Tighter Error Bound on the Inverse Midpoint Integral Approximation

Presently, we derive a tighter bound with our previous comments on equation 36 in mind. We divide up the integral into an arbitrary number of integer-width intervals, each with a lesser error bound than that given above (with the exception of those at the frame edges). The optimal case occurs when we simply choose $N$ subintervals - one for each sample - so that the bound is different (and minimal) for each sample. This way, the high error in the most rapidly varying parts of the function will only be considered in the segments in which it may occur. Having segmented the signal this way, we then sum the error bounds for each sample to obtain the overall bound. Applying equation 27 each of $N$ times gives

$$\vert\mathring{e}_m\vert \leq \sum_{n = -\ensuremath{\frac{N-1}{2}}}^{\ensuremath{\frac{N-1}{2}}} \ensuremath{\frac{M_n 1^3}{24\cdot 1^2}}$$ (37)

where $M_n$ is the bound $M$ for the subinterval whose value is approximated by the sample $n$ (obtained from each of $N$ individual values of $\vert f''(n)\vert$). The number 1 is seen twice: in the numerator as the interval width and in the denominator as the number of subintervals. Continuing the above calculation by looking at $M_n$, we have:

$$M_n \geq \sum_n \left( \left(2\alpha n -\ensuremath{\frac{2\pi k}{N}}\right)^2 + 2\alpha\right)1$$ (38)

$$= 2\alpha N + \sum_n \left(4\alpha^2 n^2 - \ensuremath{\frac{4\alpha n \pi k}{N}}+ \ensuremath{\frac{4\pi^2 k^2}{N^2}}\right)$$ (39)

$$= 2\alpha N + \ensuremath{\frac{N4\pi^2 k^2}{N^2}}+ 4\sum_n \left(\alpha^2 n^2-\ensuremath{\frac{\alpha n \pi k}{N}}\right).$$ (40)

We now observe that because of the symmetry of our sum, we may drop out the second term in the sum since it is an odd function. We also see that the remaining sum may itself be approximated using the inverse midpoint rule, generating $4(2\ensuremath{\frac{\alpha^2}{3}}(\ensuremath{\frac{N}{2}})^3) + e_m$ where $e_m \leq 2\alpha^2 N/24$. Continuing, we thus have

$$M_n \geq 2\alpha N + \ensuremath{\frac{4\pi^2 k^2}{N}}+ 4\left(2\ensuremat... ...ac{\alpha^2}{3}}\left(\ensuremath{\frac{N}{2}}\right)^3\right) + 2\alpha^2 N/24$$ (41)

$$= 2\alpha N + \ensuremath{\frac{4\pi^2 k^2}{N}}+ \ensuremath{\frac{8\alpha^2 N^3}{24}}+ \ensuremath{\frac{2\alpha^2 N}{24}}.$$ (42)

Substituting back into equation 37, we thus have

$$\fbox{ $ \displaystyle \vert\mathring{e}_m\vert = \ens... ...ha^2 N^3}{3}}+ \ensuremath{\frac{\alpha^2 N}{12}}\right). $}$$ (43)

We now verify that this bound is tighter than or equal to that given in equation 36. We start by assuming this as fact, and show that the fact is necessarily true.

$$\vert\mathring{e}_m\vert \leq \vert e_m\vert$$ (44)

$$\ensuremath{\frac{1}{24}}\left(2\alpha N + \ensuremath{\frac{4\pi... ... \ensuremath{\frac{\alpha^2 N^3}{3}}+ \ensuremath{\frac{\alpha^2 N}{12}}\right) \leq \ensuremath{\frac{1}{24}}\left(2\alpha N + \ensuremath{\frac{4\pi^2 k^2}{N}}+ N^3\alpha^2 + \vert 4\pi k N \alpha\vert\right)$$ (45)

$$\ensuremath{\frac{\alpha^2 N^3}{3}}+ \ensuremath{\frac{\alpha^2 N}{12}} \leq N^3\alpha^2 + \vert 4\pi k N \alpha\vert$$ (46)

$$\ensuremath{\frac{\alpha^2 N}{12}} \leq \ensuremath{\frac{2N^3\alpha^2}{3}}+ \vert 4\pi k N \alpha\vert$$ (47)

$$\ensuremath{\frac{\alpha^2}{4}} \leq 2N^2\alpha^2 + \vert 12\pi k \alpha\vert$$ (48)

$$0 \leq \alpha^2\left(2N^2-\ensuremath{\frac{1}{4}}+ \left\vert\ensuremath{\frac{12\pi k}{\alpha}}\right\vert \right)$$ (49)

This is necessarily true because $N\geq1$. We observe by inspecting any line above (but the last), that as $\alpha\rightarrow 0$, the two limits become the same.

Even the tighter error bound given in equation 43 may seem significant. But we recall that, due to the constraints on $\alpha$, the product $\alpha N$ will never be greater than $\pi$ (which only occurs in the diabolical Nyquist frequency case), and in the case of actual audio signals will never even approach this value, causing all $\alpha N$ terms to be less than 1. We also note that this implies that $\alpha<\ensuremath{\frac{1}{N}}$. In order to determine the relative size of our error completely, we need to know the size of our ``good data'' estimates. Looking ahead in the proof, we know that our good data will take the form of sinusoids modulated by a raised cosine of magnitude $\sqrt{\ensuremath{\frac{2\pi}{\alpha}}}$. Recalling our observation about $\alpha$ above, this modulation amplitude is always at least $\sqrt{2\pi N}$.

Armed with these observations, we consider each term in equation 43. The first and fourth terms will be less than $\ensuremath{\frac{1}{12}}$ and $\ensuremath{\frac{1}{N\cdot 288}}$, respectively, and are thus small relative to the good data. The third term will be bounded by $\ensuremath{\frac{N}{72}}$, which is in general small compared to $\sqrt{2\pi N}$. To be more specific, we only require that $\ensuremath{\frac{N}{72}}\ll \sqrt{\ensuremath{\frac{2\pi}{\alpha}}}$. This is guaranteed when $N \ll 10368\pi$, a condition that will prove easy to satisfy in general. Considering the second term (in the no zero padding case for simplicity) we know that the largest $k$ value used will be $k_{\max} = \pm \ensuremath{\frac{\alpha N^2}{4\pi}}$. Substituting this into the second term and simplifying, we find that the second term is bounded by $\ensuremath{\frac{1}{288}}$, again small relative to the good data. As before, we note that these error bounds are worst case bounds that in reality will not often be seen.

We include a practical example in figure 5. Therein, we see the the real and imaginary parts of an example $Y(k)$ and the error bound given in equation 43, also function of $k$.

Figure 5: The real and imaginary parts of an example $Y(k)$ (solid) (with $\alpha = 0.004280$, $F=8000$, $K=N=201$, and a Hann windowing function) and the inverse midpoint approximation error (dashed).