Choice of Interpolation Resolution

We now consider the error due to finite precision in the linear interpolation between stored filter coefficients. We will find that the number of bits $n_\eta$ in the interpolation factor should be about half the filter coefficient word-length $n_c$.

Quantized Interpolation Error Bound.

The quantized interpolation factor and its complement are representable as

$$\begin{eqnarray*} \eta_q&=&\eta + \nu \\ \overline{\eta}_q &=& \overline{\eta}- \nu \end{eqnarray*}$$

where, since $\eta,\overline{\eta}$ are unsigned, $\vert\nu\vert\leq 2^{-(n_\eta+1)}$. The interpolated coefficient look-up then gives

$$\begin{eqnarray*} {\hat h}_{qq}(t) &=& (\overline{\eta}-\nu)[h(t_0)+\epsilon_0] ... ...\overline{\eta}\epsilon_0 + \eta\epsilon_1 + \nu[h(t_1)-h(t_0)], \end{eqnarray*}$$

where second-order errors $\nu\epsilon_0$ and $\nu\epsilon_1$ are dropped. Since $\vert h(t_1)-h(t_0)\vert\leq M_1$, we obtain the error bound

$$\left\vert\tilde{h}_{qq}(t)\right\vert\leq 2^{-n_c} + 2^{-(n_\eta+1)}M_1 + {3\over8}M_2.$$

The three terms in Eq.$\,$(J.3.4) are caused by coefficient quantization, interpolation quantization, and linear-approximation error, respectively.

Ideal Lowpass Filter.

For the ideal lowpass, the error bound is

$$\left\vert\tilde{h}_{qq}(t)\right\vert \leq 2^{-n_c} + a 2^{-(n_l+n_\eta+1)} + {\pi^2\over 8} 2^{-n_l}.$$

Let $n_l=1+n_c/2$ and require that the added error is at most ${1\over2}2^{-n_c}$. Then we arrive at the requirement

$$n_\eta\geq {n_c\over2}.$$