Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Small losses hardly change the field pattern. The field then carries an extra factor $e^{- α z}$ , $γ = α + j τ$ , and the power falls as $P^{'} (z) = P^{'} (0) e^{- 2 α z}$ . The power lost per unit length, $P_{L}^{'} = - d P^{'} / d z = 2 α P^{'}$ , gives

Attenuation constant from the power balance

α = α_{d} + α_{c} = \frac{P _{L}^{'}}{2 P ^{'}} [Np/m], (7.2.35)

the power lost per unit length divided by twice the power carried. Multiply by $8.686$ for dB/m.

Dielectric

For a slightly lossy filling with conductivity $σ$ (or loss tangent $tan δ = σ / ω ε ≪ 1$ ) the TEM wave is the plane wave of a low-loss medium (Section 6.2), with $α = σ η /2 = k tan δ /2$ . The higher modes have the same attenuation divided by $cos θ$ , because their zig-zag waves travel a longer path for every metre of progress (Exercise 7.1.5 in Section 7.1 derives it for any guide):

α_{d}^{T E M} α_{d}^{T M, T E} = \frac{k tan δ}{2} = \frac{σ η}{2}, = \frac{σ η /2}{1 - ( f _{co, n} / f ) ^{2}} . (7.2.36)

Plates

A good conductor of conductivity $σ_{c}$ carries the surface current $J_{s} = ∣ H_{t a n} ∣$ in a skin depth and dissipates $\frac{1}{2} R_{s} ∣ J_{s} ∣^{2}$ per unit area, with the surface resistance $R_{s} = π f μ_{c} / σ_{c}$ (Section 6.2). Both plates count. For the TEM wave $∣ H_{t a n} ∣ = ∣ H_{x} ∣ = ∣ A_{0} ∣$ on each plate, so $P_{L}^{'} = 2 \times \frac{1}{2} R_{s} ∣ A_{0} ∣^{2}$ and, with $P^{'} = \frac{1}{2} η ∣ A_{0} ∣^{2} b$ , $α_{c} = R_{s} / (η b)$ . For TM $_{n}$ the tangential field on the plates is $∣ H_{x} ∣ = ω ε ∣ C ∣/ h_{n}$ , so $P_{L}^{'} = R_{s} ω^{2} ε^{2} ∣ C ∣^{2} / h_{n}^{2}$ and, with $P^{'}$ from the power subsection,

α_{c}^{T M_{n}} = \frac{R _{s} ω ^{2} ε ^{2} ∣ C ∣ ^{2} / h _{n}^{2}}{2 ω ε τ _{n} b ∣ C ∣ ^{2} / ( 4 h _{n}^{2} )} = \frac{2 R _{s} ω ε}{τ _{n} b} = \frac{2 R _{s}}{Z _{T M} b} .

For TE $_{n}$ only $H_{z}$ is tangential, and Exercise 7.2.5 carries out the same steps. Altogether

Conductor attenuation of the parallel-plate modes

α_{c}^{T E M} α_{c}^{T M_{n}} α_{c}^{T E_{n}} R_{s} = \frac{R _{s}}{η b}, = \frac{2 R _{s}}{η b 1 - ( f _{co, n} / f ) ^{2}}, = \frac{2 R _{s} ( f _{co, n} / f ) ^{2}}{η b 1 - ( f _{co, n} / f ) ^{2}}, = \frac{π f μ _{c}}{σ _{c}} . (7.2.37)

Conductor attenuation of the three lowest modes, normalised with the surface resistance at f_{c1} (so that R_s\propto\sqrt{f} is included). All modes become lossy near cutoff, where they bounce many times per metre. Far above cutoff the TE_1 loss falls as f^{-3/2}, while TEM and TM_1 grow as \sqrt f. — **Figure 7.2.12.** Conductor attenuation of the three lowest modes, normalised with the surface resistance at $f_{c 1}$ (so that $R_{s} \propto f$ is included). All modes become lossy near cutoff, where they bounce many times per metre. Far above cutoff the TE $_{1}$ loss falls as $f^{- 3/2}$ , while TEM and TM $_{1}$ grow as $f$ .

