Omega is equal to 1 / RC !

Thread Starter

kaznov

Joined Mar 10, 2012
30
Hi guys

can someone please explain to me why omega equal to 1 / RC in a RC circuit because I found some info but still not fully understand it

Thanks alot
 

Papabravo

Joined Feb 24, 2006
21,159
In an RC circuit that is the so called "corner frequency" where the magnitude of the "ideal" response begins to roll off. It is also the point where the magnitude of the response is 3 dB down.
 

daviddeakin

Joined Aug 6, 2009
207
The 'gain' or tranfer characteristic for an RC filter is:

A(jw) = 1 / (1+jwRC)

To get only the magnitude of the voltage you need to square the real and imaginary parts, then take the square root:

Av = 1 / √(1+(wRC)^2)

When w=1/RC then you have:

Av = 1/√(1+1) = 1/√2

This is the point where the *power* gain is halved, and that is called the cut-off frequency.
 

Papabravo

Joined Feb 24, 2006
21,159
And a *power* gain of one-half says EXACTLY the same thing as:

  1. The power gain is "3 dB down" or
  2. The power gain is "minus 3 dB"
because
Rich (BB code):
10 * log (1/2)
10 * (-0.301019996...)
-3.01....
 
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