# Is the sampling period arbitrary for a bilinear transform, and why?

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!

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:

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

Where:

$$f = F_d/F_s$$ 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.