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