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

$$h(t_0+\eta) = h(t_0) + \eta h^\prime (t_0) + {1\over 2} \eta^2 h^{\prime\prime}(t_0+\lambda \eta), \protect$$ (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

$$e(t) \isdef h(t) - \hat{h}(t),$$ (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

$$\hat{h}(t) \isdef \overline{\eta }h(t_0) + \eta h(t_1), \protect$$ (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 t_e denote any point at which |e(t)| reaches a maximum over the interval (t₀,t₁). Then we have

$$\vert e(t_e)\vert \ge \vert e(t)\vert, \quad\forall t\in [t_0,t_1].$$

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'(t_e)=0. Expressing e(t₀)=0 as a Taylor expansion of e(t) about t=t_e, we obtain

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

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

$$e(t_e) = - \frac{(t_0-t_e)^2}{2}e''(\xi)$$

Defining

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

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

$$\vert e(t_e)\vert \le \frac{M_2}{8} \protect$$ (9)