- #1

franck_hunter

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter franck_hunter
- Start date

- #1

franck_hunter

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

franck_hunter

Thank you very much for the complete explanation.

Share: