Uniform Plane Waves. 2.1 Uniform Plane Waves in Lossless Media. ɛ = 1 ηc, μ = η c, where c = 1 μ. , η = z = 1 c η H. z = 1 E. z = ˆx E y. c t.

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1 .. Uniform Plane Waves in Lossless Media 37. Uniform Plane Waves in Lossless Media Uniform Plane Waves The simplest electromagnetic waves are uniform plane waves propagating along some fixed direction, say the z-direction, in a lossless medium {ɛ, μ}. The assumption of uniformity means that the fields have no dependence on the transverse coordinates x, y and are functions only of z, t. Thus, we look for solutions of Maxwell s equations of the form: E(x, y, z, t)= E(z, t) and H(x, y, z, t)= H(z, t). Because there is no dependence on x, y, we set the partial derivatives x = 0 and y = 0. Then, the gradient, divergence, and curl operations take the simplified forms: =ẑ z, E = E z z, E E = ẑ z = ˆx E y z + ŷ E x z Assuming that D = ɛe and B = μh, the source-free Maxwell s equations become: E = μ H t H = ɛ E t E = 0 H = 0 ẑ E H = μ z t ẑ H z = ɛ E t E z z = 0 H z z = 0 (..) An immediate consequence of uniformity is that E and H do not have components along the z-direction, that is, E z = H z = 0. Taking the dot-product of Ampère s law with the unit vector ẑ, and using the identity ẑ (ẑ A)= 0, we have: The shorthand notation x stands for x. ( ẑ ẑ H ) = ɛ ẑ E z t = 0 E z t = 0 Because also z E z = 0, it follows that E z must be a constant, independent of z, t. Excluding static solutions, we may take this constant to be zero. Similarly, we have H z = 0. Thus, the fields have components only along the x, y directions: E(z, t) = ˆx E x (z, t)+ŷ E y (z, t) (transverse fields) (..) H(z, t) = ˆx H x (z, t)+ŷ H y (z, t) These fields must satisfy Faraday s and Ampère s laws in Eqs. (..). We rewrite these equations in a more convenient form by replacing ɛ and μ by: ɛ = ηc, μ = η c, where c = μ, η = (..3) μɛ ɛ Thus, c, η are the speed of light and characteristic impedance of the propagation medium. Then, the first two of Eqs. (..) may be written in the equivalent forms: ẑ E z = c η H t η ẑ H z = E c t (..4) The first may be solved for z E by crossing it with ẑ. Using the BAC-CAB rule, and noting that E has no z-component, we have: ( ẑ E ) ẑ = E ( (ẑ ẑ) ẑ ẑ E ) = E z z z z where we used ẑ z E = z E z = 0 and ẑ ẑ =. It follows that Eqs. (..4) may be replaced by the equivalent system: E z = (ηh ẑ) c t z (ηh ẑ)= E c t (..5) Now all the terms have the same dimension. Eqs. (..5) imply that both E and H satisfy the one-dimensional wave equation. Indeed, differentiating the first equation with respect to z and using the second, we have: E z = c ( t z (ηh ẑ)= E c t ) z E(z, t)= c t 0 (wave equation) (..6) and similarly for H. Rather than solving the wave equation, we prefer to work directly with the coupled system (..5). The system can be decoupled by introducing the socalled forward and backward electric fields defined as the linear combinations: E + = (E + ηh ẑ) E = (E ηh ẑ) (forward and backward fields) (..7) or,

2 38. Uniform Plane Waves.. Uniform Plane Waves in Lossless Media 39 Component-wise, these are: E x± = (E x ± ηh y ), E y± = (E y ηh x ) (..8) We show next that E + (z, t) corresponds to a forward-moving wave, that is, moving towards the positive z-direction, and E (z, t), to a backward-moving wave. Eqs. (..7) can be inverted to express E, H in terms of E +, E. Adding and subtracting them, and using the BAC-CAB rule and the orthogonality conditions ẑ E ± = 0, we obtain: E(z, t) = E + (z, t)+e (z, t) H(z, t) = η ẑ [ E + (z, t) E (z, t) ] (..9) In terms of the forward and backward fields E ±, the system of Eqs. (..5) decouples into two separate equations: E + z = E + c t (..0) E z =+ E c t Indeed, using Eqs. (..5), we verify: z (E ± ηh ẑ)= c t (ηh ẑ) E c t = c (E ± ηh ẑ) t Eqs. (..0) can be solved by noting that the forward field E + (z, t) must depend on z, t only through the combination z ct (for a proof, see Problem..) If we set E + (z, t)= F(z ct), where F(ζ) is an arbitrary function of its argument ζ = z ct, then we will have: Inserting these into the inverse formula (..9), we obtain the most general solution of (..5), expressed as a linear combination of forward and backward waves: E(z, t) = F(z ct)+g(z + ct) H(z, t) = η ẑ [ F(z ct) G(z + ct) ] (..) The term E + (z, t)= F(z ct) represents a wave propagating with speed c in the positive z-direction, while E (z, t)= G(z+ct) represents a wave traveling in the negative z-direction. To see this, consider the forward field at a later time t + Δt. During the time interval Δt, the wave moves in the positive z-direction by a distance Δz = cδt. Indeed, we have: E + (z, t + Δt) = F ( z c(t + Δt) ) = F(z cδt ct) E + (z Δz, t) = F ( (z Δz) ct ) = F(z cδt ct) E +(z, t + Δt)= E + (z Δz, t) This states that the forward field at time t + Δt is the same as the field at time t, but translated to the right along the z-axis by a distance Δz = cδt. Equivalently, the field at location z + Δz at time t is the same as the field at location z at the earlier time t Δt = t Δz/c, that is, E + (z + Δz, t)= E + (z, t Δt) Similarly, we find that E (z, t + Δt)= E (z + Δz, t), which states that the backward field at time t + Δt is the same as the field at time t, translated to the left by a distance Δz. Fig... depicts these two cases. E + z = ζ F(ζ) F(z ct)= z z ζ = F(ζ) ζ E + t = ζ F(ζ) F(z ct)= t t ζ = c F(ζ) ζ E + z = E + c t Vectorially, F must have only x, y components, F = ˆxF x + ŷf y, that is, it must be transverse to the propagation direction, ẑ F = 0. Similarly, we find from the second of Eqs. (..0) that E (z, t) must depend on z, t through the combination z + ct, so that E (z, t)= G(z + ct), where G(ξ) is an arbitrary (transverse) function of ξ = z + ct. In conclusion, the most general solutions for the forward and backward fields of Eqs. (..0) are: Fig... Forward and backward waves. E + (z, t) = F(z ct) E (z, t) = G(z + ct) with arbitrary functions F and G, such that ẑ F = ẑ G = 0. (..) The two special cases corresponding to forward waves only (G = 0), or to backward ones (F = 0), are of particular interest. For the forward case, we have: E(z, t) = F(z ct) H(z, t) = η ẑ F(z ct)= η ẑ E(z, t) (..3)

