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04.Multiport circuit parameters and transmission lines

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THE TRANSMISSION LINE EQUATION

61

In similar fashion the current wave equation can be found:

 

2I

1 2I

4.47

 

 

D

 

 

 

 

 

 

∂z2

v2

∂t2

The velocity of the wave is

1

 

 

 

 

 

 

 

 

 

4.48

 

v D

p

 

 

 

LC

 

The solution for these two wave equations given below in terms of the arbitray functions F1 and F2 can be verified by substitution back into Eqs. (4.46) and (4.47):

V z, t

D F1 t

z

C F2

t C

z

 

 

 

4.49

v

v

 

 

I z, t

1

F1 t

z

C F2

t C

z

 

4.50

D

 

 

 

Z0

v

v

The most useful function for F1 and F2 is the exponential function exp[j ωt š ˇz ], where ˇ D ω/v. The term Z0 is the same characteristic impedance in Eq. (4.41), which for the telegrapher’s equations is

L1

Z0 D D Lv D 4.51

C Cv

The variables L and C are given in terms of Henries and Farads per unit length and are thus distinguished from L and C used in lumped element circuit theory.

4.5THE TRANSMISSION LINE EQUATION

The transmission line equation was determined in Section 4.3 for a cascade of lumped element matched circuits. It is the input impedance of a transmission line terminated with a load, ZL, and it can also be found directly from analysis of a transmission line itself. The transmission line is characterized by its mechanical length, L, and its characteristic impedance, Z0. The characteristic impedance of a transmission line is a function only of the geometry and dielectric constant of the material between the lines and is independent of its terminating impedances. The input impedance of the transmission line depends on L, Z0, and ZL. When terminated with a nonmatching impedance, a standing wave is set up in the transmission line where the forwardand backward-going voltages and currents are as indicated in Fig. 4.10. At the load,

VL D VC C V

4.52

IL D IC I

4.53

62 MULTIPORT CIRCUIT PARAMETERS AND TRANSMISSION LINES

Since the forward current wave is IC D VC/Z0 and the reverse current wave is I D V /Z0, the current at the load is

I

VC V

 

VL

4.54

Z0

D ZL

L D

 

Replacing VL above with Eq. (4.52), the voltage reflection coefficient can be determined:

 

 

V

 

ZL

 

Z0

 

4.55

D VC

D ZL C Z0

 

 

If the transmission line is lossy, the reflection coefficient is actually

 

 

 

V

 

ZL

 

Z0

 

D

 

 

D

 

 

 

 

4.56

V

C

Z

Z

0Ł

 

 

 

 

L

C

 

 

The phase velocity of the wave is a measure of how fast a given phase moves down a transmission line. This is illustrated in Fig. 4.12 where ejωt time dependence is assumed. If time progresses from t1 to t2, then in order for ej ωt ˇz to have the same phase at each of these two times, the wave must progress in the forward direction from z1 to z2. Consequently

0 D ˇ z2 z1 ω t2 t1

 

giving the phase velocity

 

 

 

 

v D

z

D

ω

4.57

 

 

 

t

ˇ

This is to be distinguished from the group velocity,

dω

vg D

dˇ

t1 t2

z1

z 2

z

FIGURE 4.12 Forward-directed propagating wave.

THE SMITH CHART

63

which is a measure of velocity of energy flow. For low-loss media, vgv D c2, where c is the velocity of light in a vacuum. The negative-going wave of course has a phase velocity of ω/ˇ.

This traveling wave corresponds to the solution of the lossless telegrapher’s equations. The total voltage at any position, z, along the transmission line is the sum of the forwardand backward-going waves:

V z D VCe jˇz C V eCjˇz

4.58

The total current at any point z is by Kirchhoff’s law the difference of the of the two currents:

1

VCe jˇz V eCjˇz

4.59

I z D Z0

At the input to the line where z D $, the ratio of Eqs. (4.58) and (4.59) gives the input impedance:

Z

D

Z

 

VCe jˇz C V eCjˇz

 

4.60

 

 

 

 

 

 

 

 

0 VCe jˇz V eCjˇz

 

 

D

Z

 

e jˇz C eCjˇz

 

 

 

4.61

 

 

 

 

 

 

0 e jˇz eCjˇz

 

At the position z D $,

 

 

 

 

 

 

 

 

 

 

 

Z

D

Z

ZL C jZ0 tan ˇ$

 

 

4.62

 

 

in

 

0 Z0 C jZL tan ˇ$

 

If the propagation constant is the complex quantity D ˛ C jˇ, then

 

Z

 

D

Z

ZL C jZ0 tanh $

 

4.63

 

 

 

in

 

0 Z0 C jZL tanh $

 

A few special cases illustrates some basic features of the transmission line equation. If z D 0, Zin 0 D ZL no matter what Z0 is. If ZL D Z0, then Zin z D Z0 no matter what z is. For a quarter wavelength line, Zin z D &/4 D Z20/ZL. The input impedance for any length of line can be readily calculated from Eq. (4.62), or by using the Smith chart.

