The echo at the end of the line
Shout into a long, empty corridor and most of your voice travels cleanly down it — until it slams into the far wall and a fraction comes ringing back at you. A signal on a transmission line does exactly the same thing. In the last rung you met the line's [[ee-characteristic-impedance|characteristic impedance]] Z₀: the fixed ratio of voltage to current that a wave 'expects' to see as it travels — 50 ohms in nearly every radio and 75 ohms in cable TV. A travelling wave is perfectly happy as long as the world keeps offering it that same ratio. Trouble starts at the load — the antenna, the amplifier input, whatever the line is feeding. If the load's impedance Z_L is anything other than Z₀, the wave arrives expecting one voltage-to-current ratio and is handed a different one. Something has to give.
What gives is energy conservation plus Ohm's law at the boundary. The instant the wave hits the load, the voltage across that load and the current through it must obey Ohm's law for *that* load — V_L = Z_L · I_L — not the line's ratio. The only way to satisfy both the incoming wave and the stubborn load at the same time is to launch a second, reflected wave that heads back toward the source. The reflected wave carries away exactly the mismatch: it's the line's way of saying 'this much of you didn't fit, take it back.' A matched load swallows the whole wave and sends nothing home; a badly mismatched load reflects almost everything, like a mirror.
Γ: one complex number that captures the bounce
Engineers compress that whole story into a single quantity: the [[ee-reflection-coefficient|reflection coefficient]], written with the Greek capital gamma, Γ. It is the ratio of the reflected wave's voltage to the incident wave's voltage at the load — *how much bounced back, divided by how much arrived.* Because both waves are sinusoids that can be out of step, Γ is a complex number: its magnitude |Γ| (between 0 and 1) tells you what fraction of the voltage came back, and its angle tells you the phase — whether the echo returns crest-up or crest-down. The formula falls straight out of matching Ohm's law at the boundary:
Z_L - Z0
Γ = ------------- ( Z_L = load impedance )
Z_L + Z0 ( Z0 = line's characteristic impedance )
|Γ| = 0 → nothing reflects (Z_L = Z0, perfect match)
|Γ| = 1 → everything reflects (open or short — a total mirror)
Power reflected = |Γ|^2 × incident power
Power delivered = (1 - |Γ|^2) × incident power
Example: Z0 = 50 Ω, Z_L = 100 Ω
Γ = (100-50)/(100+50) = 50/150 = +0.333
|Γ|^2 = 0.111 → 11% of power bounces back,
89% reaches the load.Look hard at that formula and the three landmark cases practically introduce themselves. When Z_L = Z₀ the numerator is zero, so Γ = 0: a perfectly matched load, the wave glides in and is fully absorbed, no echo at all. When the line ends in an open circuit (Z_L → ∞, the wire just stops in mid-air), Γ = +1: the full wave reflects with no phase flip. When it ends in a dead short (Z_L = 0, the conductors touched together), Γ = −1: the full wave reflects but upside-down, phase flipped by 180°. Open and short are the two perfect mirrors — they return all the power, just with opposite signs.
Standing waves and the number on the meter
Now picture both waves living on the line at once — the forward wave streaming toward the load, the reflected wave streaming back. Where their crests meet, the voltages add and you get a tall bump; a little further along, a crest meets a trough and they cancel down to almost nothing. Critically, these high and low spots don't travel — they sit frozen at fixed positions on the line, because the two waves are locked in step. That stationary pattern of bulges and nulls is the [[ee-standing-wave|standing wave]], and it's the unmistakable fingerprint of a mismatch. A perfectly matched line has no reflected wave, so it has no ripples at all — the voltage amplitude is the same everywhere.
Matched line (Γ = 0): amplitude is flat — no ripple
|V| ───────────────────────────────── even everywhere
Mismatched line: forward + reflected = standing wave
Vmax Vmax Vmax
|V| /\ /\ /\
/ \ /\ / \ /\ / \ ← peaks & nulls
_____/ \__/ \____/ \__/ \____/ \___ sit STILL on the line
Vmin Vmin
Vmax 1 + |Γ|
VSWR = ------ = -----------
Vmin 1 - |Γ|Carrying around a complex Γ is precise but fiddly, so the bench instrument reports something simpler: the Voltage Standing Wave Ratio, or VSWR. It's just the ratio of the biggest voltage on the line to the smallest — Vmax divided by Vmin. A flat, rippleless line gives Vmax = Vmin, so VSWR = 1:1, the gold standard. As the mismatch worsens the ripples deepen and the ratio climbs: 1.5:1 is excellent, 2:1 is a common 'good enough' spec, and a 3:1 or worse usually means something is wrong. Because VSWR is built straight from |Γ|, the two carry the same information — VSWR is simply the form an engineer can read off a meter in one glance.
The Smith chart: a map of every possible impedance
In 1939 a Bell Labs engineer named Phillip Smith got tired of grinding through complex-number arithmetic by hand and drew a chart that did the algebra for him with a ruler and a compass. Eighty-odd years later, with computers everywhere, the Smith chart is *still* taped to the wall of every RF lab — because no equation gives you the at-a-glance feel for impedance that this strange circular map does. Here is its single founding idea: instead of plotting impedance on an ordinary flat grid (which runs off to infinity at an open circuit), Smith plotted the reflection coefficient Γ inside a unit circle. Every passive impedance in the universe — from a dead short to a wide-open line — maps to exactly one point *inside or on* that circle, because |Γ| never exceeds 1. The infinite plane of impedance is folded neatly into a disc you can hold in your hand.
