Summation of two IIR filters Actually my question is how to get the weighted average of two or three IIR filters given their transfer functions, which is nearly equal to this question. How to calculate the total transfer function of the summation of two IIR filters? For example, I have two filters as following
$$H_1(z) = \frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$
$$H_2(z) = \frac{b_0'+b_1'z^{-1}+b_2'z^{-2}}{1+a_1'z^{-1}+a_2'z^{-2}}$$
I can directly solve the sum of these two fractions as $H(z) = H_1(z)+H_2(z)$, but the problem is that my filter orders are generally around dozens, I have no idea how to do it.
I also know that I can put a signal through this two filters and sum the output together, but it requires double computational complexity.
Is there any other way to solve this problem? Thanks!
In theory you just do it the same way. The transfer function of a weighted sum of filters with weights $w_i$ is just
$$H_{total}(z)=\sum_iw_iH_i(z)\tag{1}$$
where $H_i(z)$ are the individual transfer functions.
However, in general the order of the total transfer function is the sum of the orders of the individual filters, unless some filters share the same poles. Consequently, the resulting filter order can become too large for a practical implementation. If the orders of the individual filters &quot;are generally around dozens&quot; then you have a problem, even using floating point arithmetic.