Can a LPF have linear phase when the impuse response is not symmetric?

On Sun, 30 Mar 2008 09:13:26 -0700, "miner_tom" <tom_nospam@yahoo.com>
wrote:

>Oppenheimer says that this figure shows a "linear phase system for which the 
>impulse response is not symmetric". I do not understand that statement.

...

>On page 309 there are figures showing the decimated impulse responses of the 
>various polyphase subfilters. The decimated impulse responses exhibit no 
>symmetry that I can see even though the original prototype filter is 
>symmetic. Each of the subfilters is described as having linear phase. 

Linear phase only means that the phase is a linear function of
frequency (so that the first derivative is a constant).  Impulse
response symmetry is sufficient to produce linear phase, but not
necessary, at least not in the discrete-time case.  The problem is in
the definition of symmetry.  If one adheres strictly to the definition
as (from page 5-34 of the Oppenheim presentation that you cited):

    h(n) = h(M-n); 0<=n<=M
                0; otherwise
or
    h(n) = -h(M-n); 0<=n<=M
                 0; otherwise

then if the envelope of the impulse response is delayed by anything
other than an integer multiple of one-half the sample period, the
symmetry definition is no longer satisfied, even though the phase is
still linear.  This can be avoided by broadening the definition of
symmetry to apply to the envelope, and not just to the discrete-time
samples themselves.

Greg

On Mar 30, 9:13 am, "miner_tom" <tom_nos...@yahoo.com> wrote:
> Hi,
>
> I had always believed that in order to have linear phase or constant group
> delay, the impulse response of a digital filter needed to have even symmetry
> or odd symmetry. In "Discrete time signal processing" by Oppenheim and
> Shafer, there is a section on "Generalized systems with linear phase". There
> is a figure, 5.32C where the impulse response is not symmetric but the
> filter is said to have the property of linear phase.
>
> Here is a link to a PPT lecture that uses this figure from the book and the
> slide in question is 5-33. The link should open to the proper slide.
>
> http://www.vcipl.okstate.edu/~glfan/ECEN5763/chapter5.ppt#289,33,Example
>
> Oppenheimer says that this figure shows a "linear phase system for which the
> impulse response is not symmetric". I do not understand that statement.
>
> There is another example of this in a very famous paper concerning Polyphase
> filtering, "Interpolation and decimation of Digital Signals" by Crochiere
> and Rabiner. The identical arguments appear in the book "Multirate Digital
> Signal processing" by the same authors.
>
> Here is a link to the paper.
>
> http://www.caip.rutgers.edu/~lrr/lrr%20papers/179_interpolation_decim...
>
> On page 309 there are figures showing the decimated impulse responses of the
> various polyphase subfilters. The decimated impulse responses exhibit no
> symmetry that I can see even though the original prototype filter is
> symmetic. Each of the subfilters is described as having linear phase. Each
> of the subfilters has a different linear phase because the samples in each
> subfilter have a different relationship to the center of symmetry of the
> original prototype filter.
>
> What is the fundamental concept that I have missed?
>
> Thank You
> Tom
> tom_cip_11...@hotmail.com

What is broken is your substitution of "exhibit symmetry I can see"
for the possession symmetry. The symmetries you have learned to  "see"
at this point are symmetries about a sample point or a point half-way
between samples. But a uniform sampling (satisfying Nyquist) of a
symmetric waveform is still symmetric. You just haven't learned to
"see" these symmetries. The set you know and have learned to "see" as
symmetric doesn't span the concept.

As Wil Rogers once said, "The problem isn't what we don't know so much
as what we know that ain't so."

Dale B Dalrymple

On Mar 30, 9:13 am, "miner_tom" <tom_nos...@yahoo.com> wrote:
> Hi,
>
> I had always believed that in order to have linear phase or constant group
> delay, the impulse response of a digital filter needed to have even symmetry
> or odd symmetry. In "Discrete time signal processing" by Oppenheim and
> Shafer, there is a section on "Generalized systems with linear phase". There
> is a figure, 5.32C where the impulse response is not symmetric but the
> filter is said to have the property of linear phase.
>
> Here is a link to a PPT lecture that uses this figure from the book and the
> slide in question is 5-33. The link should open to the proper slide.
>
> http://www.vcipl.okstate.edu/~glfan/ECEN5763/chapter5.ppt#289,33,Example
>
> Oppenheimer says that this figure shows a "linear phase system for which the
> impulse response is not symmetric". I do not understand that statement.
>
> There is another example of this in a very famous paper concerning Polyphase
> filtering, "Interpolation and decimation of Digital Signals" by Crochiere
> and Rabiner. The identical arguments appear in the book "Multirate Digital
> Signal processing" by the same authors.
>
> Here is a link to the paper.
>
> http://www.caip.rutgers.edu/~lrr/lrr%20papers/179_interpolation_decim...
>
> On page 309 there are figures showing the decimated impulse responses of the
> various polyphase subfilters. The decimated impulse responses exhibit no
> symmetry that I can see even though the original prototype filter is
> symmetic. Each of the subfilters is described as having linear phase. Each
> of the subfilters has a different linear phase because the samples in each
> subfilter have a different relationship to the center of symmetry of the
> original prototype filter.
>
> What is the fundamental concept that I have missed?
>
> Thank You
> Tom
> tom_cip_11...@hotmail.com

