Special case Lossless transmission

Analysis of a Terminated Lossless Transmission Line

The analysis of a terminated lossless transmission line begins with the general equations obtained for a lossless line, where the series resistance and shunt conductance are zero. Thus, \(R=0\) and \(G=0\). In this condition, the transmission line does not dissipate power as heat or through leakage, and the propagation constant is purely imaginary. When the line is terminated by a load impedance \(Z_L\), the incident wave arriving at the load may be completely absorbed or partially reflected depending on the relationship between \(Z_L\) and the characteristic impedance \(Z_0\). The resulting voltage and current distributions depend strongly on the nature of the termination. Three important cases are the short-circuit termination, open-circuit termination, and the quarter-wave transformer.

Basic Equations of a Lossless Transmission Line

For a lossless transmission line, the resistance and conductance are zero:

\[ R=0,\qquad G=0 \]

The propagation constant of a transmission line is generally given by:

\[ \gamma = \sqrt{(R+j\omega L)(G+j\omega C)} \]

For \(R=0\) and \(G=0\), this becomes:

\[ \gamma = j\omega\sqrt{LC} \]

Since:

\[ \gamma=\alpha+j\beta \]

comparison gives:

\[ \alpha=0 \]

and:

\[ \beta=\omega\sqrt{LC} \]

Therefore, the propagation constant of a lossless line is:

\[ \boxed{\gamma=j\beta=j\omega\sqrt{LC}} \]

The absence of the attenuation constant \(\alpha\) means that the magnitude of a traveling wave does not decrease as it propagates along the ideal lossless transmission line.

Characteristic Impedance of a Lossless Line

The general characteristic impedance of a transmission line is:

\[ Z_0 = \sqrt{\frac{R+j\omega L}{G+j\omega C}} \]

For a lossless line, \(R=0\) and \(G=0\). Therefore:

\[ Z_0 = \sqrt{\frac{j\omega L}{j\omega C}} \]

Hence:

\[ \boxed{ Z_0=\sqrt{\frac{L}{C}} } \]

The characteristic impedance of an ideal lossless transmission line is therefore purely real. It represents the impedance that must terminate the line for maximum power transfer without producing a reflected wave.

General Voltage and Current Solutions

The voltage on a lossless transmission line consists of a forward-traveling incident wave and a backward-traveling reflected wave. The general voltage solution is:

\[ \boxed{ V(z) = V_0^+e^{-j\beta z} + V_0^-e^{j\beta z} } \]

Similarly, the general current solution is:

\[ \boxed{ I(z) = \frac{V_0^+}{Z_0}e^{-j\beta z} - \frac{V_0^-}{Z_0}e^{j\beta z} } \]

Here, \(V_0^+\) represents the complex amplitude of the incident voltage wave and \(V_0^-\) represents the complex amplitude of the reflected voltage wave. The negative sign associated with the reflected current wave occurs because the reflected wave travels in the opposite direction to the incident wave.

Voltage Reflection Coefficient

Let the transmission line be terminated at \(z=0\) by a load impedance \(Z_L\). At the load, the ratio of total voltage to total current must be equal to the load impedance:

\[ Z_L=\frac{V(0)}{I(0)} \]

At \(z=0\), the voltage becomes:

\[ V(0)=V_0^++V_0^- \]

and the current becomes:

\[ I(0) = \frac{V_0^+}{Z_0} - \frac{V_0^-}{Z_0} \]

Therefore:

\[ Z_L = \frac{V_0^++V_0^-} {\dfrac{V_0^+-V_0^-}{Z_0}} \]

Rearranging the equation gives the voltage reflection coefficient:

\[ \boxed{ \Gamma = \frac{V_0^-}{V_0^+} = \frac{Z_L-Z_0}{Z_L+Z_0} } \]

This equation is fundamental to the analysis of a terminated transmission line. The value of \(\Gamma\) determines both the magnitude and phase of the reflected wave. The three important terminations considered below produce three distinct reflection-coefficient values.

Case 1: Short-Circuit Termination

In a short-circuit termination, the load impedance is zero:

\[ Z_L=0 \]

Substituting this condition into the reflection-coefficient equation gives:

\[ \Gamma = \frac{0-Z_0}{0+Z_0} \]

Therefore:

\[ \boxed{\Gamma=-1} \]

The magnitude of the reflection coefficient is unity:

\[ |\Gamma|=1 \]

This means that the entire incident wave is reflected. The negative sign indicates that the reflected voltage wave undergoes a \(180^\circ\) phase reversal with respect to the incident voltage wave.

