Analysis of co efficient quantization effects in FIR filters.

Consider an Nth order FIR transfer function.

Analysis of co efficient quantization effects in FIR filters.

Consider an $N^{th}$ order FIR transfer function.

$H(Z) = \sum^N_{n = 0} \ h(n) z^{-n}$

Quantization of the filter co efficient results in a new transfer function.

$\hat{H}(z)=\sum_{n=0}^{N} \hat{h}(n) \overline{z}^{n}=\sum_{n=0}^{N}(h(n)+e(n)) \overline{z}^{n}$

$\hat{H}(z)=H(z)+E(z)$

$E(z)=\sum_{n=0}^{N} e(n) \overline{z}^{n}$

Thus the FIR filter with quantized coefficient can be modeled as a parallel connection of two FIR filters H(z) and E(z) where H(z) represents the desired FIR filter with unquantized coefficient and E(z) is the FIR filter representing the error in the transfer function due to coefficient quantization without any loss of generally for type - 1 FIR filter the transfer function is

$H\left(e^{j \omega}\right)=\overline{e}^{j \omega N / 2}\left[h(N / 2)+\sum_{n=0}^{(N-2)/2} 2 h(n) \cos \left(\frac{N}{2}-n\right) \omega\right]$

The frequency response of the actual FIR filter with quantized coefficient $\hat{h}(n)$ can be expressed

$\hat{H}\left(e^{j \omega}\right)=H\left(e^{j \omega)}+E\left(e^{j \omega)}\right.\right.$

Where $E\left(e^{j w}\right)$ represents the error in the desired frequency response $H\left(e^{j \omega)}\right.$

$E\left(e^{j \omega}\right)=\sum_{n=0}^{N} e(n) e^{-j \omega n}$

$e(n)=\hat{h}(n)-h(n)$

$\left|E\left(e^{j \omega}\right)\right|=\left|\sum_{n=0}^{N} e(n) e^{-j \omega n}\right| \leq \sum_{n=0}^{N}|e(n)|\left|{e}^{-j \omega n}\right|$

$\leq \sum_{n=0}^{N}|e(n)|$

Assume each impulse response coefficient h(n) is a (b+1) - bit signed fraction. For example, for rounding $|e(n)| \leq \delta | 2$ $\delta={2}^{b}$ is the quantization step.

$\left|E\left(e^{jw)} | \leq \frac{(N+1) \delta}{2}\right.\right.$

The above bound is rather conservative and can be reached only if all errors are of the same sign and have the maximum value in the range.

The statistical realization is

$E\left(e^{j \omega}\right)=e^{-j \omega N / 2}\left[e\left(\frac{N}{2}\right)+\sum_{n=0}^{(N-2)/2} 2 e(n) \cos \left(\frac{N}{2}-n\right) \omega\right]$

$E\left(e^{j \omega}\right)$ is a sum of independent random variables

$\therefore \sigma_{E}^{2 (W)}=\sigma^{2}_e\left[1+4 \sum_{n=1}^{N / 2} \cos ^{2}\left(\omega_{n}\right)\right]$

$=\sigma_{e}^{2}\left[N+\frac{\sin (N+1) \omega}{\sin w}\right]$

&

$w_{N}(w)=\left[\frac{1}{2 N+1}\left(N+\frac{\sin (N+1) w}{\sin w}\right]^{1 / 2}\right.$

$\sigma_{E}(\omega)=\sigma_{e}(\sqrt{2 N+1}) W_{N}(\omega)$

For uniform distribution $\sigma_{e}=\delta / \sqrt{12}$

$=2^{-b-1} / \sqrt{3}$ $\therefore \quad \sigma_{E}(\omega) \leq \delta \sqrt{\frac{2 N+1}{12}}$

Based on above bound the word length of the FIR filter coefficient is decided.