On Mon, 25 Apr 2005 23:45:30 +1000, Kelvin Chu wrote:
: Hi Group,
:
: I'm self-reading this RF book (practical rf circuit design for modern
: wireless systems, artech house), and up to chapter 2, but stuck on this:
:
: in the book, it says 20db/decade = 6db/octave
:
: now, I understand that 1 decade = 10 fold increase, and 1 octave = 2 fold
: increase, but I can't for the life of me get why 20db/decade = 6db/octave


Assuming that the given dB versus log frequency relation is always
linear, imagine you start off at some arbitrary point on the frequency
scale, say 1 Hz. If you increase frequency by 3 octaves, you go in
the sequence 1 Hz - 2 Hz - 4 Hz - 8 Hz.

At 8 Hz, you've changed the gain / attenuation by 3 x 6 = 18 dB. The
next octave step would take you to 16 Hz, and that overshoots the
decade step of 10 Hz. If you only want to go as far as 10 Hz, then an
interpolation says you need to add 2 dB of gain / attenuation to go
from 8 Hz to 10 Hz. That is, you've changed the gain / attenuation
from 18 dB to 20 dB, and you therefore have 20 dB of change per
decade.

To do the interpolation from 8 Hz to 10 Hz properly, you have to
recall that you're really interpolating a straight line relationship
on a log-log scale. Mathematically, you'd write 6 dB x
(log(10/8)/log(16/8)), which is 1.93 dB instead of 2 dB. Apparently
the author has rounded the result of 19.93 dB per decade up to 20 dB -
a more easily remembered figure.

[Guvf fcnpr erfreirq sbe Travhf Jvfrzna gb pbzzrag.]