Problem 5.6.8: A damped wave: phasor, medium and power

Problem 5.6.8

A damped wave: phasor, medium and power

In an infinite, source-free, simple and non-magnetic medium, the electric field component of a monochromatic electromagnetic wave is measured as

E (x; t) = 20 e^{- 2 x} sin (3 x - 3 \times 1 0^{8} t) a_{z} [V/m] .

Find the frequency, phase velocity and wavelength of this wave.
Determine the phasor expression of the electric field.
Find the wavenumber $k$ , and by the help of this, determine the conductivity and relative permittivity for this medium.
Calculate the average power passing through a $5 m^{2}$ surface on $x = 0$ plane in $a_{x}$ direction.

Adapted from: M. N. O. Sadiku, Elements of Electromagnetics, Problem 10.9.

Try it yourself first.

Strategy

The field is a sine, the time enters with a minus sign, and there is a factor $e^{- 2 x}$ . Bring it to the form $A e^{- α x} cos (ω t - β x + φ)$ first: then every part is a reading. The Helmholtz equation gives $k^{2}$ , Faraday's law gives $H$ , and $\frac{1}{2} Re {E \times H^{*}}$ the power.

Bookkeeping

The phasor rules use cosines with $+ ω t$ , so first

sin (3 x - 3 \times 1 0^{8} t) = - sin (3 \times 1 0^{8} t - 3 x) = cos (3 \times 1 0^{8} t - 3 x + \frac{π}{2}) .

Then $ω = 3 \times 1 0^{8} rad/s$ , $ϕ (x) = 3 x - π /2$ so $β = 3 rad/m$ , the factor $e^{- 2 x}$ gives $α = 2 Np/m$ , and $ω t - β x$ means travel towards $+ x$ , decaying.

f v_{p} λ = \frac{3 \times 1 0 ^{8}}{2 π} = \frac{150}{π} MHz = 47.7 MHz, = \frac{ω}{∣\nabla ϕ ∣} = \frac{3 \times 1 0 ^{8}}{3} = 1 0^{8} m/s, = \frac{v _{p}}{f} = \frac{2 π}{3} = 2.09 m .

$E = Re {20 e^{- 2 x} e^{j π /2} e^{- j 3 x} a_{z} e^{j ω t}}$ , so

E (x) = 20 j e^{- (2 + 3 j) x} a_{z} V/m .

Helmholtz: $d^{2} E_{z} / d x^{2} = (2 + 3 j)^{2} E_{z} = - k^{2} E_{z}$ , so

k^{2} = - (2 + 3 j)^{2} = 5 - 12 j, k = - j (2 + 3 j) = 3 - 2 j m^{- 1} .

Of the two roots $\pm (3 - 2 j)$ , the wave that decays along $+ x$ needs $Im k < 0$ , so $k = β - j α = 3 - 2 j$ . Compare with $k^{2} = ω^{2} μ_{0} ε_{0} ε_{r} - j ω μ_{0} σ$ , where $ω^{2} μ_{0} ε_{0} = ω^{2} / c^{2} = 1$ and $ω μ_{0} = 120 π$ :

ε_{r} = 5, σ = \frac{12}{ω μ _{0}} = \frac{12}{120 π} = \frac{1}{10 π} = 0.0318 S/m .

$H$ from Faraday's law, $H = \frac{j}{ω μ _{0}} \nabla \times E$ , with

\nabla \times E = - \frac{\partial E _{z}}{\partial x} a_{y} = 20 j (2 + 3 j) e^{- (2 + 3 j) x} a_{y} = 20 (- 3 + 2 j) e^{- (2 + 3 j) x} a_{y},

H = \frac{j 20 ( - 3 + 2 j )}{120 π} e^{- (2 + 3 j) x} a_{y} = - \frac{2 + 3 j}{6 π} e^{- (2 + 3 j) x} a_{y} A/m .

The complex Poynting vector of Eq. (5.6.27), with $a_{z} \times a_{y} = - a_{x}$ :

P_{c} = \frac{1}{2} E \times H^{*} = \frac{1}{2} 20 j e^{- (2 + 3 j) x} (- \frac{2 - 3 j}{6 π}) e^{- (2 - 3 j) x} (a_{z} \times a_{y}) = \frac{5 ( 3 + 2 j )}{3 π} e^{- 4 x} a_{x},

P_{a v} W_{a v} = Re {P_{c}} = \frac{5}{π} e^{- 4 x} a_{x} W/ m^{2}, = \int_{S} P_{a v} \cdot d S = \frac{5}{π} \times 5 = \frac{25}{π} W

with $d S = d S a_{x}$ on the plane $x = 0$ .

Answer

f = 47.7 MHz, v_{p} = 1 0^{8} m/s, λ = 2.09 m, E (x) = 20 j e^{- (2 + 3 j) x} a_{z} V/m k = 3 - 2 j m^{- 1}, ε_{r} = 5, σ = 0.0318 S/m, W_{a v} = \frac{25}{π} = 7.96 W

Check

With the intrinsic impedance $η = ω μ_{0} / k = 120 π / (3 - 2 j) = 87.0 + j 58.0 Ω$ , that is $∣ η ∣ = 104.6 Ω$ at $33. 7^{\circ}$ , the relation $H = - (E_{z} / η) a_{y}$ gives the same $H$ , and $P_{a v} (0) = \frac{1}{2} ∣ E_{0} ∣^{2} Re {1/ η^{*}} = 200 \times 3/120 π = 5/ π W/ m^{2}$ .

Remark

$tan δ = σ / ω ε = 12/5 = 2.4$ : conduction dominates. $H$ lags $E$ by $33. 7^{\circ}$ , so $P_{c}$ also has a reactive part, $Im {P_{c}} = \frac{10}{3 π} e^{- 4 x} W/ m^{2}$ . Over one wavelength the amplitude falls to $e^{- 2 λ} = 1.5 %$ , strongly damped, as $tan δ = 2.4$ suggests. The power density falls as $e^{- 4 x}$ , twice the field rate: to $1/ e$ at $x = 1/2 α = 0.25 m$ and to $e^{- 4} = 1.8 %$ at $x = 1 m$ (Figure 5.6.20). The rest has turned into heat.