3 40. Uniform Plane Waves.. Uniform Plane Waves in Lossless Media 4 This solution has the following properties: (a) The field vectors E and H are perpendicular to each other, E H = 0, while they are transverse to the z-direction, (b) The three vectors {E, H, ẑ} form a right-handed vector system as shown in the figure, in the sense that E H points in the direction of ẑ, (c) The ratio of E to H ẑ is independent of z, t and equals the characteristic impedance η of the propagation medium; indeed: H(z, t)= ẑ E(z, t) E(z, t)= ηh(z, t) ẑ (..4) η The electromagnetic energy of such forward wave flows in the positive z-direction. With the help of the BAC-CAB rule, we find for the Poynting vector: P=E H = ẑ η F = c ẑ ɛ F (..5) where we denoted F = F F and replaced /η = cɛ. The electric and magnetic energy densities (per unit volume) turn out to be equal to each other. Because ẑ and F are mutually orthogonal, we have for the cross product ẑ F = ẑ F = F. Then, w e = ɛ E = ɛ F w m = μ H = μ η ẑ F = ɛ F = w e where we replaced μ/η = ɛ. Thus, the total energy density of the forward wave will be: w = w e + w m = w e = ɛ F (..6) In accordance with the flux/density relationship of Eq. (.6.), the transport velocity of the electromagnetic energy is found to be: v = P c ẑ ɛ F = = c ẑ w ɛ F As expected, the energy of the forward-moving wave is being transported at a speed c along the positive z-direction. Similar results can be derived for the backward-moving solution that has F = 0 and G 0. The fields are now: E(z, t) = G(z + ct) H(z, t) = η ẑ G(z + ct)= η ẑ E(z, t) (..7) The Poynting vector becomes P=E H = c ẑ ɛ G and points in the negative z-direction, that is, the propagation direction. The energy transport velocity is v = c ẑ. Now, the vectors {E, H, ẑ} form a right-handed system, as shown. The ratio of E to H is still equal to η, provided we replace ẑ with ẑ: H(z, t)= ( ẑ) E(z, t) E(z, t)= η H(z, t) ( ẑ) η In the general case of Eq. (..), the E/H ratio does not remain constant. The Poynting vector and energy density consist of a part due to the forward wave and a part due to the backward one: P=E H = c ẑ ( ɛ F ɛ G ) w = ɛ E + μ H = ɛ F + ɛ G (..8) Example..: A source located at z = 0 generates an electric field E(0,t)= ˆx E 0 u(t), where u(t) is the unit-step function, and E 0, a constant. The field is launched towards the positive z-direction. Determine expressions for E(z, t) and H(z, t). Solution: For a forward-moving wave, we have E(z, t)= F(z ct)= F ( 0 c(t z/c) ), which implies that E(z, t) is completely determined by E(z, 0), or alternatively, by E(0, t): E(z, t)= E(z ct, 0)= E(0,t z/c) Using this property, we find for the electric and magnetic fields: E(z, t) = E(0,t z/c)= ˆx E 0 u(t z/c) H(z, t) = η ẑ E(z, t)= ŷ E 0 u(t z/c) η Because of the unit-step, the non-zero values of the fields are restricted to t z/c 0, or, z ct, that is, at time t the wavefront has propagated only up to position z = ct. The figure shows the expanding wavefronts at time t and t + Δt. Example..: Consider the following three examples of electric fields specified at t = 0, and describing forward or backward fields as indicated: E(z, 0)= ˆx E 0 cos(kz) (forward-moving) E(z, 0)= ŷ E 0 cos(kz) (backward-moving) E(z, 0)= ˆx E cos(k z)+ŷ E cos(k z) (forward-moving) where k, k,k are given wavenumbers (measured in units of radians/m.) Determine the corresponding fields E(z, t) and H(z, t). Solution: For the forward-moving cases, we replace z by z ct, and for the backward-moving case, by z + ct. We find in the three cases: E(z, t) = ˆx E 0 cos ( k(z ct) ) = ˆx E 0 cos(ωt kz) E(z, t) = ŷ E 0 cos ( k(z + ct) ) = ŷ E 0 cos(ωt + kz) E(z, t) = ˆx E cos(ω t k z)+ŷ E cos(ω t k z) where ω = kc, and ω = k c, ω = k c. The corresponding magnetic fields are: H(z, t) = η ẑ E(z, t)= ŷ E 0 η H(z, t) = η ẑ E(z, t)= ˆx E 0 η cos(ωt kz) (forward) cos(ωt + kz) (backward) H(z, t) = η ẑ E(z, t)= ŷ E η cos(ω t k z) ˆx E η cos(ω t k z)

4 4. Uniform Plane Waves.. Monochromatic Waves 43 The first two cases are single-frequency waves, and are discussed in more detail in the next section. The third case is a linear superposition of two waves with two different frequencies and polarizations.. Monochromatic Waves Uniform, single-frequency, plane waves propagating in a lossless medium are obtained as a special case of the previous section by assuming the harmonic time-dependence: E(x, y, z, t) = E(z)e jωt H(x, y, z, t) = H(z)e jωt (..) where E(z) and H(z) are transverse with respect to the z-direction. Maxwell s equations (..5), or those of the decoupled system (..0), may be solved very easily by replacing time derivatives by t jω. Then, Eqs. (..0) become the first-order differential equations (see also Problem.3): E ± (z) = jk E ± (z), where k = ω z c = ω μɛ (..) with solutions: E + (z) = E 0+ e jkz (forward) (..3) E (z) = E 0 e jkz (backward) where E 0± are arbitrary (complex-valued) constant vectors such that ẑ E 0± = 0. The corresponding magnetic fields are: H + (z) = η ẑ E +(z)= η (ẑ E 0+)e jkz = H 0+ e jkz H (z) = η ẑ E (z)= η (ẑ E 0 )e jkz = H 0 e jkz (..4) where we defined the constant amplitudes of the magnetic fields: H 0± =± η ẑ E 0± (..5) Inserting (..3) into (..9), we obtain the general solution for single-frequency waves, expressed as a superposition of forward and backward components: E(z) = E 0+ e jkz + E 0 e jkz H(z) = η ẑ [ E 0+ e jkz E 0 e jkz] (forward + backward waves) (..6) Setting E 0± = ˆx A ± +ŷ B ±, and noting that ẑ E 0± = ẑ (ˆx A ± +ŷ B ± )= ŷ A ± ˆx B ±, we may rewrite (..6) in terms of its cartesian components: E x (z)= A + e jkz + A e jkz, H y (z)= η E y (z)= B + e jkz + B e jkz [ A+ e jkz A e jkz], H x (z)= η [ B+ e jkz B e jkz] (..7) Wavefronts are defined, in general, to be the surfaces of constant phase. A forward moving wave E(z)= E 0 e jkz corresponds to the time-varying field: E(z, t)= E 0 e jωt jkz = E 0 e jϕ(z,t), where ϕ(z, t)= kz ωt A surface of constant phase is obtained by setting ϕ(z, t)= const. Denoting this constant by φ 0 = kz 0 and using the property c = ω/k, we obtain the condition: ϕ(z, t)= ϕ 0 kz ωt = kz 0 z = ct + z 0 Thus, the wavefront is the xy-plane intersecting the z-axis at the point z = ct + z 0, moving forward with velocity c. This justifies the term plane wave. A backward-moving wave will have planar wavefronts parametrized by z = ct +z 0, that is, moving backwards. A wave that is a linear combination of forward and backward components, may be thought of as having two planar wavefronts, one moving forward, and the other backward. The relationships (..5) imply that the vectors {E 0+, H 0+, ẑ} and {E 0, H 0, ẑ} will form right-handed orthogonal systems. The magnetic field H 0± is perpendicular to the electric field E 0± and the cross-product E 0± H 0± points towards the direction of propagation, that is, ±ẑ. Fig... depicts the case of a forward propagating wave. Fig... Forward uniform plane wave. The wavelength λ is the distance by which the phase of the sinusoidal wave changes by π radians. Since the propagation factor e jkz accumulates a phase of k radians per meter, we have by definition that kλ = π. The wavelength λ can be expressed via the frequency of the wave in Hertz, f = ω/π, as follows: λ = π k = πc ω = c f (..8) If the propagation medium is free space, we use the vacuum values of the parameters {ɛ, μ, c, η}, that is, {ɛ 0,μ 0,c 0,η 0 }. The free-space wavelength and corresponding wavenumber are: λ 0 = π = c 0 k 0 f, k 0 = ω (..9) c 0 In a lossless but non-magnetic (μ = μ 0 ) dielectric with refractive index n = ɛ/ɛ 0, the speed of light c, wavelength λ, and characteristic impedance η are all reduced by a

