Understanding how to implement high-pass filter with cascaded biquad filters

Question

I'm trying to understand how to implement a high-pass filter using, quote:

a cascasded biquad direct form II IIR filter

with a cut-off of 8kHz.

DSP is a complete new field for me, and the terminologies are alien to me.

From reading the web, I roughly understand that:

1. A digital filter is roughly a function (a transfer function to be specific). Most common form of such function that I'd probably encounter are $H(z) = \frac{B(z)}{A(z)}$

2. A biquad filter is such when the order of $B(z)$ and $A(z)$ are 2. Typically, normalization is applied so that the constant term of $A(z)$ is 1.

3. cascading means feeding the output of previous output of the filter, to the input of the next iteration.

What I don't understand so far are:

1. In what domain do such filter operate? Time or Frequency domain?

2. Many other details I can't think of, but will pop up and confuse me when I notice my knowledge in such area is missing.

Answer 1

Answer to question 1 is very short: it operates in both. Filtering in the time domain will affect the frequency domain, and vice-versa.
If you meant to ask about implementation, the answer is similar: it can be implemented in both domains.

However for IIR filters the vast majority of times you’d want to operate in the time domain. Frequency domain processing comes with a set of things that one needs to be very careful about such as time-to-frequency domain computation, correct buffering, windowing and zero-padding, processing delay considerations, frequency-to-time domain computation to name a few.

Time-domain filtering is much easier and in the case of IIR filters the method of choice 99% of the time. I suggest you search this website, there are plenty of similar questions and answers on time-domain vs frequency-domain filtering.

I can’t answer question 2) because, well, there’s no question. When one does come to mind, please first use the search functionality to make sure it hasn’t been asked (and answered) before.