimum phase filter that is *much* more
> concentrated towards t=0. I would be very happy to have a conclusive
> demonstration of that.

Since the minimum-phase frequency and phase responses are a Hilbert
transform pair, it seems to me that greatly affecting one can't affect
the other only a little. Of course, the back of my envelope is rather
smudged tonight.

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

Jerry Avins <[email protected]> writes:

> Matt Timmermans wrote:
>
> > Now that's an interesting question that I would like the answer to as well.
> > In particular, knowing that a filter is minimum phase for a particlar
> > magnitude response doesn't convince me that there isn't a *nearly identical*
> > magnitude response with a minimum phase filter that is *much* more
> > concentrated towards t=0. I would be very happy to have a conclusive
> > demonstration of that.
>
> Since the minimum-phase frequency and phase responses are a Hilbert
> transform pair, it seems to me that greatly affecting one can't affect
> the other only a little. Of course, the back of my envelope is rather
> smudged tonight.

Another observation is that the overall energy of the *much* more
bunched filter cannot be different from the minimum phase original (by
Parseval's theorem).

I suspect you could probably extend that to an argument showing that
the "energy delay" (see Oppenheim and Schafer, 1989, prob 5.36) of
each cannot be significantly different.

Ciao,

Peter K.

--
Peter J. Kootsookos

"I will ignore all ideas for new works [..], the invention of which
has reached its limits and for whose improvement I see no further
hope."

- Julius Frontinus, c. AD 84

Matt Timmermans wrote:

> "Bob Cain" <[email protected]> wrote in message
> news:[email protected]..
>
>>Actually, I've always thought [...]
>> From what I'm reading in the thread I'm still not sure
>>whether that is right or dead wrong. :-)
>
>
> Yes, what you say works fine if you measure log amplitude, and the
> deformations sum.

Ah, yes, the log. That clears up another question I now
need not ask.

>
>
> The easiest explantion is in tems of impulse responses. The impulse
> response for the filter is the convolution of the impulse responses for all
> the poles and zeros. Swapping any zero outside the unit circle for the
> matching one inside the unit circle will pack more of the impulse response
> near t=0, because the impulse response for that zero will be packed more
> towards t=0, and that is easy to see, because H(z) for a single-zero FIR is
> zero at z=-h[1]/h[0].

Is there a nice interactive pole/zero placement tool (for
Win98se or Java) that will show the impulse response,
magnitude, and phase plots as you place them on the z-plane
(the ability to mirror zeros about the unit circle would be
handy too)?

>
> Now that's an interesting question that I would like the answer to as well.
> In particular, knowing that a filter is minimum phase for a particlar
> magnitude response doesn't convince me that there isn't a *nearly identical*
> magnitude response with a minimum phase filter that is *much* more
> concentrated towards t=0. I would be very happy to have a conclusive
> demonstration of that.

If you get further insight on that, please share it.


Thanks,

Bob
--

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

A. Einstein