5 44. Uniform Plane Waves.3. Energy Density and Flux 45 scale factor n compared to the free-space values, whereas the wavenumber k is increased by a factor of n. Indeed, using the definitions c = / μ 0 ɛ and η = μ 0 /ɛ, we have: c = c 0 n, η = η 0 n, λ = λ 0 n, k = nk 0 (..0) Example..: A microwave transmitter operating at the carrier frequency of 6 GHz is protected by a Plexiglas radome whose permittivity is ɛ = 3ɛ 0. The refractive index of the radome is n = ɛ/ɛ 0 = 3 =.73. The free-space wavelength and the wavelength inside the radome material are: λ 0 = c 0 f = = 0.05 m = 5cm, λ = λ 0 9 n = 5 =.9 cm.73 We will see later that if the radome is to be transparent to the wave, its thickness must be chosen to be equal to one-half wavelength, l = λ/. Thus, l =.9/ =.45 cm. Example..: The nominal speed of light in vacuum is c 0 = m/s. Because of the relationship c 0 = λf, it may be expressed in the following suggestive units that are appropriate in different application contexts: c 0 = 5000 km 60 Hz (power systems) 300 m MHz (AM radio) 40 m 7.5 MHz (amateur radio) 3m 00 MHz (FM radio, TV) 30 cm GHz (cell phones) 0 cm 3 GHz (waveguides, radar) 3cm 0 GHz (radar, satellites).5 μm 00 THz (optical fibers) 500 nm 600 THz (visible spectrum) 00 nm 3000 THz (UV) Similarly, in terms of length/time of propagation: c 0 = km/0 msec (geosynchronous satellites) 300 km/msec (power lines) 300 m/μsec (transmission lines) 30 cm/nsec (circuit boards) The typical half-wave monopole antenna (half of a half-wave dipole over a ground plane) has length λ/4 and is used in many applications, such as AM, FM, and cell phones. Thus, one can predict that the lengths of AM radio, FM radio, and cell phone antennas will be of the order of 75 m, 0.75 m, and 7.5 cm, respectively. A more detailed list of electromagnetic frequency bands is given in Appendix B. The precise value of c 0 and the values of other physical constants are given in Appendix A. Wave propagation effects become important, and cannot be ignored, whenever the physical length of propagation is comparable to the wavelength λ. It follows from Eqs. (..) that the incremental change of a forward-moving electric field in propagating from z to z + Δz is: ΔE + E + = kδz = π Δz λ (..) Thus, the change in the electric field can be ignored only if Δz λ, otherwise, propagation effects must be taken into account. For example, for an integrated circuit operating at 0 GHz, we have λ = 3 cm, which is comparable to the physical dimensions of the circuit. Similarly, a cellular base station antenna is connected to the transmitter circuits by several meters of coaxial cable. For a -GHz system, the wavelength is 0.3 m, which implies that a 30-meter cable will be equivalent to 00 wavelengths..3 Energy Density and Flux The time-averaged energy density and flux of a uniform plane wave can be determined by Eq. (.9.6). As in the previous section, the energy is shared equally by the electric and magnetic fields (in the forward or backward cases.) This is a general result for most wave propagation and waveguide problems. The energy flux will be in the direction of propagation. For either a forward- or a backward-moving wave, we have from Eqs. (.9.6) and (..5): w e = [ ] Re ɛ E ±(z) E± (z) = [ ] Re ɛ E 0±e jkz E0±e jkz = 4 ɛ E 0± w m = Re [ μ H ±(z) H ± (z) ] = 4 μ H 0± = 4 μ η ẑ E 0± = 4 ɛ E 0± = w e Thus, the electric and magnetic energy densities are equal and the total density is: w = w e + w m = w e = ɛ E 0± (.3.) For the time-averaged Poynting vector, we have similarly: P= Re[ E ± (z) H ± (z)] = η Re[ E 0± (±ẑ E 0±) ] Using the BAC-CAB rule and the orthogonality property ẑ E 0± = 0, we find: P=±ẑ η E 0± =±cẑ ɛ E 0± (.3.) Thus, the energy flux is in the direction of propagation, that is, ±ẑ. The corresponding energy velocity is, as in the previous section: v = P =±cẑ (.3.3) w In the more general case of forward and backward waves, we find: w = 4 Re[ ɛ E(z) E (z)+μ H(z) H (z) ] = ɛ E 0+ + ɛ E 0 P= Re[ E(z) H (z) ] = ẑ ( ) η E 0+ (.3.4) η E 0 Thus, the total energy is the sum of the energies of the forward and backward components, whereas the net energy flux (to the right) is the difference between the forward and backward fluxes.

6 46. Uniform Plane Waves.4 Wave Impedance For forward or backward fields, the ratio of E(z) to H(z) ẑ is constant and equal to the characteristic impedance of the medium. Indeed, it follows from Eq. (..4) that E ± (z)= ±ηh ± (z) ẑ However, this property is not true for the more general solution given by Eqs. (..6). In general, the ratio of E(z) to H(z) ẑ is called the wave impedance. Because of the vectorial character of the fields, we must define the ratio in terms of the corresponding x- and y-components: Z x (z) = Z y (z) = [ E(z) ] x [ H(z) ẑ ] x = E x(z) H y (z) [ ] E(z) y [ ] = E y(z) H(z) ẑ y H x (z) Using the cartesian expressions of Eq. (..7), we find: Z x (z) = E x(z) H y (z) = η A +e jkz + A e jkz A + e jkz A e jkz Z y (z) = E y(z) H x (z) = η B +e jkz + B e jkz B + e jkz B e jkz (wave impedances) (.4.) (wave impedances) (.4.) Thus, the wave impedances are nontrivial functions of z. For forward waves (that is, with A = B = 0), we have Z x (z)= Z y (z)= η. For backward waves (A + = B + = 0), we have Z x (z)= Z y (z)= η. The wave impedance is a very useful concept in the subject of multiple dielectric interfaces and the matching of transmission lines. We will explore its use later on..5 Polarization Consider a forward-moving wave and let E 0 = ˆx A + + ŷ B + be its complex-valued phasor amplitude, so that E(z)= E 0 e jkz = (ˆx A + + ŷ B + )e jkz. The time-varying field is obtained by restoring the factor e jωt : E(z, t)= (ˆx A + + ŷ B + )e jωt jkz The polarization of a plane wave is defined to be the direction of the electric field. For example, if B + = 0, the E-field is along the x-direction and the wave will be linearly polarized. More precisely, polarization is the direction of the time-varying real-valued field E(z, t)= Re [ E(z, t)]. At any fixed point z, the vector E(z, t) may be along a fixed linear direction or it may be rotating as a function of t, tracing a circle or an ellipse..5. Polarization 47 The polarization properties of the plane wave are determined by the relative magnitudes and phases of the complex-valued constants A +,B +. Writing them in their polar forms A + = Ae jφa and B + = Be jφb, where A, B are positive magnitudes, we obtain: E(z, t)= (ˆx Ae jφa + ŷ Be jφb) e jωt jkz = ˆx Ae j(ωt kz+φa) + ŷ Be j(ωt kz+φb) (.5.) Extracting real parts and setting E(z, t)= Re [ E(z, t) ] = ˆx E x (z, t)+ŷ E y (z, t), we find the corresponding real-valued x, y components: E x (z, t) = A cos(ωt kz + φ a ) E y (z, t) = B cos(ωt kz + φ b ) (.5.) For a backward moving field, we replace k by k in the same expression. To determine the polarization of the wave, we consider the time-dependence of these fields at some fixed point along the z-axis, say at z = 0: E x (t) = A cos(ωt + φ a ) E y (t) = B cos(ωt + φ b ) (.5.3) The electric field vector E(t)= ˆx E x (t)+ŷ E y (t) will be rotating on the xy-plane with angular frequency ω, with its tip tracing, in general, an ellipse. To see this, we expand Eq. (.5.3) using a trigonometric identity: E x (t) = A [ ] cos ωt cos φ a sin ωt sin φ a E y (t) = B [ ] cos ωt cos φ b sin ωt sin φ b Solving for cos ωt and sin ωt in terms of E x (t), E y (t), we find: cos ωt sin φ = E y(t) B sin ωt sin φ = E y(t) B sin φ a E x(t) A sin φ b cos φ a E x(t) A cos φ b where we defined the relative phase angle φ = φ a φ b. Forming the sum of the squares of the two equations and using the trigonometric identity sin ωt + cos ωt =, we obtain a quadratic equation for the components E x and E y, which describes an ellipse on the E x, E y plane: ( Ey (t) B This simplifies into: sin φ a E ) x(t) A sin φ b + E x A ( Ey (t) B cos φ a E ) x(t) A cos φ b = sin φ + E y B cos φ E xe y AB = sin φ (polarization ellipse) (.5.4) Depending on the values of the three quantities {A, B, φ} this polarization ellipse may be an ellipse, a circle, or a straight line. The electric field is accordingly called elliptically, circularly, or linearly polarized.

