NUS/ECE EE2011 Plane Propagation in Lossless Media See animation “ Viewer” 1 Plane in Lossless Media In a source free lossless medium, J = ρ = σ = 0.

Maxwell’s equations: J =current density ρ =charge density ∂H ∇× E = - σ =conductivity μ ∂t ∂E ∇× H = ε ∂t ε ∇ ⋅E = 0 μ∇ ⋅ H = 0

Hon Tat Hui 1 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Take the curl of the first equation and make use of the second and the third equations, we have: Note : μ ∇×∇×E = ∇()∇ ⋅E − ∇2E ∂ ∂ 2 ∇ 2E = ∇× H = με E ∂t ∂t 2 This is called the : ∂2 ∇2E − με E = 0 ∂t 2 A similar equation for H can be obtained:

∂2 ∇2H − με H = 0 ∂t 2

Hon Tat Hui 2 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 In free space, the wave equation for E is:

∂ 2 ∇ 2E − μ ε E = 0 0 0 ∂t 2 where 1 μ ε = 0 0 c2

c being the speed of light in free space (~ 3 × 108 (m/s)). Hence the speed of light can be derived from Maxwell’s equation.

Hon Tat Hui 3 Plane Wave Propagation in Lossless Media Signal Propagation - 시간 지연 - 시간영역 파형 유지

Ef(0)= (t ) Ef()r= ( t − rc / ) NUS/ECE EE2011 To simplify subsequent analyses, we consider a special case in which the (and the source) variation with time takes the form of a sinusoidal function: sin(ωt +φ) or cos(ωt +φ) Using complex notation, the E field, for example, can be written as: • ⎧ jωt ⎫ E()x, y, z,t = Re⎨E x, y, ()z e ⎬ ⎩ ⎭ • where E () x , y , z is called the form of E(x,y,z,t) and is in general a depending on the spatial coordinates only. Note that the phasor form also includes the initial information and is a complex number.

Hon Tat Hui 4 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 The benefits of using the phasor form are that:

n n • ω ∂ ⎧ ∂ j t ⎫ n E()x, y, z,t = Re⎨ n E()x, y, z e ⎬ ∂t ⎩∂t ⎭ • ()(⎧ n jω )t ⎫ = Re⎨ jω E x, y, z e ⎬ ⎩ ⎭

• ⎧ j t ⎫ " E()x, y, z,t dt"dt = Re⎨ " E x, y, z e ()dt"dt⎬ ∫ ∫ ⎩∫ ∫ ω ⎭ ⎧ • ⎫ 1 () jωt = Re⎨ n E()x, y, z e ⎬ ⎩ jω ⎭

Hon Tat Hui 5 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Therefore differentiation or integration with respect to time can be replaced by multiplication or division of the phasor form with the factor jω. All other field functions and source functions can be expressed in the phasor form. As all time-harmonic functions involve the common factor ejωt in their phasor form expressions, we can eliminate this factor when dealing with the Maxwell’s equation. The wave equation can now be put in phasor form as (dropping the dot on the top, same as below):

2 •• ∂ 2 ∇−22 μεEE ωμε =0 ⇒∇− E()j E = 0 00∂t 2 00 22 (dropping the dot sign) ⇒∇+ EE00 = 0

Hon Tat Hui 6 Plane Wave Propagation in Lossless Media

μεω NUS/ECE EE2011 In phasor form, Maxwell’s equations can be written as:

∇×E = - jωB ∇× H = jωD ∇ ⋅D = ρ ∇ ⋅B = 0 Using the phasor form expression, the wave equation for μ E field is also calledε theω Helmholtz’s equation, which is:

2 2 2 2 ∇ E + 0 0 E = ∇ E + k E = 0 ω μ where k = 0ε0

Hon Tat Hui 7 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 k is called the wavenumber or the propagation constant. ω μ 2πf 2π k = k0 = 0ε0 = = c λ0 where λ0 is the free space wavelength.