What is broken is your substitution of "exhibit symmetry I can see"
for the possession symmetry. The symmetries you have learned to  "see"
at this point are symmetries about a sample point or a point half-way
between samples. But a uniform sampling (satisfying Nyquist) of a
symmetric waveform is still symmetric. You just haven't learned to
"see" these symmetries. The set you know and have learned to "see" as
symmetric doesn't span the concept.

As Wil Rogers once said, "The problem isn't what we don't know so much
as what we know that ain't so."

Dale B Dalrymple

On Mar 30, 5:29 am, Greg Berchin <gberc...@comicast.net> wrote:
> On Sun, 30 Mar 2008 09:13:26 -0700, "miner_tom" <tom_nos...@yahoo.com>
> wrote:
>
> >Oppenheimer says that this figure shows a "linear phase system for which the
> >impulse response is not symmetric". I do not understand that statement.
>
> ...
>
> >On page 309 there are figures showing the decimated impulse responses of the
> >various polyphase subfilters. The decimated impulse responses exhibit no
> >symmetry that I can see even though the original prototype filter is
> >symmetic. Each of the subfilters is described as having linear phase.
>
> Linear phase only means that the phase is a linear function of
> frequency (so that the first derivative is a constant).  Impulse
> response symmetry is sufficient to produce linear phase, but not
> necessary, at least not in the discrete-time case.  The problem is in
> the definition of symmetry.  If one adheres strictly to the definition
> as (from page 5-34 of the Oppenheim presentation that you cited):
>
>     h(n) = h(M-n); 0<=n<=M
>                 0; otherwise
> or
>     h(n) = -h(M-n); 0<=n<=M
>                  0; otherwise
>
> then if the envelope of the impulse response is delayed by anything
> other than an integer multiple of one-half the sample period, the
> symmetry definition is no longer satisfied, even though the phase is
> still linear.  This can be avoided by broadening the definition of
> symmetry to apply to the envelope, and not just to the discrete-time
> samples themselves.
>
> Greg

I have proof in another book that you get linear phase rom 4 possible
cases:

Even number of weights,even symmetry
Odd number of weights,even symmetry
Even number of weights, odd symmetry
Odd numberof weights, odd symmetry.

They all have their applications too. In the case of an even number of
weights the centry point is between samples.

K.

On Mar 29, 12:55 pm, kronec...@yahoo.co.uk wrote:
> On Mar 30, 5:29 am, Greg Berchin <gberc...@comicast.net> wrote:
>
>
>
> > On Sun, 30 Mar 2008 09:13:26 -0700, "miner_tom" <tom_nos...@yahoo.com>
> > wrote:
>
> > >Oppenheimer says that this figure shows a "linear phase system for which the
> > >impulse response is not symmetric". I do not understand that statement.
>
> > ...
>
> > >On page 309 there are figures showing the decimated impulse responses of the
> > >various polyphase subfilters. The decimated impulse responses exhibit no
> > >symmetry that I can see even though the original prototype filter is
> > >symmetic. Each of the subfilters is described as having linear phase.
>
> > Linear phase only means that the phase is a linear function of
> > frequency (so that the first derivative is a constant).  Impulse
> > response symmetry is sufficient to produce linear phase, but not
> > necessary, at least not in the discrete-time case.  The problem is in
> > the definition of symmetry.  If one adheres strictly to the definition
> > as (from page 5-34 of the Oppenheim presentation that you cited):
>
> >     h(n) = h(M-n); 0<=n<=M
> >                 0; otherwise
> > or
> >     h(n) = -h(M-n); 0<=n<=M
> >                  0; otherwise
>
> > then if the envelope of the impulse response is delayed by anything
> > other than an integer multiple of one-half the sample period, the
> > symmetry definition is no longer satisfied, even though the phase is
> > still linear.  This can be avoided by broadening the definition of
> > symmetry to apply to the envelope, and not just to the discrete-time
> > samples themselves.
>
> > Greg
>
> I have proof in another book that you get linear phase rom 4 possible
> cases:
>
> Even number of weights,even symmetry
> Odd number of weights,even symmetry
> Even number of weights, odd symmetry
> Odd numberof weights, odd symmetry.
>
> They all have their applications too. In the case of an even number of
> weights the centry point is between samples.
>
> K.

