Transmission Lines

Distributed-Parameter Model

A transmission line guides electromagnetic energy from a source to a load. The distributed $RLCG$ model describes structures such as a wire pair, coaxial cable, stripline, and, approximately, microstrip; hollow waveguides require mode-dependent propagation models.

• Distributed parameters: Resistance, inductance, capacitance, and leakage conductance occur along the entire line. Voltage and current therefore depend on position as well as time.

• Electrical length: A lumped model becomes inadequate when propagation delay is appreciable compared with the sinusoidal period or a digital signal’s rise time. A length of about $\lambda/10$ is a common sinusoidal guideline, not a sharp boundary.

Equivalent circuit of a transmission-line section.

Primary Constants

Constant	Unit	Physical origin	Element in length $\Delta x$
$R$	$\Omega/\mathrm{m}$	Conductor loss, including frequency-dependent skin effect	Series resistance $R\Delta x$
$L$	$\mathrm{H/m}$	Magnetic field associated with line current	Series inductance $L\Delta x$
$C$	$\mathrm{F/m}$	Electric field between conductors	Shunt capacitance $C\Delta x$
$G$	$\mathrm{S/m}$	Leakage and dielectric loss	Shunt conductance $G\Delta x$

Note

Distributed circuit elements

• Frequency dependence: The per-unit-length values can vary with frequency, particularly conductor resistance through skin effect and conductance through dielectric loss.

• Leakage branch: A section of length $\Delta x$ has shunt conductance $G\Delta x$ and leakage resistance $1/(G\Delta x)$. If $G=0$, this resistive branch is open; the shunt capacitance remains.

Telegrapher Equations

Take $x$ from source to load and reference current in the $+x$ direction. Use sinusoidal phasors with time factor $e^{\mathrm{j}\omega t}$. To first order in $\Delta x$, KVL across the series elements and KCL through the shunt elements give:

KVL:

$$V(x)-V(x+\Delta x)=(R+\mathrm{j}\omega L)\Delta x\,I(x).$$

KCL:

$$I(x)-I(x+\Delta x)=(G+\mathrm{j}\omega C)\Delta x\,V(x).$$

Limit form:

$$\frac{dV}{dx}=-(R+\mathrm{j}\omega L)I, \qquad \frac{dI}{dx}=-(G+\mathrm{j}\omega C)V.$$

Wave equations:

$$\frac{d^2V}{dx^2}=\gamma^2V, \qquad \frac{d^2I}{dx^2}=\gamma^2I,$$

Propagation constant:

$$\gamma=\sqrt{(R+\mathrm{j}\omega L)(G+\mathrm{j}\omega C)} =\alpha+\mathrm{j}\beta.$$

Secondary Constants

Quantity	Meaning	Unit or relation
$\gamma$	Complex propagation constant	$\mathrm{m^{-1}}$
$\alpha$	Exponential attenuation per unit length	$\mathrm{Np/m}$
$\beta$	Phase change per unit length	$\mathrm{rad/m}$
$Z_0$	Characteristic impedance	$\Omega$
$v_p$	Phase velocity at the stated frequency	$v_p=\omega/\beta$
$\lambda$	Distance for a $2\pi$ phase change	$\lambda=2\pi/\beta=v_p/f$

A forward wave’s amplitude falls by a factor $e^{-\alpha\ell}$ over length $\ell$. Its one-way attenuation in decibels is

$$A_{\mathrm{dB}}=20\log_{10}(e^{\alpha\ell}) =8.686\alpha\ell.$$

Lossless Line

For $R=G=0$,

$$\alpha=0,\qquad \beta=\omega\sqrt{LC},\qquad Z_0=\sqrt{\frac{L}{C}},\qquad v_p=\frac{1}{\sqrt{LC}}.$$

Note

Real impedance without dissipation

$Z_0$ is real even when $R=G=0$: a forward wave carries energy toward the load rather than dissipating it in the line. Its voltage-to-current ratio is not the conductor’s ohmic resistance.