To make it useful, Smith overlaid a second set of curves: lines of constant resistance and constant reactance, all bent into circles and arcs by the same folding. So the chart speaks two languages at once. Read a point's distance and angle from the centre and you've got Γ (magnitude and phase). Read which resistance-circle and reactance-arc it sits on and you've got the impedance directly, normally written *normalized* to Z₀ — that is, in units of Z₀, so 50 Ω on a 50 Ω system reads as '1.0'. The dead centre of the chart is the magic spot: it's Γ = 0, which is the perfect match z = 1 (i.e. Z = Z₀). Every matching job is, quite literally, a quest to walk a point toward the centre of the Smith chart.
The Smith chart, plain-language layout
top half = inductive (+jX)
____________________________________________
/ \
| SHORT OPEN |
| (Z=0,Γ=-1) ●──────────●──────────● (Z=∞,Γ=+1)|
| left edge CENTER right edge |
| z = 1, Γ = 0 |
| PERFECT MATCH |
\____________________________________________/
bottom half = capacitive (-jX)
• horizontal axis = pure resistance (real Z)
• circles of constant R, arcs of constant X tile the disc
• distance from center = |Γ| → bigger circle = worse match
• a constant-|Γ| circle is also a constant-VSWR circleReading it: from a marker to a match
Let's actually use the thing. Suppose a network analyzer puts a marker on the Smith chart for your antenna at, say, the point reading z = 0.4 + j0.5 (normalized) on a 50 Ω system. First, *de-normalize* by multiplying by Z₀: the real load impedance is Z_L = 50 × (0.4 + j0.5) = 20 + j25 Ω. The positive imaginary part tells you the antenna looks a bit inductive at this frequency, and sits in the top half of the chart. To get Γ you don't even need the formula — you read it off the picture: the marker's distance from the center is |Γ|, and the angle of the line from center to marker is the phase. For this point |Γ| works out to about 0.5, which corresponds to a VSWR near 3:1 — clearly not matched, and worth fixing.
Matching means physically walking that marker to the center. The two classic moves: a series element slides the point along a constant-resistance circle, and a shunt (parallel) element slides it along a constant-conductance circle. The everyday recipe is the L-network — two reactive parts, one series and one shunt — chosen so the first hop lands you on the right circle and the second hop drags you home to z = 1. Because the parts are pure inductors and capacitors (which store energy rather than burning it), a good match wastes almost no power: you've simply *transformed* the stubborn load into something the line is happy to see.
- Plot the load. Put the measured load impedance (normalized to Z₀) on the chart. Note where it sits relative to the center — how far (that's your |Γ| and VSWR) and in which half (inductive up, capacitive down).
- Cancel the reactance first. Add a series reactive element to slide the point along its constant-resistance circle until it lands on the unit-resistance circle (the big circle passing through the center).
- Then fix the resistance. Add a shunt element to slide along the constant-conductance circle the rest of the way into the center — z = 1, Γ = 0.
- Read off the part values. The amount of reactance each hop required tells you the inductor and capacitor values for your matching frequency. Done — the marker is at the center and the line sees its own Z₀.
Worked example: matching an antenna feed
Time to put every piece together on a real job. You've built a small antenna for a 433 MHz remote sensor, fed by 50 Ω coax. You sweep it on a network analyzer and at 433 MHz the antenna's input impedance reads Z_L = 25 + j0 Ω — purely resistive (no reactance to cancel, lucky you), but only half the 50 Ω the coax wants. Plug it into the reflection formula and check just how bad that is before reaching for parts.
Given: Z0 = 50 Ω, Z_L = 25 + j0 Ω, f = 433 MHz
Reflection coefficient:
Γ = (25 - 50)/(25 + 50) = -25/75 = -0.333
|Γ| = 0.333 (sign is negative: Z_L < Z0, as expected)
Standing-wave ratio:
VSWR = (1 + 0.333)/(1 - 0.333) = 1.333/0.667 ≈ 2.0 : 1
Power check:
reflected = |Γ|^2 = 0.111 → 11% bounced back
delivered = 1 - 0.111 = 0.889 → 89% reaches the antenna
Verdict: VSWR 2:1 is borderline. ~11% of your transmit
power is wasted and echoing back at the source.
On a 433 MHz sensor it's livable, but on a
high-power link it's worth matching out.On the Smith chart this load sits on the horizontal axis, halfway between the center and the short — at z = 0.5, real, no reactance. The fix is a textbook L-network. First add a series inductor: it adds positive reactance, dragging the point up and around its constant-resistance circle until it hits the unit-resistance circle. Then add a shunt capacitor: it slides the point along the constant-conductance circle straight into the center. Two parts, a few minutes of arithmetic (or one click in a matching tool), and the coax now looks into a flawless 50 Ω. The reflected wave vanishes, VSWR drops to 1:1, and all of your power flies out of the antenna instead of cooking your transmitter.