# Calculating the cutoff frequency of a highpass digital filter

I am new to DSP and I am trying to find the cutoff frequency of a HP digital filter. I know the equation that describes the system, its frequency response: $$H(e^{jω})= 1 - \frac{e^{jω}}{2} - \frac{1}{2e^{jω}}$$ and of course amplitude response / phase diagrams.

I have found (p. 8) that for analog filters I can calculate $$\frac{1}{\sqrt{2}} |H(e^{jω})|$$. However I don't know how to do this for digital filters. Reading this makes it look like it's the same procedure, although this seems (to me) like a contradiction to

Digital filters are less standardized, and it is common to see 99%, 90%, 70.7%, and 50% amplitude levels defined to be the cutoff frequency.

of the first source. Am I confusing something? Please keep in mind that I am new to DSP. Thank you.

Concerning the cut-off frequency, there's really not much of a difference between analog and digital filters. $$3\textrm{ dB}$$ is common, but any other value is fine, as long as you specify it and people know what you're talking about.

In the case of the given filter, it's quite straightforward to compute any cut-off frequency. Let's choose the $$3\textrm{ dB}$$ cut-off frequency:

$$H(e^{j\omega})=1-\frac12\left(e^{j\omega}+e^{-j\omega}\right)=1-\cos(\omega)\tag{1}$$

Since $$H(e^{j\omega})$$ is real-valued and since $$H(e^{j\omega})\ge 0$$, we can simply solve the equation

$$1-\cos(\omega_c)=\frac{2}{\sqrt{2}}=\sqrt{2}\tag{2}$$

Note that the maximum of $$H(e^{j\omega})$$ is $$2$$, so the value of $$H(e^{j\omega})$$ at the $$3\textrm{ dB}$$ cut-off frequency $$\omega_c$$ is $$\sqrt{2}$$.

From $$(2)$$ we get $$\omega_c\approx 1.9979$$. The actual cut-off frequency in Hertz is given by

$$f_c=\frac{\omega_c}{2\pi}f_s\tag{3}$$

where $$f_s$$ is the sampling frequency.