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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
![\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}](img59.png) |
(6) |
for some
, where
denotes the time derivative of
h(t) evaluated at t=t0, and
is the second derivative at
t0.
The linear interpolation error is
![\begin{displaymath}
e(t) \isdef h(t) - \hat{h}(t),
\end{displaymath}](img63.png) |
(7) |
where
,
,
, and
is the interpolated value given by
![\begin{displaymath}
\hat{h}(t) \isdef \overline{\eta }h(t_0) + \eta h(t_1),
\protect
\end{displaymath}](img68.png) |
(8) |
where
and
. Thus t0 and t1 are
successive time instants for which samples of h(t) are available,
and
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
Without loss of generality, assume
. (Otherwise,
replace t0 with t1 in the following.) Since both h(t) and
are twice differentiable for all
, 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
for some
. Solving for e(te) gives
Defining
where the maximum is taken over
, and noting that
, we obtain the upper bound2
![\begin{displaymath}
\vert e(t_e)\vert \le \frac{M_2}{8}
\protect
\end{displaymath}](img79.png) |
(9) |
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