Yes K

These symmetries guarantee linear phase.

Linear phase doesn't guarantee these symmetries.

Note that for weight sets having the same delay, any linear
combination of the weight sets with these symmetries will have linear
phase.

I suggest that any weight set with linear phase can be decomposed into
no more than 4 weight sets with these symmetries and that is why you
find them in your texts.

Dale B. Dalrymple

On Mar 30, 8:27 am, dbd <d...@ieee.org> wrote:
> On Mar 29, 12:55 pm, kronec...@yahoo.co.uk wrote:
>
>
>
> > On Mar 30, 5:29 am, Greg Berchin <gberc...@comicast.net> wrote:
>
> > > On Sun, 30 Mar 2008 09:13:26 -0700, "miner_tom" <tom_nos...@yahoo.com>
> > > wrote:
>
> > > >Oppenheimer says that this figure shows a "linear phase system for which the
> > > >impulse response is not symmetric". I do not understand that statement.
>
> > > ...
>
> > > >On page 309 there are figures showing the decimated impulse responses of the
> > > >various polyphase subfilters. The decimated impulse responses exhibit no
> > > >symmetry that I can see even though the original prototype filter is
> > > >symmetic. Each of the subfilters is described as having linear phase.
>
> > > Linear phase only means that the phase is a linear function of
> > > frequency (so that the first derivative is a constant).  Impulse
> > > response symmetry is sufficient to produce linear phase, but not
> > > necessary, at least not in the discrete-time case.  The problem is in
> > > the definition of symmetry.  If one adheres strictly to the definition
> > > as (from page 5-34 of the Oppenheim presentation that you cited):
>
> > >     h(n) = h(M-n); 0<=n<=M
> > >                 0; otherwise
> > > or
> > >     h(n) = -h(M-n); 0<=n<=M
> > >                  0; otherwise
>
> > > then if the envelope of the impulse response is delayed by anything
> > > other than an integer multiple of one-half the sample period, the
> > > symmetry definition is no longer satisfied, even though the phase is
> > > still linear.  This can be avoided by broadening the definition of
> > > symmetry to apply to the envelope, and not just to the discrete-time
> > > samples themselves.
>
> > > Greg
>
> > I have proof in another book that you get linear phase rom 4 possible
> > cases:
>
> > Even number of weights,even symmetry
> > Odd number of weights,even symmetry
> > Even number of weights, odd symmetry
> > Odd numberof weights, odd symmetry.
>
> > They all have their applications too. In the case of an even number of
> > weights the centry point is between samples.
>
> > K.
>
> Yes K
>
> These symmetries guarantee linear phase.
>
> Linear phase doesn't guarantee these symmetries.
>
> Note that for weight sets having the same delay, any linear
> combination of the weight sets with these symmetries will have linear
> phase.
>
> I suggest that any weight set with linear phase can be decomposed into
> no more than 4 weight sets with these symmetries and that is why you
> find them in your texts.
>
> Dale B. Dalrymple

Ok is that what you mean by sufficent but not nescessary? I have
always been confused with these terms..

K.

kronecker@yahoo.co.uk writes:
> [...]
> Ok is that what you mean by sufficent but not nescessary? I have
> always been confused with these terms..

"p -> q" (read "p implies q", or "if p then q") is equivalent to saying
"p is sufficient for q" and "q is necessary for p". The contrapositive
(which is equivalent logicially) is ~q -> ~p (where "~" denotes "not" or
negation).

The converse (which is NOT logically equivalent) is "p -> q".

As an example, let "p" be "pregnant human" and "q" be "human female".
So p -> q means "Being a pregnant human is a sufficient condition for
being a human female" or, in the contrapositive, "If you're not a human
female, then you're not a pregnant human." Notice that the converse
is NOT true: "If you're a human female, then you're a pregnant human."

The notation "p <-> q" means "p if and only if q" ("if and only if" is
sometimes abbreviated "iff") or "p is necessary and sufficient for
q". It is a short-hand way of asserting that both "p -> q" and "q ->
p". For example, "A mapping is invertible iff the mapping is one-to-one
and onto."

[durbin] has a set of appendices that discusses this as well as other
"tricky" yet fundamental logic ideas (such as set equivalent) in his
text on modern algebra. Dated, but highly recommended if you're near
a library.

