infinite impulse response - Is there a way to derive an FIR filter using an IIR filter?

I know there's a thread similar to this one here, but the OP is asking the reverse of what I'm trying to find here. I've done some research on the web with very few sources coming up with actual solutions to this problem. What techniques are used to give an approximation of an FIR filter given one or more IIR filters with say the same order?

Answer

Approximating the frequency response of an IIR filter or physical process using an FIR filter is useful in learning control. It is quite common to do FIR filter design based on frequency response specifications. You probably want to check out two standard papers on the subject:

[1] J. H. McClellan, T. W. Parks, and L. R. Rabiner, “A computer program for designing optimum FIR linear phase digital filters,” IEEE Trans. Audio Electroacoust., vol. 21, no. 6, pp. 506–526, 1973.

[2] L. R. Rabiner, “Techniques for Designing Finite-Duration Impulse-Response Digital Filters,” IEEE Trans. Commun. Technol., vol. 19, no. 2, pp. 188–195, Apr. 1971.

Broadly, you either do windowed direct sampling of your desired frequency response, or you use one of several optimization methods to achieve similar results. If you disregard the linear phase delay in an FIR, you can practically make the IIR and FIR responses identical, if the FIR filter order is high enough.

As an elaboration on one of the other answers given; if you have an IIR filter $G(z^{-1})$, then you can do FIR filter design by frequency sampling by taking $M$ samples of the frequency response of $G(z^{-1})$, denoted $\widehat{G}(k)$, and then taking the inverse discrete Fourier transform (IDFT) of $\widehat{G}(k)$. The unit impulse response $g(n)$ of $\widehat{G}(k)$ is \begin{align*} g(n) = \frac{1}{M} \sum\limits_{k=0}^{M-1} \widehat{G}(k) \text{e}^{j \frac{2 \pi k n}{M}} \; , \end{align*} where $n \in [0, M-1] \cap \mathbb{N}_{0}$. The FIR filter is then expressed in the $z$-domain as \begin{multline*} F(z^{-1}) = g(0) + g(1) z^{-1} + ... + g(M-1) z^{-M+1} = \sum_{n=0}^{M-1} g(n) z^{-n} \; . \end{multline*}

The frequency-sampling method results in a unit impulse response which has been convoluted with a rectangular window of the same length in the frequency domain. The frequency response of $F(z^{-1})$ is therefore affected by the large side-lobes of the rectangular window. As a result, the approximation error of $F(z^{-1})$ is large between the frequency samples. This can be alleviated by the use of a window that do not contain abrupt discontinuities in the time domain, and thus have small side-lobes in the frequency domain, i.e., the window smooths the frequency response of $F(z^{-1})$.

A windowed FIR filter $\tilde{h}(n)$ is created from an un-windowed FIR filter $h(n)$ as \begin{align*} \tilde{h}(n) = w(n) h(n) \end{align*} where $w(n)$ is a window function which is non-zero only for $n \in [0, M-1] \cap \mathbb{N}_{0}$. The frequency-domain representation of the window function $W(k)$ is found as \begin{multline*} W(k) = \sum\limits_{n=0}^{M-1} w(n-M/2) \textrm{e}^{-j \frac{2 \pi k n}{M}} = \left[ \sum\limits_{n=0}^{M-1} w(n) \textrm{e}^{-j \frac{2 \pi k n}{M}} \right] \textrm{e}^{-j \frac{2 \pi k}{M} \frac{M}{2} } \; , \end{multline*} where the term $\textrm{e}^{-j (2 \pi k / M) (M/2) }$ comes from the fact that the rectangular window is not centered around $n=0$, but is time-shifted to be centered around $n=M/2$. This phase term will cause distortion of $h(n)$, unless $h(n)$ is also phase-shifted to compensate. The unit impulse response $g(n)$ is therefore phase-shifted before windowing. Due to the circular shift property of the DFT, this can be done by rearranging $g(n)$ such that \begin{equation*} \bar{g}\left( n \right) = \begin{cases} g\left( n + M/2 \right) , & \hspace{-0.6em} n = 0,1, ..., \frac{M}{2} - 1 \\ g\left( n - M/2 \right) , & \hspace{-0.6em} n = \frac{M}{2},\frac{M}{2}+1, ..., M-1 \end{cases} \end{equation*} for the case when $M$ is even. The response is then represented by the FIR filter \begin{equation*} \bar{F}(z^{-1}) = \sum_{n=0}^{M-1} \bar{g}(n) z^{-n} = z^{-M/2} F(z^{-1}) \end{equation*} which is $F(z^{-1})$ delayed by $M/2$ steps. Applying the window $w(n)$ to the time-shifted impulse response $\bar{g}(n)$, \begin{equation*} \tilde{g}(n) = w(n) \bar{g}(n) \; , \end{equation*} the filter \begin{equation*} \tilde{F}(z^{-1}) = W(z^{-1})*\left[ z^{-M/2} F(z^{-1}) \right] \end{equation*} is obtained. Now, $G^{-1}(z^{-1}) \left[ W(z^{-1})*F(z^{-1}) \right] \approx 1$ if the FIR filter is accurate. Note that the phase due to $z^{-M/2}$ is taken out.