Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

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

Status
Not open for further replies.

biolycans

Member level 2
Member level 2
Joined
Aug 24, 2013
Messages
50
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Visit site
Activity points
478
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.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top