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core30 min studyLesson 286 in Physics

Characteristic impedance and reflections

Apply a load boundary condition to forward and backward waves, track voltage and current signs, and compute reflected and delivered power.

Before you begin
On this page

The question for today

What you’ll understand

  • Derive the voltage reflection coefficient from load voltage and current.
  • Distinguish characteristic impedance from a finite line’s input impedance.
  • Calculate reflected and accepted power for a real lossless line impedance.

A traveling wave does not know the impedance of a distant load until it reaches that boundary and a reflected wave can return. Characteristic impedance is a local relation between voltage and current within one traveling component. The load impedance instead relates the total terminal voltage to the total terminal current. A mismatch between these two relations requires reflection. Keeping these meanings separate explains open-circuit voltage doubling and prevents the mistaken claim that a lossless line dissipates power like a resistor.

A backward voltage has a backward current

Use a uniform lossless line with real positive characteristic impedance, a load at z equal to zero, and the source on the negative-z side. The forward voltage phasor has negative spatial phase and its current has the same sign. The backward voltage has positive spatial phase, while its contribution to current has a minus sign under the fixed positive-z current convention. Both components can therefore carry energy in their own propagation directions without contradicting the port-current convention.

At the load, total voltage is the sum of incident and reflected voltages, but total current is their difference divided by characteristic impedance. Enforcing the load relation and solving for the reflected-to-incident voltage ratio gives the familiar reflection coefficient. The current sign is essential to the derivation. Adding both current amplitudes with the same sign would predict no distinction between open and short boundaries, which immediately fails the limiting-case check.

V(z)=V+e−jβz+V−ejβz,I(z)=V+e−jβz−V−ejβzZ0,ΓL=V−V+=ZL−Z0ZL+Z0V(z)=V_+e^{-j\beta z}+V_-e^{j\beta z},\quad I(z)=\frac{V_+e^{-j\beta z}-V_-e^{j\beta z}}{Z_0},\quad \Gamma_L=\frac{V_-}{V_+}=\frac{Z_L-Z_0}{Z_L+Z_0}
V₊ and V₋ are RMS voltage amplitudes at z = 0, I is positive toward the load, β is in rad/m, and impedances are in ohms. This form assumes a uniform lossless single-mode line with real Z₀ > 0.↔ Scroll wide equations horizontally.

Boundary limits are physical checks

A matched resistive load equals the characteristic impedance and creates no reflection. An open circuit demands zero total terminal current, so reflected voltage equals incident voltage and total load voltage doubles. A short demands zero terminal voltage, so reflected voltage is the negative of incident voltage and total current doubles. Neither open nor short absorbs average power ideally. Their large local voltage or current does not imply that energy disappears into the termination.

A finite line's input impedance depends on both the load and electrical length because incident and reflected components interfere at the input plane. A lossless line terminated in a reactive load has a reactive input impedance, except at singular limiting positions. A matched line presents its characteristic impedance at every length, but an arbitrary finite line does not. The resistance-like real value of characteristic impedance describes power transported by a traveling field, not heat generated along the line.

Zin=Z0ZL+jZ0tan⁡(βℓ)Z0+jZLtan⁡(βℓ),ZL=Z0⟹Zin=Z0Z_{\mathrm{in}}=Z_0\frac{Z_L+jZ_0\tan(\beta\ell)}{Z_0+jZ_L\tan(\beta\ell)},\qquad Z_L=Z_0\Longrightarrow Z_{\mathrm{in}}=Z_0
The input plane is at z = −ℓ, with ℓ in metres. The tangent is dimensionless, and Zin is the total voltage/current ratio at that plane. The formula assumes a lossless line and a frequency-specific load impedance.↔ Scroll wide equations horizontally.