In an arbitrary medium with ε =ε0εr and μ =μ0μ r, ω μ ε μ 2πf μ 2π μ k = 0 0 rε r = rε r = λ rε r c 0 We call, 2π λ λ = = 0 = wavelength in the medium μ k rε r Hon Tat Hui 8 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 In Cartesian coordinates, the Helmholtz’s equation can be written as three scalar equations in terms of the respective x, y, and z components of the E field. For

example, the scalar equation for the Ex component is:

⎛ ∂2 ∂2 ∂2 ⎞ ⎜ + + + k 2 ⎟E = 0 ⎜ 2 2 2 ⎟ x ⎝ ∂x ∂y ∂z ⎠

Consider a special case of the Ex in which there is no variation of Ex in the x and y directions, i.e., ∂2 ∂2 E = E = 0 ∂x2 x ∂y2 x

Hon Tat Hui 9 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

This is called the plane wave condition and Ex(z) now varies with z only. The wave equation for Ex becomes:

d 2E (z) x + k 2E ()z = 0 dz2 x

Note that a plane wave is not physically realizable because it extends to an infinite extent in the x and y directions. However, when considered over a small plane area, its propagation characteristic is very close to a spherical wave, which is a real and common form of electromagnetic wave propagating.

Hon Tat Hui 10 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Solutions to the plane wave equation take one form of the following functions, depending on the boundary conditions: + − jkz 1. Ex (z)= E0 e − + jkz 2. Ex (z)= E0 e + − jkz − + jkz 3. Ex (z)= E0 e + E0 e

+ - E0 and E0 are constants to be determined by boundary conditions.

+ − jkz − + jkz E0 e and E0 e are plane waves propagating along the +z direction and –z direction.

Hon Tat Hui 11 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 1.1 Solutions to plane wave equation In time domain, + − jkz + − jkz jωt (1) Ex ()z = E0 e → Ex (z,t)= Re{E0 e e } + = E0 cos()ωt − kz

+ Assume E0 = ω = k = 1, then Ex (z,t)= cos(t − z). We can plot this solution for several seconds to see its motion in space. We focus on one period of the function only while keeping in mind that this period repeats itself continuously in both left and right directions.

Hon Tat Hui 12 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

At t = 0, E ()z,0 = cos(− z) Ex x 1

z -2π -3π/2 -π -π/2 0 π/2 π 3π/2 2π

-1

() ( ) Ex At t = 1s, Ex z,1 = cos 1− z 1

z -2π -3π/2 -π -π/2 0 π/2 π 3π/2 2π

-1

Ex At t = 2s, Ex ()z,2 = cos(2 − z) 1

z -2π -3π/2 -π -π/2 0 π/2 π 3π/2 2π

-1 Hon Tat Hui 13 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

− + jkz − + jkz jωt (2) Ex ()z = E0 e → Ex (z,t)= Re{E0 e e } − = E0 cos()ωt + kz

− Assume E0 = ω = k = 1, then Ex (z,t)= cos(t + z). The solution is plotted on the next page for the first several seconds.

Hon Tat Hui 14 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

Ex At t = 0, Ex ()z,0 = cos(z) 1

z -2π -3π/2 -π -π/2 0 π/2 π 3π/2 2π

-1

Ex At t = 1s, Ex ()z,1 = cos(1+ z) 1

z -2π -3π/2 -π -π/2 0 π/2 π 3π/2 2π

-1

Ex At t = 2s, Ex ()z,2 = cos(2 + z) 1

z -2π -3π/2 -π -π/2 0 π/2 π 3π/2 2π

-1 Hon Tat Hui 15 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

+ − jkz E0 e is a wave propagating in the + z direction. − + jkz E0 e is a wave propagating in the − z direction.

For the wave moving in +z direction, in a time of 1 second, the wave moves in 1 unit of distance (for example, meter). Then the speed of propagation is (1 m/1 s = 1ms-1). A similar result can be obtained for the wave moving in –z direction.

Hon Tat Hui 16 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 1.2 Propagation speed of a general plane wave + If ω ≠ 1, k ≠ 1, and E 0 ≠ 1, then for the wave moving

to the right: + Ex (z,t)()= E0 cos ωt − kz .

+ (1) At t = 0, Ex (z,0)()= E0 cos − kz .

