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

On Mar 30, 6:05 pm, "Ron N." <rhnlo...@xxxxxxxxx> wrote:
On Mar 30, 5:31 pm, Rick Lyons <R.Lyons@xxxxxxxxxxxxxxx> wrote:



On Sat, 29 Mar 2008 10:45:29 -0700 (PDT), dbd <d...@xxxxxxxx> wrote:

(Snipped by Lyons)

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

Hi Dale,
Are you saying that if we use the sequence
represented by the dots in Figure 5-35(c) as
the coefficients of a tapped-delay line FIR
filter, that this filter will have linear phase?

If the sequence is a uniform sampling of a symmetric waveform
which is bandlimited to below the new sampling rate represented
by the dots, then, according to reconstruction theory, yes.
However, remember that this perfectly bandlimited waveform
can't be time limited as well, therefore any finite length FIR
filter is really a truncated approximation of some symmetric
infinite impulse response. Any phase non-linearity is thus
"really" due to finite windowing, not the apparent asymmetry
of the sample points.

Windowing often makes a difference, even if you don't notice
that some operations imply them

IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

Limits! We don't have no limits! We don't have to show you no
steeenking limits!
(with apology to those who treasure the Sierra Madre...)

Wow, Ron, does this mean someone was listening?

Dale B. Dalrymple
Dale B. Dalrymple
.