Calculating the true values of biquad filter coefficients from normalized form

Question

I have a C code in which the IIR filter(BiQuad) uses only 4 (normalized) coefficients per stage (a2/a0,a1/a0,b2/b0,b1/b0) instead of 6 coefficients (a0,a1,a2,b0,b1,b2). I know that the value of a0 is always 1. So, it is easy to find the true values of the a2,a1 coefficients (since a0 value is 1). But, i wanted to know the true values of b0, b1, b2.

So, I just want to know is there any way to retrieve/derive the true values of "b" coefficients? One of my friends told me that we can get the values by using Matlab / octave? But, I have no idea how to do it?

Answer 1

Note that a biquad has $5$ degrees of freedom (not $6$), because $a_0$ can always be chosen as $a_0=1$ without loss of generality:

$$\begin{align}H(z)&=\frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}\\&=b_0\cdot\frac{1+\hat{b}_1z^{-1}+\hat{b}_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}},\qquad \hat{b}_1=\frac{b_1}{b_0},\;\;\hat{b}_2=\frac{b_2}{b_0}\tag{1}\end{align}$$

Eq. $(1)$ shows that $b_0$ is just an overall gain (for fixed values of $\hat{b}_1$ and $\hat{b}_2$), as you've noted yourself.

If you don't know the formula that produced the filter coefficients, there is generally no way to know the intended gain. As mentioned in Dan's answer, the gain is often chosen such that quantization effects are minimized. It can often be compensated for after all computations have been performed.

In some cases, however, it is possible to estimate the intended gain. E.g., if the biquad is a low pass filter, it is reasonable to assume that the original frequency response has a value of $1$ at DC. This means that

$$b_0\frac{1+\frac{b_1}{b_0}+\frac{b_2}{b_0}}{1+a_1+a_2}=1\tag{2}$$

Similarly, for a high pass filter a common scaling makes sure that the frequency response equals $1$ at Nyquist, which implies

$$b_0\frac{1-\frac{b_1}{b_0}+\frac{b_2}{b_0}}{1-a_1+a_2}=1\tag{3}$$

You can come up with similar guesses for the intended scaling for other standard filter types, like band pass filters (unity gain at the center frequency) and band stop filters (unity gain at DC and/or Nyquist).

Answer 2

It is simply a scaling. If you use $b0=1$ for example, you have the values for $b2$ and $b1$ directly. In contrast if you use $b0=2$ then the filter would have twice the gain and the values for $b2$ and $b1$ would be double. It is that simple. Where you really need to pay attention to the actually scaling used is when you implement the filter with fixed point precision as incorrect scaling can result in much larger quantization noise, but otherwise it will be rather arbitrary similar to scaling your final output.