Reading the graph

The TE $_{1}$ loss falls because the only tangential magnetic field on its plates is $H_{z}$ , which shrinks as $θ \to 0$ : the zig-zag waves graze the plates and induce less and less current. TEM and TM modes always have their full transverse $H$ tangential to the plates. Above $f = 1.6 f_{co, 1}$ TE $_{1}$ loses less in the plates than even the TEM wave (Home exercise 7.2.4, Exercise 7.2.5).

Common mistake · Counting one plate

Taking the conductor loss of only one plate. The current flows on both plates, equal and opposite, and both dissipate. For TEM this is the difference between $R_{s} / (η b)$ and half of it.

Quick check

Far above cutoff, which mode of a parallel-plate guide has the lowest conductor loss, and how does it scale with frequency?

TE $_{1}$ : $α_{c}^{T E_{1}} \to 2 R_{s} f_{co, 1}^{2} / (η b f^{2}) \propto f / f^{2} = f^{- 3/2}$ . TEM and TM $_{1}$ grow as $R_{s} \propto f$ .

Application 7.2.2 · Why hollow waveguides at microwave frequencies

The coax problem. A coaxial cable carries its current on a thin inner conductor, where $J_{s}$ is large, so its conductor loss $α_{c} \propto R_{s} \propto f$ is high, and its dielectric adds $α_{d} \propto f$ . A bigger coax would lose less, but its first higher mode (TE $_{11}$ , with $f_{c} \approx c / [π (a + b) ε_{r}]$ , Section 7.5) then moves below the operating frequency. At $10 GHz$ the coax must stay small: a PTFE semirigid cable ( $2 a = 0.92 mm$ , $2 b = 2.98 mm$ ) loses $1.32 dB/m$ .

The hollow guide. A hollow guide has no inner conductor, no dielectric, wide walls, and for TE-type modes a wall current that decreases with frequency above cutoff (Figure 7.2.12). WR-90 ( $22.86 \times 10.16 mm$ , Section 7.3) loses $0.11 dB/m$ at $10 GHz$ and, with the air breakdown field of $3 MV/m$ (Section 3.4.4), carries about $1 MW$ peak.

A $30$ m feed at $10$ GHz

semirigid coax: 30 \times 1.32 = 39.5 dB, WR-90: 30 \times 0.108 = 3.3 dB . (7.2.38)

Application 7.2.2. Attenuation of a PTFE semirigid coax (2a=0.92\,\mathrm{mm}, 2b=2.98\,\mathrm{mm}, copper, \tan\delta=2\times10^{-4}) and of the WR-90 copper guide in its TE_{10} mode. The guide is far better in its band (shaded, 8.2 to 12.4\,\mathrm{GHz}), but it cannot be used below its cutoff at 6.56\,\mathrm{GHz}. — **Figure 7.2.13.** Application 7.2.2. Attenuation of a PTFE semirigid coax ( $2 a = 0.92 mm$ , $2 b = 2.98 mm$ , copper, $tan δ = 2 \times 1 0^{- 4}$ ) and of the WR-90 copper guide in its TE $_{10}$ mode. The guide is far better in its band (shaded, $8.2$ to $12.4 GHz$ ), but it cannot be used below its cutoff at $6.56 GHz$ .

The price is that a waveguide works only in a band (WR-90: $8.2$ to $12.4 GHz$ , between $1.25 f_{c}$ and the next mode), it is rigid and bulky, and it is dispersive (Application 7.1.1 in Section 7.1). Below about $2 GHz$ a guide would be too large, and coax wins. Radar feeds, satellite earth stations, high-power transmitters and accelerators use waveguide (Worked example 7.2.2).

7.2.10 Attenuation due to lossy dielectric and plates

Dielectric

Plates