--Randy

@book{durbin,
  title = "Modern Algebra: An Introduction",
  author = "John~R.~Durbin",
  edition = "3rd",
  year = "1992",
  publisher = "John Wiley \& Sons, Inc."}

-- 
%  Randy Yates                  % "I met someone who looks alot like you,
%% Fuquay-Varina, NC            %             she does the things you do, 
%%% 919-577-9882                %                     but she is an IBM."
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO   
http://www.digitalsignallabs.com

Randy Yates <yates@ieee.org> writes:
> [...]
> The converse (which is NOT logically equivalent) is "p -> q".

Sorry, that should be "q -> p".
-- 
%  Randy Yates                  % "Ticket to the moon, flight leaves here today 
%% Fuquay-Varina, NC            %  from Satellite 2"
%%% 919-577-9882                % 'Ticket To The Moon' 
%%%% <yates@ieee.org>           % *Time*, Electric Light Orchestra
http://www.digitalsignallabs.com

On Sat, 29 Mar 2008 12:29:29 -0500, Greg Berchin
<gberchin@comicast.net> wrote:

>Linear phase only means that the phase is a linear function of
>frequency (so that the first derivative is a constant).

Having said that, let me attempt a thought experiment.

Suppose I have a system with real, even-symmetric impulse response
(note that the value at n=0 is zero):

.                     x x
.                    x| |x
.                   x|| ||x
.                  x||| |||x
.                 x|||| ||||x
-----------------x-----x-----x------------------
<- -n*Ts               0                +n*Ts -> 

Because the system is real, the real part of the spectrum will be
even-symmetric and the imaginary part will be odd-symmetric.  In this
case the spectrum will be purely real; the phase will be everywhere
zero.  

Now suppose that I have another system with real, odd-symmetric
impulse response:

.                       x
.                       |x
.                       ||x
.                       |||x
.                       ||||x 
-----------------x-----x-----x------------------
<- -n*Ts          x||||                 +n*Ts ->
                   x|||
                    x||
                     x|
                      x

Because the system is real, the real part of the spectrum will be
even-symmetric and the imaginary part will be odd-symmetric.  In this
case the spectrum will be pure-imaginary; the phase will be -PI/2 at
positive frequencies and +PI/2 at negative frequencies (assuming the
definition of the FT having the "-j" term in the forward transform).

Now I add the two impulse responses together:

.                       x
.                       |
.                       |x
.                       ||
.                       ||x
.                       |||
.                       |||x 
.                       ||||
.                       ||||x 
.                       |||||
-----------------------x-----x------------------
<- -n*Ts               0                +n*Ts ->

Because the system is real, the real part of the spectrum will be
even-symmetric and the imaginary part will be odd-symmetric.  In this
case the phase will be -PI/4 at positive frequencies and +PI/4 at
negative frequencies (because the magnitudes of the real and imaginary
parts are equal to each other at every frequency).

Now here's what's confusing me:  Except for the discontinuity at n=0,
this system is linear phase (the first derivative of the phase is, in
fact, zero).  Yet neither the impulse response nor the envelope of the
impulse response is symmetrical about any point in time.

What is the proper way to interpret this?

Greg

Hi,

I had always believed that in order to have linear phase or constant group 
delay, the impulse response of a digital filter needed to have even symmetry 
or odd symmetry. In "Discrete time signal processing" by Oppenheim and 
Shafer, there is a section on "Generalized systems with linear phase". There 
is a figure, 5.32C where the impulse response is not symmetric but the 
filter is said to have the property of linear phase.

Here is a link to a PPT lecture that uses this figure from the book and the 
slide in question is 5-33. The link should open to the proper slide.

http://www.vcipl.okstate.edu/~glfan/ECEN5763/chapter5.ppt#289,33,Example

Oppenheimer says that this figure shows a "linear phase system for which the 
impulse response is not symmetric". I do not understand that statement.


There is another example of this in a very famous paper concerning Polyphase 
filtering, "Interpolation and decimation of Digital Signals" by Crochiere 
and Rabiner. The identical arguments appear in the book "Multirate Digital 
Signal processing" by the same authors.

Here is a link to the paper.

http://www.caip.rutgers.edu/~lrr/lrr%20papers/179_interpolation_decimation.pdf

On page 309 there are figures showing the decimated impulse responses of the 
various polyphase subfilters. The decimated impulse responses exhibit no 
symmetry that I can see even though the original prototype filter is 
symmetic. Each of the subfilters is described as having linear phase. Each 
of the subfilters has a different linear phase because the samples in each 
subfilter have a different relationship to the center of symmetry of the 
original prototype filter.

What is the fundamental concept that I have missed?

Thank You
Tom
tom_cip_11551@hotmail.com