Terminated Lossless Transmission Lines

Total Voltage and Current on the Line

In the previous article we derived the general solution to the telegrapher's equations for a lossless line (R' = G' = 0). The result was a superposition of a forward-travelling wave and a backward-travelling wave. Now we attach a load impedance Z_L at the end of the line and determine what the boundary condition at the load forces those two wave amplitudes to be.

We place the load at z = 0 and the source at z = −l. This is the standard convention for terminated line analysis: it makes the load the reference plane and measures distance back toward the source as positive l. The general phasor solutions for voltage and current on a lossless line (γ = jβ) are:

Total voltage on the lossless line

V (z) = V_{0}^{+} (e^{- j β z} + Γ e^{+ j β z})

V₀⁺ is the complex amplitude of the forward wave. Γ = V₀⁻/V₀⁺ is the voltage reflection coefficient at the load (z = 0), defined in Section 2.

Total current on the lossless line

I (z) = \frac{V_{0}^{+}}{Z_{0}} (e^{- j β z} - Γ e^{+ j β z})

The minus sign on the reflected current term reflects that the backward current wave flows in the −z direction. The characteristic impedance Z₀ = √(L'/C') is real for a lossless line.

These two expressions are the complete description of the fields anywhere on the line. All the quantities we care about — input impedance, reflection coefficient at any point, power delivered to the load, standing wave envelope — are derived from them directly. The load boundary condition pins down Γ; the line length l then determines everything at the source end.

Fig 1 — Coordinate convention: load at z = 0, source at z = −l, distance measured toward source

The load sets the ratio of backward to forward wave amplitude at z = 0. The line of length l transforms this ratio into the input impedance seen at z = −l.

📌 Why put the load at z = 0?

With the load at z = 0, evaluating V(0)/I(0) gives Z_L directly — no exponentials to evaluate. The input plane at z = −l then has e^(jβl) terms, which are the familiar "rotation" factors used in Smith chart analysis. This convention is universal in microwave engineering textbooks and matches the Smith chart's direction of rotation.

Reflection Coefficients

The Voltage Reflection Coefficient at the Load

The load imposes a boundary condition: at z = 0, the ratio of total voltage to total current must equal Z_L. Evaluating V(0) and I(0) from the general solutions:

Boundary condition at z = 0

Z_{L} = \frac{V (0)}{I (0)} = \frac{V_{0}^{+} (1 + Γ)}{\frac{V_{0}^{+}}{Z_{0}} (1 - Γ)} = Z_{0} \frac{1 + Γ}{1 - Γ}

At z = 0, e^(0) = 1 so both exponentials evaluate to unity. V₀⁺ cancels from numerator and denominator.

Solving for Γ:

Voltage reflection coefficient at the load — Γ_L

Γ = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}}

Γ is in general complex. Its magnitude |Γ| ≤ 1 for a passive load. Its phase angle ∠Γ describes the phase shift on reflection.

This is one of the most important equations in RF engineering. It tells us the fraction of the incident voltage wave that is reflected at the load. A few key special cases are worth memorising:

Load Z_L	Γ	\|Γ\|	Physical meaning
Z_L = Z₀ (matched)	0	0	No reflection. All power absorbed by load. Perfect match.
Z_L = 0 (short circuit)	−1	1	Total reflection. Voltage inverted on reflection. V = 0 at load.
Z_L = ∞ (open circuit)	+1	1	Total reflection. Voltage doubles at load. I = 0 at load.
Z_L = jX (pure reactance)	e^(jθ)	1	Total reflection with phase shift. No power absorbed.
Z_L = R + jX (general)	\|Γ\|e^(jφ)	0 < \|Γ\| < 1	Partial reflection. Some power absorbed, some reflected.

The Input Reflection Coefficient Γ_in

The reflection coefficient is not fixed at the value Γ_L everywhere on the line. At an arbitrary position z on the line, the ratio of backward to forward wave amplitude is:

Reflection coefficient at position z

Γ (z) = \frac{V_{0}^{-} e^{+ j β z}}{V_{0}^{+} e^{- j β z}} = Γ e^{+ j 2 β z}

The magnitude |Γ(z)| = |Γ| is constant everywhere on a lossless line. Only the phase of Γ changes with position, rotating at rate 2β rad/m.

At the input plane z = −l, the input reflection coefficient Γ_in is:

Input reflection coefficient

Γ_{in} = Γ (- l) = Γ e^{- j 2 β l} = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} e^{- j 2 β l}

Moving from load toward source by distance l rotates Γ clockwise in the complex plane by 2βl radians. On the Smith chart this is exactly the "rotate toward generator" operation.

