Wave Propagation and Polarization Normal Incidence: Reflection and Transmission Summary of Key Equations Reflection Coefficient (Γ \Gamma Γ :
Perpendicular Polarization: Γ < e m > ⊥ = η < / e m > 2 cos ( θ < e m > t ) − η < / e m > 1 cos ( θ < e m > i ) η < / e m > 2 cos ( θ < e m > t ) + η < / e m > 1 cos ( θ i ) \Gamma<em>{\perp} = \frac{\eta</em>2 \cos(\theta<em>t) - \eta</em>1 \cos(\theta<em>i)}{\eta</em>2 \cos(\theta<em>t) + \eta</em>1 \cos(\theta_i)} Γ < e m > ⊥ = η < / e m > 2 c o s ( θ < e m > t ) + η < / e m > 1 c o s ( θ i ) η < / e m > 2 c o s ( θ < e m > t ) − η < / e m > 1 c o s ( θ < e m > i ) Parallel Polarization: Γ < e m > ∣ ∣ = η < / e m > 2 cos ( θ < e m > i ) − η < / e m > 1 cos ( θ < e m > t ) η < / e m > 2 cos ( θ < e m > i ) + η < / e m > 1 cos ( θ t ) \Gamma<em>{||} = \frac{\eta</em>2 \cos(\theta<em>i) - \eta</em>1 \cos(\theta<em>t)}{\eta</em>2 \cos(\theta<em>i) + \eta</em>1 \cos(\theta_t)} Γ < e m > ∣∣ = η < / e m > 2 c o s ( θ < e m > i ) + η < / e m > 1 c o s ( θ t ) η < / e m > 2 c o s ( θ < e m > i ) − η < / e m > 1 c o s ( θ < e m > t ) Normal Incidence: Γ = η < e m > 2 − η < / e m > 1 η < e m > 2 + η < / e m > 1 \Gamma = \frac{\eta<em>2 - \eta</em>1}{\eta<em>2 + \eta</em>1} Γ = η < e m > 2 + η < / e m > 1 η < e m > 2 − η < / e m > 1 Transmission Coefficient (τ \tau τ :
Perpendicular Polarization: τ < e m > ⊥ = 2 η < / e m > 2 cos ( θ < e m > i ) η < / e m > 2 cos ( θ < e m > t ) + η < / e m > 1 cos ( θ i ) \tau<em>{\perp} = \frac{2\eta</em>2 \cos(\theta<em>i)}{\eta</em>2 \cos(\theta<em>t) + \eta</em>1 \cos(\theta_i)} τ < e m > ⊥ = η < / e m > 2 c o s ( θ < e m > t ) + η < / e m > 1 c o s ( θ i ) 2 η < / e m > 2 c o s ( θ < e m > i ) Parallel Polarization: τ < e m > ∣ ∣ = 2 η < / e m > 2 cos ( θ < e m > i ) η < / e m > 2 cos ( θ < e m > i ) + η < / e m > 1 cos ( θ t ) \tau<em>{||} = \frac{2\eta</em>2 \cos(\theta<em>i)}{\eta</em>2 \cos(\theta<em>i) + \eta</em>1 \cos(\theta_t)} τ < e m > ∣∣ = η < / e m > 2 c o s ( θ < e m > i ) + η < / e m > 1 c o s ( θ t ) 2 η < / e m > 2 c o s ( θ < e m > i ) Normal Incidence: τ = 2 η < e m > 2 η < / e m > 2 + η 1 \tau = \frac{2\eta<em>2}{\eta</em>2 + \eta_1} τ = η < / e m > 2 + η 1 2 η < e m > 2 Relation between Reflection and Transmission Coefficients :
τ = 1 + Γ \tau = 1 + \Gamma τ = 1 + Γ τ < e m > ∣ ∣ = ( 1 + Γ < / e m > ∣ ∣ ) cos ( θ < e m > i ) cos ( θ < / e m > t ) \tau<em>{||} = (1 + \Gamma</em>{||}) \frac{\cos(\theta<em>i)}{\cos(\theta</em>t)} τ < e m > ∣∣ = ( 1 + Γ < / e m > ∣∣ ) c o s ( θ < / e m > t ) c o s ( θ < e m > i ) Reflectivity (R) :
