Attenuation

kerem1

Feb 18, 2005
42
Joined
Feb 18, 2005
Messages
42
Hi, I am very confused about this subject of attenuation. I have a circuit posted, and it is working at a center frequency of 25Mhz. And I need to calculate the attenuation at 2nd, 3rd, 4th and 5th harmonics of 25Mhz. From internet I found this formula:
A(db) = -10 log(1+((w/wc)^2n)) where w is the desired freq. and wc is the center frequency, n is the number of elements (inductors and capacitors) in the circuit.

 
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aartak

Oct 13, 2004
21
Joined
Oct 13, 2004
Messages
21
This formula is wrong
it will not vork with your circuit
attenuation of your circuit is 40db/decade or 12db/octave
it means when you double frecuency it passes 4 time less voltage
for example its module of transfer function is
~1 at 25MHz
0.25 at 50Mhz
1/16 at 100Mhz
in your formula you must substitute 2 for n to get the correct result

 

ante1

Jan 24, 2004
4,138
Joined
Jan 24, 2004
Messages
4,138
Hi sergen,

I have changed your image to a more suitable format for posting (PNG).
The size is now 10k instead of 2019.4k (200 times smaller) and it is easier to see to.
Please use this format, the pages loads much faster and members using dialup will be more comfortable.

View attachment 39930

 
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