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Chapter 2: Transmission lines and waveguides

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Chapter 2: Transmission lines and waveguides

2.1 Generation solution for TEM, TE and TM waves

2.2 Parallel plate waveguide

2.3 Rectangular waveguide

2.4 Circular waveguide

2.5 Coaxial line

2.6 Surface waves on a grounded dielectric slab

2.7 Stripline

2.8 Microstrip

2.9 Wave velocities and dispersion

2.10 Summary of transmission lines and waveguids

Closed waveguide

Two-conductor TL

Electromagnetic fields (time harmonic eiωt and propagating along z axis):

where e(x,y) and h(x,y) represent the transverse (x,y) E and H components, while ezand hzare the longitudinal E and H components.

In the case of source free, Maxwell’s equations can be written as:

With the e-iβz dependence, the above vector equations can be divided into six component equations and then solve the transverse fields in terms of the longitudinal components Ez amd Hz:

The cutoff frequency:

(1) TEM waves (Ez = Hz = 0)

0

0

- Propagation constant:

(kc = 0, no cutoff)

- The Helmholtz equation for Ex:

For a e-jz dependence, and the above equation can be simplified

Laplace equations, equal to static fields

Similarly,

- Transverse magnetic field:

0

- Wave impedance:

0

- Note:
- Wave impedance, Z
- relates transverse field components and is dependent only on the material constant.
- For TEM wave
- Characteristic impedance of a transmission line, Z0:
- relates an incident voltage and current and is a function of the line geometry as well as the material filling the line.
- For TEM wave: Z0= V/I

(V incident wave voltage

I incident wave current)

(2) TE waves (Ez = 0 and Hz0)

- The field components can be simplified as:

- is a function of frequency and TL/WG structure

- Solve Hz from the Helmholtz equation

Because , then

where . Boundaries conditions will be used to solve the above equation.

- TE wave impedance:

(3) TM waves (Hz = 0 and Ez0)

- The field components can simplified:

- is a function of frequency and TL/WG structure

- Solve Ez from Helmholtz equation:

Because , then

where . Boundaries conditions will be used to solve the above equation.

- TE wave impedance:

(4) Attenuation due to dielectric loss

Total attenuation constant in TL or WG = c + d.

c: due to conductive loss; calculated using the perturbation method; must be evaluated separately for each type.

d: due to the dielectric loss; calculated from the propagation constant.

Taylor expansion (tan << 1)

2.2 Parallel plate waveguide

w >> d (fringing fields and any x variation could be ignored)

- Formed from two flat plates or strips
- Probably the simplest type of guide
- Support TEM, TE and TM modes
- Important for practical reasons.

(a) TEM modes (Ez = Hz = 0)

- Laplace equation for the electric potential (x,y)

for

Boundary conditions:

- The transverse field , so that we have

- Characteristic impedance:

- Phase velocity:

(b) TM modes (Hz = 0)

- The transverse ez(x,y) satisfies

Bn = 0

Boundary conditions: ez(x,y) = 0 at y = 0, d.

k > kc traveling wave

k = kc ?

k < kcevanescent wave

Cutoff frequency

- Propagation constant:

- The field components:

Wave impedance:

Power flow:

(n > 0)

(c) TE modes (Ez = 0)

An = 0

The transverse hz(x,y) satisfies

And boundary Ex(x,y) = 0 at y = 0, d.

where the propagation constant

- Wave impedance:

- Cutoff frequency :

Parallel plate waveguide

TEM

TM

TE