7 48. Uniform Plane Waves.5. Polarization 49 To get linear polarization, we set φ = 0orφ = π, corresponding to φ a = φ b = 0, or φ a = 0,φ b = π, so that the phasor amplitudes are E 0 = ˆx A ± ŷ B. Then, Eq. (.5.4) degenerates into: E x A representing the straight lines: E y =± B A E x + E y B E xe y AB = 0 ( Ex A E ) y = 0 B and φ b = π/, we have complex amplitude E 0 = A(ˆx + jŷ). Then, Eq. (.5.3) becomes: E x (t) = A cos ωt E y (t) = A cos(ωt + π/)= A sin ωt The tip of the electric field vector rotates clockwise on the xy-plane. Since the wave is moving forward, this will represent left-circular polarization. Fig..5. depicts the four cases of left/right polarization with forward/backward waves. The fields (.5.) take the forms, in the two cases φ = 0 and φ = π: E x (t)= A cos ωt E y (t)= B cos ωt and E x (t)= A cos ωt E y (t)= B cos(ωt π)= B cos ωt To get circular polarization, we set A = B and φ =±π/. In this case, the polarization ellipse becomes the equation of a circle: E x A + E y A = The sense of rotation, in conjunction with the direction of propagation, defines leftcircular versus right-circular polarization. For the case, φ a = 0 and φ b = π/, we have φ = φ a φ b = π/ and complex amplitude E 0 = A(ˆx jŷ). Then, Fig..5. Left and right circular polarizations. E x (t) = A cos ωt E y (t) = A cos(ωt π/)= A sin ωt Thus, the tip of the electric field vector rotates counterclockwise on the xy-plane. To decide whether this represents right or left circular polarization, we use the IEEE convention [5], which is as follows. Curl the fingers of your left and right hands into a fist and point both thumbs towards the direction of propagation. If the fingers of your right (left) hand are curling in the direction of rotation of the electric field, then the polarization is right (left) polarized. Thus, in the present example, because we had a forward-moving field and the field is turning counterclockwise, the polarization will be right-circular. If the field were moving backwards, then it would be left-circular. For the case, φ = π/, arising from φ a = 0 To summarize, the electric field of a circularly polarized uniform plane wave will be, in its phasor form: E(z)= A(ˆx jŷ)e jkz (right-polarized, forward-moving) E(z)= A(ˆx + jŷ)e jkz (left-polarized, forward-moving) E(z)= A(ˆx jŷ)e jkz (left-polarized, backward-moving) E(z)= A(ˆx + jŷ)e jkz (right-polarized, backward-moving) If A B, but the phase difference is still φ =±π/, we get an ellipse with major and minor axes oriented along the x, y directions. Eq. (.5.4) will be now: E x A + E y B = Most engineering texts use the IEEE convention and most physics texts, the opposite convention. Finally, if A B and φ is arbitrary, then the major/minor axes of the ellipse (.5.4) will be rotated relative to the x, y directions. Fig..5. illustrates the general case.

8 50. Uniform Plane Waves.5. Polarization 5 Example.5.: Determine the real-valued electric and magnetic field components and the polarization of the following fields specified in their phasor form (given in units of V/m): Fig..5. General polarization ellipse. It can be shown (see Problem.5) that the tilt angle θ is given by: tan θ = AB cos φ (.5.5) A B The ellipse semi-axes A,B, that is, the lengths OC and OD, are given by: A = B = (A + B )+ s (A + B ) s (A B ) +4A B cos φ (A B ) +4A B cos φ (.5.6) where s = sign(a B). These results are obtained by defining the rotated coordinate system of the ellipse axes: E x =E x cos θ +E y sin θ E y =E (.5.7) y cos θ E x sin θ and showing that Eq. (.5.4) transforms into the standardized form: E x A + E y B = (.5.8) The polarization ellipse is bounded by the rectangle with sides at the end-points ±A, ±B, as shown in the figure. To decide whether the elliptic polarization is left- or right-handed, we may use the same rules depicted in Fig..5.. The angle χ subtended by the major to minor ellipse axes shown in Fig..5. is given as follows and is discussed further in Problem.5: sin χ = AB A + B sin φ, π 4 χ π 4 that is, it can be shown that tan χ = B /A or A /B, whichever is less than one. (.5.9) a. E(z)= 3j ˆx e jkz b. E(z)= ( 3 ˆx + 4 ŷ ) e +jkz c. E(z)= ( 4 ˆx + 3 ŷ ) e jkz d. E(z)= ( 3e jπ/3 ˆx + 3 ŷ ) e +jkz e. E(z)= ( 4 ˆx + 3e jπ/4 ŷ ) e jkz f. E(z)= ( 3e jπ/8 ˆx + 4e jπ/8 ŷ ) e +jkz g. E(z)= ( 4e jπ/4 ˆx + 3e jπ/ ŷ ) e jkz h. E(z)= ( 3e jπ/ ˆx + 4e jπ/4 ŷ ) e +jkz Solution: Restoring the e jωt factor and taking real-parts, we find the x, y electric field components, according to Eq. (.5.): a. E x (z, t)= 3 cos(ωt kz π/), E y (z, t)= 0 b. E x (z, t)= 3 cos(ωt + kz), E y (z, t)= 4 cos(ωt + kz) c. E x (z, t)= 4 cos(ωt kz + π), E y (z, t)= 3 cos(ωt kz) d. E x (z, t)= 3 cos(ωt + kz + π/3), E y (z, t)= 3 cos(ωt + kz) e. E x (z, t)= 4 cos(ωt kz), E y (z, t)= 3 cos(ωt kz π/4) f. E x (z, t)= 3 cos(ωt + kz π/8), E y (z, t)= 4 cos(ωt + kz + π/8) g. E x (z, t)= 4 cos(ωt kz + π/4), E y (z, t)= 3 cos(ωt kz π/) h. E x (z, t)= 3 cos(ωt + kz π/), E y (z, t)= 4 cos(ωt + kz + π/4) Since these are either forward or backward waves, the corresponding magnetic fields are obtained by using the formula H(z, t)= ±ẑ E(z, t)/η. This gives the x, y components: (cases a, c, e, g): H x (z, t)= η E y(z, t), H y (z, t)= η E x(z, t) (cases b, d, f, h): H x (z, t)= η E y(z, t), H y (z, t)= η E x(z, t) To determine the polarization vectors, we evaluate the electric fields at z = 0: a. E x (t)= 3 cos(ωt π/), E y (t)= 0 b. E x (t)= 3 cos(ωt), E y (t)= 4 cos(ωt) c. E x (t)= 4 cos(ωt + π), E y (t)= 3 cos(ωt) d. E x (t)= 3 cos(ωt + π/3), E y (t)= 3 cos(ωt) e. E x (t)= 4 cos(ωt), E y (t)= 3 cos(ωt π/4) f. E x (t)= 3 cos(ωt π/8), E y (t)= 4 cos(ωt + π/8) g. E x (t)= 4 cos(ωt + π/4), E y (t)= 3 cos(ωt π/) h. E x (t)= 3 cos(ωt π/), E y (t)= 4 cos(ωt + π/4) The polarization ellipse parameters A, B, and φ = φ a φ b, as well as the computed semi-major axes A,B, tilt angle θ, sense of rotation of the electric field, and polarization