Voltage Distribution for a Short Circuit

For a short circuit:

\[ V_0^-=\Gamma V_0^+ \]

Since \(\Gamma=-1\):

\[ V_0^-=-V_0^+ \]

Substituting this into the general voltage equation:

\[ V(z) = V_0^+e^{-j\beta z} - V_0^+e^{j\beta z} \]

Taking \(V_0^+\) common:

\[ V(z) = V_0^+ \left( e^{-j\beta z} - e^{j\beta z} \right) \]

Using:

\[ e^{-j\theta}-e^{j\theta} = -2j\sin\theta \]

the voltage distribution becomes:

\[ \boxed{ V(z) = -2jV_0^+\sin(\beta z) } \]

Thus, the voltage varies sinusoidally along the transmission line and has zero value at the short-circuited load.

Current Distribution for a Short Circuit

The general current equation is:

\[ I(z) = \frac{V_0^+}{Z_0}e^{-j\beta z} - \frac{V_0^-}{Z_0}e^{j\beta z} \]

Since \(V_0^-=-V_0^+\):

\[ I(z) = \frac{V_0^+}{Z_0}e^{-j\beta z} + \frac{V_0^+}{Z_0}e^{j\beta z} \]

Therefore:

\[ I(z) = \frac{V_0^+}{Z_0} \left( e^{-j\beta z} + e^{j\beta z} \right) \]

Using:

\[ e^{-j\theta}+e^{j\theta} = 2\cos\theta \]

we obtain:

\[ \boxed{ I(z) = \frac{2V_0^+}{Z_0}\cos(\beta z) } \]

Therefore, the voltage and current standing-wave patterns are \(90^\circ\) apart in space. Where the voltage is zero, the current reaches a maximum, and where the voltage reaches a maximum magnitude, the current reaches a minimum.

Voltage and Current at the Short-Circuited Load

At the load, \(z=0\). Therefore:

\[ V(0) = -2jV_0^+\sin(0) = 0 \]

Hence, the short circuit produces a voltage node.

For current:

\[ I(0) = \frac{2V_0^+}{Z_0}\cos(0) \]

Therefore:

\[ \boxed{ I(0)=\frac{2V_0^+}{Z_0} } \]

Thus, the short-circuited termination corresponds to a current antinode at the load. This agrees with the physical condition of a short circuit, where the voltage must be zero while current can reach a maximum.

Input Impedance for a Short-Circuited Line

The input impedance of a lossless transmission line terminated by \(Z_L\) is:

\[ Z_{\mathrm{in}} = Z_0 \frac{ Z_L+jZ_0\tan(\beta l) }{ Z_0+jZ_L\tan(\beta l) } \]

For a short-circuit termination, \(Z_L=0\). Therefore:

\[ Z_{\mathrm{in}} = Z_0 \frac{ jZ_0\tan(\beta l) }{ Z_0 } \]

Hence:

\[ \boxed{ Z_{\mathrm{in}} = jZ_0\tan(\beta l) } \]

The input impedance is therefore purely reactive. Its value changes periodically with the electrical length \(l\), alternating between inductive and capacitive reactance as the observation point moves along the line.

Fig: Voltage, Current and Input Reactance for a Short-Circuited Line

Case 2: Open-Circuit Termination

In an open-circuit termination, the load impedance approaches infinity:

\[ Z_L\rightarrow\infty \]

The reflection coefficient is:

\[ \Gamma = \lim_{Z_L\rightarrow\infty} \frac{Z_L-Z_0}{Z_L+Z_0} \]

Dividing the numerator and denominator by \(Z_L\):

\[ \Gamma = \lim_{Z_L\rightarrow\infty} \frac{1-\dfrac{Z_0}{Z_L}} {1+\dfrac{Z_0}{Z_L}} \]

As \(Z_L\rightarrow\infty\), the ratio \(Z_0/Z_L\) approaches zero. Therefore:

\[ \boxed{\Gamma=+1} \]

The reflected voltage wave has the same magnitude and the same phase as the incident voltage wave. Therefore, the voltage waves add at the open-circuited load.

Voltage and Current at an Open Circuit

For an open circuit:

\[ V_0^-=V_0^+ \]

Substituting into the voltage equation gives:

\[ V(z) = V_0^+ \left( e^{-j\beta z} + e^{j\beta z} \right) \]

Therefore:

\[ \boxed{ V(z) = 2V_0^+\cos(\beta z) } \]

For current:

\[ I(z) = \frac{V_0^+}{Z_0} \left( e^{-j\beta z} - e^{j\beta z} \right) \]

Hence:

\[ \boxed{ I(z) = -\frac{2jV_0^+}{Z_0}\sin(\beta z) } \]

At the open-circuited load \(z=0\):

\[ V(0) = 2V_0^+ \]

while:

\[ I(0) = 0 \]

Thus, an open circuit produces a voltage antinode and a current node at the load. This is consistent with the physical condition of an open circuit, where current cannot flow through the termination.