Low-Loss and Distortionless Lines

For $R\ll\omega L$ and $G\ll\omega C$,

$$Z_0\approx\sqrt{\frac{L}{C}},\qquad \alpha\approx\frac{R}{2Z_0}+\frac{GZ_0}{2},\qquad \beta\approx\omega\sqrt{LC}.$$

An idealised distortionless line satisfies the Heaviside condition

$$\frac{R}{L}=\frac{G}{C}.$$

Under this condition, with frequency-independent parameters,

$$Z_0=\sqrt{\frac{L}{C}},\qquad \alpha=\sqrt{RG},\qquad \beta=\omega\sqrt{LC}.$$

Note

Distortionless-line result

• Attenuation is frequency-independent.

• Phase constant is proportional to frequency.

• Waveform shape is preserved except for attenuation and delay.

Characteristic Impedance

• Wave ratio: Characteristic impedance $Z_0$ is the voltage-to-current ratio of a single forward travelling wave.

• Input impedance: An infinitely long uniform line has input impedance $Z_0$ because no wave returns from its far end. A finite uniform line terminated in $Z_0$ has the same input impedance.

For a forward wave, $dV/dx=-\gamma V$:

$$Z_0=\frac{V^+}{I^+} =\frac{R+\mathrm{j}\omega L}{\gamma} =\sqrt{\frac{R+\mathrm{j}\omega L}{G+\mathrm{j}\omega C}}.$$

For a lossy line, $Z_0$ is generally complex and frequency-dependent. For a lossless line, it reduces to $\sqrt{L/C}$.

Matched Termination

• No reflection: With $Z_L=Z_0$, the incident wave already satisfies the load’s voltage-to-current ratio. No reflected wave is required, and the line’s input impedance is $Z_0$ at any length.

• Source matching: Eliminating load reflection is distinct from maximising power drawn from a source. Maximum available source power requires a conjugate match at the source port; the source impedance and any intervening matching network must also be considered.

Incident and Reflected Waves

An impedance discontinuity generates a reflected wave that travels toward the source. For a uniform line from $x=0$ to $x=\ell$, with current referenced in the $+x$ direction, the total voltage and current are the sums of forward and backward waves.

Incident and reflected waves on a terminated transmission line.

$$V(x)=V_0^+e^{-\gamma x}+V_0^-e^{\gamma x},$$ $$I(x)=\frac{V_0^+}{Z_0}e^{-\gamma x} -\frac{V_0^-}{Z_0}e^{\gamma x}.$$

$V_0^+$ and $V_0^-$ are the wave amplitudes at $x=0$. The reflected-current term has a minus sign because both currents use the same $+x$ reference, while the reflected wave carries energy in the opposite direction.

Load Reflection Coefficient

Let $V_L^+$ and $V_L^-$ be the incident and reflected voltage-wave amplitudes at the load. Their total voltage and current must satisfy the load impedance:

$$Z_L=\frac{V_L^++V_L^-}{(V_L^+-V_L^-)/Z_0}.$$

With $\Gamma_L=V_L^-/V_L^+$:

$$\boxed{\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}}.$$

$\Gamma_L$ is dimensionless and generally complex. Its magnitude is the reflected-to-incident voltage amplitude ratio; its angle is the reflected wave’s phase relative to the incident wave at the load.

At distance $d$ from load toward source:

$$\Gamma(d)=\Gamma_L e^{-2\gamma d}.$$

The factor 2 accounts for travel to the load and back. On a lossless line, moving the reference plane changes only the phase of $\Gamma$; on a lossy line, its magnitude toward the source decreases by $e^{-2\alpha d}$.

