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

There is an example where he takes the ratio of the two (10/2 = 5) and then
justifies it by saying 20dB - 6dB = 14dB, with a table of other values,
which make me more confused)

So, if anyone can answer my question, which is
1. why is 20db/decade = 6db/octave and
2. what's the significance? (a short pointer would be good, motivate me to
keep reading :-P),
I'd be eternally grateful (and have more hairs left on my head...)

Thanks,

Kelv

My heavens. An actual TECHNICAL question in this ng. What is this world
coming to?

This is a common way to specify filters. With a single "pole" (reactive
element) the slope of the curve in the attenuation range is 6 dB/octave or
20 dB/decade. Here is what it means.


6 dB = 2:1 in voltage and 1 octave = 2:1 in frequency.

20 dB = 10:1 in voltage and 1 decade - 10:1 in frequency.

Thus, a single pole falls just as fast in voltage as you increase the
frequency. If you double (or triple, or quadruple) the number of reactive
elements, the "fall" is that multiple times the dB fall for a single
element. A three-pole filter will fall 18 dB/octave or 60 dB/decade.

This is a relatively simple-minded explanation; I could post Lesson G in my
freshman electronics class which takes a little over an hour and a half to
explain the same phenomenon, with charts and graphs all a-writing.

{;-)


Jim





"Kelvin Chu" wrote in message
...
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

There is an example where he takes the ratio of the two (10/2 = 5) and
then
justifies it by saying 20dB - 6dB = 14dB, with a table of other values,
which make me more confused)

So, if anyone can answer my question, which is
1. why is 20db/decade = 6db/octave and
2. what's the significance? (a short pointer would be good, motivate me to
keep reading :-P),
I'd be eternally grateful (and have more hairs left on my head...)

Thanks,

Kelv

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.]