📌 The Factor of 2 in 2βl

The phase of Γ_in rotates by twice the electrical length βl of the line, not once. This is because the wave travels the length l twice — once going forward to the load, and once coming back as the reflected wave. A quarter-wave line (βl = π/2) therefore rotates Γ by π radians (180°), mapping an open circuit at the load into a short circuit at the input, and vice versa. This is the quarter-wave transformer principle.

Return Loss and Mismatch Loss

Two additional quantities are commonly specified in datasheets and measurements. Return Loss (RL) is the ratio of reflected to incident power, expressed in dB:

Return loss

RL = - 20 {log}_{10} | Γ | dB

Return loss is defined as a positive number (the minus sign makes it so). A perfect match gives RL = ∞ dB; a total reflection gives RL = 0 dB. Typical antenna specifications require RL ≥ 10 dB (|Γ| ≤ 0.316).

Mismatch Loss (ML) is the fraction of incident power lost to reflection, i.e. not delivered to the load:

Mismatch loss

ML = - 10 {log}_{10} (1 - {| Γ |}^{2}) dB

For |Γ| = 1/3 (VSWR 2:1): ML = −10 log(1 − 1/9) = −10 log(8/9) ≈ 0.51 dB. Less than half a dB lost — often acceptable in practice.

Input Impedance of a Lossless Terminated Line

The input impedance Z_in is what the source "sees" looking into the transmission line. It is the ratio of total voltage to total current at z = −l. Substituting z = −l into the general solutions:

V(−l) and I(−l)

V (- l) = V_{0}^{+} (e^{+ j β l} + Γ e^{- j β l}) I (- l) = \frac{V_{0}^{+}}{Z_{0}} (e^{+ j β l} - Γ e^{- j β l})

At z = −l, e^(−jβ(−l)) = e^(+jβl) for the forward term, and e^(+jβ(−l)) = e^(−jβl) for the backward term.

Forming the ratio V(−l)/I(−l) and substituting Γ = (Z_L − Z₀)/(Z_L + Z₀):

Step 1 — Ratio before substituting Γ

Z_{in} = Z_{0} \frac{e^{j β l} + Γ e^{- j β l}}{e^{j β l} - Γ e^{- j β l}}

Substituting Γ, multiplying through by (Z_L + Z₀), and collecting terms using Euler's identities e^(±jβl) = cos(βl) ± j sin(βl):

Input impedance — general lossless line formula

Z_{in} = Z_{0} \frac{Z_{L} + j Z_{0} tan (β l)}{Z_{0} + j Z_{L} tan (β l)}

This is the fundamental transmission line impedance transformation formula. It maps any load impedance Z_L through a lossless line of electrical length βl into the input impedance Z_in seen by the source.

This single equation encodes all the impedance behaviour of a terminated lossless line. Because tan(βl) is periodic with period π (i.e. every half-wavelength), Z_in repeats with period λ/2. A few special cases of this formula are central to RF design:

Special Cases of Input Impedance

Half-Wavelength Line: βl = π (l = λ/2)

tan(π) = 0, so:

Input impedance of a half-wave line

Z_{in} = Z_{0} \frac{Z_{L} + j Z_{0} \cdot 0}{Z_{0} + j Z_{L} \cdot 0} = Z_{L}

A half-wave line is transparent — it reproduces the load impedance exactly at the input, regardless of Z₀. This also holds for any integer multiple of λ/2.

Quarter-Wavelength Line: βl = π/2 (l = λ/4)

tan(π/2) → ∞. Dividing numerator and denominator by tan(βl) and taking the limit:

Input impedance of a quarter-wave line

Z_{in} = \frac{Z_{0}^{2}}{Z_{L}}

The quarter-wave transformer. Z_in × Z_L = Z₀². To match a load Z_L to a source Z_S using a λ/4 section, choose Z₀ = √(Z_S × Z_L).

Short-Circuit Stub: Z_L = 0

Input impedance — short-circuit stub

Z_{in} = Z_{0} \frac{0 + j Z_{0} tan (β l)}{Z_{0} + j \cdot 0 \cdot tan (β l)} = j Z_{0} tan (β l)

A shorted stub is purely reactive. For 0 < βl < π/2 it looks inductive (Z_in = +jX); for π/2 < βl < π it looks capacitive (Z_in = −jX). This is the basis of stub matching networks.

