EE-335 - EM Wave Propagation

4 Lecture pp: 287-295 7-1:2

This is a typical wireless communication scheme:

Radiation Pattern Receiving Antenna Receiver Transmission Line

Transmission Generator Line Transmitting Antenna Free Space Propagation

We will now look at free space propagation

∂ 2 Ψ ∂ 2 Ψ Maxwell’s equations imply wave equations: − u 2 = 0 1-D wave equation ∂t 2 ∂z 2 Where Ψ is an electromagnetic field (E or H) and u is the wave velocity ( a function of medium). The solutions are going to be a function of (z ± ut) where

z – ut is propagation in the +Z direction z + ut is propagation in the –Z direction

d 2 Ψ if assume sinusoidal time dependence (ejωt) then + β Ψ = 0 where β = ω/u and the dz 2 solution will look like a wave traveling in the (+) direction and a wave traveling in the (-) direction:

Ψ = Ψ + + Ψ − = A+ e j()ω t−β z + A−e j(ω t+β z) amplitude: A angular frequency: ω = 2πf phase constant: β = 2π/λ

4-1 The frequency determines the EM spectrum

AM Radio 535-1605 KHz FM Radio/VHF TV 54-806 MHz EM UHF TV 470-806 MHz Microwaves 3-300 GHz IR 103 – 104 GHz Visible 105 – 106 GHz Optical UV 106 – 108 GHZ X-ray 108 – 109 GHZ

EM wave radiate from source, if you are far enough away the curved wave fronts look planar. Plane wave approximation.

Assume sinusoidal time dependence of the electric field

r ~ ~ E()()()x, y, z,t = E x, y, z f t = ℜe{E(x, y, z)e jω t }

Maxwell’s Equations then become:

~ ρ~ ∇ ⋅ E = V ε ~ ~ ∇ × E = − jω µ H ~ ∇ ⋅ H = 0 ~ ~ ~ ∇ × H = J + jω ε E

Now apply the point form of Ohm’s Law to Ampere’s Equation

4-2 ~ ~ J = σ E ~ ~ ~ ~ ∇ × H = σ E + jω ε E = ()σ + jω ε E ⎛ σ ⎞ ~ = jω⎜ε − j ⎟E ⎝ ω ⎠ ~ = jω ε c E

σ Where the complex permittivity is defined as ε = ε − j = ε ′ − jε ′′ c ω

The imaginary part of the complex permittivity is the loss term due to the conductivity of the medium

~ Now if we assume no free charge then ρv = 0 and the Maxwell’s equations become:

~ ∇ ⋅ E = 0 ~ ~ ∇ × E = − jω µ H ~ ∇ ⋅ H = 0 ~ ~ ∇ × H = jω ε c E

If we take the curl of the curl equations and then apply the other corresponding curl equation gives:

~ ~ ~ ~ ∇ × (∇ × E) = ∇ × (− jω µ H ) = − jω µ (∇ × H ) = − jω µ(jω ε E) c ~ 2 ~ ∇ × ()∇ × E = ω µ ε c E now apply the curl of the curl identity

~ ~ ~ ~ ~ ∇(∇ ⋅ E)− ∇ 2 E = ω 2 µ ε E ⇒ ∇ 2 E − γ 2 E = 0 c 2 2 γ = −ω µ ε c

A similar results for the magnetic field

These are the homogeneous wave equations with gamma being the propagation constant

If the medium is nonconducting, (lossless) then σ = 0, then

The permittivity is no longer complex and the propagation constant gets replaced with the wave number k (lossless)

γ 2 = −k 2 ⇒ k = ω µ ε

4-3 the phase velocity of the velocity of propagation becomes:

ω 1 1 1 c u p = = = = ()m / s k µ ε µ rε r µ 0ε 0 µ rε r

The phase velocity is the speed of light scale by the inverse of the square root of the relative permittivity and permeability product.

And the wavelength of the radiation is given by:

2π u λ = = p ()m k f and the intrinsic impedance of the medium is:

ω µ µ η ≡ = ()Ω k ε

8 in vacuum up = c = 3x10 (m/s) and η = η0 = 377 Ω or 120π Ω

Look at the 1-D case of the wave equation:

The solution is a plane wave that has no electric or magnetic field components in the direction of propagation.

