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1

A zero at $z = e^{j\theta_1}$ corresponds to the transfer function $H_1(z) = 1 - e^{j\theta_1}z^{-1}$. You have an implicit plus sign (rather than minus) in your vectors b1 and b2.

3

Given your hardware constrains mentioned in the comments, your best shot is probably to do this as parallel second order section. Since the parallel sections are independent of each other, it's pretty straight forward to vectorize and it's also a little cheaper: each section has a complex conjugate pole pair but only one real zero. Things get a bit more ...

5

First of all, what is the order of your IIR filter? The highest order I have ever used was an order-10 IIR filter for a control loop application. I feel like it is unlikely that you need more that this. Second, it is a good idea to split your filter in second-order-sections (SOS) and cascade them , this usually fix most issues. https://www.dsprelated.com/...

5

The Jacobian is not computed numerically but analytically and then just evaluated. The frequency response of the IIR filter is $$H(e^{j\omega})=\frac{b_0+b_1e^{-j\omega}+\ldots+b_Me^{-jM\omega}}{1+a_1e^{-j\omega}+\ldots+a_Ne^{-jN\omega}}=\frac{B(e^{j\omega})}{A(e^{j\omega})}\tag{1}$$ Now you need the derivative with respect to the filter coefficients: \...

4

You're right, there usually is a better IIR filter (if you have enough data). The discrete-time Wiener filter is not "by definition" FIR. It is common to constrain the filter to the FIR case because it's often more straightforward to implement, and because such a filter can be made adaptive more easily. Also, in practice you often want to consider only a ...

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