# How to set the frequency cut-off for a low pass RC filter?

How do I set the frequency for a low pass filter illustrated below? • Note that there is a separate StackExchange site for electronics questions: electronics.stackexchange.com May 28, 2014 at 21:35

You probably know that the frequency response of this first order low-pass filter is

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

Now compute its magnitude

$$|H(\omega)|=\frac{1}{\sqrt{1+\omega^2R^2C^2}}$$

In order to find the 3-dB cut-off point $\omega_0$, you have to solve $|H(\omega_0)|=1/\sqrt{2}$:

$$\frac{1}{\sqrt{1+\omega_0^2R^2C^2}}=\frac{1}{\sqrt{2}}\Rightarrow \omega_0=\frac{1}{RC}$$

And since the frequency in Hertz is $f=\omega/2\pi$, you get

$$f_0=\frac{1}{2\pi RC}$$

To set cut-off frequency, design $R$ and $C$ values such that $$f_c = \frac{1}{2\pi RC}$$$$\small{\text{(see Matt's answer to see how this equation is obtained)}}$$

Assuming a standard value for capacitor and then calculating the value for $R$ as, $$R=\frac{1}{2\pi f_cC}$$ is the usual method followed.