Matched, Open and Short Circuits

Termination	$\Gamma_L$	Voltage at load	Current at load
$Z_L=Z_0$	$0$	Incident voltage only	Incident current only
Open circuit	$+1$	Incident and reflected voltages add	Incident and reflected currents cancel
Short circuit	$-1$	Incident and reflected voltages cancel	Incident and reflected currents add

Note

Termination note

• Open or short on a real-$Z_0$ lossless line reflects all incident power.

• Matched load absorbs the power arriving at it.

Standing Waves and VSWR

• Standing waves: Interference between incident and reflected waves creates stationary maxima and minima in the voltage envelope along the line.

• VSWR: The ratio of maximum to minimum voltage amplitude measures mismatch on a lossless line. A matched line has a uniform envelope and VSWR of 1.

For a lossless line,

$$V_{\max}=|V^+|(1+|\Gamma_L|),\qquad V_{\min}=|V^+|(1-|\Gamma_L|).$$

The voltage standing-wave ratio is

$$\boxed{S=\mathrm{VSWR}=\frac{V_{\max}}{V_{\min}} =\frac{1+|\Gamma_L|}{1-|\Gamma_L|}}.$$

Therefore,

$$|\Gamma_L|=\frac{S-1}{S+1}.$$

Voltage and current standing waves for matched, open and shorted lines.

• Voltage maxima are separated by $\lambda/2$, and voltage minima are separated by $\lambda/2$.

• A voltage maximum and its nearest voltage minimum are separated by $\lambda/4$.

• On a lossless line, a voltage maximum coincides with a current minimum and vice versa.

• At an open load, voltage is maximum and current is zero; at a shorted load, voltage is zero and current is maximum.

• A matched lossless line has uniform voltage/current amplitudes and $S=1$; complete reflection gives $S\to\infty$.

Note

VSWR cautions

• Plots show amplitude envelopes, not instantaneous voltage/current versus time.

• Simple VSWR formula assumes a lossless or low-loss measured section.

• Incident/reflected amplitudes must be nearly constant across the measured pattern.

Return Loss and Reflected Power

Return loss compares incident and reflected power at the same reference plane. For a real, positive reference impedance $Z_0$, the reflected power fraction is $|\Gamma_L|^2$, giving

$$\boxed{RL=10\log_{10}\frac{P_i}{P_r} =-20\log_{10}|\Gamma_L|\ \mathrm{dB}}.$$

Useful relationships:

$$\frac{P_r}{P_i}=|\Gamma_L|^2=10^{-RL/10}, \qquad |\Gamma_L|=10^{-RL/20},$$ $$RL=20\log_{10}\frac{S+1}{S-1},\qquad S>1.$$

Variation of return loss and VSWR with reflection coefficient.

Condition	$\lvert\Gamma_L\rvert$	VSWR	Return loss	Reflected power
Perfect match	$0$	$1$	$\infty$	$0$
Good match example	$0.1$	$1.222$	$20\,\mathrm{dB}$	$1\%$
Moderate mismatch	$0.2$	$1.5$	$13.98\,\mathrm{dB}$	$4\%$
VSWR of 2	$1/3$	$2$	$9.54\,\mathrm{dB}$	$11.11\%$
Open or short	$1$	$\infty$	$0\,\mathrm{dB}$	$100\%$

Note

Reflection and delivered power

• Match quality: Small $|\Gamma_L|$, VSWR near 1, and high return loss all indicate little reflection.

• Insertion loss: Measures the reduction in through-path power caused by inserting a component. Return loss instead measures power reflected toward the source.

• Load power: On a lossless line with real $Z_0$, $P_L=P_i(1-|\Gamma_L|^2)$, where $P_i$ is the power incident at the load. If incident power is specified at the input of a lossy line, propagation loss must also be included.

• Complex reference impedance: A voltage-wave reflection coefficient does not generally give a reflected power fraction of $|\Gamma|^2$ when $Z_0$ is complex; power must be calculated with the appropriate voltage and current relations.

