Linear Interpolation Error Bound

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 [#!Goldstein!#, p. 119], we have

[画像:\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 [画像:$\lambda \in[0,1]$], where [画像:$h^\prime (t_0)$] denotes the time derivative of h(t) evaluated at t=t₀, and [画像:$h^{\prime\prime}(t_0)$] 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 $t=t_0+\eta,ドル [画像:$t_0=\lfloor{t}\rfloor $], $\eta=t-t_0,ドル and [画像:$\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 [画像:$\overline{\eta }\isdef 1-\eta$] and [画像:$t_1\isdef t_0+1$]. Thus t₀ and t₁ are successive time instants for which samples of h(t) are available, and [画像:$\eta \in[0,1)$] is the linear interpolation factor.

By definition, e(t₀)=e(t₁)=0. That is, the interpolation error is zero at the known samples. Let t_e 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 [画像:$t_e \le t_0 + 1/2$]. (Otherwise, replace t₀ with t₁ in the following.) Since both h(t) and [画像:$\hat{h}(t)$] are twice differentiable for all [画像:$t\in(t_0,t_1)$], then so is e(t), and therefore e'(t_e)=0. Expressing e(t₀)=0 as a Taylor expansion of e(t) about t=t_e, we obtain

[画像:\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 [画像:$\xi\in(t_0,t_1)$]. Solving for e(t_e) gives

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

Defining

[画像:\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 [画像:$t\in(t_0,t_1)$], and noting that [画像:$\vert t_0-t_e\vert\le 1/2$], we obtain the upper bound²

[画像:\begin{displaymath} \vert e(t_e)\vert \le \frac{M_2}{8} \protect \end{displaymath}] (9)

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 © 2020年09月17日 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA), Stanford University
CCRMA

Linear Interpolation Error Bound

Linear Interpolation Error Bound

``The Digital Audio Resampling Home Page'', by Julius O. Smith III. Copyright © 2020年09月17日 by Julius O. Smith III Center for Computer Research in Music and Acoustics (CCRMA), Stanford University CCRMA

``The Digital Audio Resampling Home Page'', by Julius O. Smith III.
Copyright © 2020年09月17日 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA), Stanford University
CCRMA