4.6THE SMITH CHART

The Smith chart, as shown in Fig. 4.13, is merely a plot of the transmission line equation on a set of polar coordinates. The reflection coefficient is really an alternate way of expressing the input impedance relative to some standard value (Z0), which is typically 50 '. The reflection coefficient, , has a magnitude

64 MULTIPORT CIRCUIT PARAMETERS AND TRANSMISSION LINES

FIGURE 4.13 The Smith chart.

between 0 and 1 and a phase angle between 0° and 360°. The equations describing the radii and centers of the circles of the coordinates of the Smith chart are found by solving the normalized version of (4.60):

(

 

r

 

jx

 

Zin

 

1 C e 2jˇ$

4.64

D

C

D Z0

D 1 e 2jˇ$

 

 

 

 

Solution of the real part of Eq. (4.64) gives the center of the resistance circles asr/ 1 C r , 0 with a radius of 1/ 1 C r . Solution of the imaginary part gives the center of the reactance circles as 1, 1/x with a radius of 1/x [2, pp. 121–129].

COMMONLY USED TRANSMISSION LINES

65

The Smith chart can be used as a computational tool, and it often gives insight where straight equation solving will not. It is also a convenient plotting tool of measured or calculated data, since any passive impedance will fall within its boundaries.

4.7COMMONLY USED TRANSMISSION LINES

Because TEM transmission lines have neither an electric nor a magnetic field component in the direction of propagation, the characteristic impedance can be found from Eq. (4.51) and electrostatics. Since the velocity of propagation in the given media is presumably known all that is necessary is to calculate the electrostatic capacitance between the conductors. When the geometry is particularly nasty and the solution is needed quickly, the field-mapping approach described in Chapter 2 can be used.

4.7.1Two-Wire Transmission Line

The two-wire transmission line, commonly used, for example, between a TV antenna and the receiver, consists of two round conductors each with a radius of a and separated by a distance b (Fig. 4.14). The dielectric surrounding the wires has a dielectric constant of ε. The field theory analysis, such as that given in [2], shows that the characteristic impedance of the two wire line is

+b

Z0 D

 

Arccosh

 

4.65

,

2a

where

 

 

 

 

 

 

 

+ D

 

ε

 

 

4.66

 

 

 

-

 

 

 

 

ε r

b

2a

FIGURE 4.14 The two-wire transmission line.

66 MULTIPORT CIRCUIT PARAMETERS AND TRANSMISSION LINES

While the field analysis for a given structure may bring some challenges, the good news is that once Z0 is known, the rest of the problem can be solved from circuit theory. Losses in the two-wire transmission line stem from the lossy dielectric between the conductors and the resistive losses experienced by the current as it flows along the conductor. Since a wave is attenuated as it goes down a line by exp ˛z , the power loss is proportional to exp 2˛z , where

˛ D ˛d C ˛c

4.67

The dielectric and conductor losses are

˛d D

.d+

 

 

 

4.68

2

 

 

 

 

 

˛c D

2a

 

 

+Arccosh b/2a

4.69

2.c

 

 

1

 

 

ω-

1

 

where .d and .c are the conductivities of the dielectric and conductor, respectively. The two-wire line is inexpensive and widely used in UHF applications.

4.7.2Two Parallel Plate Transmission Line

The parallel plate transmission line consists of two separate conductors of width b and separated by a distance a (Fig. 4.15). This is a rectangular waveguide without the side walls. It is fundamentally distinct from the rectangular waveguide, which is not a TEM transmission line. The Maxwell equations for a plane wave of this system are an exact analog to the telegrapher’s equations:

 

∂Ex

D -

∂Hy

4.70

 

∂z

∂t

∂Hy

D ε

∂Ex

 

4.71

 

∂z

 

∂t

a

x

 

z

y

b

FIGURE 4.15 The parallel plate transmission line.

COMMONLY USED TRANSMISSION LINES

67

The voltage between the plates is the integral of the electric field:

a

V D Exdx D aEx 4.72

0

The magnetic field in the y direction will produce a current in the conductor that will travel in the z direction according to Ampere’s` law:

b

I D Hydy D Hyb 4.73

0

Substitution of Eqs. (4.72) and (4.73) into Eq. (4.44) gives

∂Ex

D

Lb ∂Hy

4.74

∂z

a

 

∂t

Comparison of this with Eq. (4.70) indicates that

-a

L D

 

4.75

 

 

b

A similar substitution of Eqs. (4.72) and (4.73) into Eq. (4.45) and comparison with Eq. (4.71) indicates that

εb

C D

 

4.76

 

