Plane Wave Propagation Concepts Unbounded EM Waves Unbounded EM waves and Guided EM Waves. Plane wave: Constant phase front on a planar surface.Uniform: Electric field (E) and Magnetic field (H) are not functions of x and y (E , H ≠ f ( x , y ) E, H \neq f(x, y) E , H = f ( x , y ) Maxwell's Equations - General Maxwell's contribution to electromagnetism:Gauss's law for electric field: ∇ ⋅ D = ρ v \nabla \cdot D = \rho_v ∇ ⋅ D = ρ v Gauss's law for magnetic field: ∇ ⋅ B = 0 \nabla \cdot B = 0 ∇ ⋅ B = 0 Faraday's law of induction: ∇ × E = − ∂ B ∂ t \nabla \times E = -\frac{\partial B}{\partial t} ∇ × E = − ∂ t ∂ B Ampere-Maxwell's law: ∇ × H = J + ∂ D ∂ t \nabla \times H = J + \frac{\partial D}{\partial t} ∇ × H = J + ∂ t ∂ D Where: D = ϵ E D = \epsilon E D = ϵ E B = μ H B = \mu H B = μ H J J J ρ v \rho_v ρ v Continuity equation:∇ ⋅ J = − ∂ ρ v ∂ t \nabla \cdot J = -\frac{\partial \rho_v}{\partial t} ∇ ⋅ J = − ∂ t ∂ ρ v For a conducting medium, J = σ E J = \sigma E J = σ E σ \sigma σ Therefore, ∇ ⋅ E = ρ v ϵ \nabla \cdot E = \frac{\rho_v}{\epsilon} ∇ ⋅ E = ϵ ρ v Concept of Time-Harmonic Fields Time-harmonic fields assume a sinusoidal time dependence:e j ω t e^{j\omega t} e j ω t ω \omega ω j j j This implies that ∂ ∂ t ⇒ j ω \frac{\partial}{\partial t} \Rightarrow j\omega ∂ t ∂ ⇒ j ω Complex Permittivity Complex permittivity ϵ c \epsilon_c ϵ c ϵ c = ϵ ′ − j ϵ ′ ′ \epsilon_c = \epsilon' - j\epsilon'' ϵ c = ϵ ′ − j ϵ ′′ ϵ ′ \epsilon' ϵ ′ ϵ ′ ′ \epsilon'' ϵ ′′ Loss tangent (tan δ c \delta_c δ c t a n δ c = ϵ ′ ′ ϵ ′ = σ ω ϵ tan \delta_c = \frac{\epsilon''}{\epsilon'} = \frac{\sigma}{\omega \epsilon} t an δ c = ϵ ′ ϵ ′′ = ω ϵ σ σ \sigma σ Time Harmonic Waves - Source Free Region (Radiated Fields) In a source-free region (ρ v = 0 \rho_v = 0 ρ v = 0 E ( x , y , z ) = x ^ E < e m > x ( x , y , z ) + y ^ E < / e m > y ( x , y , z ) + z ^ E z ( x , y , z ) E(x, y, z) = \hat{x}E<em>x(x, y, z) + \hat{y}E</em>y(x, y, z) + \hat{z}E_z(x, y, z) E ( x , y , z ) = x ^ E < e m > x ( x , y , z ) + y ^ E < / e m > y ( x , y , z ) + z ^ E z ( x , y , z ) Wave Equations – Source Free Region Separation of variables leads to three Scalar Helmholtz equations for the electric field components E < e m > x , E < / e m > y , E z E<em>x, E</em>y, E_z E < e m > x , E < / e m > y , E z Propagation constant γ \gamma γ γ = α + j β = j ω μ ( σ + j ω ϵ ) \gamma = \alpha + j\beta = \sqrt{j\omega \mu(\sigma + j\omega \epsilon)} γ = α + j β = j ω μ ( σ + j ω ϵ ) α \alpha α β \beta β If the medium is non-conducting (σ = 0 \sigma = 0 σ = 0 Plane Wave Direction Direction of wave propagation:Cartesian coordinates: + z ^ +\hat{z} + z ^ Spherical coordinates: + R ^ +\hat{R} + R ^ For a uniform plane wave:∂ E < e m > x ∂ x = ∂ E < / e m > x ∂ y = 0 \frac{\partial E<em>x}{\partial x} = \frac{\partial E</em>x}{\partial y} = 0 ∂ x ∂ E < e m > x = ∂ y ∂ E < / e m > x = 0 ∂ H < e m > x ∂ x = ∂ H < / e m > x ∂ y = 0 \frac{\partial H<em>x}{\partial x} = \frac{\partial H</em>x}{\partial y} = 0 ∂ x ∂ H < e m > x = ∂ y ∂ H < / e m > x = 0 General form of the solution:E ~ ( z ) = x ^ E < e m > x 0 + e − j k z + y ^ E < / e m > y 0 + e − j k z \tilde{E}(z) = \hat{x}E<em>{x0}^+e^{-jkz} + \hat{y}E</em>{y0}^+e^{-jkz} E ~ ( z ) = x ^ E < e m > x 0 + e − j k z + y ^ E < / e m > y 0 + e − j k z Intrinsic Impedance Intrinsic impedance η \eta η η = μ ϵ ≈ 120 π ≈ 377 Ω \eta = \sqrt{\frac{\mu}{\epsilon}} \approx 120\pi \approx 377 \Omega η = ϵ μ ≈ 120 π ≈ 377Ω Directional Relation Between E and H For any Transverse ElectroMagnetic (TEM) wave:∇ × E ~ = − j ω μ H ~ \nabla \times \tilde{E} = -j\omega \mu \tilde{H} ∇ × E ~ = − j ω μ H ~ ∇ × H ~ = j ω ϵ c E ~ \nabla \times \tilde{H} = j\omega \epsilon_c \tilde{E} ∇ × H ~ = j ω ϵ c E ~ H ~ = 1 η k ^ × E ~ \tilde{H} = \frac{1}{\eta} \hat{k} \times \tilde{E} H ~ = η 1 k ^ × E ~ E ~ = − η k ^ × H ~ \tilde{E} = -\eta \hat{k} \times \tilde{H} E ~ = − η k ^ × H ~ Right-hand rule: Point the fingers of the right hand from E to H, and the thumb points in the direction of wave travel, k. E and H Fields Electric and magnetic fields for a wave traveling in the +z direction:E ~ ( z ) = x ^ E < e m > x 0 + e − j k z + y ^ E < / e m > y 0 + e − j k z \tilde{E}(z) = \hat{x}E<em>{x0}^+e^{-jkz} + \hat{y}E</em>{y0}^+e^{-jkz} E ~ ( z ) = x ^ E < e m > x 0 + e − j k z + y ^ E < / e m > y 0 + e − j k z H ~ ( z ) = 1 η ( − x ^ E < e m > y 0 + e − j k z + y ^ E < / e m > x 0 + e − j k z ) \tilde{H}(z) = \frac{1}{\eta}(-\hat{x}E<em>{y0}^+e^{-jkz} + \hat{y}E</em>{x0}^+e^{-jkz}) H ~ ( z ) = η 1 ( − x ^ E < e m > y 0 + e − j k z + y ^ E < / e m > x 0 + e − j k z ) Wave Decomposition A uniform plane wave traveling in the +z-direction with both x- and y-components:E ~ = x ^ E ~ < e m > x + ( z ) + y ^ E ~ < / e m > y + ( z ) \tilde{E} = \hat{x}\tilde{E}<em>x^+(z) + \hat{y}\tilde{E}</em>y^+(z) E ~ = x ^ E ~ < e m > x + ( z ) + y ^ E ~ < / e m > y + ( z ) H ~ = x ^ H ~ < e m > x + ( z ) + y ^ H ~ < / e m > y + ( z ) \tilde{H} = \hat{x}\tilde{H}<em>x^+(z) + \hat{y}\tilde{H}</em>y^+(z) H ~ = x ^ H ~ < e m > x + ( z ) + y ^ H ~ < / e m > y + ( z ) H ~ = 1 η ( z ^ × E ~ ) \tilde{H} = \frac{1}{\eta}(\hat{z} \times \tilde{E}) H ~ = η 1 ( z ^ × E ~ ) H ~ < e m > x + ( z ) = − E ~ < / e m > y + ( z ) η \tilde{H}<em>x^+(z) = -\frac{\tilde{E}</em>y^+(z)}{\eta} H ~ < e m > x + ( z ) = − η E ~ < / e m > y + ( z ) H ~ < e m > y + ( z ) = E ~ < / e m > x + ( z ) η \tilde{H}<em>y^+(z) = \frac{\tilde{E}</em>x^+(z)}{\eta} H ~ < e m > y + ( z ) = η E ~ < / e m > x + ( z ) Wave's Phase Velocity Phase velocity (v p v_p v p v p = ω k = 1 μ ϵ v_p = \frac{\omega}{k} = \frac{1}{\sqrt{\mu\epsilon}} v p = k ω = μ ϵ 1 Wavelength (λ \lambda λ λ = 2 π k = v p f \lambda = \frac{2\pi}{k} = \frac{v_p}{f} λ = k 2 π = f v p In vacuum:v < e m > p = c = 1 μ < / e m > 0 ϵ 0 = 3 × 10 8 v<em>p = c = \frac{1}{\sqrt{\mu</em>0\epsilon_0}} = 3 \times 10^8 v < e m > p = c = μ < / e m > 0 ϵ 0 1 = 3 × 1 0 8 η = η < e m > 0 = μ < / e m > 0 ϵ 0 = 377 Ω ≈ 120 π \eta = \eta<em>0 = \sqrt{\frac{\mu</em>0}{\epsilon_0}} = 377 \Omega \approx 120\pi η = η < e m > 0 = ϵ 0 μ < / e m > 0 = 377Ω ≈ 120 π Example 7-1: EM Plane Wave in Air Given: 1-MHz plane wave traveling in the +z-direction in air, polarized along the x-direction.Peak value of the electric field = 1.2 π 1.2\pi 1.2 π Wavelength in air:λ = 3 × 10 8 1 × 10 6 = 300 \lambda = \frac{3 \times 10^8}{1 \times 10^6} = 300 λ = 1 × 1 0 6 3 × 1 0 8 = 300 Wavenumber:k = 2 π 300 k = \frac{2\pi}{300} k = 300 2 π General expression for the electric field:E ( z , t ) = x ^ ∣ E x 0 ∣ c o s ( ω t − k z + ϕ ) E(z, t) = \hat{x} |E_{x0}| cos(\omega t - kz + \phi) E ( z , t ) = x ^ ∣ E x 0 ∣ cos ( ω t − k z + ϕ ) E ( z , t ) = x ^ 1.2 π c o s ( 2 π × 10 6 t − 2 π 300 z + ϕ ) E(z, t) = \hat{x} 1.2\pi cos(2\pi \times 10^6 t - \frac{2\pi}{300}z + \phi) E ( z , t ) = x ^ 1.2 π cos ( 2 π × 1 0 6 t − 300 2 π z + ϕ ) At t = 0 and z = 50 m:2 π × 50 300 + ϕ = 0 \frac{2\pi \times 50}{300} + \phi = 0 300 2 π × 50 + ϕ = 0 π + ϕ = 0 \pi + \phi = 0 π + ϕ = 0 Example 7-1 Continued Phase constant:ϕ = − π 3 \phi = -\frac{\pi}{3} ϕ = − 3 π Electric field expression:E ( z , t ) = x ^ 1.2 π c o s ( 2 π × 10 6 t − 2 π 300 z − π 3 ) E(z, t) = \hat{x} 1.2\pi cos(2\pi \times 10^6 t - \frac{2\pi}{300}z - \frac{\pi}{3}) E ( z , t ) = x ^ 1.2 π cos ( 2 π × 1 0 6 t − 300 2 π z − 3 π ) Magnetic field expression:H ( z , t ) = y ^ E ( z , t ) η 0 = y ^ 1.2 π 377 c o s ( 2 π × 10 6 t − 2 π 300 z − π 3 ) H(z, t) = \hat{y} \frac{E(z, t)}{\eta_0} = \hat{y} \frac{1.2\pi}{377} cos(2\pi \times 10^6 t - \frac{2\pi}{300}z - \frac{\pi}{3}) H ( z , t ) = y ^ η 0 E ( z , t ) = y ^ 377 1.2 π cos ( 2 π × 1 0 6 t − 300 2 π z − 3 π ) H ( z , t ) = y ^ 10 c o s ( 2 π × 10 6 t − 2 π 300 z − π 3 ) H(z, t) = \hat{y} 10 cos(2\pi \times 10^6 t - \frac{2\pi}{300}z - \frac{\pi}{3}) H ( z , t ) = y ^ 10 cos ( 2 π × 1 0 6 t − 300 2 π z − 3 π ) CD Module 7.1 Plane Wave Observe a plane wave propagating along the z-direction; note the temporal and spatial variations of E and H, and examine how the wave properties change as a function of the values selected for the wave parameters-frequency and E-field amplitude and phase-and the medium's constitutive parameters (ϵ , μ , σ \epsilon, \mu, \sigma ϵ , μ , σ 2012 Test 1 Q3 The electric field of a uniform plane wave propagating in a homogeneous medium is given byE ~ = ( y ^ − j z ^ ) E o e − j 0.4 x ( V / m ) \tilde{E}=(\hat{y}-j\hat{z})E_oe^{-j0.4x} (V/m) E ~ = ( y ^ − j z ^ ) E o e − j 0.4 x ( V / m ) The phase velocity of the wave is v < e m > p = 1.5 × 10 8 m / s v<em>p = 1.5 \times 10^8 m/s v < e m > p = 1.5 × 1 0 8 m / s μ < / e m > r = 2.4 \mu</em>r = 2.4 μ < / e m > r = 2.4 2012 Test 1 Q3 Solution Wavelength:k = 0.4 = 2 π λ ⇒ λ = 2 π 0.4 = 5 π = 15.7 m k = 0.4 = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2\pi}{0.4} = 5\pi = 15.7 m k = 0.4 = λ 2 π ⇒ λ = 0.4 2 π = 5 π = 15.7 m Frequency:v < e m > p = f λ ⇒ f = v < / e m > p λ = 1.5 × 10 8 15.7 = 9.55 M H z v<em>p = f\lambda \Rightarrow f = \frac{v</em>p}{\lambda} = \frac{1.5 \times 10^8}{15.7} = 9.55 MHz v < e m > p = f λ ⇒ f = λ v < / e m > p = 15.7 1.5 × 1 0 8 = 9.55 M H z Relative permittivity:v < e m > p = c μ < / e m > r ϵ < e m > r ⇒ ϵ < / e m > r = c 2 v < e m > p 2 μ < / e m > r = ( 3 × 10 8 1.5 × 10 8 ) 2 1 2.4 = 4 2.4 = 1.67 v<em>p = \frac{c}{\sqrt{\mu</em>r\epsilon<em>r}} \Rightarrow \epsilon</em>r = \frac{c^2}{v<em>p^2\mu</em>r} = (\frac{3 \times 10^8}{1.5 \times 10^8})^2 \frac{1}{2.4} = \frac{4}{2.4} = 1.67 v < e m > p = μ < / e m > r ϵ < e m > r c ⇒ ϵ < / e m > r = v < e m > p 2 μ < / e m > r c 2 = ( 1.5 × 1 0 8 3 × 1 0 8 ) 2 2.4 1 = 2.4 4 = 1.67 2012 Test 1 Q3 Continued Intrinsic impedance:η = μ < e m > r ϵ < / e m > r η o \eta = \sqrt{\frac{\mu<em>r}{\epsilon</em>r}} \eta_o η = ϵ < / e m > r μ < e m > r η o η = 2.4 1.67 120 π = 451.94 \eta = \sqrt{\frac{2.4}{1.67}}120\pi = 451.94 η = 1.67 2.4 120 π = 451.94 Phasor magnetic field:H ~ = 1 η k ^ × E ~ = 1 451.94 ( − y ^ − j z ^ ) E o e − j 0.4 x \tilde{H} = \frac{1}{\eta} \hat{k} \times \tilde{E} = \frac{1}{451.94} (-\hat{y} - j\hat{z}) E_o e^{-j0.4x} H ~ = η 1 k ^ × E ~ = 451.94 1 ( − y ^ − j z ^ ) E o e − j 0.4 x H ~ = E o 451.94 ( y ^ + j z ^ ) e − j 0.4 x \tilde{H} = \frac{E_o}{451.94} (\hat{y} + j\hat{z}) e^{-j0.4x} H ~ = 451.94 E o ( y ^ + j z ^ ) e − j 0.4 x 2012 Test 1 Q3 - Time-Domain Expression Time-domain expressions for E and H:E ( x , t ) = y ^ E < e m > o c o s ( ω t − 0.4 x ) + z ^ E < / e m > o s i n ( ω t − 0.4 x ) E(x, t) = \hat{y}E<em>o cos(\omega t - 0.4x) + \hat{z}E</em>o sin(\omega t - 0.4x) E ( x , t ) = y ^ E < e m > ocos ( ω t − 0.4 x ) + z ^ E < / e m > os in ( ω t − 0.4 x ) H ( x , t ) = y ^ E < e m > o 451.94 c o s ( ω t − 0.4 x ) − z ^ E < / e m > o 451.94 s i n ( ω t − 0.4 x ) H(x, t) = \hat{y}\frac{E<em>o}{451.94} cos(\omega t - 0.4x) - \hat{z}\frac{E</em>o}{451.94} sin(\omega t - 0.4x) H ( x , t ) = y ^ 451.94 E < e m > o cos ( ω t − 0.4 x ) − z ^ 451.94 E < / e m > o s in ( ω t − 0.4 x ) General Wave Propagation General form for a wave propagating in an arbitrary direction:E ~ = ( x ^ E < e m > x + y ^ E < / e m > y ) e − j k ⃗ ⋅ r ⃗ \tilde{E} = (\hat{x}E<em>x + \hat{y}E</em>y) e^{-j\vec{k} \cdot \vec{r}} E ~ = ( x ^ E < e m > x + y ^ E < / e m > y ) e − j k ⋅ r k ⃗ = k ( c o s ϕ x ^ − s i n ϕ y ^ ) \vec{k} = k(cos\phi \hat{x} - sin\phi \hat{y}) k = k ( cos ϕ x ^ − s in ϕ y ^ ) r ⃗ = x x ^ + y y ^ + z z ^ \vec{r} = x\hat{x} + y\hat{y} + z\hat{z} r = x x ^ + y y ^ + z z ^ E ~ = ( x ^ c o s ϕ − y ^ s i n ϕ ) E o e − j k ( x c o s ϕ − y s i n ϕ ) \tilde{E} = (\hat{x} cos\phi - \hat{y} sin\phi) E_o e^{-jk(x cos\phi - y sin\phi)} E ~ = ( x ^ cos ϕ − y ^ s in ϕ ) E o e − j k ( x cos ϕ − y s in ϕ ) H ~ = z ^ E o η e − j k ( x c o s ϕ − y s i n ϕ ) \tilde{H} = \hat{z} \frac{E_o}{\eta} e^{-jk(x cos\phi - y sin\phi)} H ~ = z ^ η E o e − j k ( x cos ϕ − y s in ϕ )