Input Impedance

For a uniform line of length $\ell$ terminated in $Z_L$,

$$Z_{\mathrm{in}}=Z_0 \frac{Z_L+Z_0\tanh(\gamma\ell)} {Z_0+Z_L\tanh(\gamma\ell)}.$$

For a lossless line, $\gamma=\mathrm{j}\beta$ and $\tanh(\mathrm{j}\beta\ell)=\mathrm{j}\tan(\beta\ell)$, giving

$$Z_{\mathrm{in}}=Z_0 \frac{Z_L+\mathrm{j}Z_0\tan(\beta\ell)} {Z_0+\mathrm{j}Z_L\tan(\beta\ell)}.$$

Lossless-line condition	Input impedance
Matched load	$Z_{\mathrm{in}}=Z_0$
Short-circuited load	$Z_{\mathrm{in}}=\mathrm{j}Z_0\tan(\beta\ell)$
Open-circuited load	$Z_{\mathrm{in}}=-\mathrm{j}Z_0\cot(\beta\ell)$
Half-wavelength line	$Z_{\mathrm{in}}=Z_L$
Quarter-wavelength line	$Z_{\mathrm{in}}=Z_0^2/Z_L$

Note

Special line lengths

• Quarter-wave line transforms open $\leftrightarrow$ short.

• Half-wave line repeats the load impedance.

• Quarter-wave section can be used as an impedance transformer.

Impedance Matching

Impedance matching reduces reflections, standing waves, and power returned to the transmitter. By limiting repeated reflections, it also reduces response ripple and reflection-related echo.

Method	Principle	Main limitation
Matched termination	Set the terminating impedance equal to the line’s $Z_0$	Dissipates received power in the terminating load
Transformer	Reflect impedance through the square of the turns ratio	Practical bandwidth, loss and parasitics
Quarter-wave section	Transform a real resistance using a specified line impedance and electrical length	Frequency-sensitive; simplest form matches real resistances
Reactive network or stub	Cancel reactance and transform the resistive part	Frequency dependence and tuning requirements

Quarter-Wave Transformer

A lossless section of characteristic impedance $Z_t$ and length $\lambda_t/4$ transforms a real load resistance $R_L$ into

$$Z_{\mathrm{in}}=\frac{Z_t^2}{R_L}.$$

To match it to a line of real characteristic impedance $Z_0$,

$$\boxed{Z_t=\sqrt{Z_0R_L}},\qquad \ell=\frac{\lambda_t}{4}=\frac{v_{p,t}}{4f_0}.$$

Quarter-wave impedance transformer.

Note

Quarter-wave transformer notes

• $f_0$ is the design frequency.

• $v_{p,t}$ is phase velocity in the matching section.

• Use guided wavelength in the matching section, not automatically free-space wavelength.

• Away from $f_0$, electrical length changes and matching degrades.

Worked Problems

Lossless Line Constants

A lossless line has $L=0.25\,\mu\mathrm{H/m}$, $C=100\,\mathrm{pF/m}$, and $R=G=0$. At $f=10\,\mathrm{MHz}$, find $Z_0$, $v_p$, $\lambda$, and $\beta$.

$$Z_0=\sqrt{\frac{0.25\times10^{-6}}{100\times10^{-12}}} =50\,\Omega.$$ $$v_p=\frac{1}{\sqrt{(0.25\times10^{-6})(100\times10^{-12})}} =2.00\times10^8\,\mathrm{m/s}.$$ $$\lambda=\frac{2.00\times10^8}{10\times10^6}=20.0\,\mathrm{m}, \qquad \beta=\frac{2\pi}{20}=0.3142\,\mathrm{rad/m}.$$

Lossy Line from RLCG

At $1\,\mathrm{MHz}$, a line has $R=0.2\,\Omega/\mathrm{m}$, $L=0.25\,\mu\mathrm{H/m}$, $G=10\,\mu\mathrm{S/m}$ and $C=100\,\mathrm{pF/m}$.

