I have been given the specifications for a digital highpass filter (stopband, passband, stopband attenuation and maximum passband ripple).

I am expected to design a prototype lowpass filter in the analogue domain, use frequency transformation to convert it to a highpass, then use bilinear transformation to convert it into a digital filter. I am using a Butterworth filter for prototype.

However I have not been given the sampling frequency(or period) for pre-warping. I have seen textbooks take period = 1 as it is arbitrary, however I don't understand why. Any confirmation that this is okay, and explanations will be appreciated!


1 Answer 1


The sampling frequency is not arbitrary and important to specify for pre-warping. This is clear when reviewing how each analog frequency in the bilinear transform is mapped one-one to a digital frequency according to:

$$f = \frac{F_d}{F_s} = \frac{1}{\pi}\tan^{-1} \bigg(\pi\frac{F_a}{F_s}\bigg) \tag{1} \label{1}$$


$f = F_d/F_s$ is the normalized digital frequency in units of cycles/sample,
$F_a$ is an analog frequency in Hz
$F_s$ is the sampling rate in Hz

Observe that $tan^{-1}(\infty) = \pi$ and thus the Bilinear Transform "warps" the entire analog frequency domain from $F=0$ to $F= \infty$ to the digital frequency domain of $f=0$ to $f=1$ (where $f=1$ is the sampling rate in units of normalized frequency), but the specific mapping of each analog frequency over that range is dependent on the value of $F_s$ that is used, as given by \ref{1}.

Normalizing the discrete-time frequency domain is commonly done such that the resulting spectrum associated with DC to the sampling rate extends from $f =0 \text{ to } 1$ cycles/sample, which may be the source of the OP's confusion.

  • 1
    $\begingroup$ ya know, i am a big believer in ordinary frequency (instead of angular frequency) in the continuous-time domain. but for some reason (i think the unit circle) i think normalizing angular frequency is more handy. for me it's always $$ \omega = \frac{2 \pi F_d}{F_s} = 2 \arctan\left( \pi \frac{F_a}{F_s} \right) $$ so it would be in the unitless radians per sample and would occur at the angle of $\omega$ on the unit circle. $\endgroup$ Jul 15, 2021 at 0:13
  • $\begingroup$ @robertbristow-johnson agreed, for digital signals I also find angular frequency to be the simplest path. $\endgroup$ Jul 15, 2021 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.