As descipbed in this great answer (thank you Matt L.), the suggested exact formula was very precise. I need a first order high-pass with similar precision.

EDIT: (clarification) The low-pass filter discussed in the link above was this one, derived from the simple RC filter. What I'm looking for is the discrete version of the simple high-pass RC filter with a formula for getting the exact alpha term (that's what Matt L. did) for an amplitude response of -3dB at a given cutoff frequency.

I've tried and measured the low pass version and it worked as expected.

  • $\begingroup$ could you put numbers to what you mean with "similar precision"? What kind of high-pass filter, i.e. cutting of where exactly? You don't really have a lot of degrees of freedom in a first-order system. To be exact, you have exactly two parameters you can adjust, at all. Now, I trust you have looked up the formulas describing the frequency response resulting from the choice of these two numbers, so could you try to specify what the question is? $\endgroup$ Commented Jun 28, 2020 at 0:21
  • $\begingroup$ Thank you Mr Müller, I've edited my question. $\endgroup$ Commented Jun 28, 2020 at 5:02

2 Answers 2


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:

$$y[n]=\frac{1+\alpha}{2}\big(x[n]-x[n-1]\big)+\alpha y[n-1],\qquad -1<\alpha<1\tag{1}$$

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 $\omega_c$:

$$\begin{align}\big|H(e^{j\omega_c})\big|^2&=\frac{(1+\alpha)^2}{4}\frac{\left|1-e^{-j\omega_c}\right|^2}{\left|1-\alpha e^{-j\omega_c}\right|^2}\\&=\frac{(1+\alpha)^2}{4}\frac{2\big(1-\cos(\omega_c)\big)}{1-2\alpha\cos(\omega_c)+\alpha^2}\stackrel{!}{=}\frac12\tag{3}\end{align}$$

Eq. $(3)$ results in a quadratic equation for $\alpha$ with the solution

$$\alpha=\begin{cases}\displaystyle\frac{1-\sin(\omega_c)}{\cos(\omega_c)},&\omega_c\in(0,\pi)\setminus \frac{\pi}{2}\\0,&\omega_c=\frac{\pi}{2}\end{cases}\tag{4}$$

(where the requirement $|\alpha|<1$) 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. $(4)$.

enter image description here

  • $\begingroup$ When taking the quadratic, both roots give identical magnitude, but inverted phase response. Aside from an awareness of the filter type (ie. LP, HP), is there a guide on which root to pick? $\endgroup$
    – bazz
    Commented Mar 17, 2023 at 19:48
  • $\begingroup$ @bazz: The correct root is determined by the stability condition $|\alpha|<1$. The other root always results in an unstable solution. $\endgroup$
    – Matt L.
    Commented Mar 18, 2023 at 7:47
  • $\begingroup$ Thank you. By putting eq. 2 into factored form we can observe how $\alpha$ plays a role on the poles and zeros. For this equation we have a pole directly at $\alpha$ which is why alpha must be in [-1, 1], right? It's to be kept within the unit circle. $\endgroup$
    – bazz
    Commented Mar 18, 2023 at 19:00

Here are couple examples:

% R is the resistance value (in ohms)
% C is the capacitance value (in farrads)
% fs is the digital sample rate (in Hz)

% Constants
RC = R * C;
T  = 1 / fs;

% Analog Cutoff Fc
w = 1 / (RC);

% Prewarped coefficient for Bilinear transform
A = 1 / (tan((w*T) / 2));

% using Bilinear transform of
%             1          ( 1 - z^-1 )
% s -->  -----------  *  ------------
%         tan(w*T/2)     ( 1 + z^-1 )

b(1) = (A)/(1+A);
b(2) = -b(1);
a(2) = (1-A)/(1+A);

and an alternative implementation could be:

w = 2.0 * pi * fc/fs;
cx = cos(w);
sx = sin(w);
b0 =   cx + 1;
b1 = -(cx + 1);
a0 =   cx + sx + 1;
a1 =   sx - cx - 1;
  • $\begingroup$ Great answer! It was really hard to decide the accepted one between yours and Matt's but I had to choose one. Matt's answer was complemented with clear explanations and well presented. BTW I evaluated and plotted all three propositions and they were correct and identical. $\endgroup$ Commented Jun 28, 2020 at 10:48

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