Analog Intuition

Part 2 · Impedance Matching

Impedance \(Z = R + jX\)

Key point: at one frequency a two-terminal port is one complex number \(Z = R + jX\). \(R\) dissipates; \(X\) stores. Frequency moves \(X\) — that is why a lumped match is narrowband.

Series R with L or C

\(Z(j\omega) = R + j\omega L\) or \(R - j/(\omega C)\) · phasor + sweep

R (real) X (imag) Z phasor this f

Phasor · Re = R · Im = X

R
X
|Z|
∠Z
\[ \omega = 2\pi f \]
=
\[ X = +\omega L \]
=
\[ Z = R + jX \]
=
\[ |Z| = \sqrt{R^2 + X^2} \]
=
\[ \angle Z = \mathrm{atan2}(X,R) \]
=

For series L: \(X = +\omega L\) (inductive). For series C: \(X = -1/(\omega C)\) (capacitive).

\(|Z|(f)\) · amber line = this frequency

\(\angle Z(f)\) · + = inductive · − = capacitive

Try it: “50 + j25 @ 1 GHz”, then raise \(f\) — the phasor climbs and \(|Z|(f)\) keeps rising. Switch to series RC and raise \(f\) — \(|X|\) collapses toward \(R\). Same resistor, opposite frequency story.