Shanks methods for least-mean squares filter design
Given the desired impulse response $h_d(n), n \geq 0$ for an IIR filter, the IIR filter to be designed can be written as
$$
H(z)=\frac{\sum_{k=0}^M b_k z^{-k}}{1+\sum_{k=1}^M a_k z^{-k}}=\sum_{k=0}^{\infty} h(k) z^{-k} .
$$
Find $b_k$ and $a_k$ such that the sum of squared errors between $h_d(n)$ and $h(n)$ is minimized.
(a) From Padé approximation, we can set
$$
\tilde{h}_d(n)=-\sum_{k=1}^N a_k h_d(n-k)
$$
Show that the equations determine $a_k$ by minimizing the following squared errors:
$$
E_1=\sum_{n=M+1}^{\infty}\left[h_d(n)-\tilde{h}_d(n)\right]^2
$$
That is, for $m=1,2, \cdots, N$
$$
\sum_{k=1}^N a_k \sum_{n=M+1}^{\infty} h_d(n-k) h_d(n-m)=-\sum_{n=M+1}^{\infty} h_d(n) h_d(n-m) .
$$
(b) To determine $b_k$, we first split $H(z)$ into $H_1(z)$ and $H_2(z)$, that is, $H(z)=H_1(z) H_2(z)$ where
$$
H_1(z) \frac{1}{1+\sum_{k=1}^N \tilde{a}_k z^{-k}} \text { and } H_2(z)=\sum_{k=0}^M b_k z^{-k}
$$
Let $v(n)$ be the impulse response of $H_1(z)$, that is,
$$
v(n)=-\sum_{k=1}^N \tilde{a}_k v(n-k)+\delta(n)
$$
Then
$$
\hat{h}_d(n)=\sum_{k=0}^M b_k v(n-k)
$$
After minimizing the sum of squared error
$$
E_2=\sum_{n=0}^{\infty}\left[h_d(n)-\hat{h}_d(n)\right]^2,
$$
Show that for $m=0,1, \cdots, M$
$$
\sum_{k=0}^M b_k \sum_{n=0}^{\infty} v(n-k) v(n-m)=\sum_{n=0}^{\infty} h_d(n) v(n-m) .
$$