9 5. Uniform Plane Waves.6. Uniform Plane Waves in Lossy Media 53 type are given below: case A B φ A B θ rotation polarization a o o linear/forward b o o linear/backward c o o linear/forward d o o left/backward e o o right/forward f o o right/backward g o o right/forward h o o right/backward In the linear case (b), the polarization ellipse collapses along its A -axis (A = 0) and becomes a straight line along its B -axis. The tilt angle θ still measures the angle of the A - axis from the x-axis. The actual direction of the electric field will be 90 o o = 53.3 o, which is equal to the slope angle, atan(b/a)= atan(4/3)= 53.3 o. In case (c), the ellipse collapses along its B -axis. Therefore, θ coincides with the angle of the slope of the electric field vector, that is, atan( B/A)= atan( 3/4)= o. With the understanding that θ always represents the slope of the A -axis (whether collapsed or not, major or minor), Eqs. (.5.5) and (.5.6) correctly calculate all the special cases, except when A = B, which has tilt angle and semi-axes: θ = 45 o, A = A + cos φ, B = A cos φ (.5.0) The MATLAB function ellipse.m calculates the ellipse semi-axes and tilt angle, A, B, θ, given the parameters A, B, φ. It has usage: [a,b,th] = ellipse(a,b,phi) % polarization ellipse parameters For example, the function will return the values of the A,B,θ columns of the previous example, if it is called with the inputs: A = [3, 3, 4, 3, 4, 3, 4, 3] ; B = [0, 4, 3, 3, 3, 4, 3, 4] ; phi = [-90, 0, 80, 60, 45, -45, 35, -35] ; To determine quickly the sense of rotation around the polarization ellipse, we use the rule that the rotation will be counterclockwise if the phase difference φ = φ a φ b is such that sin φ>0, and clockwise, if sin φ<0. This can be seen by considering the electric field at time t = 0 and at a neighboring time t. Using Eq. (.5.3), we have: E(0) = ˆx A cos φ a + ŷ B cos φ b E(t) = ˆx A cos(ωt + φ a )+ŷ B cos(ωt + φ b ) The sense of rotation may be determined from the cross-product E(0) E(t). If the rotation is counterclockwise, this vector will point towards the positive z-direction, and otherwise, it will point towards the negative z-direction. It follows easily that: E(0) E(t)= ẑ AB sin φ sin ωt (.5.) Thus, for t small and positive (such that sin ωt > 0), the direction of the vector E(0) E(t) is determined by the sign of sin φ..6 Uniform Plane Waves in Lossy Media We saw in Sec..4 that power losses may arise because of conduction and/or material polarization. A wave propagating in a lossy medium will set up a conduction current J cond = σe and a displacement (polarization) current J disp = jωd = jωɛ d E. Both currents will cause ohmic losses. The total current is the sum: J tot = J cond + J disp = (σ + jωɛ d )E = jωɛ c E where ɛ c is the effective complex dielectric constant introduced in Eq. (.4.): jωɛ c = σ + jωɛ d ɛ c = ɛ d j σ (.6.) ω The quantities σ,ɛ d may be complex-valued and frequency-dependent. However, we will assume that over the desired frequency band of interest, the conductivity σ is realvalued; the permittivity of the dielectric may be assumed to be complex, ɛ d = ɛ d jɛ d. Thus, the effective ɛ c has real and imaginary parts: ( ɛ c = ɛ jɛ = ɛ d j ɛ d + σ ) (.6.) ω Power losses arise from the non-zero imaginary part ɛ. We recall from Eq. (.4.5) that the time-averaged ohmic power losses per unit volume are given by: dp loss dv = Re[ J tot E ] = ωɛ E = (σ + ωɛ d ) E (.6.3) Uniform plane waves propagating in such lossy medium will satisfy Maxwell s equations (.9.), with the right-hand side of Ampère s law given by J tot = J + jωd = jωɛ c E. The assumption of uniformity ( x = y = 0), will imply again that the fields E, H are transverse to the direction ẑ. Then, Faraday s and Ampère s equations become: E = jωμh H = jωɛ c E ẑ z E = jωμh ẑ z H = jωɛ c E (.6.4) These may be written in a more convenient form by introducing the complex wavenumber k c and complex characteristic impedance η c defined by: k c = ω μɛ c, η c = μ ɛ c (.6.5) They correspond to the usual definitions k = ω/c = ω μɛ and η = μ/ɛ with the replacement ɛ ɛ c. Noting that ωμ = k c η c and ωɛ c = k c /η c, Eqs. (.6.4) may

10 54. Uniform Plane Waves.6. Uniform Plane Waves in Lossy Media 55 be written in the following form (using the orthogonality property ẑ E = 0 and the BAC-CAB rule on the first equation): [ ] [ ][ ] E 0 jkc E = (.6.6) z η c H ẑ jk c 0 η c H ẑ To decouple them, we introduce the forward and backward electric fields: E + = ( E + ηc H ẑ) E = E + + E E = ( E ηc H ẑ) H = ẑ [ ] E + E η c Then, Eqs. (.6.6) may be replaced by the equivalent system: [ ] [ ][ ] E+ jkc 0 E+ = z E 0 jk c E with solutions: (.6.7) (.6.8) E ± (z)= E 0± e jkcz, where ẑ E 0± = 0 (.6.9) Thus, the propagating electric and magnetic fields are linear combinations of forward and backward components: E(z) = E 0+ e jkcz + E 0 e jkcz H(z) = η c ẑ [ E 0+ e jkcz E 0 e jkcz] (.6.0) In particular, for a forward-moving wave we have: E(z)= E 0 e jkcz, H(z)= H 0 e jkcz, with ẑ E 0 = 0, H 0 = η c ẑ E 0 (.6.) Eqs. (.6.0) are the same as in the lossless case but with the replacements k k c and η η c. The lossless case is obtained in the limit of a purely real-valued ɛ c. Because k c is complex-valued, we define the phase and attenuation constants β and α as the real and imaginary parts of k c, that is, k c = β jα = ω μ(ɛ jɛ ) (.6.) We may also define a complex refractive index n c = k c /k 0 that measures k c relative to its free-space value k 0 = ω/c 0 = ω μ 0 ɛ 0. For a non-magnetic medium, we have: n c = k c ɛc ɛ jɛ = = n r jn i (.6.3) k 0 ɛ 0 ɛ 0 where n r,n i are the real and imaginary parts of n c. The quantity n i is called the extinction coefficient and n r, the refractive index. Another commonly used notation is the propagation constant γ defined by: γ = jk c = α + jβ (.6.4) It follows from γ = α + jβ = jk c = jk 0 n c = jk 0 (n r jn i ) that β = k 0 n r and α = k 0 n i. The nomenclature about phase and attenuation constants has its origins in the propagation factor e jkcz. We can write it in the alternative forms: e jkcz = e γz = e αz e jβz = e k0niz e jk0nrz (.6.5) Thus, the wave amplitudes attenuate exponentially with the factor e αz, and oscillate with the phase factor e jβz. The energy of the wave attenuates by the factor e αz,as can be seen by computing the Poynting vector. Because e jkcz is no longer a pure phase factor and η c is not real, we have for the forward-moving wave of Eq. (.6.): P(z) = Re[ E(z) H (z) ] [ ] = Re η E 0 (ẑ E 0 )e (α+jβ)z e (α jβ)z c = ẑ Re( ) η c E0 e αz = ẑ P(0)e αz = ẑ P(z) Thus, the power per unit area flowing past the point z in the forward z-direction will be: P(z)= P(0)e αz (.6.6) The quantity P(0) is the power per unit area flowing past the point z = 0. Denoting the real and imaginary parts of η c by η,η, so that, η c = η + jη, and noting that E 0 = η c H 0 ẑ = η c H 0, we may express P(0) in the equivalent forms: P(0)= Re( ) η c E0 = η H 0 (.6.7) The attenuation coefficient α is measured in nepers per meter. However, a more practical way of expressing the power attenuation is in db per meter. Taking logs of Eq. (.6.6), we have for the db attenuation at z, relative to z = 0: [ ] P(z) A db (z)= 0 log 0 = 0 log P(0) 0 (e)αz = αz (.6.8) where we used the numerical value 0 log 0 e = Thus, the quantity α db = α is the attenuation in db per meter: α db = α (db/m) (.6.9) Another way of expressing the power attenuation is by means of the so-called penetration or skin depth defined as the inverse of α: δ = α Then, Eq. (.6.8) can be rewritten in the form: A db (z)= z δ (skin depth) (.6.0) (attenuation in db) (.6.)

