Linkwitz-Riley transient response

On Wed, 06 Dec 2006 14:04:54 -0600, "jeff227" <rocksonics@earthlink.net>
wrote:

>So if the phase BETWEEN the LP and HP outputs is always constant why
>doesn't summing the LP and HP outputs reproduce a square wave?  What am I
>not understanding here?

Because a Linkwitz-Riley pair does not sum to unity, it sums to allpass.
In other words, when you sum the transfer functions of the lowpass
section and the highpass section, you don't get "1", you get the
transfer function of an allpass filter.  Allpass filters have unity
magnitude response but nonlinear phase response.  That nonlinear phase
response is what causes the poor summed transient response that you
observe.

-- Greg

>Because a Linkwitz-Riley pair does not sum to unity, it sums to allpass.
>In other words, when you sum the transfer functions of the lowpass
>section and the highpass section, you don't get "1", you get the
>transfer function of an allpass filter.  Allpass filters have unity
>magnitude response but nonlinear phase response.  That nonlinear phase
>response is what causes the poor summed transient response that you
>observe.


Greg thank you for turning on the light!!  Now I understand!

Thanks everyone for your comments.  Sometimes it takes me a while to "get
it".

On Wed, 06 Dec 2006 15:30:41 -0500, Jerry Avins <jya@ieee.org> wrote:

>Sharp corners in frequency make for ringing in time. It's harder to show 
>the reverse.
>
>Jerry

It's not that bad, even for a single-stage FIR.   This was pointed out
by a really insightful DSP guru early in my career:  The steepness of
the step response depends on the width of the main lobe in the filter.

It isn't too hard to imagine a simple FIR convolving with a step
function and that the steepness of the output step is dependent almost
completely on the width of the main lobe.   The "ringing" is then just
an artifact of the sidelobes.   That was a really useful insight for
me as it gives yet another meaningful connection between the impulse
response and other filter characteristics.   It's not a big leap to
realize that a narrow main lobe is also a sharp filter in frequency,
so there's an intuitive connection between ringing in time and
sharpness in frequency (as well as the steepness of the step
response).   They're all related.


Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org

On Wed, 06 Dec 2006 18:33:45 -0500, Jerry Avins <jya@ieee.org> wrote:

>Vladimir Vassilevsky wrote:
>> 
>> 
>> Jerry Avins wrote:
>> 
>> 
>>> Basically, it's not phase error that causes the ringing, but the sharp 
>>> cutoff.
>> 
>> To be more exact, the ringing is caused by the cutoff sharpness vs delay 
>> in the filter. One can build a filter as sharp as he like and without 
>> any ringing. That can be either FIR or IIR structure, it does not matter.
>> 
>>> Sharp corners in frequency make for ringing in time. It's harder to 
>>> show the reverse.
>> 
>> 
>> Imagine a cascade of moving average filters for FIR or a cascade of 1st 
>> order filters for IIR. The cutoff can be as sharp as you like depending 
>> on the number of stages, the phase will be nonlinear for IIR, however 
>> there will be no ringing at all.
>
>Interesting. A cascade of moving average filters is a binomial filter, a 
>digital approximation to a Gaussian. Do Gaussians not ring? As 2-D 
>filters, they are separable. What other remarkable properties do they have?
>
>Jerry

Hi Jer,
  well, if we have a Gaussian time-domain sequence 
whose duration is finite, the Fourier transform of 
that sequence certainly does have ringing in 
the form of "sidelobes" in the freq domain.

The Fourier transform of an infinitely-long 
Gaussian function is itself a Gaussian function.
However we can't have infinitely-long functions 
(sequences) in our computers. 

See Ya,
[-Rick-]