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



On Mar 30, 5:25 am, Greg Berchin <gberc...@xxxxxxxxxxxx> wrote:
On Sat, 29 Mar 2008 12:29:29 -0500, Greg Berchin

<gberc...@xxxxxxxxxxxx> 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).

I'm not sure where you got the last claim. The magnitude of
the spectrum of the symmetric FIR IR should be positive and
relatively large around the DC bin, whereas the magnitude of
the spectrum of the antisymmetric FIR IR should be zero at
the DC bin and small in the neighborhood . Around Fs/2, the
two magnitudes should also be different for similar reasons.

Since the magnitudes of the two spectrum aren't equal, the
phase of the vector sum will have some non-linear "twist",
depending on the changing magnitude ratios, probably more
apparent around Fs/2.




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



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