How Do You Get the Reflection Coefficient and the Load From Measured Vmax and Vmin?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•8 min read

Divide to get the VSWR, S = Vmax ÷ Vmin, convert it with |Γ| = (S − 1) ÷ (S + 1), and use the position of the first voltage minimum to find the phase of Γ; then ZL = Z0(1 + Γ) ÷ (1 − Γ). On a 50 Ω line with Vmax = 1.5 V, Vmin = 0.5 V and the first minimum 0.1λ from the load, that gives S = 3, |Γ| = 0.5, a return loss of 6.02 dB and ZL = 24.05 − j30.50 Ω.

What you'll learn

Why the ratio of the maximum to the minimum voltage fixes the magnitude of the reflection coefficient
How return loss, mismatch loss and delivered power follow from |Γ|
How the position of the first minimum gives the phase of Γ
Two independent routes to the load impedance, worked to the same answer
Quick sanity checks for a load that is purely resistive

Conventions used here

A lossless line of characteristic impedance Z0 = 50 Ω, phasors with e^(jωt) time dependence, and distance d measured from the load towards the generator. The load reflection coefficient is ΓL = (ZL − Z0) ÷ (ZL + Z0) = |Γ|e^(jθ), and the voltage on the line is V(d) = V+e^(jβd)[1 + ΓLe^(−j2βd)], with β = 2π/λ.

The measurement: a probe slides along a 50 Ω line ending in an unknown load. It finds Vmax = 1.5 V, Vmin = 0.5 V, and the first minimum 0.1λ from the load.

Why do Vmax and Vmin give |Γ|?

The bracket 1 + ΓLe^(−j2βd) is the forward wave plus the reflected wave. As you move along the line the reflected part rotates relative to the forward part. Where they line up the magnitude is |V+|(1 + |Γ|), and where they oppose it is |V+|(1 − |Γ|). Their ratio is the voltage standing wave ratio, S = (1 + |Γ|) ÷ (1 − |Γ|), and solving for |Γ| gives |Γ| = (S − 1) ÷ (S + 1). The magnitude of the incident wave cancels, so you do not need to know the source voltage.

Worked example, step 1: the magnitude

S = Vmax ÷ Vmin = 1.5 ÷ 0.5 = 3.
|Γ| = (3 − 1) ÷ (3 + 1) = 0.5.
Return loss = −20 log10 |Γ| = 6.02 dB. It is quoted as a positive number of decibels; larger means a better match.
Fraction of incident power reflected = |Γ|² = 0.25, so 75% reaches the load. The mismatch loss is −10 log10(1 − |Γ|²) = 1.25 dB.

Worked example, step 2: the phase from the first minimum

At a voltage minimum the reflected wave is exactly opposite the forward wave, so the angle of ΓLe^(−j2βd) is −180° there. With θ the angle of ΓL, that means θ − 2βdmin = −180°, so θ = 2βdmin − 180° = 4π(dmin/λ) − 180°. For dmin = 0.1λ, 4π × 0.1 rad is 72°, so θ = 72° − 180° = −108°. The load reflection coefficient is ΓL = 0.5∠−108° = −0.1545 − j0.4755.

Worked example, step 3: the load impedance, two ways

Route 1, from Γ: ZL = 50 × (1 + ΓL) ÷ (1 − ΓL) = 50 × (0.8455 − j0.4755) ÷ (1.1545 + j0.4755) = 24.05 − j30.50 Ω.
Route 2, from the minimum: the impedance at a voltage minimum is real and equal to Z0 ÷ S = 50 ÷ 3 = 16.67 Ω. Move 0.1λ back towards the load with the line equation Z = Z0(Zmin − jZ0 tan βl) ÷ (Z0 − jZmin tan βl), where βl = 36°. The result is again 24.05 − j30.50 Ω.
The negative reactance means the load is capacitive at this frequency.

Why is the impedance at a minimum exactly Z0 ÷ S?

At a voltage minimum the reflected voltage subtracts from the forward voltage, while the reflected current, which carries the opposite sign, adds to the forward current. Both totals are in phase with the forward wave, so their ratio is real: V ÷ I = Z0(1 − |Γ|) ÷ (1 + |Γ|), which is Z0 ÷ S. At a maximum the roles swap and the impedance is Z0 × S. Here that is 16.67 Ω at each minimum and 150 Ω at each maximum, a quick check on any Smith chart reading.

The standing wave rebuilt in code from ZL = 24.05 − j30.50 Ω. It reproduces the measurement: a minimum of 0.5 V at 0.1λ and a maximum of 1.5 V at 0.35λ, repeating every half wavelength.

How can you sanity-check the answer?

Maxima and minima alternate every quarter wavelength, so the first maximum here is at 0.1λ + 0.25λ = 0.35λ, matching the plot. Two special cases are worth remembering. If the first minimum were right at the load, the load would be real and smaller than Z0: ZL = Z0 ÷ S = 16.7 Ω. If the first maximum were at the load, which is the same as the first minimum at 0.25λ, it would be real and larger: ZL = Z0 × S = 150 Ω. A minimum anywhere in between, like 0.1λ, means a complex load. On a Smith chart the same calculation is a rotation of 0.1λ from the minimum towards the load, which is counterclockwise.

Common mistakes

Treating Vmax ÷ Vmin as a power ratio. It is a voltage ratio; |Γ|² is the power fraction reflected.
Measuring d from the generator instead of the load, which reverses the sign of the phase step.
Using the second or third minimum without accounting for the extra half wavelengths. Minima repeat every λ/2, so subtract whole multiples of 0.5λ first.
Giving return loss as a negative number of decibels in a question that expects the usual positive convention.
Rotating towards the generator (clockwise) on the Smith chart when moving from the minimum back to the load.

Check your understanding

Q1. A line shows a VSWR of 2. What are |Γ|, the return loss and the fraction of power reflected? Answer: |Γ| = 1/3, return loss = 9.54 dB, and 1/9 = 11.1% of the power is reflected.
Q2. Same 50 Ω line and Vmax ÷ Vmin = 3, but the first minimum is 0.25λ from the load. What is ZL? Answer: θ = 0, so ZL = Z0 × S = 150 Ω, purely resistive.
Q3. What do you measure if the load is a short circuit? Answer: Γ = −1, so Vmin = 0 and the VSWR is infinite, with minima at the load and every half wavelength from it.

Official source

GATE 2027 EC syllabus (PDF), Section 8: Electromagnetics

Last verified

The GATE 2027 EC syllabus lists transmission lines, characteristic impedance, impedance matching and transformation, S-parameters and the Smith chart under Section 8, Electromagnetics (checked against the official PDF from IIT Madras on 10 October 2026). The load was computed two independent ways in code, and the standing wave rebuilt from it reproduces the stated Vmax, Vmin and minimum position.