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I want to calculate the cutoff frequency for a specific filter, but I can't find any formula for that.

I know the formula for the cutoff frequency of a low pass filter:

$$f_c=\frac{1}{2\pi RC}$$

But how is that derived in the first place? I don't have a regular low pass filter, but something similar that I want to calculate the cutoff frequency of.

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    \$\begingroup\$ In your question You speak about a "specific filter". Please note that (a) the formula for fc you have given applies to a first order lowpass only and (b) that the -3dB criterion does NOT apply for all kind of filters. Therefore, please tell us how your "specific filter" looks like. \$\endgroup\$
    – LvW
    Nov 28 '14 at 8:44
  • \$\begingroup\$ Can you specify the filter by providing a circuit diagram. I can explain the cut off frequency for a simple RC low pass filter and others already have but without knowing what filter you have in mind its difficult to be specific for your case. \$\endgroup\$ Nov 28 '14 at 9:03
  • \$\begingroup\$ stack dsp forum does a lot of filter maths. \$\endgroup\$ Dec 10 '17 at 21:17
  • \$\begingroup\$ analogzoo.com/2015/12/deriving-the-rc-filter-transfer-function \$\endgroup\$
    – 1110101001
    Apr 27 '18 at 1:49
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The specific formula applies only for a first order RC low pass filter. This is derived from its frequency response:

$$H(j\omega)=\frac{1}{1+j\omega RC}$$

The cutoff frequency is defined as the frequency where the amplitude of \$H(j\omega)\$ is \$1\over\sqrt2\$ times the DC amplitude (approximately -3dB, half power point).

$$|H(j\omega_c)|=\frac{1}{\sqrt{1^2+\omega_c^2R^2C^2}}=\frac{1}{\sqrt{2}}\cdot|H(j0)|=\frac{1}{\sqrt{2}}$$

Solve it for \$\omega_c\$ (cutoff angular frequency), you'll get \$1\over RC\$. Divide that by \$2\pi\$ and you get the cutoff frequency \$f_c\$.

If you know the frequency response of your filter, you can apply this method (given that the cutoff frequency is defined as above). Obviously, for high-pass filters for example, you calculate with the value for \$\omega\to \infty\$ as opposed to the DC value (always the maximum of the amplitude response, relative to which there is a 3dB decrease in amplitude at the cutoff frequency.)

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  • \$\begingroup\$ Haven't you calculated the "corner" frequency above ? \$\endgroup\$
    – Vedanshu
    Nov 12 '17 at 8:08
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    \$\begingroup\$ @AnshKumar As far as I know, these two terms have the same meaning, at least in case of a first-order LPF. \$\endgroup\$
    – hryghr
    Nov 16 '17 at 23:40
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For a simple RC low pass filter, cut-off (3dB point) is defined as when the resistance is the same magnitude as the capacitive reactance: -

\$R = \dfrac{1}{2\pi f C}\$

It's a simple math trick to say: -

\$f = \dfrac{1}{2\pi R C}\$

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  • \$\begingroup\$ It is not only a pure definition. The main conception of definition cut-off frequency is that the output power becomes half of the input power or equally since P = V^2 / Z the amplitude of output voltage becomes square root of 0.5 of input voltage. Then you have a simple voltage divider circuit as Vout / Vin = Z1 / Z1 + Z2 and can easily find the cut-off frequency by equaling it to square root of 0.5. \$\endgroup\$
    – Pana
    Dec 12 '19 at 10:37
  • \$\begingroup\$ It's a circular argument, a chicken or egg situation. I regard both as equally valid. \$\endgroup\$
    – Andy aka
    Dec 12 '19 at 10:41
  • \$\begingroup\$ Yes, right. But please pay attention that OP needs to know the concept of obtaining cut-off frequency to be able to calculate the cut-off frequency of their own filter: I don't have a regular low pass filter, but something similar that I want to calculate the cutoff frequency \$\endgroup\$
    – Pana
    Dec 12 '19 at 10:45
  • \$\begingroup\$ I believe I have answered the OP's questions and he/she has accepted a different answer (back in 2014) i.e. 5 years ago. \$\endgroup\$
    – Andy aka
    Dec 12 '19 at 10:50

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