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

The square root in Eq. (6.2.6) can be taken once and for all. Square $γ = α + j β$ and compare it with $γ^{2} = - ω^{2} μ ε + j ω μ σ$ from Eq. (6.2.4):

α^{2} - β^{2} + j 2 α β = - ω^{2} μ ε + j ω μ σ . (6.2.8)

The real and the imaginary parts give two equations,

β^{2} - α^{2} = ω^{2} μ ε, 2 α β = ω μ σ, (6.2.9)

and a third follows from the modulus, $∣ γ^{2} ∣ = ∣ γ ∣^{2} = α^{2} + β^{2}$ :

β^{2} + α^{2} = (ω^{2} μ ε)^{2} + (ω μ σ)^{2} = ω^{2} μ ε 1 + (\frac{σ}{ω ε})^{2} . (6.2.10)

Adding the first equation of Eq. (6.2.9) to Eq. (6.2.10) gives $2 β^{2}$ , subtracting it gives $2 α^{2}$ . Taking the positive roots ( $α, β \geq 0$ for a wave that travels and decays along $+ z$ ) gives the exact constants.

The exact attenuation and phase constants

α β = ω \frac{μ ε}{2} [1 + (\frac{σ}{ω ε})^{2} - 1]^{1/2} Np/m, = ω \frac{μ ε}{2} [1 + (\frac{σ}{ω ε})^{2} + 1]^{1/2} rad/m . (6.2.11)

The medium enters only through $ω μ ε$ and the ratio $σ / ω ε$ .

In plain words

Measure both constants in units of the lossless wave number $k_{0} = ω μ ε$ and they depend on one number only, $σ / ω ε$ . A single pair of curves, Figure 6.2.2, therefore describes every medium at every frequency.

The exact attenuation and phase constants of Eq. (6.2.11), in units of k_0=\omega\sqrt{\mu\varepsilon}, against the ratio \sigma/\omega\varepsilon. For \sigma/\omega\varepsilon\ll1 the exact \alpha follows the green line and \beta\approx k_0. For \sigma/\omega\varepsilon\gg1 it follows the grey line, on which \alpha=\beta. In between, roughly from 0.1 to 10, only the exact formulas are reliable. — **Figure 6.2.2.** The exact attenuation and phase constants of Eq. (6.2.11), in units of $k_{0} = ω μ ε$ , against the ratio $σ / ω ε$ . For $σ / ω ε ≪ 1$ the exact $α$ follows the green line and $β \approx k_{0}$ . For $σ / ω ε ≫ 1$ it follows the grey line, on which $α = β$ . In between, roughly from $0.1$ to $10$ , only the exact formulas are reliable.

Two checks are worth making at once. For $σ = 0$ the bracket of $α$ vanishes and the bracket of $β$ is $2$ , so $α = 0$ and $β = ω μ ε$ , the lossless wave of Section 6.1. And whatever the medium, $β > ω μ ε$ : losses always shorten the wavelength and slow the phase down, because the two equations of Eq. (6.2.9) force $β^{2} = ω^{2} μ ε + α^{2}$ .

Common mistake · Using an approximation outside its range

Reaching for $α = π f μ σ$ because the medium "conducts", or for $α = \frac{σ}{2} μ / ε$ because it "is a dielectric". The class of a medium is decided by $σ / ω ε$ at the working frequency, not by its name. Wet soil at $100 MHz$ has $σ / ω ε = 0.225$ (Home exercise 6.2.1) and brackish water at $112.5 MHz$ has exactly $1$ (Exercise 6.2.10): there the good-conductor formula is $55 %$ wrong and only Eq. (6.2.11) will do.

6.2.2 The exact attenuation and phase constants