Why is the frequency cut-off 0.707 in filters

  • Thread starter franck_hunter
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  • #1
franck_hunter

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Hi Guys,

There is a formula which shows that the frequency cut-off (0.707) is found from 1/(2*Pi*RC). But why?

Could someone show me the proof, so that I could understand?


Thx in advance
 

Answers and Replies

  • #2
turin
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This is for a very simple RC filter, I imagine. Soe you've got the low pass and the high pass. If you want a deep understanding, you should know how to deal with the impedances and the transfer fucntion.

The impedences come from the Laplace transform of the characteristic equation of the circuit. It turns out that the impedance for a capacitor is ZC = 1/sC and the impedance for a resistor is R. That s is a complex frequency. The real part is a characteristic decay and the imaginary part is a characteristic frequency.

To get the transfer function, you use the simple voltage divider model. For the low pass filter, the capacitor is the shunt, so the voltage divider gives the transfer function as:

H(s) = ZC/(ZC + R) = 1/(1 + sRC)

If the input doesn't introduce any inherent decay (it is steady state), the system responds as:

H(jw) = 1/(1 + jwRC) = 1/(1 + j2πfRC)

This is a complex valued function of the variable, f, which is the steady-state frequency of the input. It has a power transfer associated with it that is the square of the magnitude of this amplitude transfer:

|H(f)|2 = 1/(1 + 4π2f2R2C2)

When this equals 1/2, that is considered the cufoff point (3 dB point), and the frequency associated with it is the cufoff frequency (3 dB frequency). Notice that the squareroot of 1/2 = 0.707.

The freqency that satisfies the condition is easily seen from the power transfer function as:

f3dB = 1/(2πRC)

It is a very similar treatment for the high pass.
 
  • #3
franck_hunter
Thank you very much for the complete explanation.:wink:
 

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