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 $\vert f\vert \gt\gt \vert\epsilon\vert$,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 way, and it permeates the subject of signal analysis.

To apply this method in the Fourier domain, suppose 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,  

$$X(\omega) \eq { \overline{F(\omega)} \ Y(\omega) \over \overline{F(\omega)} F(\omega) \ +\ \epsilon^2}$$ (12)

Now the denominator must always be a positive number greater than zero, so division is always safe.

In preparing figures with equation (12), I learned that it is helpful to recast the equation in a scaled form. First replace $\epsilon^2$,which has physical units of |F|², by $\epsilon^2 = \lambda \sigma_F^2$, where $\lambda$ is a dimensionless parameter and $\sigma_F^2$ is the average value of $\overline{F}F$.Then I rescaled equation (12) to  

$$X(\omega) \eq { \overline{F(\omega)} \ Y(\omega) \over \ove... ...F(\omega) \ +\ \lambda \sigma_F^2} \ \ (2+\lambda /2) \sigma_F$$ (13)

The result is that the scale of X is independent of the scale of F and the scale of $\lambda$.This facilitates plotting X over a range of those parameters. I found the 2s in the expression by experimentation. Of course, if the plotting software you are using adjusts a scale factor to fill a defined area, then the scaling may be unimportant. Equation (13) 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 all these filters have the same phase.

The filter $\overline{F}$ is called the ``matched filter." If nature created Y by random bursts of energy into F, then building X from Y and F by choosing $\lambda=\infty$ in equation (13) amounts to $X=Y\overline{F}$ which crosscorrelates F with the randomly placed copies of F that are in Y.