Impedance Z = R + jX — Impedance Matching

Part 2 · Impedance Matching

Impedance \(Z = R + jX\)

Key point: at one frequency a port is one number, \(Z = R + jX\). Frequency moves \(X\), so a lumped match is narrowband.

Emerald is \(R\). Rose is \(X\). The amber mark on the sweeps is this frequency.

1
Almost a pure resistor.
50 Ω and 0.01 nH at 1 GHz. \(X\) is +j0.06 Ω. \(|Z|\) is 50.00 Ω. Angle is 0.1°. The phasor lies on the real axis.
2
Series L, +j25.
3.98 nH at 1 GHz. \(X\) is +j25.01 Ω. \(|Z|\) is 55.90 Ω. Angle is 26.6°. Inductive, above the axis.
3
Series C, the other sign.
5.3 pF at 1 GHz. \(X\) is −j30.03 Ω. \(|Z|\) is 58.32 Ω. Angle is −31.0°. Same \(R\), capacitive.
4
A large L at 2 GHz.
20 Ω and 40 nH. \(X\) is +j502.65 Ω. \(|Z|\) is 503.05 Ω. Angle is 87.7°. The resistor is a small piece of a big reactance.
5
Same L, lower frequency.
Still 20 Ω and 40 nH. \(f\) is 0.5 GHz. \(X\) falls to +j125.66 Ω. \(|Z|\) is 127.25 Ω. Angle is 81.0°. That is why the match will not hold across a band.

Series RL or RC only. Part 3 draws the same \(Z\) as a point \(\Gamma\) on the unit disk.

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

Type R (Ω) L (nH) f (GHz)

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

Phasor · Re = R · Im = X

R—

X—

|Z|—

∠Z—

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

\(Z\) from \(R\), \(X\), and \(f\)

\[ \omega = 2\pi f \]

= —

\[ X = +\omega L \]

= —

\[ Z = R + jX \]

= —

\[ |Z| = \sqrt{R^2 + X^2} \]

= —

\[ \angle Z = \mathrm{atan2}(X,R) \]

= —