- #1

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

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- Thread starter franck_hunter
- Start date

- #1

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

- #2

turin

Homework Helper

- 2,323

- 3

The impedences come from the Laplace transform of the characteristic equation of the circuit. It turns out that the impedance for a capacitor is Z

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) = Z

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)|

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:

f

It is a very similar treatment for the high pass.

- #3

Thank you very much for the complete explanation.

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