[Sonnova]

On Sat, 31 May 2008 20:13:25 -0700, jwvm wrote
(in article ):

On May 31, 11:39 am, Sonnova wrote:
On Fri, 30 May 2008 07:49:25 -0700, jwvm wrote
(in article ):



On May 29, 10:56 am, Sonnova wrote:
On Wed, 28 May 2008 07:49:55 -0700, Dave wrote
(in article ):

"Arny Krueger" wrote in message
...
I thought that the AudioEngine 2 review begged comparison with this one
for
the classic NHT SZ:

http://www.stereophile.com/loudspeak...04/index9.html

I am curious about the high-pass filter employed in the article. The
author
notes that the R in the RC filter is in parallel with the amplifier's
output
impedance of 51K. Given that the amp's impedance is upstream of the
capacitor, is this right? Or did he actually construct a filter with a
rolloff frequency of 106Hz?

Vin Vout
o------------||-----------------o
0.1uF "C" of |
\ RC filter \
/ /
51K \ Amp \ 15K "R" of RC filter
/ /

o-------------------------------o

That's a classic single-pole high-pass filter "Pi" network. But such a
filter
is only 3dB/octave - not very steep.

I am sure that you meant 6 dB/octave. :-)

Doesn't that rather depend upon whether your talking about voltage or power?
It's been years since I've dealt with this stuff, but it seems to me that
the slope of a single-pole filter gives 3dB with voltage and 6dB with power.
I could be misremembering though.

Sorry but that is not correct. Measurements in decibels are power
ratio measurements. The 20log(Vout/Vin) type calculation can be
rewritten as 10log((Vout/Vin)^2) again showing the power relationship.
Note also that 6 dB/octave is the same slope as 20 dB/decade. The 3 dB
figure is the loss at the corner frequency of the first-order filter.

That must have been where I got the 3dB figure from, then. Sorry. A single
pole filter is 6dB/octave