Consider a zero point (z coordinate = z0) of the wave (for example the first one to the right of the origin), then π Ex ()z0 ,0 = 0 ⇒ cos(− kz0 )= 0 π ⇒ −kz = − ⇒ z = 0 2 0 2k Hon Tat Hui 17 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 + (2) At t = 1s, Ex (z,0)= E0 cos(ω − kz). Consider the same zero point of the wave as in (1) but now its z coordinateω has move from z to z , then π 0 1 Ex ()z1,1 = 0 ⇒ cos(ω − kz1 )= 0 ω π ⇒ − kz = − ⇒ z = + 1 2 1 k 2k ω π π ω Distance change in 1s = z − z = + − = 1 0 k 2k 2k k ω ω Thus propagation speed = /1s = ms−1 k k A same result can be obtained for the wave moving to the left. Hon Tat Hui 18 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 1.3 Solution for the Once the is known, the accompanying magnetic field H can be found from the Maxwell’s equation ∇× E = - jωμH For example, if the solution for E is, + − jkz E(z)= xˆEx = xˆE0 e , then the solution for H is:

ωμ+ − jkz E ∂e k H()z = yˆ 0 = yˆ E +e− jkz = yˆH − j ∂z ωμ 0 y Note that H is ⊥ to E and they are shown on next page. Hon Tat Hui 19 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

H and E propagate in free space

See animation “Plane Wave E and H Vector Motions ” Hon Tat Hui 20 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

The ratioη of Ex to Hy is called the intrinsic impedance of the medium, η. ωμ ωμ ( ) ω μ Ex z με = ()= = = (Ω) H y z k ε

Note that η is independent of z. In free space, η μ

ε0 0 = =120π ≈ 377 Ω 0

and Ex and Hy are in phase (as η0 is a real number). Hon Tat Hui 21 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 The (propagation speed of a constant-

phase point) of the wave up is given by: ω ω ω 1 u = = με = (m/s) p k με

See animation “Plane Wave in 3D”

The plane wave is also called the TEM wave (TEM =

Transverse ElectroMagnetic) in which Ez = Hz = 0 where z is the direction of propagation.

Hon Tat Hui 22 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 1.4 Expressions for a general plane wave The general expressions of a plane wave are:

− jk⋅r E = E0e − jk⋅r H = H0e

E0 and H0 are vectors in arbitrary directions. k is the vector propagation constant whose magnitude is k and whose direction is the direction of propagation of the wave. r is the observation position vector.

2 2 2 k = kxxˆ + k yyˆ + kzzˆ, k = kx + k y + kz r = xxˆ + yyˆ + zzˆ

Hon Tat Hui 23 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

k Right-hand rule: r E (index finger) (thumb) E k H H (middle finger)

Hon Tat Hui 24 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Using Maxwell’s equations ∇×E = - jωμH ∇× H = jωεE it can be shown that we have the following relations for the field vectors and the propagation direction. E ⊥ H ⊥ k η 1 H = kˆ × E, E = ηH ×kˆ

Hon Tat Hui 25 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Example 1

A uniform plane wave with E = xˆ E x propagates in the +z- direction in a lossless medium with εr = 4 and μr = 1. Assume that Ex is sinusoidal with a of 100 MHz and that it has a positive maximum value of 10-4 V/m at t = 0 and z = 1/8 m.

(a) Calculate the wavelength λ and the phase velocity up, and find expressions for the instantaneous electric and magnetic field intensities.

(b) Determine the positions where Ex is a positive maximum at the time instant t = 10-8s.

Hon Tat Hui 26 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Solutions eˆ = unit vector of E (a) kˆ , kkkˆˆ ⋅r = kz,

k ω 1 c u = = = = 1.5×108 (m/s) p k με 4 kˆ (phasor form)

(instantaneous form)

Hon Tat Hui 27 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 The cosine function has a positive maximum when its argument equals zero (ignoring the 2nπ ambiguity). Thus at t = 0 and z = 1/8,

kˆ

kˆ

Hon Tat Hui 28 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

kˆ = zˆ

(m)

See animation “Plane Wave Simulator”

Hon Tat Hui 29 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 2 of Plane Waves The polarization of a plane wave is the figure the tip of the instantaneous electric-field vector E traces out with time at a fixed observation point. There are three types of polarizations: the linear, circular, and elliptical polarizations.