11 56. Uniform Plane Waves.6. Uniform Plane Waves in Lossy Media 57 This gives rise to the so-called 9-dB per delta rule, that is, every time z is increased by a distance δ, the attenuation increases by db. A useful way to represent Eq. (.6.6) in practice is to consider its infinitesimal version obtained by differentiating it with respect to z and solving for α : P (z)= αp(0)e αz = αp(z) α = P (z) P(z) The quantity P loss = P represents the power lost from the wave per unit length (in the propagation direction.) Thus, the attenuation coefficient is the ratio of the power loss per unit length to twice the power transmitted: α = P loss P transm (attenuation coefficient) (.6.) If there are several physical mechanisms for the power loss, then α becomes the sum over all possible cases. For example, in a waveguide or a coaxial cable filled with a slightly lossy dielectric, power will be lost because of the small conduction/polarization currents set up within the dielectric and also because of the ohmic losses in the walls of the guiding conductors, so that the total α will be α = α diel + α walls. Next, we verify that the exponential loss of power from the propagating wave is due to ohmic heat losses. In Fig..6., we consider a volume dv = lda of area da and length l along the z-direction. On the other hand, according to Eq. (.6.3), the ohmic power loss per unit volume will be ωɛ E(z) /. Integrating this quantity from z = 0toz = l will give the total ohmic losses within the volume lda of Fig..6.. Thus, we have: dp ohmic = l [ ] ωɛ E(z) dz da = l 0 ωɛ E 0 e αz dz da, or, 0 dp ohmic da = ωɛ 4α E 0 ( e αl) (.6.4) Are the two expressions in Eqs. (.6.3) and (.6.4) equal? The answer is yes, as follows from the following relationship among the quantities η c,ɛ,α (see Problem.7): Re ( ) η ωɛ c = (.6.5) α Thus, the power lost from the wave is entirely accounted for by the ohmic losses within the propagation medium. The equality of (.6.3) and (.6.4) is an example of the more general relationship proved in Problem.5. In the limit l, we have P(l) 0, so that dp ohmic /da =P(0), which states that all the power that enters at z = 0 will be dissipated into heat inside the semi-infinite medium. Using Eq. (.6.7), we summarize this case: dp ohmic da = Re( ) η c E0 = η H 0 (ohmic losses) (.6.6) This result will be used later on to calculate ohmic losses of waves incident on lossy dielectric or conductor surfaces, as well as conductor losses in waveguide and transmission line problems. Fig..6. Power flow in lossy dielectric. Example.6.: The absorption coefficient α of water reaches a minimum over the visible spectrum a fact undoubtedly responsible for why the visible spectrum is visible. Recent measurements [36] of the absorption coefficient show that it starts at about 0.0 nepers/m at 380 nm (violet), decreases to a minimum value of nepers/m at 48 nm (blue), and then increases steadily reaching the value of 0.5 nepers/m at 600 nm (red). Determine the penetration depth δ in meters, for each of the three wavelengths. Determine the depth in meters at which the light intensity has decreased to /0th its value at the surface of the water. Repeat, if the intensity is decreased to /00th its value. From the definition of P(z) as power flow per unit area, it follows that the power entering the area da at z = 0 will be dp in =P(0)dA, and the power leaving that area at z = l, dp out =P(l)dA. The difference dp loss = dp in dp out = [ P(0) P(l) ] da will be the power lost from the wave within the volume lda. Because P(l)= P(0)e αl,we have for the power loss per unit area: dp loss da =P(0) P(l)= P(0)( e αl) = Re( ) η c E0 ( e αl) (.6.3) Solution: The penetration depths δ = /α are: δ = 00, 7.3, m for α = 0.0, , 0.5 nepers/m Using Eq. (.6.), we may solve for the depth z = (A/8.9696)δ. Since a decrease of the light intensity (power) by a factor of 0 is equivalent to A = 0 db, we find z = (0/8.9696)δ =.8 δ, which gives: z =.8, 56.3,.3 m. A decrease by a factor of 00 = 0 0/0 corresponds to A = 0 db, effectively doubling the above depths.