λ

Eo Ex x z,k

The x-component of the electric field will propagate in the z-direction

The wave equation becomes:

~ d 2 E ~ x + k 2 E = 0 and the solution will have the form: dz 2 x

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E0 cos()ω t ± kz + φ0 xˆ

sign gives direction of propagation (-)Îpositive z direction, (+) Î negative z direction

Amplitude of the wave Reference phase, like initial conditions at t=0, z=0

Direction of traveling wave Field direction

ω = the angular frequency = 2πf

the results are a sinusoidal dependence on phase that depends on time (ωt) and on position (kz)

So far we have found that the plane waves are solutions to the wave equation of the electric and magnetic fields, this occurs simultaneously both the electric field and the magnetic field will propagate in the same direction. There is a relationship between the field direction, magnitude of the fields and the propagation direction. Given by:

~ 1 ~ ~ ~ H = kˆ × E E = −η kˆ × H η

Example: given the electric field in free space find the corresponding magnetic field.

r E = 103 sin()()ω t − β z yˆ V / m What is the direction of propagation? (+z)

What is the direction of the E-Field? (y)

~ 1 ~ 1 ~ E H = kˆ × E = ()zˆ × E yˆ = ()− xˆ η η η 0 0 r −103 H = sin()ω t − β z xˆ 120π

4-5 x

z

y E H

Example:

A E-field plane wave at 1Mhz in the x-direction is traveling in the z direction in air with a peak amplitude of 1.2π (mV/m) Emax is t = 0 and z = 50m fine E(z,t) and H(z,t)

1. Find the frequency: f = 1×106 (Hz) ⇒ ω = 2π f = 2π ×106 (rad / sec) 8 2. Find the velocity u p = c = 3×10 (m / sec) c 3×108 3. Find the wavelength: λ = = = 300()m f 106 2π 2π 4. Find the wave number k = = λ 300

Combine to form the field as a function of time:

r E()z,t = E peak cos (ω t − kz + φ0 )xˆ

⎡ 6 ⎛ 2π ⎞ ⎤ = 1.2π cos⎢()2π ×10 t − ⎜ ⎟z + φ0 ⎥xˆ()mV / m ⎣ ⎝ 300 ⎠ ⎦

Need to solve for the arb. Phase by using the initial conditions, cosine is a maximum when the argument is zero or at t=0, z=50

2π 100π π − ()50 + φ = 0 ⇒ φ = = 300 0 0 300 3

r ⎡ 6 ⎛ 2π ⎞ π ⎤ E()z,t = 1.2π cos⎢()2π ×10 t − ⎜ ⎟z + ⎥xˆ()mV / m ⎣ ⎝ 300 ⎠ 3 ⎦

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Now find the magnetic field

r 1 r H = kˆ × E η 1 ⎛ ⎡ 2π π ⎤ ⎞ ⎜ 6 ⎛ ⎞ ⎟ = ()zˆ × ⎜1.2π cos⎢()2π ×10 t − ⎜ ⎟z + ⎥xˆ⎟ 120π ⎝ ⎣ ⎝ 300 ⎠ 3 ⎦ ⎠

1 ⎡ 6 ⎛ 2π ⎞ π ⎤ = cos⎢()2π ×10 t − ⎜ ⎟z + ⎥ yˆ()mA/ m 100 ⎣ ⎝ 300 ⎠ 3 ⎦

µ 0 for free space: η = η0 = = 377Ω ≅ 120π Ω ε 0

r ⎡1.2π ⎛ 6 2π π ⎞⎤ Therefore: H ()z,t = yˆ⎢ cos⎜2π ×10 t − z + ⎟⎥()µ A/ m ⎣120π ⎝ 300 3 ⎠⎦

Note: E and H are in phase in lossless medium.

What happens if we have a lossy medium? Resistance or conductivity

Opaque fog for visible light, but transparent to RF waves Why?

Function of frequency, wavelength and size of the objects. (Attenuation, Scattering, diffraction)

CASE I: Free Space:

σ = 0 ε = ε0 µ = µ0

1 2π α = 0 β = ω/c c = λ = ε 0 µ 0 β

4-7 CASE II: Lossless Dielectric:

σ ≈ 0 ε = ε0εr µ = µ0µr

1 c α = 0 β = ω µ ε u = = µ ε µ rε r

µ µ r µ 0 µ r η = = = η0 impedance gets scaled by sqrt of ratio of relative ε ε r ε 0 ε r permeability and relative permittivity.

EXAMPLE For a relative permittivity of 2.56 and nonmagnetic medium r E = yˆ20cos()8π ×109 t − β z ()V / m Find f,u,λ,β,η,H

ω 8π ×109 ω = 2π f ⇒ f = = = 4×109 Hz 2π 2π c 3×108 u = = = 1.875×108 ()m / s µ rε r ()(1 2.56 ) u 1.875×108 λ = = = .047m f 4×109

2π β = = 134m −1 λ µ µ 1 η = = r 377 = 377 = 235.6Ω ε ε r 2.56 r 1 r 20 H = kˆ × E = cos()()()ωt − β z − xˆ A/ m = −0.85cos()8π ×109 t −134z xˆ()A/ m η 235.6

CASE III Good Conductors (σ >> ωε)

σ Rule of thumb: good conductor if > 100 This is frequency dependent some materials ω ε will be good conductors at some frequencies and not at others.

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