3 dB cut-off frequency of first-order IIR high-pass filter
A discrete-time first-order high pass filter with unity gain at Nyquist and a zero at DC is described by the following difference equation:
$$y[n]=\frac{1+\alpha}{2}\big(x[n]-x[n-1]\big)+\alpha y[n-1],\qquad -1&lt;\alpha&lt;1\tag{1}$$
Its transfer function is given by
$$H(z)=\frac{1+\alpha}{2}\frac{1-z^{-1}}{1-\alpha z^{-1}}\tag{2}$$
Evaluating the squared magnitude of $(2)$ on the unit circle $z=e^{j\omega}$ and equating it to $\frac12$ ($-3$ dB) results in the following relation between $\alpha$ and the $3$ dB cut-off frequency $\omega_c$:
$$\begin{align}\big|H(e^{j\omega_c})\big|^2&amp;=\frac{(1+\alpha)^2}{4}\frac{\left|1-e^{-j\omega_c}\right|^2}{\left|1-\alpha e^{-j\omega_c}\right|^2}\\&amp;=\frac{(1+\alpha)^2}{4}\frac{2\big(1-\cos(\omega_c)\big)}{1-2\alpha\cos(\omega_c)+\alpha^2}\stackrel{!}{=}\frac12\tag{3}\end{align}$$
Eq. $(3)$ results in a quadratic equation for $\alpha$ with the solution
$$\alpha=\begin{cases}\displaystyle\frac{1-\sin(\omega_c)}{\cos(\omega_c)},&amp;\omega_c\in(0,\pi)\setminus \frac{\pi}{2}\\0,&amp;\omega_c=\frac{\pi}{2}\end{cases}\tag{4}$$
(where the requirement $|\alpha|&lt;1$) has been taken into account).
For $\omega_c=\pi/2$ we obtain $\alpha=0$ and the corresponding filter is a simple $2$-tap FIR filter. All other cut-off frequencies $\omega_c\in(0,\pi)$ result in IIR filters.
The figure below shows the magnitude responses of $9$ high-pass filters with specified cut-off frequencies $0.1\pi,0.2\pi, \ldots,0.9\pi$. The corresponding values for $\alpha$ were computed according to Eq. $(4)$.