8.2 Standing Waves, Reflection Coefficient, SWR, and Return Loss

8.2 Standing Waves, Reflection Coefficient, SWR, and Return Loss

When a radio transmitter generates alternating electromagnetic energy, it launches a forward-travelling wave of voltage and current along the transmission line toward the antenna. If the impedance presented by the load ($Z_L$) exactly matches the characteristic impedance of the feedline ($Z_0$), all incident energy is smoothly absorbed by the load and converted into radiated electromagnetic waves or heat. In this matched state ($Z_L = Z_0$), the transmission line operates in a pure travelling wave regime.

However, if the load impedance differs from the line's characteristic impedance ($Z_L \neq Z_0$), boundary conditions at the termination dictate that not all incident energy can be absorbed. The electrical mismatch forces a portion—and in extreme cases, all—of the incident energy to reflect back from the load toward the transmitter. The superposition of the forward-travelling wave and the reverse-travelling reflected wave creates a stationary interference pattern known as a standing wave.

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The Mechanism of Standing Waves: Wave Superposition

Let distance $x$ be measured along a lossless line from the load ($x=0$) toward the generator ($x>0$). The complex voltage and current phasors are the sums of the incident and reflected waves below. The instantaneous quantities are $\operatorname{Re}\{V(x)e^{j\omega t}\}$ and $\operatorname{Re}\{I(x)e^{j\omega t}\}$.

$$V(x) = V^+ e^{+j\beta x} + V^- e^{-j\beta x}$$ $$I(x) = I^+ e^{+j\beta x} - I^- e^{-j\beta x} = \frac{V^+}{Z_0} e^{+j\beta x} - \frac{V^-}{Z_0} e^{-j\beta x}$$

where:

• $V^+$ and $V^-$ are complex voltage-wave phasors at the load, using peak rather than RMS magnitudes.
• $I^+=V^+/Z_0$ and $I^-=V^-/Z_0$ are the corresponding wave coefficients; the reflected term has the explicit minus sign shown.
• $\beta = 2\pi / \lambda$ is the phase constant.

Example: Gamma = -0.5, so the load is a voltage minimum.

Distance from load:    0     lambda/4    lambda/2    3lambda/4    lambda
Voltage magnitude:   Vmin     Vmax        Vmin         Vmax       Vmin
Current magnitude:   Imax     Imin        Imax         Imin       Imax

Vmax = 1.5 |V+|       Vmin = 0.5 |V+|       VSWR = 3:1
Adjacent voltage extrema: lambda/4; repeated maxima: lambda/2.
Other reflection phases shift this pattern relative to the load.

Interference Nodes and Antinodes

As the forward and reflected waves travel in opposite directions, their relative phase shift changes continuously with position along the line:

1. Voltage Antinodes (Voltage Maxima, $V_{\text{max}}$): At points where the forward and reflected voltage phasors align in phase, they interfere constructively, creating a voltage peak: $V_{\text{max}} = |V^+| + |V^-|$
2. Voltage Nodes (Voltage Minima, $V_{\text{min}}$): At points located a quarter-wavelength ($\lambda/4$, or $90^\circ$ electrically) away from the antinodes, the forward and reflected voltage phasors are exactly $180^\circ$ out of phase, interfering destructively to create a voltage trough: $V_{\text{min}} = |V^+| - |V^-|$
3. Current Distribution: Because the reflected current is inverted in sign, current antinodes ($I_{\text{max}}$) occur precisely at voltage nodes ($V_{\text{min}}$), and current nodes ($I_{\text{min}}$) occur precisely at voltage antinodes ($V_{\text{max}}$): $I_{\text{max}} = \frac{|V^+| + |V^-|}{Z_0} = \frac{V_{\text{max}}}{Z_0}$ $I_{\text{min}} = \frac{|V^+| - |V^-|}{Z_0} = \frac{V_{\text{min}}}{Z_0}$
4. Spatial Periodicity: Adjacent nodes and antinodes are separated by $\lambda/4$ ($90^\circ$). Successive peaks (or successive troughs) are separated by $\lambda/2$ ($180^\circ$).

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The Voltage Reflection Coefficient ($\Gamma$)

The voltage reflection coefficient (denoted by $\Gamma$, the uppercase Greek letter gamma, or occasionally $\rho$, rho) is the complex ratio of the reflected voltage wave to the forward incident voltage wave at the load termination ($x = 0$):

$$\Gamma = \frac{V^-}{V^+}$$

At the load, the total voltage is $V_L = V^+ + V^-$ and the total current is $I_L = (V^+ - V^-) / Z_0$. By Ohm's law, the load impedance is $Z_L = V_L / I_L$:

$$Z_L = Z_0 \left( \frac{V^+ + V^-}{V^+ - V^-} \right) = Z_0 \left( \frac{1 + \Gamma}{1 - \Gamma} \right)$$

Solving explicitly for $\Gamma$ yields the fundamental Reflection Coefficient Formula:

$$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$$

Because load impedance $Z_L = R_L + jX_L$ is generally a complex number containing both resistance and reactance, $\Gamma$ is also a complex quantity characterized by a scalar magnitude ($|\Gamma|$) and a phase angle ($\theta$):

$$\Gamma = |\Gamma| e^{j\theta} = |\Gamma| \angle \theta$$

• The magnitude $|\Gamma|$ represents the proportion of incident voltage reflected, bounded strictly between zero and unity: $0 \le |\Gamma| \le 1$.
• The phase angle $\theta$ represents the phase shift between the reflected voltage wave and the incident voltage wave at the load boundary.