Power requires an amplitude convention

For RMS voltage waves on a real-impedance lossless line, incident power is incident voltage magnitude squared divided by characteristic impedance. Reflected power uses the same denominator and the reflected magnitude. Consequently the reflected power fraction is the magnitude squared of the voltage reflection coefficient. A negative reflection coefficient reverses voltage phase but still carries a positive backward power magnitude. Net accepted load power is incident power minus reflected power.

For a passive load with nonnegative real impedance, the reflection magnitude cannot exceed one under these assumptions. A purely reactive termination has unit magnitude and exchanges energy without net absorption. A lossy line with complex characteristic impedance requires more careful power accounting; one cannot blindly reuse real-impedance voltage-wave formulas. Also distinguish power accepted at the load from generator available power, since source mismatch and multiple returns can change the incident amplitude that actually reaches the load.

P+=∣V+∣2Z0,P−=∣ΓL∣2P+,PL=(1−∣ΓL∣2)P+P_+=\frac{|V_+|^2}{Z_0},\quad P_-=|\Gamma_L|^2P_+,\quad P_L=(1-|\Gamma_L|^2)P_+
P₊, P₋, and PL are average powers in watts for RMS wave voltages and real positive Z₀. Peak voltage amplitudes require an additional factor 1/2. This relation is evaluated at the load reference plane.↔ Scroll wide equations horizontally.

Work it through

A resistive mismatch with power balance

A 50 Ω lossless line carries a 3 V RMS incident voltage at a 100 Ω resistive load. Find reflection coefficient, total load voltage, total load current, and accepted power.

  1. The voltage coefficient is ΓL=(100−50)/(100+50)=1/3\Gamma_L=(100-50)/(100+50)=1/3. The reflected RMS voltage is therefore 1 V, in phase with the incident voltage at the load.

  2. Total load voltage is 3 + 1 = 4 V RMS. Current is the wave difference divided by 50 Ω: (3 − 1)/50 = 0.040 A RMS, consistent with 4/100.

  3. Incident power is 32/50=0.18 W3^2/50=0.18\,\mathrm W and reflected power is 12/50=0.020 W1^2/50=0.020\,\mathrm W. Their difference is 0.160 W, also equal to 42/1004^2/100.

The result

The reflection coefficient is +1/3; the load receives 4 V RMS, 40 mA RMS, and 0.160 W. One ninth of incident power is reflected.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Boundary phase addition at one location

Reset and set BOTH Phase difference to 0 degrees and Amplitude ratio to 1; predict a doubled sum. Then Reset and set BOTH Phase difference to 180 degrees and Amplitude ratio to 1; predict zero sum. Treat this only as voltage addition at a boundary. These are co-propagating equal-frequency waves, so the lab cannot show a reflected wave, standing pattern, current sign, or line power.

Boundary phase addition at one locationTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe open and short voltage limits are addition identities, but physical reflection additionally requires opposite propagation and the correct current relation.

Pause & reason

Check your understanding

An ideal short terminates a lossless line with real Z₀. Which load-plane voltage reflection coefficient follows?

Your turn

Try it yourself

A 75 Ω lossless line is terminated by 25 Ω and receives a 2 V RMS incident wave. Find reflected voltage, reflected power fraction, and load power.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The voltage coefficient is signed; its squared magnitude determines the backward power fraction.

Compare with the solution

The coefficient is (25−75)/(25+75)=−0.5(25-75)/(25+75)=-0.5, so the reflected voltage is −1 V RMS relative to the incident phase. The reflected power fraction is 0.25. Incident power is 4/75 W4/75\,\mathrm W, so load power is 0.75(4/75)=0.040 W0.75(4/75)=0.040\,\mathrm W. At the load, total voltage is 1 V RMS and total current is (2−(−1))/75=0.040 A(2-(-1))/75=0.040\,\mathrm A, confirming 40 mW in the 25 Ω resistance.

Keep the idea

Take it with you

  • Characteristic impedance relates individual traveling voltage and current components.
  • The backward current sign makes open and short boundary conditions consistent.
  • For real line impedance, reflected power fraction is the squared voltage-reflection magnitude.

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