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

About the further exercises of Section 6.2

Thirteen further exercises in the spirit of the problems, the worked examples and the home exercises, graded from easy to harder within each group. Exercises 6.2.1–6.2.3 are about lossy media and the loss tangent, Exercises 6.2.4–6.2.5 about good conductors and the skin effect, Exercises 6.2.6–6.2.8 about plasmas and the group velocity, and Exercises 6.2.9–6.2.13 come from the exercise lecture and mix all three. They use only the equations listed at the head of this subsection, together with $v_{g} = d ω / d β$ and $γ η_{c} = j ω μ$ , $γ / η_{c} = σ + j ω ε$ .

Exercise 6.2.10

Neither a dielectric nor a conductor

Brackish estuary water has $σ = 0.5 S/m$ , $ε_{r} = 80$ , $μ_{r} = 1$ . (a) Find the frequency $f_{t}$ at which $p = σ / ω ε = 1$ . (b) At $f_{t}$ find exactly $α$ , $β$ , $λ$ , $η_{c}$ and the depth at which the field is $20 dB$ down. Show that $α / β = tan 22. 5^{\circ}$ . (c) Evaluate the low-loss and the good-conductor approximations of $α$ , $β$ , $η_{c}$ at $f_{t}$ and their errors. (d) Show that the low-loss $α$ is too large by the factor $\approx 1 + p^{2} /8$ and the good-conductor $α$ by $\approx 1 + 1/ (2 p)$ . For which $p$ is each in error by $5 %$ ? Translate into frequencies for this water.

Try it yourself first.

Strategy

At $p = 1$ the factor $1 - j p$ in $γ = j ω μ ε 1 - j p$ has the angle $- 4 5^{\circ}$ , its square root $- 22. 5^{\circ}$ . So $γ$ has the angle $67. 5^{\circ}$ and $η_{c}$ the angle $22. 5^{\circ}$ , exactly halfway between a dielectric ( $0^{\circ}$ ) and a conductor ( $4 5^{\circ}$ ). Neither approximation is designed for this point.

$ε = 80 ε_{0} = 7.074 \times 1 0^{- 10} F/m$ and $f_{t} = σ /2 π ε = 112.5 MHz$ .

$ω μ ε = 2 π f_{t} 80 / c = 21.07 rad/m$ and $1 - j = 2^{1/4} ∠ - 22. 5^{\circ}$ :

γ = 21.07 \times 1.1892∠ (9 0^{\circ} - 22. 5^{\circ}) = 25.06∠67. 5^{\circ} = 9.59 + j 23.15 m^{- 1},

so $α = 9.59 Np/m$ , $β = 23.15 rad/m$ , $α / β = tan 22. 5^{\circ} = 2 - 1 = 0.414$ , $λ = 2 π / β = 27.1 cm$ , and

η_{c} = \frac{η _{0} / 80}{2 ^{1/4}} ∠22. 5^{\circ} = \frac{42.15}{1.1892} ∠22. 5^{\circ} = 35.4∠22. 5^{\circ} Ω.

The field is $20 dB$ down at $ln 10/ α = 0.240 m$ , less than one wavelength.

Low-loss: $α \approx \frac{σ}{2} μ / ε = 0.25 \times 42.15 = 10.54 Np/m$ ( $+ 9.9 %$ ), $β \approx 21.07 rad/m$ ( $- 9.0 %$ ), $η_{c} \approx 42.15 (1 + j 0.5) = 47.1∠26. 6^{\circ} Ω$ . Good conductor: $α = β = π f μ_{0} σ = 14.90 m^{- 1}$ ( $+ 55 %$ and $- 36 %$ ), $η_{c} = (1 + j) π f μ_{0} / σ = 42.1∠4 5^{\circ} Ω$ .

From the exact formula, $α_{LL} / α = p / 2 (1 + p^{2} - 1)$ . For small $p$ , $1 + p^{2} - 1 \approx \frac{p ^{2}}{2} - \frac{p ^{4}}{8}$ , so the ratio is $(1 - p^{2} /4)^{- 1/2} \approx 1 + p^{2} /8$ . Similarly $α_{GC} / α = p / 1 + p^{2} - 1$ , and for large $p$ , $1 + p^{2} - 1 \approx p - 1$ , so the ratio is $(1 - 1/ p)^{- 1/2} \approx 1 + 1/ (2 p)$ . The estimates give $5 %$ at $p \approx 0.63$ and $p \approx 10$ , solved exactly $p = 0.672$ and $p = 10.2$ . Since $p = f_{t} / f$ , for this water the low-loss $α$ is good to $5 %$ above $112.5/0.672 = 167 MHz$ and the good-conductor $α$ below $112.5/10.2 = 11.0 MHz$ .

Answer

f_{t} = 112.5 MHz, γ = 9.59 + j 23.15 m^{- 1}, η_{c} = 35.4∠22. 5^{\circ} Ω, α errors: low-loss + 9.9 %, good conductor + 55 %, 5 % for p < 0.67 and p > 10.2

Check

$∣ γ ∣ ∣ η_{c} ∣ = 25.06 \times 35.44 = 888.2 = ω μ_{0}$ , and the angles add to $9 0^{\circ}$ , as $γ η_{c} = j ω μ$ requires. The rule of thumb "low-loss for $p < 0.1$ , good conductor for $p > 10$ " is therefore a statement about $5 %$ accuracy in $α$ at the conductor end and much better accuracy at the dielectric end. Between about $p = 0.6$ and $p = 10$ (here $11$ to $170 MHz$ , a decade and a half that includes the VHF band) only the exact formulas will do.

cf. Home exercise 6.2.1

Exercise 6.2.10: Neither a dielectric nor a conductor