Ey Ey Ey

Ex Ex Ex

Eectric-field vector Eectric-field vector Eectric-field vector Linearly polarized Circularly polarized Elliptically polarized

See animation “Polarization of a Plane Wave - 2D View” Hon Tat Hui 30 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

See animation “Polarization of a Plane Wave - 3D View”

Polarization of a Plane Wave Hon Tat Hui 31 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 (a) A plane wave is linearly polarized at a fixed observation point if the tip of the electric-field vector at that point moves along the same straight line at every instant of time. (b) A plane wave is circularly polarized at a a fixed observation point if the tip of the electric-field vector at that point traces out a circle as a function of time. Circular polarization can be either right-handed or left- handed corresponding to the electric-field vector rotating clockwise (right-handed) or anti-clockwise (left-handed). Hon Tat Hui 32 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 (c) Elliptical Polarization A plane wave is elliptically polarized at a a fixed observation point if the tip of the electric-field vector at that point traces out an ellipse as a function of time. Elliptically polarization can be either right-handed or left-handed corresponding to the electric-field vector rotating clockwise (right-handed) or anti-clockwise (left-handed).

Hon Tat Hui 33 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

For example, consider a plane wave: Ex0 and Ey0 are both real numbers − jkz E = xˆEx + yˆEy Ex = Ex0e − jkz − jkz E = − jE e− jkz = xˆEx0e − yˆjEy0e y y0

Note that the phase difference between Ex and Ey is 90º. The instantaneous expressionω for E is: ω j t− jkz jωt− jkz E()z,t = Re{xˆEx0e − yˆjEy0e }

= xˆEx0 cos()t − kz + yˆEy0 sin ωt − kz () Let:

X =−==−EExx=cos00(ωω tkzY) , E y E y sin( tkz) Hon Tat Hui 34 Plane Wave Propagation in Lossless Media NUS/ECE EE2011

Case 1: EExo== 0 or yo 0, then XY==0 or 0 Both are straight lines. Hence the wave is linearly polarized.

Case 2: EECxo== yo , then 22 2 2 2 2 XYC+=⎣⎦⎡⎤cos(ωω tkz −+) sin ( tkzC −) = X and Y describe a circle. Hence the wave is circularly polarized.

Case 3: EExo≠ yo , then 22 XY 22 22+=cos()()ωωtkz −+ sin tkz −= 1 EExy00 X and Y describe an ellipse. Hence the wave is elliptically polarized. Hon Tat Hui 35 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Example 2 Two circularly polarized plane waves watched at z = 0 are given by: ω E1()t = xˆ5cos(ωt + 53.1°)+ yˆ5sin(ωt + 53.1°) ()()() E2 t = xˆ5cos t − 53.1° − yˆ5sin ωt − 53.1° Show that they combine together to form a linearly polarized wave. Solutions: ω E = E1 + E2 = xˆ[5cos(ωt + 53.1°)+ 5cos(ωt − 53.1°)] ω + yˆ[]5sin()()t + 53.1° − 5sinω t − 53.1° = xˆ10cosω()t cos 53 (.1° + yˆ10 )cos t sin ()53.1° ( ) = xˆ6cos()t + yˆ8cos ωt ()ω Hon Tat Hui 36 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Now,

Ex = 6cos()ωt , Ey = 8cos(ωt)

Let X = Ex , Y = Ey Y 8cosω()t 4 Then = = X 6cos()ωt 3 4 4 Y = X ⇒ equation of a straight line with slope = 3 3 Hence the locus of the combined electric field falls on a straight line and the polarization of the combined wave is thus linear.

Hon Tat Hui 37 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Generic Polarization Description Method In general, the polarization state of an EM wave is characterized by two parameters.

− jkz jδ − jkz Ex = Ex0e , Ey = e Ey0e γ 1. Ratio of Ey0 to Ex0

−1⎛ Ey0 ⎞ ⇒ = tan ⎜ ⎟, 0 ≤ γ ≤ 90° ⎝ Ex0 ⎠ δ 2. Phase difference between Ex and Ey , i.e., , -180° ≤ δ ≤ 180°

Hon Tat Hui 38 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 For example: γ = 0 or 90º and for any value of δ ⇒ linearly polarized γ = 45° and δ = 90° ⇒ (right - hand) circularly polarized

Hon Tat Hui 39 Plane Wave Propagation in Lossless Media NUS/ECE EE2011 Plane Wave Propagation in Lossy Media 1 Plane Waves in Lossy Media In a lossy medium, J = σE, ρ ≠ 0, σ ≠ 0.