Open-Circuit Stub: Z_L = ∞

Input impedance — open-circuit stub

Z_{in} = Z_{0} \frac{Z_{L}}{j Z_{L} tan (β l)} = \frac{Z_{0}}{j tan (β l)} = - j Z_{0} cot (β l)

An open stub is also purely reactive but with opposite sign to the shorted stub at the same length. For 0 < βl < π/2 it looks capacitive. Open stubs are preferred on microstrip where drilling for a via is undesirable.

💡 Stubs as Lumped Reactances

Shorted and open stubs are the transmission line equivalents of lumped inductors and capacitors. Because they are distributed structures, they are not limited by component parasitics and can be fabricated directly in PCB copper at microwave frequencies. Single-stub and double-stub matching, as well as stub-based filters, exploit this property extensively. The input impedance formula above is all you need to design them.

Standing Waves and the Voltage Standing Wave Ratio

When a forward wave and a backward wave coexist on the line, they superpose to form a standing wave pattern. Unlike a pure travelling wave (which has uniform amplitude everywhere), a standing wave has fixed spatial maxima and minima of voltage and current magnitude. These are a direct observable consequence of the mismatch.

The Voltage Magnitude Envelope

The magnitude of the total voltage phasor at position z = −l′ (where l′ = −z ≥ 0 measures distance from the load) is found by factoring V₀⁺ out of the general solution and computing |V|:

Voltage magnitude as a function of distance from load l′

| V (- l') | = | V_{0}^{+} | | 1 + Γ e^{- j 2 β l'} |

Writing Γ = |Γ|e^(jφ), this becomes |V₀⁺| |1 + |Γ|e^(j(φ − 2βl′))|. The exponential rotates as l′ increases.

The term inside the absolute value is a phasor of magnitude 1 plus a phasor of magnitude |Γ|. As l′ varies, the phase of the second phasor rotates. The maximum and minimum values of |V| occur when the two phasors are aligned (add constructively) and anti-aligned (add destructively):

Maximum and minimum voltage magnitudes

|_{V |}_{\max} = | V_{0}^{+} | (1 + | Γ |) |_{V |}_{\min} = | V_{0}^{+} | (1 - | Γ |)

|V|_max occurs at planes where φ − 2βl′ = 0, 2π, 4π … (phasors add). |V|_min occurs where φ − 2βl′ = π, 3π … (phasors oppose).

The Voltage Standing Wave Ratio (VSWR)

The VSWR (sometimes written SWR) is the ratio of the maximum to minimum voltage magnitude on the line:

VSWR definition

VSWR = \frac{{| V |}_{\max}}{{| V |}_{\min}} = \frac{1 + | Γ |}{1 - | Γ |}

VSWR ≥ 1 always. VSWR = 1 for a perfect match (|Γ| = 0). VSWR = ∞ for a total reflection (|Γ| = 1, open or short circuit).

The inverse relationship — finding |Γ| from VSWR — is equally important:

Reflection coefficient magnitude from VSWR

| Γ | = \frac{VSWR - 1}{VSWR + 1}

For VSWR = 2: |Γ| = 1/3. For VSWR = 3: |Γ| = 1/2. For VSWR = 1.5: |Γ| ≈ 0.2. These are numbers worth knowing by feel.

Positions of Maxima and Minima

Writing Γ = |Γ|e^(jφ), voltage maxima occur at distances l′_max from the load satisfying φ − 2βl′_max = 2nπ, i.e.:

Distance from load to n-th voltage maximum

l_{\max}^{'} = \frac{φ - 2 n π}{2 β} = \frac{φ λ g}{4 π} + \frac{n λ_{g}}{2} (n = 0, 1, 2, \dots)

Maxima are separated by λg/2. The first maximum is at l′ = φλg/4π from the load (taking the first positive value). Voltage minima are offset by λg/4 from each maximum.

A critical observation: current maxima occur at voltage minima, and vice versa. Wherever |V| is maximum on the line, |I| = |V|_max / Z₀ × (1 − |Γ|)/(1 + |Γ|) is at its minimum, and the local impedance seen at that plane is purely real and equal to Z₀ × VSWR. At a voltage minimum, the local impedance is Z₀ / VSWR — also purely real.

Standing Wave Pattern Visualised

Fig 2 — Standing wave envelope |V(l′)| for |Γ| = 0.5 (VSWR = 3). Maxima at λg/2 intervals, minima midway between.

The envelope oscillates between |V₀⁺|(1 + |Γ|) and |V₀⁺|(1 − |Γ|). Adjacent maxima and minima are separated by exactly λg/4. Measuring the ratio of any maximum to the adjacent minimum directly gives the VSWR.

