i am design low pass filter RC filter for attenuate the incoming signal above 3000hz to 300 times of the input(i.e 1/300 of the input at 0hz) so i need to know the calculation for this attenuation i know cut of freq = 1/(2 pi RC) so i can calculate the rc but how about attenuation 1/300
Is it passive RC Yaseer !!!
If it is active with opamp then Av = 1 +(Rf/R1) can be set to the factor you want and recalculate the R's
So it is with oscillators , A is the gain and beta the phase shifting factor , condition : A > beta
For a simple first order RC low-pass filter: Voltage gain G = 1/sqrt(1 + f2/f02)
where f0 is the cut-off frequency.
You need f0 = 3000 Hz / sqrt(3002 - 1) which is approximately 10 Hz
Dear Yaser, what is your cut off frequency?. The cut off frequency is that frequency such that the output voltage= 0,707 of the input voltage. It is the -3dB point of the filter.
If you want to build your filter using one rc filter section, then the frequency response of the filter will be mag. (Vo/Vi)= 1/ sq.root ( 1+( w/wc)^2) with wc=1/rc, with r is the resistance and c the capacitance of the filter.
What is the requires stop frequency which is defined as the frequency at which the required stop band attenuation must be achieved.
Please specify your filter in a more elaborate form as i introduced in my post.
Single rc section may not able to satisfy your requirement.
best wishes
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