Analysis of Boundary Termination Conditions

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Voltage Standing Wave Ratio (VSWR)

The Voltage Standing Wave Ratio (VSWR, or simply SWR) is defined as the ratio of the maximum RF envelope voltage ($V_{\text{max}}$) to the minimum RF envelope voltage ($V_{\text{min}}$) along the transmission line:

$$\text{VSWR} = \frac{V_{\text{max}}}{V_{\text{min}}} = \frac{|V^+| + |V^-|}{|V^+| - |V^-|}$$

Factoring out $|V^+|$ from numerator and denominator, and recognizing that $|V^-| / |V^+| = |\Gamma|$:

$$\text{VSWR} = \frac{1 + |\Gamma|}{1 - |\Gamma|}$$

Conversely, given a known VSWR, the magnitude of the reflection coefficient is uniquely determined:

$$|\Gamma| = \frac{\text{VSWR} - 1}{\text{VSWR} + 1}$$

Because $0 \le |\Gamma| \le 1$, SWR values range from a minimum of $1.0$ (expressed as $1:1$ or $1.0:1$) for a perfectly matched line, up to infinity ($\infty:1$) for total reflection (short, open, or pure reactance).

SWR for Purely Resistive Loads

When the load is non-reactive ($Z_L = R_L + j0$), the reflection coefficient formula simplifies to:

$$\Gamma = \frac{R_L - Z_0}{R_L + Z_0}$$

Substituting this into the VSWR equation yields two straightforward rules:

$$\text{SWR} = \frac{R_L}{Z_0} \quad (\text{when } R_L > Z_0)$$ $$\text{SWR} = \frac{Z_0}{R_L} \quad (\text{when } R_L < Z_0)$$

Practical Example: The Duality of Resistive Mismatches

Consider a standard $50\ \Omega$ transmission line:

• Terminating in a $150\ \Omega$ resistor: $\text{SWR} = 150 / 50 = 3.0:1$. The reflection coefficient is $\Gamma = (150 - 50)/(150 + 50) = +0.50$.
• Terminating in a $16.67\ \Omega$ resistor: $\text{SWR} = 50 / 16.67 = 3.0:1$. The reflection coefficient is $\Gamma = (16.67 - 50)/(16.67 + 50) = -0.50 = 0.50 \angle 180^\circ$.

Both loads produce an identical standing wave ratio ($3.0:1$) and reflect the exact same percentage of incident power ($|\Gamma|^2 = 0.25 = 25\%$). However, at the load boundary, the $150\ \Omega$ termination creates a voltage antinode ($V_{\text{max}}$), whereas the $16.67\ \Omega$ termination creates a voltage node ($V_{\text{min}}$) and current antinode ($I_{\text{max}}$).

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Return Loss ($RL$) and Reflected Power Ratios

In RF laboratory testing and professional telecommunications, engineers frequently express impedance match quality using Return Loss ($RL$) and power ratios rather than VSWR.

Reflected Power Ratio

The power carried by a forward-travelling wave is $P_{\text{fwd}} = |V^+|^2 / (2Z_0)$, and the power carried by the reflected wave is $P_{\text{ref}} = |V^-|^2 / (2Z_0)$. The ratio of reflected power to forward incident power is:

$$\frac{P_{\text{ref}}}{P_{\text{fwd}}} = \left| \frac{V^-}{V^+} \right|^2 = |\Gamma|^2$$

The fraction of incident power accepted by the load is the difference between forward and reflected powers at that load plane. Antenna radiation is only part of accepted power if the antenna also dissipates energy as heat:

$$\frac{P_{\text{absorbed}}}{P_{\text{fwd}}} = 1 - |\Gamma|^2 = 1 - \left( \frac{\text{VSWR} - 1}{\text{VSWR} + 1} \right)^2$$

Return Loss Formula

Return Loss ($RL$), expressed in decibels ($\text{dB}$), is defined as the logarithmic ratio of forward incident power to reflected power:

$$RL = -20 \log_{10}|\Gamma| = -10 \log_{10}\left( \frac{P_{\text{ref}}}{P_{\text{fwd}}} \right)$$

Master Conversion Table: SWR, Reflection Coefficient, Return Loss, and Power

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Practical Realities and Common SWR Myths

High SWR means a mismatch at the measurement plane. Its effect on a transmitter depends on the impedance presented to the final stage and the amplifier’s ruggedness and protection. Both electrical breakdown and excessive dissipation are possible; reflected-power heating is not forbidden by electromagnetic theory.

What Actually Happens to Reflected Power?

