The day a node stops being a node
Everything you learned in basic circuits rested on one quiet assumption you were probably never told about: that a wire is a single node. Connect two ends of a copper wire and you assumed the voltage was the same everywhere along it, instantly. Flip a switch here, and the bulb over there responds — for all practical purposes — at the very same instant. That assumption is a beautiful lie. It works only because, at the frequencies of a flashlight or a doorbell, the lie is invisible.
Here is the truth the lie hides. A change in voltage does not appear everywhere at once — it propagates, as a wave, at a speed close to the speed of light. In a typical cable that speed is roughly 2 × 10⁸ metres per second, about two-thirds of c. That sounds instantaneous, and at human scale it is. But signals don't live at human scale. A 1 GHz clock edge changes meaningfully every nanosecond, and in one nanosecond a wave travels only about 20 centimetres. Suddenly the length of your cable is the same order as the distance the signal moves between updates — and the two ends genuinely disagree about what the voltage is right now.
Wavelength on a typical PCB (effective dielectric, vp ~ 1.5e8 m/s):
lambda = vp / f
f lambda lambda/10 "long" wire is...
------ ---------- ---------- ------------------
1 MHz 150 m 15 m basically never
100 MHz 1.5 m 15 cm a long backplane
1 GHz 15 cm 1.5 cm a short trace!
10 GHz 1.5 cm 1.5 mm a via, a bond wire
At 1 GHz a 1.5 cm trace is already "electrically long."Slicing the wire into a ladder
If a single node won't do, what replaces it? The answer is one of the most elegant moves in all of engineering: stop treating the wire as one component, and treat it as an infinite ladder of tiny components. Take any short slice of length Δx. That slice has a little series inductance (the magnetic field around the current), a little series resistance (the copper isn't perfect), a little shunt capacitance (the two conductors form a capacitor), and a little shunt conductance (the insulator between them leaks a bit). Chain millions of these slices end to end and you have rebuilt the wire — but now as a distributed system.
We describe a transmission line not by total values but by per-unit-length parameters, written with a prime: L′ (henries per metre), R′ (ohms per metre), C′ (farads per metre), and G′ (siemens per metre). A length of RG-58 coax, for example, has roughly L′ ≈ 250 nH/m and C′ ≈ 100 pF/m. These four numbers, plus the operating frequency, capture everything the line does. The R′ here is the same physical resistance you already know, just smeared continuously along the conductor.
One slice of the distributed model (length dx):
o---[ R'dx ]---[ L'dx ]---+---o --> toward load
in |
[G'dx] [C'dx]
| |
o-------------------------+------+---o
ground
Series arm: R' dx (loss) + L' dx (magnetic energy)
Shunt arm: C' dx (electric energy) + G' dx (dielectric loss)
Chain an infinite number of these -> a transmission line.The telegrapher's equations
Now apply the two laws you already trust to a single slice. Kirchhoff's voltage law around the series arm says the voltage drops a little as you cross R′ and L′; the drop depends on how fast the current is changing. Kirchhoff's current law at the shunt node says current leaks off a little into C′ and G′; that leak depends on how fast the voltage is changing. Write those two statements, shrink Δx to zero, and you get a pair of coupled equations linking how voltage changes along the line (∂V/∂x) to how current changes in time (∂I/∂t), and vice versa.
The telegrapher's equations (1880s, named for the telegraph):
dV/dx = -( R' I + L' dI/dt ) voltage falls along x
dI/dx = -( G' V + C' dV/dt ) current leaks along x
Differentiate and combine -> a WAVE EQUATION:
d2V/dx2 = L'C' * d2V/dt2 (for a lossless line, R'=G'=0)
Solutions are forward + backward travelling waves:
V(x,t) = f(x - vp t) + g(x + vp t)
\___________/ \___________/
toward load reflected back
propagation speed vp = 1 / sqrt(L' C')This is the punchline of the whole rung. The telegrapher's equations are nothing but KVL and KCL applied to an infinitesimal slice — no new physics, just honest bookkeeping at the scale where the lie breaks down. And out of them falls a wave equation, identical in form to the equation for a vibrating guitar string or a sound wave in air. Voltage and current on a transmission line genuinely travel, with a definite speed, and they can move in both directions at once. That second, backward-travelling wave is the seed of everything in the next rung.