Input Impedance for an Open-Circuited Line

Starting with the general input-impedance equation:

\[ Z_{\mathrm{in}} = Z_0 \frac{ Z_L+jZ_0\tan(\beta l) }{ Z_0+jZ_L\tan(\beta l) } \]

Divide the numerator and denominator by \(Z_L\):

\[ Z_{\mathrm{in}} = Z_0 \frac{ 1+j\frac{Z_0}{Z_L}\tan(\beta l) }{ \frac{Z_0}{Z_L}+j\tan(\beta l) } \]

For an open circuit, \(Z_L\rightarrow\infty\), so:

\[ \frac{Z_0}{Z_L}\rightarrow0 \]

Therefore:

\[ Z_{\mathrm{in}} = \frac{Z_0}{j\tan(\beta l)} \]

Since:

\[ \frac{1}{j}=-j \]

we obtain:

\[ \boxed{ Z_{\mathrm{in}} = -jZ_0\cot(\beta l) } \]

Thus, the input impedance of an open-circuited lossless line is also purely imaginary and varies periodically with the electrical length of the line.

Fig: Voltage, Current and Input Reactance for an Open-Circuited Line

Case 3: Quarter-Wave Transformer

A quarter-wave transformer is a lossless transmission-line section whose electrical length is one-quarter wavelength. It is widely used for impedance transformation and matching because a quarter-wavelength section transforms the load impedance according to a simple inverse relationship. Unlike the short-circuit and open-circuit cases, the quarter-wave transformer is primarily important because of its ability to transform one impedance into another.

Fig: Quarter-Wave Transformer with Terminated Load

Electrical Length

For a quarter-wave transmission-line section:

\[ l=\frac{\lambda}{4} \]

The phase constant is:

\[ \beta=\frac{2\pi}{\lambda} \]

Therefore, the electrical length is:

\[ \beta l = \frac{2\pi}{\lambda} \frac{\lambda}{4} \]

Hence:

\[ \boxed{ \beta l=\frac{\pi}{2} } \]

Thus, a physical length of \(\lambda/4\) corresponds to an electrical length of \(90^\circ\).

Input Impedance of a Quarter-Wave Transformer

The general input impedance of a lossless transmission line is:

\[ Z_{\mathrm{in}} = Z_0 \frac{ Z_L+jZ_0\tan(\beta l) }{ Z_0+jZ_L\tan(\beta l) } \]

For a quarter-wave section:

\[ \beta l=\frac{\pi}{2} \]

Therefore:

\[ \tan(\beta l) = \tan\left(\frac{\pi}{2}\right) \rightarrow\infty \]

To simplify the expression, divide the numerator and denominator by \(\tan(\beta l)\):

\[ Z_{\mathrm{in}} = Z_0 \frac{ \dfrac{Z_L}{\tan(\beta l)}+jZ_0 }{ \dfrac{Z_0}{\tan(\beta l)}+jZ_L } \]

As \(\tan(\beta l)\rightarrow\infty\):

\[ \frac{Z_L}{\tan(\beta l)} \rightarrow0 \]

and:

\[ \frac{Z_0}{\tan(\beta l)} \rightarrow0 \]

Therefore:

\[ Z_{\mathrm{in}} = Z_0 \frac{jZ_0}{jZ_L} \]

Canceling \(j\):

\[ \boxed{ Z_{\mathrm{in}} = \frac{Z_0^2}{Z_L} } \]

Quarter-Wave Transformer Relation

The quarter-wave transformer relation can also be written as:

\[ \boxed{ Z_{\mathrm{in}}Z_L=Z_0^2 } \]

This equation shows that a quarter-wave transmission-line section acts as an impedance inverter. A high load impedance is transformed into a low input impedance, while a low load impedance is transformed into a high input impedance. The characteristic impedance of the quarter-wave section determines the relationship between the load impedance and the impedance observed at its input.

For example, if a quarter-wave section is required to match a real load \(Z_L\) to a real source impedance \(Z_S\), the characteristic impedance of the quarter-wave section can be selected such that:

\[ Z_{\mathrm{in}}=Z_S \]

Using the quarter-wave relation:

\[ Z_S = \frac{Z_0^2}{Z_L} \]

Therefore:

\[ \boxed{ Z_0=\sqrt{Z_SZ_L} } \]

This property makes the quarter-wave transformer an important technique for impedance matching in RF and microwave transmission-line systems.