1. Source-End Behavior: A returning wave may be absorbed or re-reflected, depending on the impedance looking toward the source. A matched termination absorbs an incoming wave. An ATU transforms the feedline input impedance to a suitable transmitter load; it does not automatically eliminate standing waves on the antenna side or guarantee that all returning power is re-reflected.
2. Energy Conservation: Separate transmitter output, line dissipation, power accepted by the antenna and power radiated. For a low-loss line with approximately real $Z_0$, a matched-line loss of $0.2\text{ dB}$ and load SWR of $10:1$ give about $0.91\text{ dB}$ total line loss, or $81\%$ transfer of net line-input power to the load. Antenna losses reduce the radiated fraction further. Settling time depends on line delay, reflections and circuit response.
3. The SWR Loss Penalty on Lossy Coax: Why, then, is high SWR undesirable on coaxial feedlines? Because on a lossy transmission line (such as RG-58 at $144\text{ MHz}$), every time the reflected wave traverses back and forth across the line, it suffers the cable's normal attenuation. Higher standing wave current antinodes ($I_{\text{max}}$) dramatically increase $I^2 R'$ conductor skin heating, and voltage antinodes ($V_{\text{max}}$) increase $V^2 G'$ dielectric heating. This compounding effect is known as the additional SWR loss penalty.

Transmitter Protection Circuits: Power Foldback

How does automatic power foldback reduce the risk from an antenna mismatch?

• Peak Transistor Voltage and Current Ratings: Solid-state bipolar junction transistors (BJTs) and laterally diffused MOSFETs (LDMOS) have strict absolute maximum ratings for collector-emitter breakdown voltage ($BV_{\text{ceo}}$) or drain-source breakdown voltage ($V_{\text{dss}}$), as well as maximum safe operating current ($I_{\text{max}}$).
• Adverse Load Impedance: Feedline length transforms a mismatch to different impedances at the transmitter. The resulting RF voltage/current swings can exceed device ratings, while reduced amplifier efficiency can increase heat dissipation. The severity depends on reflection phase, power level and circuit design.
• Power Foldback: A protection circuit senses reflected power or related device stress and reduces drive, gain or supply voltage. This reduces available output power and helps protect the final stage. Thresholds, response speed and the control path depend on the transceiver; follow its specified mismatch limits.

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Step-by-Step Worked Calculation: Complex Load Mismatch

An amateur antenna system operating on the 20-meter band ($14.15\text{ MHz}$) uses a $50\ \Omega$ coaxial feedline. An antenna analyzer connected to the line measures a complex load impedance of:

$$Z_L = 75 + j50\ \Omega$$

Calculate:

1. The complex voltage reflection coefficient ($\Gamma$)
2. The magnitude ($|\Gamma|$) and phase angle ($\theta$)
3. The Voltage Standing Wave Ratio (VSWR)
4. The Return Loss ($RL$) in decibels
5. The percentage of forward incident power reflected

Step 1: Calculate the Complex Reflection Coefficient ($\Gamma$)

$$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{(75 + j50) - 50}{(75 + j50) + 50} = \frac{25 + j50}{125 + j50}$$

To divide complex numbers, multiply numerator and denominator by the complex conjugate of the denominator ($125 - j50$):

$$\Gamma = \frac{(25 + j50)(125 - j50)}{(125 + j50)(125 - j50)} = \frac{(25 \times 125) - j(25 \times 50) + j(50 \times 125) - j^2(50 \times 50)}{125^2 + 50^2}$$ $$\Gamma = \frac{3125 - j1250 + j6250 + 2500}{15625 + 2500} = \frac{5625 + j5000}{18125}$$ $$\Gamma \approx 0.3103 + j0.2759$$

Step 2: Determine Magnitude ($|\Gamma|$) and Phase Angle ($\theta$)

$$|\Gamma| = \sqrt{(0.3103)^2 + (0.2759)^2} = \sqrt{0.0963 + 0.0761} = \sqrt{0.1724} \approx 0.4152$$ $$\theta = \arctan\left(\frac{0.2759}{0.3103}\right) = \arctan(0.8891) \approx 41.64^\circ$$

In polar form: $\Gamma \approx 0.4152 \angle 41.64^\circ$.

Step 3: Calculate the Voltage Standing Wave Ratio (VSWR)

$$\text{VSWR} = \frac{1 + |\Gamma|}{1 - |\Gamma|} = \frac{1 + 0.4152}{1 - 0.4152} = \frac{1.4152}{0.5848} \approx 2.42:1$$

Step 4: Calculate Return Loss ($RL$)

$$RL = -20 \log_{10}|\Gamma| = -20 \log_{10}(0.4152) = -20 \times (-0.3818) \approx 7.64\text{ dB}$$

Step 5: Determine Reflected Power Percentage

$$\frac{P_{\text{ref}}}{P_{\text{fwd}}} = |\Gamma|^2 = (0.4152)^2 \approx 0.1724 = 17.24\%$$

If the transmitter delivers $100\text{ W}$ forward incident power, approximately $17.2\text{ W}$ is reflected by the load mismatch, leaving $82.8\text{ W}$ absorbed by the antenna system on the initial wave passage.