12 58. Uniform Plane Waves.7. Propagation in Weakly Lossy Dielectrics 59 Example.6.: A microwave oven operating at.45 GHz is used to defrost a frozen food having complex permittivity ɛ c = (4 j)ɛ 0 farad/m. Determine the strength of the electric field at a depth of cm and express it in db and as a percentage of its value at the surface. Repeat if ɛ c = (45 5j)ɛ 0 farad/m. Solution: The free-space wavenumber is k 0 = ω μ 0 ɛ 0 = πf/c 0 = π( )/(3 0 8 )= 5.3 rad/m. Using k c = ω μ 0 ɛ c = k 0 ɛc /ɛ 0, we calculate the wavenumbers: k c = β jα = j = 5.3(.0 0.5j)= j m k c = β jα = j = 5.3(6.80.0j)= j m The corresponding attenuation constants and penetration depths are: α =.73 nepers/m, α = 56.6 nepers/m, δ = 7.86 cm δ =.77 cm It follows that the attenuations at cm will be in db and in absolute units: A = z/δ =. db, 0 A/0 = 0.88 A = z/δ = 4.9 db, 0 A/0 = 0.57 Thus, the fields at a depth of cm are 88% and 57% of their values at the surface. The complex permittivities of some foods may be found in [37]. A convenient way to characterize the degree of ohmic losses is by means of the loss tangent, originally defined in Eq. (.4.8). Here, we set: τ = tan θ = ɛ ɛ = σ + ωɛ d ωɛ d Then, ɛ c = ɛ jɛ = ɛ ( jτ)= ɛ d ( jτ). Therefore, k c,η c may be written as: μ k c = ω μɛ d ( jτ)/, η c = (.6.7) ɛ ( jτ) / (.6.8) d The quantities c d = / μɛ d and η d = μ/ɛ d would be the speed of light and characteristic impedance of an equivalent lossless dielectric with permittivity ɛ d. In terms of the loss tangent, we may characterize weakly lossy media versus strongly lossy ones by the conditions τ versus τ, respectively. These conditions depend on the operating frequency ω : σ + ωɛ d ωɛ d versus σ + ωɛ d ωɛ d The expressions (.6.8) may be simplified considerably in these two limits. Using the small-x Taylor series expansion (+x) / +x/, we find in the weakly lossy case ( jτ) / jτ/, and similarly, ( jτ) / + jτ/. On the other hand, if τ, we may approximate ( jτ) / ( jτ) / = e jπ/4 τ /, where we wrote ( j) / = (e jπ/ ) / = e jπ/4. Similarly, ( jτ) / e jπ/4 τ /. Thus, we summarize the two limits: j τ ( jτ) / =, if τ τ e jπ/4 τ / = ( j), if τ (.6.9) + j τ ( jτ) /, if τ = e jπ/4 τ / = ( + j) τ, if τ (.6.30).7 Propagation in Weakly Lossy Dielectrics In the weakly lossy case, the propagation parameters k c,η c become: ( k c = β jα = ω μɛ d j τ ) ( = ω μɛ d j σ + ) ωɛ d ωɛ d ( μ η c = η + jη = ɛ + j τ ) ( μ = d ɛ + j σ + ) (.7.) ωɛ d d ωɛ d Thus, the phase and attenuation constants are: β = ω μɛ d = ω, α = μ c d ɛ (σ + ωɛ d )= d η d(σ + ωɛ d ) (.7.) For a slightly conducting dielectric with ɛ d = 0 and a small conductivity σ, Eq. (.7.) implies that the attenuation coefficient α is frequency-independent in this limit. Example.7.: Seawater has σ = 4 Siemens/m and ɛ d = 8ɛ 0 (so that ɛ d = 8ɛ 0, ɛ d = 0.) Then, n d = ɛ d /ɛ 0 = 9, and c d = c 0 /n d = m/sec and η d = η 0 /n d = 377/9 = 4.89 Ω. The attenuation coefficient (.7.) will be: α = η dσ = = nepers/m α db = α = 78 db/m The corresponding skin depth is δ = /α =.9 cm. This result assumes that σ ωɛ d, which can be written in the form ω σ/ɛ d,orf f 0, where f 0 = σ/(πɛ d ). Here, we have f 0 = 888 MHz. For frequencies f f 0, we must use the exact equations (.6.8). For example, we find: f = khz, α db =.09 db/m, δ = 7.96 m f = MHz, α db = db/m, δ = 5.8 cm f = GHz, α db = db/m, δ =.9 cm Such extremely large attenuations explain why communication with submarines is impossible at high RF frequencies.

13 60. Uniform Plane Waves.8 Propagation in Good Conductors A conductor is characterized by a large value of its conductivity σ, while its dielectric constant may be assumed to be real-valued ɛ d = ɛ (typically equal to ɛ 0.) Thus, its complex permittivity and loss tangent will be: ɛ c = ɛ j σ ( ω = ɛ j σ ), ωɛ τ = σ ωɛ (.8.) A good conductor corresponds to the limit τ, or, σ ωɛ. Using the approximations of Eqs. (.6.9) and (.6.30), we find for the propagation parameters k c,η c : k c = β jα = ω τ μɛ η c = η + jη = Thus, the parameters β, α, δ are: ωμσ β = α = = πfμσ, ωμσ ( j)= μ ( + j)= ɛ τ ( j) ωμ ( + j) σ (.8.) δ = α = ωμσ = (.8.3) πfμσ where we replaced ω = πf. The complex characteristic impedance η c can be written in the form η c = R s ( + j), where R s is called the surface resistance and is given by the equivalent forms (where η = μ/ɛ ):.9. Propagation in Oblique Directions 6 J s = J(z)dz = σe 0 e γz dz = σ 0 0 γ E 0, or, J s = Z s E 0 (.8.5) where we defined the surface impedance Z s = γ/σ. In the good-conductor limit, Z s is equal to η c. Indeed, it follows from Eqs. (.8.3) and (.8.4) that: Z s = γ σ = α + jβ σ = α σ ( + j)= R s( + j)= η c Because H 0 ẑ = E 0 /η c, it follows that the surface current will be related to the magnetic field intensity at the surface of the conductor by: J s = H 0 ẑ = ˆn H 0 (.8.6) where ˆn = ẑ is the outward normal to the conductor. The meaning of J s is that it represents the current flowing in the direction of E 0 per unit length measured along the perpendicular direction to E 0, that is, the H 0 -direction. It has units of A/m. The total amount of ohmic losses per unit surface area of the conductor may be calculated from Eq. (.6.6), which reads in this case: dp ohmic da = R s H 0 = R s J s (ohmic loss per unit conductor area) (.8.7) ωɛ ωμ R s = η σ = σ = α σ = σδ (.8.4).9 Propagation in Oblique Directions Example.8.: For copper we have σ = Siemens/m. The skin depth at frequency f is: δ = = πfμσ π 4π f / = f / 7 We find at frequencies of khz, MHz, and GHz: f = khz, f = MHz, f = GHz, δ =.09 mm δ = 0.07 mm δ =.09 μm Thus, the skin depth is extremely small for good conductors at RF. ( f in Hz) Because δ is so small, the fields will attenuate rapidly within the conductor, depending on distance like e γz = e αz e jβz = e z/δ e jβz. The factor e z/δ effectively confines the fields to within a distance δ from the surface of the conductor. This allows us to define equivalent surface quantities, such as surface current and surface impedance. With reference to Fig..6., we define the surface current density by integrating the density J(z)= σe(z)= σe 0 e γz over the top-side of the volume lda, and taking the limit l : So far we considered waves propagating towards the z-direction. For single-frequency uniform plane waves propagating in some arbitrary direction in a lossless medium, the propagation factor is obtained by the substitution: e jkz j k r e where k = kˆk, with k = ω μɛ = ω/c and ˆk is a unit vector in the direction of propagation. The fields take the form: jωt j k r E(r,t)= E 0 e H(r,t)= H 0 e jωt j k r (.9.) where E 0, H 0 are constant vectors transverse to ˆk, that is, ˆk E 0 = ˆk H 0 = 0, such that: H 0 = ωμ k E 0 = η ˆk E 0 (.9.) where η = μ/ɛ. Thus, {E, H, ˆk} form a right-handed orthogonal system. The solutions (.9.) can be derived from Maxwell s equations in a straightforward fashion. When the gradient operator acts on the above fields, it can be simplified into jk. This follows from:

