Next  |  Prev  |  Up  |  Top  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Application to the Ideal Lowpass Filter

For the ideal lowpass filter, we have

\begin{displaymath}
h(t) = \mbox{sinc}(\omega_L t/ \pi) \isdef {\sin(\omega_L t)\over \omega_L t}
= {1\over \omega_L}\int_0^{\omega_L}\cos(\omega t)\do ,
\protect
\end{displaymath} (10)

where $\omega_L=\pi/L$, and L=2nl is the number of table entries per zero-crossing. Note that the rightmost form in Eq.$\,$(10) is simply the inverse Fourier transform of the ideal lowpass-filter frequency response. Twice differentiating with respect to t, we obtain
\begin{displaymath}
h^{\prime\prime}(t) = -{1\over\omega_L}\int_0^{\omega_L} \omega^2\cos(\omega t) \do,
\protect
\end{displaymath} (11)

from which it follows that the maximum magnitude is
\begin{displaymath}
M_2 = {\omega_L^2\over 3} = {\pi^2\over 3L^2}.
\protect
\end{displaymath} (12)

Note that this bound is attained at t=0. Substituting Eq.$\,$(12) into Eq.$\,$(9), we obtain the error bound
\begin{displaymath}
\left\vert e(t_e)\right\vert \leq {\pi^2\over 24 L^2} <
{0.412\over L^2} = 0.412\cdot2^{-2n_l}.
\protect
\end{displaymath} (13)

Thus for the ideal lowpass filter $h(t)=\mbox{sinc}(t/L)$, the pointwise error in the interpolated lookup of h(t) is bounded by 0.412/L2. This means that nl must be about half the coefficient word-length nc used for the filter coefficients. For example, if h(t) is quantized to 16 bits, L must be on the order of 216/2=256. In contrast, we will show that without linear interpolation, nl must increase proportional to nc for nc-bit samples of h(t). In the 16-bit case, this gives $L\sim 2^{16}=65536$. The use of linear interpolation of the filter coefficients reduces the memory requirements considerably.

The error bounds obtained for the ideal lowpass filter are typically accurate also for lowpass filters used in practice. This is because the error bound is a function of M2, the maximum curvature of the impulse response h(t), and most lowpass designs will have a value of M2 very close to that of the ideal case. The maximum curvature is determined primarily by the bandwidth of the filter since, generalizing equations Eq.$\,$(10) and Eq.$\,$(11),

\begin{displaymath}
h^{\prime\prime}(0) = -{1\over\pi}\int_0^{\pi} \omega^2 H(\omega) \do,
\end{displaymath}

which is just the second moment of the lowpass-filter frequency response $H(\omega)$ (which is real for symmetric FIR filters obtained by symmetrically windowing the ideal sinc function [13]). A lowpass-filter design will move the cut-off frequency slightly below that of the ideal lowpass filter in order to provide a ``transition band'' which allows the filter response to give sufficient rejection at the ideal cut-off frequency which is where aliasing begins. Therefore, in a well designed practical lowpass filter, the error bound M2 should be lower than in the ideal case.


Next  |  Prev  |  Up  |  Top  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Download resample.pdf
[How to cite and copy this work]  [Comment on this page via email]

``The Digital Audio Resampling Home Page'', by Julius O. Smith III.
Copyright © 2014-01-10 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA