Distributed Parameter Equivalents and Models

(_b, _b = _o)

(_a, _a = _o)

LC \not=

_a = _b

z

z C^-1

z L^-1

Transverse Electromagnetic Waves

_l

(a)	Show that, defined as zero on the outer conductor, Az
(b)	Using these expressions, show that the L and C needed to complete the transmission line equations are and hence that LC = .

14.2.2A transmission line consists of a conductor having the cross-section shown in Fig. P4.7.5 adjacent to an L-shaped return conductor comprised of ``ground planes'' in the planes x = 0 and y = 0, intersecting at the origin. Assuming that the region between these conductors is free space, what are the transmission line parameters L and C?  

Transients on Infinite Transmission Lines

= 2.5 _o

= _o

14.3.4On an infinite line, when t = 0, V = V_o \exp (-z²/2a²), and I = 0, determine analytical expressions for V(z, t) and I(z, t).

14.3.5^*In the energy conservation theorem for a transmission line, (14.2.19), VI is the power flow. Show that at any location, z, and time, t, it is correct to think of power flow as the superposition of power carried by the + wave in the +z direction and - wave in the -z direction.

Transients on Bounded Transmission Lines

14.4.2The transient is to be determined as in Prob. 14.4.1, except the line is now terminated at z = l in a ``short circuit.'' 14.4.3The transmission line of Fig. 14.4.1 is terminated in a resistance R_L = Z_o. Show that, provided that the voltage and current over the length of the line are initially zero, the line has the same effect on the circuit connected at z = 0 as would a resistance Z_o.

14.4.4A transmission line having characteristic impedance Z_a is terminated at z = l + L in a resistance R_a = Z_a. At the other end, where z = l, it is connected to a second transmission line having the characteristic impedance Z_b. This line is driven at z = 0 by a voltage source V_g(t) in series with a resistance R_b = Z_b. With V_g = 0 for t < 0, the driving voltage makes a step change to V_g = V_o, a constant voltage. Determine the voltage V(0, t).

14.4.5A pair of transmission lines is connected as in Prob. 14.4.4. However, rather than being turned on when t = 0, the voltage source has been on for a long time and when t = 0 is suddenly turned off. Thus, V_g = V_o for t < 0 and V_g = 0 for 0 < t. The lines have the same wave velocity c. Determine V(0, t). (Note that, by contrast with the situation in Prob. 14.4.4, the line having characteristic impedance Z_a now has initial values of voltage and current.)

14.4.6A transmission line is terminated at z = l in a ``short'' and driven at z = 0 by a current source I_g(t) in parallel with a resistance R_g. For 0 < t < T, I_g = I_o = constant, while for t < 0 and T < t, I_g = 0. For R_g = Z_o, determine V(0, t).

14.4.7With R_g not necessarily equal to Z_o, the line of Prob. 14.4.6 is driven by a step in current; for t < 0, I_g = 0, while for 0 < t, I_g = I_o = constant. the current I(0, t).

(a)	Using an approach suggested by Example 14.4.3, determine
(b)	If the transmission line is MQS, the system can be represented by a parallel inductor and resistor. Find I(0, t) assuming such a model.
(c)	Show that in the limit where the round-trip transit time 2l/c is short compared to the time = lL/Rg, the current I(0, t) found in (a) approaches that predicted by the MQS model.

14.4.8The transmission line shown in Fig. P14.4.8 is terminated in a series load resistance, R_L, and capacitance C_L.

reflected wave at z = l, given by (8) for the load resistance alone, is replaced by the differential equation at z = l

(a)	Show that the algebraic relation between the incident and
(b)	Show that if the capacitor voltage is Vc when t = 0, then
(c)	Given that Vg(t) = 0 for t < 0, Vg(t) = Vo = constant for 0 < t, and that Rg = Zo, determine V(0, t).

Transmission Lines in the Sinusoidal Steady State

14.5.2A line having length l is terminated in an open circuit.

function of l/c.

(a)	Determine the line admittance Y(-l) and sketch it as a
(b)	Show that the low-frequency admittance is that of a capacitor lC.

14.5.3^*A line is matched at z = 0 and driven at z = -l by a voltage source V_g(t) = V_o sin ( t) in series with a resistance equal to the characteristic impedance of the line. Thus, the line is as shown in Fig. 14.4.5 with R_g = Z_o. Show that in the sinusoidal steady state,

_g \equiv -jV_o

14.5.4In Prob. 14.5.3, the drive is zero for t < 0 and suddenly turned on when t = 0. Thus, for 0 < t, V_g(t) is as in Prob. 14.5.3. With the solution written in the form of (1), where V_s(z, t) is the sinusoidal steady state solution found in Prob. 14.5.3, what are the initial and boundary conditions on the transient part of the solution? Determine V(z, t) and I(z, t).

 Lines 14.6.1^*The normalized load impedance is Z_L/Z_o = 2 + j2. Use the Smith chart to show that the impedance of a quarter-wave line with this termination is Z/Z_o = (1 - j)/4. Check this result using (20).

14.6.2For a normalized load impedance Z_L/Z_o = 2 + j2, use (3) to evaluate the reflection coefficient, |\Gamma |, and hence the VSWR, (10). Use the Smith chart to check these results.

14.6.3For the system shown in Fig. 14.6.6a, the load admittance is Y_L = 2Y_o. Determine the position, l, and length, l_s, of a shorted stub, also having the characteristic admittance Y_o, that matches the load to the line.

14.6.4In practice, it may not be possible or convenient to control the position l of the stub, as required for single stub matching of a load admittance Y_L to a line having characteristic admittance Y_o. In that case, a ``double stub'' matching approach can be used, where two stubs at arbitrary locations but with adjustable lengths are used. At the price of restricting the range of loads that can be matched, suppose that the first stub is attached in parallel with the load and shorted at length l₁, and that the second stub is shorted at length l₂ and connected in parallel with the line at a given distance l from the load. The stubs have the same characteristic admittance as the line. Describe how, given the load admittance and the distance l to the second stub, the lengths l₁ and l₂ would be designed to match the load to the line. (Hint: The first stub can be adjusted in length to locate the effective load anywhere on the circle on the Smith chart having the normalized conductance g_L of the load.) Demonstrate for the case where Y_L = 2Y_o and l = 0.042 .

14.6.5Use the Smith chart to obtain the VSWR on the line to the left in Fig. 14.5.3 if the load resistance is R_L/Z_o = 2 and Z_o^a = 2Z₀. (Hint: Remember that the impedance of the Smith chart is normalized to the characteristic impedance at the position in question. In this situation, the lines have different characteristic impedances.)  Dissipation 14.7.1Following the steps exemplified in Section 14.1, derive (1) and (2).

14.7.2For Example 14.7.1,

(a)	Determine I(z, t).
(b)	Find the impedance at z = -l.
(c)	In the long wave limit, | l| \ll 1, what is this impedance and what equivalent circuit does it imply?

14.7.3The configuration is as in Example 14.7.1 except that the line is shorted at z = 0. Determine V(z, t) and I(z, t), and hence the impedance at z = -l. In the long wave limit, | l| \ll 1, what is this impedance and what equivalent circuit does it imply? 14.7.4^*Following steps suggested by the derivation of (14.2.19),

(a)	Use (1) and (2) to derive the power theorem
(b)	The product of two sinusoidally varying quantities is a constant (time average) part plus a part that varies sinusoidally at twice the frequency. In complex notation, Use (11.5.7) to prove this identity.
(c)	Show that, in describing the sinusoidal steady state, the time average of the power theorem becomes Show that for Example 14.7.1, it follows that the time average power input is equal to the integral over the length of the time average power dissipation per unit length.
(d)	Evaluate the time average input power on the left in this relation and the integral of the time average dissipation per unit length on the right and show that they are indeed equal.

Uniform and TEM Waves in Ohmic Conductors

_g \exp (j t)

14.8.3The terminations and material between the conductors of a transmission line are as described in Prob. 14.8.2. However, rather than being coaxial, the perfectly conducting transmission line conductors are in the parallel wire configuration of Fig. 14.2.2a. In terms of (x, y, z, t) and A_z(x, y, z, t), determine the electric and magnetic fields over the length of the line, including their dependencies on the transverse coordinates. What are L, C, and G and hence and Z_o?

14.8.4^*The transmission line model for the strip line of Fig. 14.8.4a is derived in Prob. 14.1.1. Because the permittivity is not uniform over the cross-section of the line, the waves represented by the model are not exactly TEM. The approximation is valid as long as the wavelength is long enough so that (25) is satisfied. In the approximation, E_x is taken as being uniform with x in each of the dielectrics, E^a and E^b, respectively. To estimate the longitudinal field E_z and compare it to E^a, an incremental surface between z + z and z and between the perfect conductors to derive Faraday's transmission line equation written in terms of E^a.

(a)	Use the integral form of the law of induction applied to
(b)	Then carry out this same procedure using a surface that again has edges at z + z and z on the upper perfect conductor, but which has its lower edge at the interface between dielectrics. With the axial electric field at the interface defined as Ez, show that
(c)	Now show that in order for this field to be small compared to Ea, (25) must hold.

Quasi-One-Dimensional Models (G = 0)

R a

14.9.2In the coaxial transmission line of Fig. 14.2.2b, the outer conductor has a thickness . Each conductor has the conductivity . What are C, L, and R, and over what frequency range are (7) and (10) valid? Give a condition on the transverse dimensions that insures the model being valid into the frequency range where the inductive reactance dominates the resistance.

14.9.3Find V(z, t) on the charge diffusion line of Fig. 14.9.4 in the case where the applied voltage has been zero for t < 0 and suddenly becomes V_p = constant for 0 < t and the line is shorted at z = 0. (Note Example 10.6.1.) 14.9.4Find V(z, t) under the conditions of Prob. 14.9.3 but with the line ``open circuited'' at z = 0.

\bye

14.11 Quasi-One-Dimensional Models (G = 0)

The transmission-line model of Sec. 14.7 is useful for representing the effects of losses in the parallel conductors. With the conductors having a finite conductivity, currents in the z direction imply that there is also a component of E in that direction. Because the tangential E is continuous at the surfaces of the conductors, this axial electric field extends into the insulating region between the conductors as well. We conclude that, with finite conductor losses, the fields are no longer exactly TEM

.

Under what circumstances can the series distributed resistance R be used to represent the conductor losses? We will find that the conductivity must be sufficiently low that the skin depth is large compared to the conductor thickness. Interestingly, we find that this ``constant resistance'' model can remain valid even under circumstances where line losses are small, in the sense that the effect of the distributed inductance is much larger than that of the series resistance. In the opposite extreme, where the effect of the series resistance is large compared to that of the inductance, the model represents EQS charge diffusion. A demonstration is used to exemplify physical situations modeled by this distributed R-C line. These include solid-state electronic devices and physiological systems.

We conclude this section with a model that is appropriate if the skin depth is much less than the conductor thickness. By restricting the model to the sinsusoidal steady state, the series distributed resistance R can be replaced by a ``frequency dependent'' resistance. This approximate model is typical of those used for representing losses in metallic conductors at radio frequencies and above.

We assume conductors in which the conduction current dominates the displacement current. In the sinusoidal steady state, this is true if

Thus, as the frequency is raised, the distribution of current density in the conductors is at first determined by quasistationary conduction (first half of Chap. 7) and then by the magnetic diffusion processes discussed in Secs. 10.3-10.7. That is, with the frequency low enough that magnetic diffusion is essentially instantaneous, the current density is uniformly distributed over the conductor cross-sections. Then, as the frequency is raised, the current tends to be redistributed so as to null the magnetic flux density normal to the conductor surfaces. Intuitively, we should expect that the constant resistance R only represents conductor losses at sufficiently low frequencies that the distribution of current density in the conductors does not depend on rates of change.

The circuit equations used to describe the incremental circuit in Sec. 14.7 express the integral laws of Faraday and Ampère for incremental lengths of the transmission-line. The ``current loop equation'' for loop C₁ in the circuit of Fig. 14.9.1a can be derived by applying Faraday's law to the surface S₁ enclosed by the contour C₁ also shown in that figure.

With the line integrals between conductors defined as the voltages and the flux through the surface as z LI, this expression becomes and in the limit where z 0 we obtain In the limit where the conductor conductivities are infinite, E_z 0 and this is the same expression as found for the incremental circuit model, (14.7.2). The field-equivalent of requiring charge conservation for the circuit node enclosed by the surface S₂ in Fig. 14.9.1b is Ampère's integral law applied to the surface S₂ enclosed by the contour C₂ also shown in that figure. Note that C₂ almost encircles one of the conductors with oppositely directed adjacent segments completing the z directed parts of the contour. For a surface S₂ of incremental length z, Ampère's integral law requires that where the contributions from the oppositely directed legs where the contour integral is in the z direction cancel. Ampère's integral law requires that the integral of H ds on the contours essentially surrounding the conductor be the enclosed current I. Gauss' integral law requires that the surface integral of E da be equal z CV. Thus, (6) becomes

and in the limit, the second transmission-line equation.

If the current density is uniformly distributed over the cross-sectional areas A₁ and A₂ of the respective conductors, it follows that the current densities are related to the total current by In each conductor, J_z = E_z, so the axial electric fields required to complete (4) are related to I by and indeed, the voltage equation is the same as for the distributed line, where the resistance per unit length has been found to be

Example 14.11.1. Low-Frequency Losses on Parallel Plate Line

In the parallel plate transmission-line shown in Fig. 14.9.2, the conductor thickness is b and the cross-sectional areas are A₁ = A₂ = bw. It follows from (11) that the resistance is Under the assumption that the conductor thickness, b, is much less than the plate spacing

---

² So that the magnetic energy stored in the plates themselves is negligible compared to that between the plates., a, the inductance per unit length is the same as found in Example 14.1.1, as is also the capacitance per unit length. As the frequency is raised, the current distribution over the cross-sections of the conductors becomes nonuniform when the skin depth (10.7.2) gets to be on the order of the plate thickness. Thus, for the model to be valid using the resistance given by (12), With this inequality we require that the effects of magnetic induction in determining the distribution of current in the conductors be negligible. Under what conditions are we justified in ignoring this effect of magnetic induction but nevertheless keeping that represented by the distributed inductance? Put another way, we ask if the inductive reactance j L can be large compared to the resistance R and still satisfy the condition of (14).

Combined, these last two conditions require that We conclude that, so long as the conductor thicknesses are small compared to their spacing, R represents the loss over the full frequency range from d-c to a high enough frequency that the line begins to behave as the ideal loss-free transmission-line. This is true because the time-constant _m \equiv L/R = ab that determines the frequency at which the resistance is equal to the inductive reactance

³ Familiar from Sec. 10.3.

---

is much longer than the magnetic diffusion time b² based on the thickness of the conductors.

Charge Diffusion Transmission-line

If the resistance is large enough that the inductance has little effect, the lossy transmission line becomes an EQS model. The line is simply composed of the series resistance shunted by the distributed capacitance of Fig. 14.9.3. To see that the voltage (and hence charge) and current on this line are governed by the diffusion equation, (10) is solved for I and that expression substituted into the z derivative of (7).