Maxwell’s equations: Usually,if dampingloss is small : ε σ ε '= 0ε r ∇ × E = - jωμH ε σ ⎛ ⎞ ε ∇ × H = E + jωD = jω⎜ '+ ⎟E = jωεcE ε σ ⎝ jω ⎠ ε ⎛ σ ⎞ c = '− j = '⎜1− j ⎟ = complex ω ⎝ ωε' ⎠

Hon Tat Hui 1 Plane Wave Propagation in Lossy Media NUS/ECE ε EE2011 σ = loss tangent = tanδ ω ' c The complex permittivity can be written as: ε ε c = ε0ε rc ⎛ε σ ⎞ ⎜ ⎟ rc = ⎜ 'r − j ⎟ = complex relative permittivity ⎝ ωε0 ⎠

Current terms: Usually ε'r = ε r σE = conduction current ω j D = jωε'E = displacement current Note that conduction current and displacement current are out of phase by π/2. Hon Tat Hui 2 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 Helmholtz’s equation:

2 2 ∇ E + kc E = 0 ω kc = μεc a complex number

By replacing k with kc, all the previous results derived for lossless media are applicable to lossy media. But

since kc is a complex number, the plane wave will experience loss when it propagates.

Hon Tat Hui 3 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 The solutions for the Helmholtz equation in lossy media

are same as those for lossless media if k is replaced by kc.

E (z)= E ±e∓ jkc z x 0 γ ± ∓ z = E0 e α ± − z ∓ jβz = E0 e e

where γ = jkc = α + jβ = complex propagation constant.

Note that solutions with the +α constant have been discarded as they imply waves with increasing which is impossible. Hon Tat Hui 4 Plane Wave Propagation in Lossy Media NUS/ECE EE2011

α = constant ω με 2 ' ⎛ ⎛σ ⎞ ⎞ = ⎜ 1+ ⎜ ⎟ −1⎟ (Np/m) 2 ⎜ ωε' ⎟ ⎝ ⎝ ⎠ ⎠

β = propagation constant ω με 2 ' ⎛ ⎛σ ⎞ ⎞ = ⎜ 1+ ⎜ ⎟ +1⎟ (rad/m) 2 ⎜ ωε' ⎟ ⎝ ⎝ ⎠ ⎠

Hon Tat Hui 5 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 As in the lossless case, the general form of a plane wave in a lossy medium is: ˆ − jk c ⋅r jφ − jkc (k c ⋅r) k = k xˆ + k yˆ + k zˆ = k kˆ E = E0e = eˆ E0 e e c cx cy cz c c r = xxˆ + yyˆ + zzˆ E − jk ()krˆ ⋅ −⋅jkrc ˆ 0 jφ cc φ = initial phase HH==0eee h ηc ˆ ˆ ˆ E ⊥ H ⊥ k k c = e×h η c ˆ ˆ ˆ 1 h = k c ×e H = kˆ × E, E = η H ×kˆ c c c eˆ = hˆ ×kˆ c c

ηc = complexμ intrinsic impedance ε η = = r + jηi c Hon Tat Hui 6 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 The real and imaginary parts of η can be derived as: η μ c ⎛ σ 2 ⎞ ε ⎜ ⎛ ⎞ ⎟ ε ⎜ 1+ωε⎜ ⎟ +1 ⎟ 0 1 ⎝ ' ⎠ r = ⎜ 2 ⎟ Ω 0 2 ' ⎜ ⎛σ ⎞ ⎟ r 1+ ⎜ ⎟ ⎜ ⎝ωε' ⎠ ⎟ ⎝ ⎠ μ ⎛ σ 2 ⎞ ε ⎜ ⎛ ⎞ ⎟ ε ⎜ 1+ωε⎜ ⎟ −1 ⎟ 0 1 ⎝ ' ⎠ ηi = ⎜ 2 ⎟ Ω 0 2 ' ⎜ ⎛σ ⎞ ⎟ r 1+ ⎜ ⎟ ⎜ ⎝ωε' ⎠ ⎟ ⎝ ⎠ ηc =ηr + jηi Hon Tat Hui 7 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 2 Approximation Analysis for Two Special Cases

The formulas for kc and ηc are complicated. For some special cases depending on the loss tangent σ/ωε’, approximation formulas can be derived. We study two special cases in which the loss tangent is either much greater than 1 or much smaller than 1.

