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Damped solution

Another way to say $x=y/f$ is to say $fx-y$ is small, or $(fx-y)^2$ is small. This does not solve the problem of $f$ going to zero, so we need the idea that $x^2$ does not get too big. To find $x$, we minimize the quadratic function in $x$.
\begin{displaymath}
Q(x) \eq (f x-y)^2 + \epsilon^2 x^2
\end{displaymath} (2)

The second term is called a ``damping factor,'' because it prevents $x$ from going to $\pm \infty$ when $f\rightarrow 0$. Set $dQ/dx=0$, which gives:
\begin{displaymath}
0 \eq f(f x-y) + \epsilon^2 x
\end{displaymath} (3)

Equation (3) yields our earlier common-sense guess $x=fy/(f^2+\epsilon^2)$. It also leads us to wider areas of application in which the elements are complex vectors and matrices.

With Fourier transforms, the signal $X$ is a complex number at each frequency $\omega$. Therefore we generalize equation (2) to:

\begin{displaymath}
Q(\bar X, X) \eq
(\overline{FX-Y}) (FX-Y) + \epsilon^2 \bar X X \eq
(\bar X \bar F - \bar Y) (FX-Y) + \epsilon^2 \bar X X
\end{displaymath} (4)

To minimize $Q$, we could use a real-values approach, where we express $X=u+iv$ in terms of two real values $u$ and $v$, and then set $\partial Q/\partial u=0$ and $\partial Q/\partial v=0$. The approach we take, however, is to use complex values, where we set $\partial Q/\partial X=0$ and $\partial Q/\partial \bar X=0$. Let us examine $\partial Q/\partial \bar X$:
\begin{displaymath}
{\partial Q(\bar X, X)\over \partial \bar X} \eq
\bar F (FX-Y) + \epsilon^2 X \eq 0
\end{displaymath} (5)

The derivative $\partial Q/\partial X$ is the complex conjugate of $\partial Q/\partial \bar X$. Therefore, if either is zero, the other is also zero. Thus, we do not need to specify both $\partial Q/\partial X=0$ and $\partial Q/\partial \bar X=0$. I usually set $\partial Q/\partial \bar X$ equal to zero. Solving equation (5) for $X$ gives equation (1).

Equation (1) solves $Y=XF$ for $X$, giving the solution for what is called ``the deconvolution problem with a known wavelet $F$.'' Analogously, we can use $Y=XF$ when the filter $F$ is unknown, but the input $X$ and output $Y$ are given. Simply interchange $X$ and $F$ in the derivation and result.


next up previous [pdf]

Next: Smoothing the denominator spectrum Up: UNIVARIATE LEAST SQUARES Previous: Dividing by zero smoothly

2014-12-01