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Dividing by zero smoothly

Think of any real numbers $x$, $y$, and $f$ and any program containing $x=y/f$. How can we change the program so that it never divides by zero? A popular answer is to change $x=y/f$ to $x=yf/(f^2+\epsilon^2)$, where $\epsilon$ is any tiny value. When $\vert f\vert >> \vert\epsilon\vert$, then $x$ is approximately $y/f$ as expected. But when the divisor $f$ vanishes, the result is safely zero instead of infinity. The transition is smooth, but some criterion is needed to choose the value of $\epsilon$. This method may not be the only way or the best way to cope with zero division, but it is a good method, and permeates the subject of signal analysis.

To apply this method in the Fourier domain, suppose that $X$, $Y$, and $F$ are complex numbers. What do we do then with $X=Y/F$? We multiply the top and bottom by the complex conjugate $\overline{F}$, and again add $\epsilon^2$ to the denominator. Thus,

\begin{displaymath}
X(\omega) \eq
\frac{ \overline{F(\omega)} \ Y(\omega) }{ \overline{F(\omega)} F(\omega) \ +\ \epsilon^2}
\end{displaymath} (1)

Now, the denominator must always be a positive number greater than zero, so division is always safe. Equation (1) ranges continuously from inverse filtering, with $X=Y/F$, to filtering with $X=\overline{F}Y$, which is called ``matched filtering.'' Notice that for any complex number $F$, the phase of $1/F$ equals the phase of $\overline{F}$, so the filters have the same phase.
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Next: Damped solution Up: UNIVARIATE LEAST SQUARES Previous: UNIVARIATE LEAST SQUARES

2014-12-01