σ >> 1 ⇒ a good conductor ωε ' σ << 1 ⇒ a low - loss dielectric ωε'

Hon Tat Hui 8 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 2.1 Good conductors Condition:

σ ⎛ σ ⎞ loss tangent = >>1 ⎜for example >100⎟ γ ωε ' α ωε ' ⎝ ⎠ β Then, = + j ≈ (1+ j) πfμσ α β = ≈ πfμσ η μ ε α c = ≈ ()1+ j c σ

Hon Tat Hui 9 Plane Wave Propagation in Lossy Media NUS/ECE EE2011

In a good conductor, the intrinsic impedance ηc is a complex number, meaning that the electric and magnetic fields are not in phase as in the case of a lossless medium. ω λ ω2 u = phase velocity = β ≈ p π μσ β π 2 π2μσ π = wavelength = ≈ = 2 f fμσ

Note that H is still ⊥ to E as shown on next page. Hon Tat Hui 10 Plane Wave Propagation in Lossy Media NUS/ECE EE2011

H and E inside a good conductor See animation “Plane Wave Simulator” Hon Tat Hui 11 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 In a good conductor, because of the attenuation constant α, the wave becomes smaller when it propagates. The distance δ through which the amplitude of a travelling plane wave decreases by a factor of e-1 = 0.368 = 36.8% is called the skin depth or depth of penetration of the conductor.

Field amplitude

E0

-1 E0e

z 0 δ 2δ …….

Hon Tat Hui 12 Plane Wave Propagation in Lossy Media αδ NUS/ECE βδ EE2011 E ()z = δ E +e− e− j x = 0 = e−αδ = e−1 + Ex ()z = 0 E0 αδ =1 δ 1 = α

7 For copper, σ = 5.8 × 10 S/m and μr = 1. δ -3 α1 2 6.61×10−2 at 60Hz, δ = 8.5 × 10 m ωμ at 1MHz, δ =6.6 × 10-5 m = = μ = 0 rσ f at 30GHz, δ = 3.8 × 10-7 m

Hon Tat Hui 13 Plane Wave Propagation in Lossy Media NUS/ECE EE2011

Skin depth of some common materials

Hon Tat Hui 14 Plane Wave Propagation in Lossy Media γ α NUS/ECE β EE2011 2.2 Low-loss ω με 1/ 2 ⎛ σ ⎞ = + j = jkc = j '⎜1− j ⎟ ⎝ ωε' ⎠ Condition: σ Binomial expansion (when x << 1) : loss tangent = <<1 1 γ ωε ' ()1+ x n ≈ 1+ nx + n()n −1 x2 + ω 2 ⎛ σ ⎞ με ⎜for example ≤ 0.01⎟ ε ⎝ ωε ' ⎠ 2 ⎡ '' 1 ⎛ε '' ⎞ ⎤ Then, ≈ j '⎢1− j ε + ⎜ ⎟ ⎥ ⎣⎢ 2 ' 8 ⎝ ε' ⎠ ⎦⎥ σ ε'' σ ε''= , = = loss tangent ω ε' ωε' Hon Tat Hui 15 Plane Wave Propagation in Lossy Media α ωε NUS/ECE μ EE2011 Hence, β ε ω με 2 '' ⎡ 1 ⎛ε '' ⎞ ⎤ ≈ , ≈ ' ⎢1+ ⎜ ⎟ ⎥ 2 ' ⎣⎢ 8 ⎝ ε' ⎠ ⎦⎥ η μ ε ⎛ ε '' ⎞ c ≈ ⎜1+ j ⎟ ' ⎝ 2ε' ⎠ ω β 2 με1 ⎡ 1 ⎛ε '' ⎞ ⎤ u p = ≈ ⎢1− ⎜ ⎟ ⎥ ' ⎣⎢ 8 ⎝ ε' ⎠ ⎦⎥

Hon Tat Hui 16 Plane Wave Propagation in Lossy Media λ π NUS/ECE EE2011 β 2 2 1με ⎡ 1 ⎛ε '' ⎞ ⎤ = wavelength = ≈ ⎢1− ⎜ ⎟ ⎥ f ' ⎣⎢ 8 ⎝ ε' ⎠ ⎦⎥

A skin depth δ can be similarly defined as in the good conductor case: δ α1 2 ε ε σ 0 r = = μ 0μr

Hon Tat Hui 17 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 Example 1 The electric field intensity of a linearly polarised uniform plane wave propagating in the +z direction in seawater is Ex=πˆ 100cos(107 t ) at z = 0. The constitutive parameters of seawater are εr = 72, μr = 1, and σ = 4 S/m. (a) Determine the attenuation constant, intrinsic impedance, phase velocity, wavelength, and skin depth. (b) Write expressions for H(z,t) and E(z,t).