$$R+\mathrm{j}\omega L=0.2+\mathrm{j}1.5708\ \Omega/\mathrm{m},$$ $$G+\mathrm{j}\omega C=10^{-5}+\mathrm{j}6.2832\times10^{-4}\ \mathrm{S/m}.$$

Using the passive-line square-root branch,

$$Z_0=\sqrt{\frac{0.2+\mathrm{j}1.5708} {10^{-5}+\mathrm{j}6.2832\times10^{-4}}} \approx(50.121-\mathrm{j}2.778)\,\Omega.$$ $$\gamma=\sqrt{(0.2+\mathrm{j}1.5708) (10^{-5}+\mathrm{j}6.2832\times10^{-4})} \approx0.0022465+\mathrm{j}0.0314644\ \mathrm{m^{-1}}.$$

Thus $\alpha=0.0022465\,\mathrm{Np/m}$, $\beta=0.0314644\,\mathrm{rad/m}$, and $\lambda=2\pi/\beta\approx199.69\,\mathrm{m}$. The one-way travelling-wave attenuation over 100 m is

$$A_{\mathrm{dB}}=8.686(0.0022465)(100)\approx1.95\,\mathrm{dB}.$$

Real Load Mismatch

A lossless $50\,\Omega$ line is terminated in $75\,\Omega$.

$$\Gamma_L=\frac{75-50}{75+50}=0.2, \qquad S=\frac{1.2}{0.8}=1.5.$$ $$RL=-20\log_{10}(0.2)=13.98\,\mathrm{dB}, \qquad \frac{P_r}{P_i}=0.2^2=0.04.$$

For incident power $P_i=10\,\mathrm{W}$, the reflected power is $0.4\,\mathrm{W}$ and the load receives $9.6\,\mathrm{W}$.

Complex Load Mismatch

A lossless $50\,\Omega$ line is terminated in $Z_L=100+\mathrm{j}50\,\Omega$.

$$\Gamma_L=\frac{50+\mathrm{j}50}{150+\mathrm{j}50} =\frac{(50+\mathrm{j}50)(150-\mathrm{j}50)}{150^2+50^2} =0.4+\mathrm{j}0.2.$$ $$|\Gamma_L|=\sqrt{0.4^2+0.2^2}=0.4472, \qquad \angle\Gamma_L=\tan^{-1}(0.2/0.4)=26.565^\circ.$$ $$S=\frac{1+0.4472}{1-0.4472}=2.618, \qquad RL=-20\log_{10}(0.4472)=6.99\,\mathrm{dB}.$$

The reflected power fraction is $|\Gamma_L|^2=0.20=20\%$. Both VSWR and return loss depend on the magnitude of the complex reflection coefficient, not its real part alone.

Return Loss from VSWR

A lossless line has $S=2$. Its reflection magnitude and return loss are

$$|\Gamma_L|=\frac{2-1}{2+1}=\frac13, \qquad RL=-20\log_{10}\left(\frac13\right)=9.54\,\mathrm{dB}.$$ $$\frac{P_r}{P_i}=\left(\frac13\right)^2=0.1111.$$

Note

VSWR-to-power caution

• $|\Gamma|=1/3$ means voltage reflection magnitude is $33.33\%$.

• Reflected power fraction is $|\Gamma|^2=11.11\%$.

Quarter-Wave Matching

Match a $100\,\Omega$ load to a $50\,\Omega$ line at $100\,\mathrm{MHz}$ using a lossless quarter-wave section. The section’s phase velocity is $2.0\times10^8\,\mathrm{m/s}$. Its required impedance and length are

$$Z_t=\sqrt{50\times100}=70.71\,\Omega,$$ $$\ell=\frac{2.0\times10^8}{4(100\times10^6)}=0.50\,\mathrm{m}.$$

At the design frequency,

$$Z_{\mathrm{in}}=\frac{70.71^2}{100}\approx50\,\Omega,$$

so the section presents $50\,\Omega$ to the main line at the design frequency.