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Linear Interpolation Error Bound

Let h(t) denote the lowpass filter impulse response, and assume it is twice continuously differentiable for all t. By Taylor's theorem [6, p. 119], we have

\begin{displaymath}
h(t_0+\eta) = h(t_0) + \eta h^\prime (t_0) + {1\over 2} \eta^2 h^{\prime\prime}(t_0+\lambda \eta),
\protect
\end{displaymath} (6)

for some $\lambda \in[0,1]$, where $h^\prime (t_0)$ denotes the time derivative of h(t) evaluated at t=t0, and $h^{\prime\prime}(t_0)$ is the second derivative at t0.

The linear interpolation error is

\begin{displaymath}
e(t) \isdef h(t) - \hat{h}(t),
\end{displaymath} (7)

where $t=t_0+\eta$, $t_0=\lfloor{t}\rfloor $, $\eta=t-t_0$, and $\hat{h}(t)$ is the interpolated value given by
\begin{displaymath}
\hat{h}(t) \isdef \overline{\eta }h(t_0) + \eta h(t_1),
\protect
\end{displaymath} (8)

where $\overline{\eta }\isdef 1-\eta$ and $t_1\isdef t_0+1$. Thus t0 and t1 are successive time instants for which samples of h(t) are available, and $\eta \in[0,1)$ is the linear interpolation factor.

By definition, e(t0)=e(t1)=0. That is, the interpolation error is zero at the known samples. Let te denote any point at which |e(t)| reaches a maximum over the interval (t0,t1). Then we have

\begin{displaymath}
\vert e(t_e)\vert \ge \vert e(t)\vert, \quad\forall t\in [t_0,t_1].
\end{displaymath}

Without loss of generality, assume $t_e \le t_0 + 1/2$. (Otherwise, replace t0 with t1 in the following.) Since both h(t) and $\hat{h}(t)$ are twice differentiable for all $t\in(t_0,t_1)$, then so is e(t), and therefore e'(te)=0. Expressing e(t0)=0 as a Taylor expansion of e(t) about t=te, we obtain

\begin{displaymath}
0 = e(t_0)
= e(t_e) + (t_0-t_e)e'(t_e) + \frac{(t_0-t_e)^2}{2!}e''(\xi)
= e(t_e) + \frac{(t_0-t_e)^2}{2}e''(\xi)
\end{displaymath}

for some $\xi\in(t_0,t_1)$. Solving for e(te) gives

\begin{displaymath}
e(t_e) = - \frac{(t_0-t_e)^2}{2}e''(\xi)
\end{displaymath}

Defining

\begin{displaymath}
M_2 \isdef \max_{t} \left\vert h^{\prime\prime}(t)\right\vert = \max_{t} \left\vert e^{\prime\prime}(t)\right\vert,
\end{displaymath}

where the maximum is taken over $t\in(t_0,t_1)$, and noting that $\vert t_0-t_e\vert\le 1/2$, we obtain the upper bound2
\begin{displaymath}
\vert e(t_e)\vert \le \frac{M_2}{8}
\protect
\end{displaymath} (9)


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``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
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