Characteristic impedance: what the wave sees
Here is the question that unlocks practical engineering. A wave launches into a long line and starts marching forward. At the instant it leaves the source, the source has no way of knowing where the far end is or what's connected there — the news of the load hasn't arrived yet. So what current flows for a given voltage? The wave must obey a fixed ratio, set entirely by the line's own L′ and C′. That ratio is the characteristic impedance Z₀.
For a forward-travelling wave, V and I are locked in lockstep:
Z0 = V_forward / I_forward = sqrt( L' / C' ) (lossless)
RG-58 coax: L' = 250 nH/m, C' = 100 pF/m
Z0 = sqrt( 250e-9 / 100e-12 )
= sqrt( 2500 )
= 50 ohms <-- why coax is "50-ohm cable"
Note: Z0 does NOT depend on length. A 1 m and a 1 km
coax of the same type both look like 50 ohms to a wave
just entering them.Sit with how strange this is. The characteristic impedance is measured in ohms, yet a lossless line contains no resistors at all — only inductance and capacitance, both of which store energy rather than burn it. Z₀ is not a resistance that dissipates power; it is the ratio of voltage to current that a travelling wave is forced to maintain. Yet a battery connected to an infinitely long lossless 50 Ω line really does push out a steady current as if it were feeding a 50 Ω resistor — because the energy keeps marching off down the line and never comes back. The line *acts* resistive while storing, not burning, the energy.
Three worked lines, and why reflections matter
Let's make this concrete with the three lines you'll actually meet. First, coax. Connect a 50 Ω coax to an oscilloscope and the wave takes real time to arrive — about 5 ns per metre, since vp ≈ 2×10⁸ m/s. That delay is why a long scope probe shifts your measured edges, and why oscilloscope inputs offer a '50 Ω' setting: it terminates the cable so the wave is absorbed at the end rather than bounced back.
- PCB trace at high speed. A 10 cm microstrip on FR-4 carries a wave at vp ≈ 1.5×10⁸ m/s, so it adds ≈ 0.67 ns of delay one-way. For a 500 MHz signal (λ ≈ 30 cm) the trace is a third of a wavelength — firmly a transmission line, and it must be routed at a controlled 50 Ω.
- Geometry sets Z₀. Widen the trace and C′ rises while L′ falls, so Z₀ = √(L′/C′) drops; narrow it and Z₀ rises. PCB houses publish exactly the width-over-a-given-stackup that yields 50 Ω. This is why 'impedance-controlled' boards cost more — the fab guarantees the geometry.
- The mismatch. Now drive that 50 Ω trace into a CMOS gate whose input looks like a few-picofarad, near-open circuit. The wave arrives, finds an impedance that is nothing like 50 Ω, and cannot simply continue — so part of it bounces back toward the source.
That bounce is a reflection, and it is the whole reason characteristic impedance is worth caring about. When a forward wave hits a load that doesn't match Z₀, the line can't satisfy both 'V/I = Z₀ for the wave' and 'V/I = Z_load at the load' at once. Its only escape is to launch a second, backward-travelling wave whose job is to fix up the books. If Z_load = Z₀, the books already balance and nothing reflects — the wave is absorbed cleanly. Any mismatch, and an echo races back up the line.
A 1 V step launched into 50 ohm line, OPEN at the far end:
t=0 __________________________ 1V edge leaves source
|
t=T |--------> reaches open end
reflects FULLY, same sign
t=2T <--------| echo returns, V -> 2V
|
source sees a clean 1V, then a step UP to 2V one round
trip (2T) later -- "ringing" / overshoot.
Far end OPEN -> reflection coeff = +1 (voltage doubles)
Far end SHORT -> reflection coeff = -1 (voltage cancels)
Far end = Z0 -> reflection coeff = 0 (no echo: matched!)This is the failure mode that turns a working schematic into a flaky board. A digital edge launched down an unterminated 50 Ω trace reflects off the high-impedance receiver, races back, reflects again off the low-impedance driver, and rings back and forth — overshooting the rails, undershooting, and possibly making the receiver see a single edge as several. The fix is to match: terminate the line in its own Z₀, or source-match the driver, so the reflection coefficient is zero. *How* much reflects, *why* a quarter-wave of line can transform one impedance into another, and how the Smith chart tames all of it — that is exactly where the next rung begins.