 

a

so that the characteristic impedance for the parallel plate guide is

La -

Z0 D D 4.77

C b ε

4.7.3Coaxial Transmission Line

Coaxial transmission line comes in the form of rigid, semirigid, and flexible forms. The end view of a coaxial line, which is shown in Fig. 4.16, consists of an inner conductor and the outer conductor, which is normally grounded. The electric field points from the outer to the inner conductor, and the longitudinal current on the center conductor produces a magnetic field concentric to the inner conductor. The potential between the two conductors is a solution of the transverse form of Laplace’s equation in cylindrical coordinates where there is no potential difference in the longitudinal z direction. The notation for the divergence and curl operators follows that given in [3]:

0 D rt

 

1 2

 

0 D

1

r

C

4.78

r

 

∂r

∂r

r2

 

∂22

 

68 MULTIPORT CIRCUIT PARAMETERS AND TRANSMISSION LINES

 

a

b

Φ = 0

Φ = V0

 

FIGURE 4.16 The coaxial transmission line.

Because there is no potential variation in the z direction, the z derivative ofis zero. Because of symmetry there is no variation of in the 2 direction either. Thus Eq. (4.78) simplifies to an ordinary second-order differential equation subject to the boundary conditions that D 0 on the outer conductor and D V0

on the inner conductor:

r dr

4.79

0 D r dr

 

1 d

 

d

 

Integration of Eq. (4.79) twice gives

 

 

 

D C1 ln r C C2

4.80

which upon applying the boundary conditions gives the potential anywhere between the two conductors:

r D

V0

ln

r

 

ln a/b

b

The electric field is easily obtained by differentiation.

V e jˇz rO

E D rt D 0 ln b/a r

The magnetic field is then

H D zO ð E

D

V0e jˇz

2O

r+ ln b/a

4.81

4.82

4.83

COMMONLY USED TRANSMISSION LINES

69

r

H2

H1

FIGURE 4.17 The continuity of the magnetic field along the center conductor.

The outward normal vector of the center conductor, rO, is shown in Fig. 4.17. The surface current on the center conductor is determined by the boundary condition for the tangential magnetic field:

Js D rO ð H2 H1 D rO ð H2

4.84

The later result occurs because the magnetic field is zero inside the conductor. The total current flowing in the center conductor is

0 O

2,

 

a+ ln b/a

 

D 0

 

 

I z

 

 

 

 

 

 

zVO

0

 

ad2

 

 

 

zO2,V0

 

 

 

4.85

 

 

 

 

 

 

 

 

 

D a+ ln b/a

 

 

 

 

 

 

 

 

 

 

so that

D I0

D 2, ln

a

4.86

Z0

 

 

 

V0

 

 

+

 

 

 

b

 

 

Coaxial Dielectric Loss. The differential form of Ampere’s` law relates the magnetic field to both the conduction current and the displacement current. In the absence of a conductor,

H D J C

D

4.87

∂t

³ jωεE

4.88

By taking the curl of Eq. (4.88), the Helmholtz wave equation for H can be found. Solution of the wave equation would give the propagation constant, :

 

 

p

 

 

jk

p

 

 

4.89

D

ε

r

 

 

 

D

 

0

 

 

where εr is the relative dielectric constant and k0 is the propagation constant in free space. A lossy dielectric is typically represented as the sum of the lossless

70 MULTIPORT CIRCUIT PARAMETERS AND TRANSMISSION LINES

(real) and lossy (imaginary) parts:

εr D εr0 r00

4.90

The revised propagation constant is found by substituting this into Eq. (4.89). The result can be simplified by taking the first two terms of the Taylor series expansion since ε00r − ε0r:

 

 

 

 

 

1 j

ε00

 

 

D ˛ C jˇ D jk0 εr0

r

4.91

2εr0

so that

 

 

 

 

 

 

˛d D

k0εr00

 

 

 

 

 

4.92

2

 

 

 

 

 

εr0

 

 

 

 

ˇ D k0

 

 

 

4.93

εr0

 

The power loss is proportional to exp 2˛z .

Coaxial Conductor Loss. The power loss per unit length, P$, is obtained by taking the derivative of the power at a given point along a transmission line:

P D P0e 2˛cz

4.94

 

dP

 

P$ D

 

D 2˛cP

4.95

dz

For a low-loss conductor where the dielectric losses are negligible, Eq. (4.87) becomes, with the help of Ohm’s law,

H D .E

4.96

where . is the metal conductivity. This would be the same as Eq. (4.88) if

ε )

.

4.97

 

 

With this substitution the wave impedance becomes the metal surface impedance:

-

) Zm D 1 C j

ω-0

4.98

ε

2.

At the surface there will be a longitudinal electric field of ZmJs directed in the zO direction. Thus, in a lossy line, the fields will no longer be strictly TEM. This longitudinal electric field produces energy flow into the conductor proportional