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equation frequency in - 3 dB. Bode help !!!!!!

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biolycans

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Hi,

I found an equation related to -3 dB

F = 1 / (2 * pi * CL * Ro)

Can you explain me how can I find the deduction ?

Regards,
 

In a simple RC filter, the -3dB point is just the frequency where the resistance of the resistor equals the reactance of the capacitor R = [1 / (2 * pi * f* C)]. Then, for example, for an RC high-pass filter the output would then be Vo = Vin [R / (R+ 1/(2*pi*f*C))] where f is the -3dB frequency.
 

Is your deduction meaning attenuation? If yes, please see below.

When you have a simple transfer function like H(s) = 1/(1+sRC). That means |H(jw)|=1/sqrt(1+(wRC)^2).
At DC (f=0), |H(jw)| = 1.
At -3dB frequency, which w = 1/RC, you can deduct that |H(jw)| = 1/sqrt(2), which is -3dB.
 

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