kerem Posted October 5, 2006 Report Share Posted October 5, 2006 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. Quote Link to comment Share on other sites More sharing options...
aartak Posted October 15, 2006 Report Share Posted October 15, 2006 This formula is wrongit will not vork with your circuitattenuation of your circuit is 40db/decade or 12db/octaveit means when you double frecuency it passes 4 time less voltagefor example its module of transfer function is~1 at 25MHz0.25 at 50Mhz1/16 at 100Mhzin your formula you must substitute 2 for n to get the correct result Quote Link to comment Share on other sites More sharing options...
ante Posted October 15, 2006 Report Share Posted October 15, 2006 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. Quote Link to comment Share on other sites More sharing options...
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