(c) Find the distance z1 at which the amplitude of the electric field is 1% of its value at z = 0. (d) Compute the skin depth at a frequency of 1 GHz.

Hon Tat Hui 18 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 Solutionsω (a) = 107π rad/s ⇒ f = 5×106 Hz σ Here ≈ 200 >> 1 . We may therefore approximate η ωε’ seawater as aπ good conductor at this frequency. μ μ α=β= πf μ μ σ=8.89 Np/m or rad/m r 0σ π f = (1+ j) r 0 = (1+ j) = π e jπ / 4 Ω c 2 ω 2π u ==3.53 × 106 m/s λ= =0.707 m p β β δ=1/ α= 0.112 m

Hon Tat Hui 19 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 φ (b) Phasor fields: eˆ = unit vector of E(r) j − jk (kˆ ⋅r) In a lossy medium, E(r) = eˆE e φ e c c φ 0 α ˆ ˆ j − (kc ⋅r) − jβ (kc ⋅r) = eˆE0e e e α ˆ Here, eˆ = xˆ, = 0, E0 = 100, = β = 8.89, k c ⋅r = z Therefore, E(z) = xˆ 100e−8.89z e− j8.89z π 1 ˆ H(r) = k cπ× E ηc 100 100 ⇒ H(z) = yˆ e−8.89z e− j8.89z = yˆ e−8.89z e− j(8.89z+π / 4) e j / 4 π

Hon Tat Hui 20 Plane Wave Propagation in Lossy Media NUS/ECE ω EE2011 Instantaneous fields: E(z,t) = Re[E(z)e j t ]

7 = Re[]xˆ 100ωe−8.89z e− j8.89z e j10 π t = xˆ 100 e−8.89z cos(107π t − 8.89z) H(z,t) = Re[H(z)e j t ] π ⎡ 100 −8.89z − j(8.89z+ / 4) j107 π t ⎤ = Re⎢yˆ e e e ⎥ ⎣ π π ⎦ 100 = yˆ e−8.89z cos(107 t − 8.89z −π / 4) π

(b) exp()− αz1 = 0.01 ⇒ z1 = 0.518 m Hon Tat Hui 21 Plane Wave Propagation in Lossy Media ω NUS/ECE EE2011 π σ μ 9 μ −7 −12 = 2 ×10 , = 4, π= 0 = 4 ×10 , '= 0 r = 8.854×10 ×72 ε ε α ε Usingω με the general formulasα for and on page 5, we have : β ⎛ σ 2 ⎞ ' ⎜ ⎛ ⎞ ⎟ ∴ = 1ωε+ ⎜ ⎟ −1 = 80.837 (Np/m) 2 ⎜ ⎝ ' ⎠ ⎟ β ⎝ ⎠ ω με 2 ' ⎛ ⎛σ ⎞ ⎞ = ⎜ 1+ ⎜ ⎟ +1⎟ = 195.35 (rad/m) 2 ⎜ ωε' ⎟ ⎝ ⎝ ⎠ ⎠

Hence at 1 GHz, α = 80.837 Np/m and δ = 1/α = 12.37 mm.

Hon Tat Hui 22 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 3 Power Flow and Poynting Vector The cross product of E and H has the dimension of power per unit area. It is called the Poynting vector, S and it represents the power carried by the electromagnetic field through a unit area (see Note 1* below). The direction of the Poynting vector indicates the direction of power flow. Instantaneous Poynting vector: ω S = E()x, y, z,t × H(x, y, z,t) =()Re{}E x, y, z e j t × Re{}H()x, y, z e jωt (W/m2 )

(*Note 1: for a rigorous derivation of the Poynting vector, pls see Intensive Reading notes, section 2.5.) Hon Tat Hui 23 Plane Wave Propagation in Lossy Media NUS/ECE EE2011

Average Poynting vector (see Intensive Reading notes, section 2.5): T 2 1 1 * 2 Sav = lim S dt = Re{}E()()x, y, z × H x, y, z (W/m ) T →∞ ∫ T −T 2 2

The magnitude of Sav gives the average power density (per unit area) of the EM wave. Vector magnitude, not absolute value 3.1 In a lossless medium 1 P ==SEHRe ×* 1 av av {} H = kˆ × E 2 η 1 22η ==ηEH (W/m2 ) E =ηH ×kˆ 2200