Power Flow on a Mismatched Line

The time-average power delivered to the load is the difference between the forward and backward wave powers:

Net power delivered to the load

P_{avg} = \frac{{| V_{0}^{+} |}^{2}}{2 Z_{0}} (1 - {| Γ |}^{2})

For a lossless line this power is constant along the entire line — it equals the power delivered to the load. The factor (1 − |Γ|²) is the power transmission coefficient. No energy is stored in the standing wave pattern itself; the reactive energy oscillates back and forth.

Note that power is constant along a lossless line. Even though |V| and |I| vary dramatically from point to point, the product Re{V × I*}/2 is the same everywhere. This is a powerful consistency check: if your analysis gives position-dependent power on a lossless line, something has gone wrong.

📌 What Comes Next

We now have the complete toolkit for terminated line analysis: the reflection coefficient at the load and at any point on the line, the input impedance formula, the standing wave pattern, and VSWR. The natural next step is to represent all of this graphically — the Smith chart is a conformal mapping of the complex Γ plane that makes impedance transformation, stub design, and matching network synthesis visual and immediate. That is the subject of the next article in this series.

Questions & Worked Examples

Apply the theory from this article to realistic RF engineering scenarios. Work through each problem fully before reading the solution.

01

Reflection Coefficient and VSWR for a Complex Load

Problem: A 50 Ω lossless coaxial cable is connected to an antenna modelled as Z_L = 35 − j20 Ω at 2.4 GHz. Calculate: (a) the reflection coefficient Γ at the load, (b) the return loss in dB, (c) the VSWR, and (d) the mismatch loss in dB.

(a) Reflection coefficient Γ

Γ = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} = \frac{(35 - j 20) - 50}{(35 - j 20) + 50} = \frac{- 15 - j 20}{85 - j 20}

Multiply numerator and denominator by conjugate of denominator

Γ = \frac{(- 15 - j 20) (85 + j 20)}{(85 - j 20) (85 + j 20)} = \frac{- 1275 - j 300 - j 1700 + 400}{7225 + 400} = \frac{- 875 - j 2000}{7625}

Result

Γ = - 0.1148 - j 0.2623 | Γ | = \sqrt{{0.1148}^{2} + {0.2623}^{2}} = 0.286

∠Γ = arctan(−0.2623/−0.1148) — third quadrant ≈ −113.6°

(b) Return loss

RL = - 20 {log}_{10} (0.286) = 10.87 dB

Just above the typical 10 dB threshold — this antenna is marginally matched at 2.4 GHz.

VSWR = \frac{1 + 0.286}{1 - 0.286} = \frac{1.286}{0.714} = 1.80

(d) Mismatch loss

ML = - 10 {log}_{10} (1 - {0.286}^{2}) = - 10 {log}_{10} (0.918) = 0.37 dB

Only 8.2% of incident power is reflected (|Γ|² = 0.082), and the mismatch loss is a modest 0.37 dB. In a practical system this is usually acceptable without a matching network, especially given that antenna impedance varies with proximity to the user's body. The complex phase of Γ tells us the impedance is on the capacitive side of the Smith chart — adding a small series inductor or shunt capacitor can reduce |Γ| further.

02

Input Impedance of a Terminated 50 Ω Line

Problem: A 50 Ω lossless microstrip line is 3λg/8 long (electrical length βl = 3π/4 = 135°) and is terminated in Z_L = 100 + j50 Ω. Find the input impedance Z_in.

tan(βl) = tan(135°)

tan (135 °) = - 1

Substitute into Z_in formula

Z_{in} = 50 \frac{(100 + j 50) + j (50) (- 1)}{50 + j (100 + j 50) (- 1)} = 50 \frac{100 + j 50 - j 50}{50 - j 100 + 50}

Simplify numerator and denominator

Z_{in} = 50 \times \frac{100}{100 - j 100} = 50 \times \frac{100}{100 (1 - j)} = \frac{50}{1 - j}

Multiply by conjugate: (1 + j)/(1 + j)

Z_{in} = \frac{50 (1 + j)}{2} = 25 + j 25 Ω

Answer: Z_in = 25 + j25 Ω

The 3λg/8 line has transformed 100 + j50 Ω into 25 + j25 Ω. Notice that the imaginary part of Z_L (j50) completely cancelled in the numerator — a coincidence of this particular electrical length. The resulting Z_in = 25 + j25 Ω has |Γ| = |(25 + j25 − 50)/(25 + j25 + 50)| = |(−25 + j25)/(75 + j25)| = √(1250)/√(6250) ≈ 0.447, which is the same as at the load — as expected for a lossless line. Only the phase of Γ has changed, not its magnitude.

