Paper V visual guide · Visual 25 of 40 · LOCAL DERIVATIVES / FINITE INCREMENTS

Sensitivity grows before the pole is reached

Traditional monograph: Section 14 (p. 29).

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Away from \(\Gamma=1\), the inverse Cayley map satisfies

\[ \frac{dz}{d\Gamma}=\frac2{(1-\Gamma)^2}. \]

Along the real axis, three exact derivatives already show the scale change:

\[ \Gamma=\tfrac12:\ 8,\qquad \Gamma=\tfrac9{10}:\ 200,\qquad \Gamma=\tfrac{99}{100}:\ 20000. \]
Cayley sensitivity near the boundary
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

The curve plots the labeled sensitivity weight against reflection coefficient on a logarithmic vertical scale. A table gives exact derivative values approaching the pole.

Drawing labels and data 20 labels

Read the original labels as selectable text. Tables preserve grid relationships; other short labels are grouped compactly.

  • \(-.9\)
  • \(-.5\)
  • 0
  • \(.5\)
  • \(.9\)
  • \(1\)
  • \(10\)
  • \(100\)
  • \(1000\)

symmetric sensitivity \(w_{\rm Smith}\) (log scale)

  • \(1/2\)
  • 8
  • \(9/10\)
  • 200
  • \(99/100\)
  • 20000
  • \(\Gamma\)
  • \(|dz/d\Gamma|\)
  • \(\Delta z=20/9\ne2\)
  • real reflection coefficient \(\Gamma\)

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A differential is not the full finite-error formula

For a finite increment \(\varepsilon\), with both denominators nonzero,

\[ z(\Gamma+\varepsilon)-z(\Gamma) =\frac{2\varepsilon}{(1-\Gamma-\varepsilon)(1-\Gamma)}. \]

At \(\Gamma=9/10\) and \(\varepsilon=1/100\), the actual increment is \(20/9\), while derivative times increment is \(2\). Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.

What a proposed weight can and cannot certify

Impedance is sensitive near \(\Gamma=1\), and admittance near \(\Gamma=-1\). The symmetric weight

\[ w(\Gamma)=|1-\Gamma|^{-2}+|1+\Gamma|^{-2} \]

reflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point comes to either pole.

Read this as

Use the exact finite difference for an error budget. A large derivative diagnoses conditioning. Choosing an optimal algorithm requires further input.

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