Hon Tat Hui 24 Plane Wave Propagation in Lossy Media φ

NUS/ECE EE2011 φ 3.2 In a lossy medium α ˆ ˆ j − jk c ⋅r j − (k c ⋅r)()− jβ k c ⋅r E = E e e = eˆ E eφ e e 0 0 α E ˆ ˆ − jk c ⋅r ˆ 0 j − ()k c ⋅r − jβ k c ⋅r () H = H0e = h e e e ηc 2 ⎧⎫E ˆ 11* ⎪⎪ˆ 0 −⋅2α()krc Peav==SEHkav Re{} ×= Re⎨⎬ c 22η ⎩⎭⎪⎪c

α 2 −⋅2 krˆ ⎧⎫ 11ˆ ()c = k c Ee0 Re⎨⎬ 2 ⎩⎭c α ˆ 112 −⋅2 ()krc ⎧⎫η 2 = Ee0 Re⎨⎬ (W/m ) 2 ⎩⎭c Hon Tat Hui 25 Plane Wave Propagationη in Lossy Media NUS/ECE EE2011 Example 2 For the plane wave in Example 1, find the power densities at distances of skin depth z = δ and z = 0.

Solutions From Example 1: E(z) = xˆ 100e−8.89z e− j8.89z 100 H(z) = yˆ e−8.89z e− j(8.89z+π / 4) π skin depth δ = 0.112 m η jπ / 4 c = π e Ω α = 8.89 Np/m Hon Tat Hui 26 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 Power density at skin depth δ: δ δ ˆ 112 −⋅2α()krc ⎧⎫ PEe= 0 Re⎨⎬ 2 ⎩⎭ηc

112 −×2 8.89 × 0.112 ⎧⎫ = 100e Re⎨⎬j /4 2 ⎩⎭e π = 153. 63 (W/m2 ) π Power density at z = 0:

112 ⎧ ⎫ P0 = 100 Re⎨ jπ /4 ⎬ 2 ⎩⎭π e = 1125.4 (W/m2 ) Hon Tat Hui 27 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 Example 3

Compute the average power density Pav of a uniform sinusoidal plane wave propagating in air which has the following expression for the instantaneous magnetic field: π ⎛ 1 1 π ⎞ H(x, z,t) = ⎜− xˆ + zˆ ⎟ cos(ωt − 6x − 8z) A/m ⎝ 15 20 ⎠ Solutions In phasor form: ⎛ 1 1 ⎞ H(r) = H(x, z) = ⎜− π xˆ + zˆ ⎟ e− j(6 x+8z) A/m ⎝ 15 20π ⎠ Hon Tat Hui 28 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 hˆ = unit vector of H(r) ˆ − j k (kˆ⋅r) H(r) = h H0 e

22 ⎛⎞⎛⎞111111 H0 =−⎜⎟⎜⎟ + = + = ⎝⎠⎝⎠15π 20ππ1522 20 12 π ˆ ⎛⎞111 hxz=−⎜⎟ˆˆ +/ =− ( 0.8, 0,0.6) ⎝⎠15πππ 20 12 ˆ k(k ⋅r) = kx x + k y y + kz z = 6x + 8z

⇒ kx = 6, k y = 0, kz = 8

2 2 k = kx + kz = 10 ˆ k = ()kx ,0,kz k = (0.6,0,0.8) Hon Tat Hui 29 Plane Wave Propagation in Lossy Media NUS/ECE EE2011 ˆ Using, E(r) = −η0 k × H E(r) = E(x, z) = yˆ 10 e− j(6 x+8z) V/m 1 S = Re[]E()r × H* r () av 2

1 ⎡ − j(6 x+8z) ⎛ 1 1 ⎞ j(6 x+8z) ⎤ = Re⎢yˆ 10 e ×⎜− xˆ + zˆ ⎟ e ⎥ 2 ⎣ ⎝ 15 20 ⎠ ⎦ π 1 ⎡ 10 10 ⎤ = Re π xˆ + zˆ π 2 ⎣⎢20 15π ⎦⎥ 1 1 5 = xˆ + zˆ (W/m 2 ) ∴=P S = (W/m2 ) 4π 3π av av 12π

Hon Tat Hui 30 Plane Wave Propagation in Lossy Media