14 6. Uniform Plane Waves.9. Propagation in Oblique Directions 63 ( e j k r) = jk ( j e k r) After canceling the common factor e jωt j k r, Maxwell s equations (..) take the form: jk E 0 = jωμh 0 k E 0 = ωμh 0 jk H 0 = jωɛe 0 k E 0 = 0 k H 0 = ωɛe 0 k E 0 = 0 (.9.3) k H 0 = 0 k H 0 = 0 The last two imply that E 0, H 0 are transverse to k. The other two can be decoupled by taking the cross product of the first equation with k and using the second equation: Fig..9. TM and TE waves. k (k E 0 )= ωμ k H 0 = ω μɛ E 0 (.9.4) The left-hand side can be simplified using the BAC-CAB rule and k E 0 = 0, that is, k (k E 0 )= k(k E 0 ) E 0 (k k)= (k k)e 0. Thus, Eq. (.9.4) becomes: Thus, we obtain the consistency condition: (k k)e 0 = ω μɛ E 0 k k = ω μɛ (.9.5) e j k r = e j(kzz+kxx) = e jk(z cos θ+x sin θ) = e jkz Because {E 0, H 0, k } form a right-handed vector system, the electric field may have components along the new transverse (with respect to z ) axes, that is, along x and y. Thus, we may resolve E 0 into the orthogonal directions: E 0 = ˆx A + ŷ B = (ˆx cos θ ẑ sin θ)a + ŷ B (.9.7) The corresponding magnetic field will be H 0 = ˆk E 0 /η = ẑ (ˆx A+ŷ B)/η. Using the relationships ẑ ˆx = ŷ and ẑ ŷ = ˆx, we find: Defining k = k k = k, we have k = ω μɛ. Using the relationship ωμ = kη and defining the unit vector ˆk = k/ k =k/k, the magnetic field is obtained from: H 0 = [ŷ A ˆx B ] = [ŷ ] A (ˆx cos θ ẑ sin θ)b η η (.9.8) H 0 = k E 0 ωμ = k E 0 kη = η ˆk E 0 The constant-phase (and constant-amplitude) wavefronts are the planes k r = constant, or, ˆk r = constant. They are the planes perpendicular to the propagation direction ˆk. As an example, consider a rotated coordinate system {x,y,z } in which the z x axes are rotated by angle θ relative to the original zx axes, as shown in Fig..9.. Thus, the new coordinates and corresponding unit vectors will be: z = z cos θ + x sin θ, x = x cos θ z sin θ, y = y, ẑ = ẑ cos θ + ˆx sin θ ˆx = ˆx cos θ ẑ sin θ ŷ = ŷ (.9.6) We choose the propagation direction to be the new z-axis, that is, ˆk = ẑ, so that the wave vector k = k ˆk = k ẑ will have components k z = k cos θ and k x = k sin θ : The propagation phase factor becomes: k = k ˆk = k(ẑ cos θ + ˆx sin θ)= ẑ k z + ˆx k x The complete expressions for the fields are then: E(r,t)= [ (ˆx cos θ ẑ sin θ)a + ŷ B ] jωt jk(z cos θ+x sin θ) e H(r,t)= η [ŷ A (ˆx cos θ ẑ sin θ)b ] e jωt jk(z cos θ+x sin θ) (.9.9) Written with respect to the rotated coordinate system {x,y,z }, the solutions become identical to those of Sec..: E(r,t)= [ˆx A + ŷ B ] e jωt jkz H(r,t)= [ŷ A ˆx B ] (.9.0) e jωt jkz η They are uniform in the sense that they do not depend on the new transverse coordinates x,y. The constant-phase planes are z = ẑ r = z cos θ + x sin θ = const. The polarization properties of the wave depend on the relative phases and amplitudes of the complex constants A, B, with the polarization ellipse lying on the x y plane. The A- and B-components of E 0 are referred to as transverse magnetic (TM) and transverse electric (TE), respectively, where transverse is meant here with respect to

15 64. Uniform Plane Waves.0. Complex or Inhomogeneous Waves 65 the z-axis. The TE case has an electric field transverse to z; the TM case has a magnetic field transverse to z. Fig..9. depicts these two cases separately. This nomenclature arises in the context of plane waves incident obliquely on interfaces, where the xz plane is the plane of incidence and the interface is the xy plane. The TE and TM cases are also referred to as having perpendicular and parallel polarization vectors with respect to the plane of incidence, that is, the E-field is perpendicular or parallel to the xz plane. We may define the concept of transverse impedance as the ratio of the transverse (with respect to z) components of the electric and magnetic fields. In particular, by analogy with the definitions of Sec..4, we have: η TM = E x = A cos θ = η cos θ H y η A η TE = E y H x = B = η B cos θ η cos θ (.9.) Such transverse impedances play an important role in describing the transfer matrices of dielectric slabs at oblique incidence. We discuss them further in Chap Complex or Inhomogeneous Waves The steps leading to the wave solution (.9.) do not preclude a complex-valued wavevector k. For example, if the medium is lossy, we must replace {η, k} by {η c,k c }, where k c = β jα, resulting from a complex effective permittivity ɛ c. If the propagation direction is defined by the unit vector ˆk, chosen to be a rotated version of ẑ, then the wavevector will be defined by k = k c ˆk = (β jα)ˆk. Because k c = ω μɛ c and ˆk ˆk =, the vector k satisfies the consistency condition (.9.5): The propagation factor will be: k k = k c = ω μɛ c (.0.) e j k r = e jkc ˆk r = e (α+jβ) ˆk r = e α ˆk r e jβ ˆk r The wave is still a uniform plane wave in the sense that the constant-amplitude planes, α ˆk r = const., and the constant-phase planes, β ˆk r = const., coincide with each other being the planes perpendicular to the propagation direction. For example, the rotated solution (.9.0) becomes in the lossy case: E(r,t)= [ˆx A + ŷ B ] e jωt jkcz = [ˆx A + ŷ B ] e jωt (α+jβ)z H(r,t)= η c [ŷ A ˆx B ] e jωt jkcz = η c [ŷ A ˆx B ] e jωt (α+jβ)z (.0.) In this solution, the real and imaginary parts of the wavevector k = β jα are collinear, that is, β = β ˆk and α = α ˆk. More generally, there exist solutions having a complex wavevector k = β jα such that β, α are not collinear. The propagation factor becomes now: e j k r = e (α+jβ) r = e α r e jβ r (.0.3) If α, β are not collinear, such a wave will not be a uniform plane wave because the constant-amplitude planes, α r = const., and the constant-phase planes, β r = const., will be different. The consistency condition k k = k c = (β jα) splits into the following two conditions obtained by equating real and imaginary parts: (β jα) (β jα)= (β jα) β β α α = β α β α = αβ (.0.4) With E 0 chosen to satisfy k E 0 = (β jα) E 0 = 0, the magnetic field is computed from Eq. (.9.), H 0 = k E 0 /ωμ = (β jα) E 0 /ωμ. Let us look at an explicit construction. We choose β, α to lie on the xz plane of Fig..9., and resolve them as β = ẑ β z + ˆx β x and α = ẑ α z + ˆx α x. Thus, k = β jα = ẑ (β z jα z )+ ˆx (β x jα x )= ẑ k z + ˆx k x Then, the propagation factor (.0.3) and conditions (.0.4) read explicitly: e j k r = e (αzz+αxx) e j(βzz+βxx) β z + β x α z α x = β α β z α z + β x α x = βα (.0.5) Because k is orthogonal to both ŷ and ŷ k, we construct the electric field E 0 as the following linear combination of TM and TE terms: E 0 = (ŷ ˆk)A + ŷ B, where ˆk = k = β jα (.0.6) k c β jα This satisfies k E 0 = 0. Then, the magnetic field becomes: H 0 = k E 0 ωμ = η c [ŷ A (ŷ ˆk)B ] (.0.7) The vector ˆk is complex-valued and satisfies ˆk ˆk =. These expressions reduce to Eq. (.0.), if ˆk = ẑ. Waves with a complex k are known as complex waves, or inhomogeneous waves. In applications, they always appear in connection with some interface between two media. The interface serves either as a reflecting/transmitting surface, or as a guiding surface. For example, when plane waves are incident obliquely from a lossless dielectric onto a planar interface with a lossy medium, the waves transmitted into the lossy medium are of such complex type. Taking the interface to be the xy-plane and the lossy medium to be the region z 0, it turns out that the transmitted waves are characterized by attenuation only in the z-direction. Therefore, Eqs. (.0.5) apply with α z > 0 and α x = 0. The parameter β x is fixed by Snel s law, so that Eqs. (.0.5) provide a system of two equations in the two unknowns β z and α z. We discuss this further in Chap. 7.