By contrast with the charge relaxation process undergone by charge in a uniform conductor, the charge in this heterogeneous system diffuses. The distributed R-C line is used to model EQS processes that range from those found in nervous conduction to semiconductor dynamics. We can either view the solution of (17) and (18) as a special case from Sec. 14.7 or exploite the complete analogy to the magnetic diffusion processes described in Secs. 10.6 and 10.7.

\begindemo1(Charge Diffusion Line

A simple demonstration of the charge diffusion line is shown in Fig. 14.9.4. A thin insulating sheet is sandwiched between a resistive sheet on top (the same Teledeltos paper used in Demonstration 7.6.2) and a metal plate on the bottom. With sinusoidal steady state conditions established by means of a voltage source at z = l and a short circuit at the right, the voltage distribution is the analogue of that described for magnetic diffusion in Example 10.7.1. The skin depth for the charge diffusion process is given by (10.7.2) with RC.

With this the new definition of , the magnitude of the voltage measured by means of the high impedance voltmeter can be compared to the theory, plotted on the inset to Fig. 10.7.1. Typical values are = 3.5 _o, b = 4.5 x 10^-4 mhos (where b is the surface conductivity of the conducting sheet) and a = 25 m, in which case RC = /(b )a = 2.7 x 10^-3 sec/m² and \simeq 0.5 m at a frequency of 500 Hz.

n B = 0

n B = 0

l

Skin-depth Short Compared to All Dimensions of Interest

In this case, the axial conduction currents are confined to within a few skin-depths of the conductor surfaces. Within a few skin-depths, the tangential magnetic field decays from its value at the conductor surface to zero. Because the magnetic field decays so rapidly with respect to a coordinate perpendicular to a given point on the conductor surface, the effects on the magnetic diffusion of spatial variations in the axial direction are negligible. For this reason, fields in the conductors can be approximated by the one-dimensional magnetic diffusion process described in Sec. 10.7. The following example illustrates.

Example 14.11.2. High-Frequency Losses on Parallel Plate Line

The parallel plate transmission-line is shown again by Fig. 14.9.5, this time with the axial current distribution in the conductors in thin regions on the inner surfaces of the conductors rather than uniform. In the conductors, the displacement current is negligible, so that the magnetic field is governed by the magnetic diffusion equation, (10.5.8). In the sinusoidal steady state, the y component of this equation requires that

The first term on the left is of the order of H_y/( )² while the second is of the order of H_y k² = H_y (2 / )² (where is the wavelength in the axial (z) direction). Thus, the derivative with respect to y can be ignored compared to that with respect to x provided that

In this case, (20) becomes the one-dimensional magnetic diffusion equation studied in Sec. 10.7. In the lower conductor, the magnetic field diffuses in the -x direction, so the appropriate solution to (20) is where H_o is the magnetic field intensity at the surface of the lower conductor [see (10.7.7)]. Ampère's law gives the current density associated with this field distribution It follows from either integrating this expression over the cross-section of the lower conductor or appealing to Ampère's integral law that the total current in the lower conductor is The axial electric field intensity at the surface of the lower conductor can now be written in terms of this total current by first using Ohm's law and the current density of (23) evaluated at the surface and then using (24) to express this field in terms of the total current.

A similar derivation gives an axial electric field at the surface of the upper conductor that is the negative of this result. Thus, we can complete the sinusoidal steady state version of the voltage transmission line equation, (4).

Because the magnetic energy stored within the conductor is usually negligible compared to that in the region between conductors, and (26) becomes the first of the two sinusoidal steady state transmission line equations.

The second follows directly from (7).

Comparison of these expressions with those describing the line operating with the conductor thickness much less than the skin depth, (10) and (7), shows that here there is an equivalent distributed resistance. (Here, is the permeability of the conductor, not of the region between conductors.) Note that this is the series dc resistance of conductors having width w and thickness . Because is inversely proportional to the square root of the frequency, this equivalent resistance increases with the square root of the frequency.