# How do I calculate the cutoff frequency of a low pass rc circuit?

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.

• 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.
– LvW
Nov 28 '14 at 8:44
• 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. Nov 28 '14 at 9:03
• stack dsp forum does a lot of filter maths. Dec 10 '17 at 21:17
• analogzoo.com/2015/12/deriving-the-rc-filter-transfer-function Apr 27 '18 at 1:49

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.)

• Haven't you calculated the "corner" frequency above ? Nov 12 '17 at 8:08
• @AnshKumar As far as I know, these two terms have the same meaning, at least in case of a first-order LPF. Nov 16 '17 at 23:40

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: -

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

• 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.
– Pana
Dec 12 '19 at 10:37
• It's a circular argument, a chicken or egg situation. I regard both as equally valid. Dec 12 '19 at 10:41
• 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
– Pana
Dec 12 '19 at 10:45
• 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. Dec 12 '19 at 10:50