03

Quarter-Wave Transformer Design

Problem: A 75 Ω antenna (purely resistive) must be matched to a 50 Ω system using a single quarter-wave transformer at 1.575 GHz (GPS L1). The substrate is Rogers RO4350B with εeff = 3.48. Find: (a) the required Z₀ of the matching section, (b) its physical length, and (c) verify that Z_in = 50 Ω.

(a) Characteristic impedance of matching section

Z_{0} = \sqrt{Z_{S} \times Z_{L}} = \sqrt{50 \times 75} = \sqrt{3750} \approx 61.24 Ω

The matching section must have a characteristic impedance that is the geometric mean of source and load impedances.

(b) Physical length — guided wavelength at 1.575 GHz on RO4350B

λ_{g} = \frac{c}{f \sqrt{ε_{eff}}} = \frac{3 \times 10^{8}}{1.575 \times 10^{9} \times \sqrt{3.48}} = \frac{3 \times 10^{8}}{1.575 \times 10^{9} \times 1.866} \approx 102.0 mm

Quarter-wave physical length

l = \frac{λ_{g}}{4} = \frac{102.0}{4} = 25.5 mm

Z_{in} = \frac{Z_{0}^{2}}{Z_{L}} = \frac{{61.24}^{2}}{75} = \frac{3750}{75} = 50 Ω ✓

Answer: Z₀ = 61.24 Ω | l = 25.5 mm | Z_in = 50 Ω ✓

A 61.24 Ω microstrip trace, 25.5 mm long on RO4350B, perfectly matches 75 Ω to 50 Ω at 1.575 GHz. In practice, you would use the Microstrip Width Calculator to find the trace width that gives 61.24 Ω for your substrate thickness, then lay out the 25.5 mm section. Note that the match is perfect only at exactly 1.575 GHz — the bandwidth of a quarter-wave transformer is finite, and wideband applications may require a multi-section design.

04

Standing Wave Maxima and Minima Positions

Problem: A 50 Ω line is terminated in Z_L = 25 Ω (real). At 3 GHz with εeff = 2.2, find: (a) Γ, (b) VSWR, (c) the position of the first voltage minimum from the load, (d) the position of the first voltage maximum from the load, and (e) the impedance at the first voltage minimum.

(a) Reflection coefficient — real load, real Z₀

Γ = \frac{25 - 50}{25 + 50} = \frac{- 25}{75} = - \frac{1}{3} | Γ | = \frac{1}{3} φ = 180 °

Γ is real and negative (φ = π) because Z_L < Z₀. Voltage is inverted on reflection — the load looks like it's on the short-circuit side of the Smith chart.

(b) VSWR

VSWR = \frac{1 + 1 / 3}{1 - 1 / 3} = \frac{4 / 3}{2 / 3} = 2.0

λ_{g} = \frac{c}{f \sqrt{ε_{eff}}} = \frac{3 \times 10^{8}}{3 \times 10^{9} \times \sqrt{2.2}} \approx 67.4 mm

Voltage minima occur when φ − 2βl′ = π (mod 2π). With φ = π:

Condition for first voltage minimum (n = 0)

π - 2 β l_{\min}^{'} = π ⟹ l_{\min}^{'} = 0

The first voltage minimum is right at the load (l′ = 0). This makes physical sense: Γ = −1/3 is real and negative, so the reflected wave opposes the incident wave in phase at z = 0.

(d) First voltage maximum — λg/4 from the minimum

l_{\max}^{'} = l_{\min}^{'} + \frac{λ_{g}}{4} = 0 + \frac{67.4}{4} = 16.85 mm from load

(e) Impedance at voltage minimum

Z_{in} (at V-min) = \frac{Z_{0}}{VSWR} = \frac{50}{2.0} = 25 Ω

At a voltage minimum the impedance is real and equal to Z₀/VSWR = 25 Ω — which is exactly Z_L. This confirms that the minimum is at the load itself: Z_in(l′=0) = Z_L, as it must be.

Answer: Γ = −1/3, φ = 180° | VSWR = 2.0 | V-min at l′ = 0 (load) | V-max at l′ = 16.85 mm | Z at V-min = 25 Ω

When Z_L is real and less than Z₀, the voltage minimum always falls at the load — Γ is real and negative so the two waves are already in anti-phase at z = 0. Conversely, when Z_L is real and greater than Z₀ (as in Q3 of the previous article), Γ is real and positive (φ = 0°) and the voltage maximum falls at the load. These two cases are the starting points for single-stub matching design: you add a stub at the first real-impedance point on the line to cancel the residual reactance.