Approximation and IIR Filters

Question

Consider the discrete-time time-invariant system with input $x[n]$ and output $y[n]$ satisfying

$$y[n] = \sum_{k=1}^5{x[n-k]}$$

Consider approximating the desired system with a second-order IIR system with system function

$$H^{'}(z)= \frac{1}{1+a_1z^{-1}+a_2z^{-2}}$$

Use the following error criterion:

where $h_d$ is the desired impulse response. $$E = \sum_{n=-\infty}^{\infty}\left\lvert h_d[n]+a_1h_d[n-1]+a_2h_d[n-2]\right\rvert^2$$

How can the particular error function be useful in solving the system of equations or the desired impulse response?

Do I need to take derivative w.r.t both $a_1$ and $a_2$ and make it 0 to get the system of equations?

Now this is what I am getting: $$0 = \sum_{n=-\infty}^{\infty}( h_d[n]h_d[n-1]+a_1h_d[n-1]h_d[n-1]+a_2h_d[n-2]h_d[n-1])$$ and
$$0 = \sum_{n=-\infty}^{\infty}( h_d[n]h_d[n-2]+a_1h_d[n-1]h_d[n-2]+a_2h_d[n-2]h_d[n-2])$$

Now how can I solve these system of two equations and how can I approximate the desired impulse response from that?

Answer 1

I think it's instructive to try to understand how this error function is derived and why it makes sense. First, the desired impulse response $h_d[n]$ is given implicitly by the input-output relation of the desired system:

$$y[n]=\sum_{k=1}^{5}x[n-k]=\sum_{k=-\infty}^{\infty}h_d[k]x[n-k]\tag{1}$$

The given error function is the error function minimized by Prony's method for the design of IIR filters. We try to approximate a given transfer function $H_d(z)$ by an IIR filter $H(z)=B(z)/A(z)$:

$$B(z)\stackrel{!}{=}H_d(z)A(z)\tag{2}$$

For the given example we get (with $B(z)=1$)

$$1\stackrel{!}{=}\left(h_d[0]+h_d[1]z^{-1}+\ldots+h_d[N-1]z^{-(N-1)}\right)\left(1+a_1z^{-1}+z_2z^{-2}\right)\tag{3}$$

For equation $(3)$ to be satisfied, we need all coefficients associated with negative powers of $z$ to vanish:

$$\begin{align}h_d[1]+h_d[0]a_1&\stackrel{!}{=}0\\ h_d[2]+h_d[1]a_1+h_d[0]a_2&\stackrel{!}{=}0\\ h_d[3]+h_d[2]a_1+h_d[1]a_2&\stackrel{!}{=}0\\\vdots\end{align}\tag{4}$$

In practice, we can solve $(4)$ in an approximate way by minimizing the sum of the squares of the left-hand side, leading to the given error function

$$E=\sum_{n=-\infty}^{\infty}\big(h_d[n]+h_d[n-1]a_1+h_d[n-2]a_2\big)^2\tag{5}$$

Finally, the optimal coefficients $a_1$ and $a_2$ are obtained by taking the derivative of $(5)$ w.r.t. $a_1$ and $a_2$ and equating them to zero. This results in two linear equations with two unknowns.