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

On Mar 30, 8:27 am, dbd <d...@xxxxxxxx> wrote:
On Mar 29, 12:55 pm, kronec...@xxxxxxxxxxx wrote:



On Mar 30, 5:29 am, Greg Berchin <gberc...@xxxxxxxxxxxx> wrote:

On Sun, 30 Mar 2008 09:13:26 -0700, "miner_tom" <tom_nos...@xxxxxxxxx>
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.
.

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