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www.ece.rutgers.edu/∼orfanidi/ewa 263<br />

264 Electromagnetic Waves & Antennas – S. J. Orfanidis – June 21, 2004<br />

Second, the magnetic fields on the conductor surfaces are determined and the corresponding<br />

induced surface currents are calculated by J s = ˆn × H, where ˆn is the outward<br />

normal to the conductor.<br />

Third, the ohmic losses per unit conductor area are calculated by Eq. (2.8.7). Figure<br />

8.2.1 shows such an infinitesimal conductor area dA = dl dz, where dl is along the<br />

cross-sectional periphery of the conductor. Applying Eq. (2.8.7) to this area, we have:<br />

dP loss<br />

dA<br />

= dP loss<br />

dldz = 1 2 R s|J s | 2 (8.2.6)<br />

where R s is the surface resistance of the conductor given by Eq. (2.8.4),<br />

√ √ √<br />

ωµ ωɛ<br />

R s =<br />

2σ = η 2σ = 1 2 δωµ , δ = 2<br />

= skin depth (8.2.7)<br />

ωµσ<br />

Integrating Eq. (8.2.6) around the periphery of the conductor gives the power loss per<br />

unit z-length due to that conductor. Adding similar terms for all the other conductors<br />

gives the total power loss per unit z-length:<br />

P ′ loss = dP loss<br />

dz<br />

∮<br />

∮<br />

= 1<br />

C a 2 R s|J s | 2 1<br />

dl +<br />

C b 2 R s|J s | 2 dl (8.2.8)<br />

One common property of all three types of modes is that the transverse fields E T , H T<br />

are related to each other in the same way as in the case of uniform plane waves propagating<br />

in the z-direction, that is, they are perpendicular to each other, their cross-product<br />

points in the z-direction, and they satisfy:<br />

H T = 1<br />

η T<br />

ẑ × E T (8.3.1)<br />

where η T is the transverse impedance of the particular mode type, that is, η, η TE ,η TM<br />

in the TEM, TE, and TM cases.<br />

Because of Eq. (8.3.1), the power flow per unit cross-sectional area described by the<br />

Poynting vector P z of Eq. (8.2.2) takes the simple form in all three cases:<br />

TEM modes<br />

P z = 1 2 Re(E T × H ∗ T )·ẑ = 1<br />

2η T<br />

|E T | 2 = 1 2 η T|H T | 2 (8.3.2)<br />

In TEM modes, both E z and H z vanish, and the fields are fully transverse. One can set<br />

E z = H z = 0 in Maxwell equations (8.1.5), or equivalently in (8.1.16), or in (8.1.17).<br />

From any point view, one obtains the condition k 2 c = 0, or ω = βc. For example, if<br />

the right-hand sides of Eq. (8.1.17) vanish, the consistency of the system requires that<br />

η TE = η TM , which by virtue of Eq. (8.1.13) implies ω = βc. It also implies that η TE ,η TM<br />

must both be equal to the medium impedance η. Thus, the electric and magnetic fields<br />

satisfy:<br />

H T = 1 η ẑ × E T (8.3.3)<br />

Fig. 8.2.1<br />

Conductor surface absorbs power from the propagating fields.<br />

These are the same as in the case of a uniform plane wave, except here the fields<br />

are not uniform and may have a non-trivial x, y dependence. The electric field E T is<br />

determined from the rest of Maxwell’s equations (8.1.5), which read:<br />

where C a and C b indicate the peripheries of the conductors. Finally, the corresponding<br />

attenuation coefficient is calculated from Eq. (2.6.22):<br />

α c = P′ loss<br />

(conductor losses) (8.2.9)<br />

2P T<br />

Equations (8.2.1)–(8.2.9) provide a systematic methodology by which to calculate the<br />

transmitted power and attenuation losses in waveguides. We will apply it to several<br />

examples later on.<br />

8.3 TEM, TE, and TM modes<br />

The general solution described by Eqs. (8.1.16) and (8.1.19) is a hybrid solution with nonzero<br />

E z and H z components. Here, we look at the specialized forms of these equations<br />

in the cases of TEM, TE, and TM modes.<br />

∇ T × E T = 0<br />

∇ T · E T = 0<br />

(8.3.4)<br />

These are recognized as the field equations of an equivalent two-dimensional electrostatic<br />

problem. Once this electrostatic solution is found, E T (x, y), the magnetic field<br />

is constructed from Eq. (8.3.3). The time-varying propagating fields will be given by<br />

Eq. (8.1.1), with ω = βc. (For backward moving fields, replace β by −β.)<br />

We explore this electrostatic point of view further in Sec. 9.1 and discuss the cases<br />

of the coaxial, two-wire, and strip lines. Because of the relationship between E T and H T ,<br />

the Poynting vector P z of Eq. (8.2.2) will be:<br />

P z = 1 2 Re(E T × H ∗ T )·ẑ = 1<br />

2η |E T| 2 = 1 2 η|H T| 2 (8.3.5)

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