Transmission-Line Reflections

Signal integrity

A step behind a source resistance $R_s$ enters a lossless line of impedance $Z_0=\sqrt{L/C}$ , drawn as its distributed $L$ – $C$ ladder. Each end reflects a fraction $\Gamma$ of every arriving wave; the lattice sum gives $V(x,t)$ at every point and instant, so the probes read a real time-domain reflectometer.

Live circuit matched source, one echo, then quiet

Drag a value tag to change that part. Click a node on the line to clip a probe on; the scope shows ten one-way delays.

Causal signal trace paused

Launch the step The source sees only the line impedance $Z$ ₀, so it launches V₁ = Vs· $Z$ ₀/(Rs + $Z$ ₀) whatever the load is.

t_{\mathrm{stage}}=\htmlData{math-slot=value}{\mathtt{0.00}}\,\mathrm{s}

Step 1 / 5 Launch the step

Notebook: 16 parts and 15 wires, modelled as lattice (bounce) diagram of a lossless line

Recommended lab 13 of 18

Read a reflection coefficient from a near-end echo and terminate a line.

Builds on: Dynamic RC Low-Pass Filter

Predict, then test

A $50\ \Omega$ line ends in an open circuit. When the step first reaches the far end, what does the far-end voltage do?

Commit before the explanation appears. This is a prediction, not a quiz score.

Governing model

\Gamma = \dfrac{Z - Z_0}{Z + Z_0} \;\cdot\; V_1 = V_s\,\dfrac{Z_0}{R_s + Z_0}

Parts

step source $V_s$ + $R_s$
$3$ -section $L$ – $C$ ladder $(Z_0,\ t_d)$
load $Z_L$
near / far probes

Limits of the model

$Z_L: 5\ \Omega$ – $5\ \mathrm{k\Omega}$
$Z_0: 25$ – $150\ \Omega$
$t: 0$ – $10\,t_d$

Diagnosis challenge The fault is hidden. Work it out from the changed outputs first.

Notebook: 16 parts and 15 wires, modelled as lattice (bounce) diagram of a lossless line

A 50 Ω50\ \Omega50 Ω line ends in an open circuit. When the step first reaches the far end, what does the far-end voltage do?

A $50\ \Omega$ line ends in an open circuit. When the step first reaches the far end, what does the far-end voltage do?