R = ∣ Γ ∣ 2 R = |\Gamma|^2 R = ∣Γ ∣ 2 Perpendicular Polarization: R < e m > ⊥ = ∣ Γ < / e m > ⊥ ∣ 2 R<em>{\perp} = |\Gamma</em>{\perp}|^2 R < e m > ⊥ = ∣Γ < / e m > ⊥ ∣ 2 Parallel Polarization: R < e m > ∣ ∣ = ∣ Γ < / e m > ∣ ∣ ∣ 2 R<em>{||} = |\Gamma</em>{||}|^2 R < e m > ∣∣ = ∣Γ < / e m > ∣∣ ∣ 2 Transmissivity (T) :
T = ∣ τ ∣ 2 η < e m > 1 η < / e m > 2 T = |\tau|^2 \frac{\eta<em>1}{\eta</em>2} T = ∣ τ ∣ 2 η < / e m > 2 η < e m > 1 Perpendicular Polarization: T < e m > ⊥ = ∣ τ < / e m > ⊥ ∣ 2 η < e m > 2 cos ( θ < / e m > t ) η < e m > 1 cos ( θ < / e m > i ) T<em>{\perp} = |\tau</em>{\perp}|^2 \frac{\eta<em>2 \cos(\theta</em>t)}{\eta<em>1 \cos(\theta</em>i)} T < e m > ⊥ = ∣ τ < / e m > ⊥ ∣ 2 η < e m > 1 c o s ( θ < / e m > i ) η < e m > 2 c o s ( θ < / e m > t ) Parallel Polarization: T < e m > ∣ ∣ = ∣ τ < / e m > ∣ ∣ ∣ 2 η < e m > 2 cos ( θ < / e m > t ) η < e m > 1 cos ( θ < / e m > i ) T<em>{||} = |\tau</em>{||}|^2 \frac{\eta<em>2 \cos(\theta</em>t)}{\eta<em>1 \cos(\theta</em>i)} T < e m > ∣∣ = ∣ τ < / e m > ∣∣ ∣ 2 η < e m > 1 c o s ( θ < / e m > i ) η < e m > 2 c o s ( θ < / e m > t ) Relation between Reflectivity and Transmissivity :
T = 1 − R T = 1 - R T = 1 − R Perpendicular Polarization: T < e m > ⊥ = 1 − R < / e m > ⊥ T<em>{\perp} = 1 - R</em>{\perp} T < e m > ⊥ = 1 − R < / e m > ⊥ Parallel Polarization: T < e m > ∣ ∣ = 1 − R < / e m > ∣ ∣ T<em>{||} = 1 - R</em>{||} T < e m > ∣∣ = 1 − R < / e m > ∣∣ Additional Notes :
sin ( θ < e m > t ) = μ < / e m > 1 ϵ < e m > 1 μ < / e m > 2 ϵ < e m > 2 sin ( θ < / e m > i ) \sin(\theta<em>t) = \sqrt{\frac{\mu</em>1 \epsilon<em>1}{\mu</em>2 \epsilon<em>2}} \sin(\theta</em>i) sin ( θ < e m > t ) = μ < / e m > 2 ϵ < e m > 2 μ < / e m > 1 ϵ < e m > 1 sin ( θ < / e m > i ) η < e m > 1 = μ < / e m > 1 ϵ 1 \eta<em>1 = \sqrt{\frac{\mu</em>1}{\epsilon_1}} η < e m > 1 = ϵ 1 μ < / e m > 1 η < e m > 2 = μ < / e m > 2 ϵ 2 \eta<em>2 = \sqrt{\frac{\mu</em>2}{\epsilon_2}} η < e m > 2 = ϵ 2 μ < / e m > 2 For non-magnetic media: η < e m > 2 η < / e m > 1 = n < e m > 1 n < / e m > 2 \frac{\eta<em>2}{\eta</em>1} = \frac{n<em>1}{n</em>2} η < / e m > 1 η < e m > 2 = n < / e m > 2 n < e m > 1 2014 Test 2 Q1 Problem Setup A uniform plane wave in air is incident on a dielectric plane boundary at z = 0 z = 0 z = 0 The electric field is given by: E ⃗ ( x , z , t ) = y ^ E < e m > 0 cos ( ω t + k < / e m > x x − k z z ) V/m \vec{E}(x, z, t) = \hat{y} E<em>0 \cos(\omega t + k</em>x x - k_z z) \text{ V/m} E ( x , z , t ) = y ^ E < e m > 0 cos ( ω t + k < / e m > xx − k z z ) V/m Dielectric constant of the second medium: ϵ r = 2.25 \epsilon_r = 2.25 ϵ r = 2.25 Incident angle: θ i = 60 ∘ \theta_i = 60^\circ θ i = 6 0 ∘ Transmitted angle: θ t = 35.26 ∘ \theta_t = 35.26^\circ θ t = 35.2 6 ∘ Phase constant in air: k 1 = 5 rad/m k_1 = 5 \text{ rad/m} k 1 = 5 rad/m Task Find the incident phasor magnetic field, H ⃗ i \vec{H}_i H i Find the reflected phasor electric field, E ⃗ r \vec{E}_r E r Find the frequency of the wave, f f f Solution The wave vector is given by: k ⃗ = − x ^ k < e m > x + z ^ k < / e m > z \vec{k} = -\hat{x} k<em>x + \hat{z} k</em>z k = − x ^ k < e m > x + z ^ k < / e m > z Given values:k < e m > x = k < / e m > i sin ( θ i ) = 5 sin ( 60 ∘ ) = 4.33 rad/m k<em>x = k</em>i \sin(\theta_i) = 5 \sin(60^\circ) = 4.33 \text{ rad/m} k < e m > x = k < / e m > i sin ( θ i ) = 5 sin ( 6 0 ∘ ) = 4.33 rad/m k < e m > z = k < / e m > i cos ( θ i ) = 5 cos ( 60 ∘ ) = 2.5 rad/m k<em>z = k</em>i \cos(\theta_i) = 5 \cos(60^\circ) = 2.5 \text{ rad/m} k < e m > z = k < / e m > i cos ( θ i ) = 5 cos ( 6 0 ∘ ) = 2.5 rad/m The incident magnetic field is: H ⃗ = 1 η k ^ × E ⃗ \vec{H} = \frac{1}{\eta} \hat{k} \times \vec{E} H = η 1 k ^ × E In air, η = η 0 = 377 Ω \eta = \eta_0 = 377 \Omega η = η 0 = 377Ω H ⃗ < e m > i = E < / e m > 0 η < e m > 0 ( − x ^ cos ( θ < / e m > i ) − z ^ sin ( θ i ) ) e − j ( 4.3 x + 2.5 z ) \vec{H}<em>i = \frac{E</em>0}{\eta<em>0} (-\hat{x} \cos(\theta</em>i) - \hat{z} \sin(\theta_i)) e^{-j(4.3x + 2.5z)} H < e m > i = η < e m > 0 E < / e m > 0 ( − x ^ cos ( θ < / e m > i ) − z ^ sin ( θ i )) e − j ( 4.3 x + 2.5 z ) H ⃗ < e m > i = E < / e m > 0 377 ( − 0.5 x ^ − 0.866 z ^ ) e − j ( 4.3 x + 2.5 z ) \vec{H}<em>i = \frac{E</em>0}{377} (-0.5 \hat{x} - 0.866 \hat{z}) e^{-j(4.3x + 2.5z)} H < e m > i = 377 E < / e m > 0 ( − 0.5 x ^ − 0.866 z ^ ) e − j ( 4.3 x + 2.5 z ) The reflection coefficient for perpendicular polarization is:Γ < e m > ⊥ = η < / e m > 2 cos ( θ < e m > t ) − η < / e m > 1 cos ( θ < e m > i ) η < / e m > 2 cos ( θ < e m > t ) + η < / e m > 1 cos ( θ i ) \Gamma<em>{\perp} = \frac{\eta</em>2 \cos(\theta<em>t) - \eta</em>1 \cos(\theta<em>i)}{\eta</em>2 \cos(\theta<em>t) + \eta</em>1 \cos(\theta_i)} Γ < e m > ⊥ = η < / e m > 2 c o s ( θ < e m > t ) + η < / e m > 1 c o s ( θ i ) η < / e m > 2 c o s ( θ < e m > t ) − η < / e m > 1 c o s ( θ < e m > i ) Since η = μ ϵ \eta = \sqrt{\frac{\mu}{\epsilon}} η = ϵ μ μ \mu μ η ∝ 1 ϵ \eta \propto \frac{1}{\sqrt{\epsilon}} η ∝ ϵ 1 η < e m > 1 η < / e m > 2 = ϵ < e m > 2 ϵ < / e m > 1 = 2.25 = 1.5 ⇒ η < e m > 2 = η < / e m > 1 1.5 = 377 1.5 = 251.33 \frac{\eta<em>1}{\eta</em>2} = \sqrt{\frac{\epsilon<em>2}{\epsilon</em>1}} = \sqrt{2.25} = 1.5 \Rightarrow \eta<em>2 = \frac{\eta</em>1}{1.5} = \frac{377}{1.5} = 251.33 η < / e m > 2 η < e m > 1 = ϵ < / e m > 1 ϵ < e m > 2 = 2.25 = 1.5 ⇒ η < e m > 2 = 1.5 η < / e m > 1 = 1.5 377 = 251.33 Γ ⊥ = 251.33 cos ( 35.26 ∘ ) − 377 cos ( 60 ∘ ) 251.33 cos ( 35.26 ∘ ) + 377 cos ( 60 ∘ ) = 205.4 − 188.5 205.4 + 188.5 = 16.9 393.9 = 0.042 \Gamma_{\perp} = \frac{251.33 \cos(35.26^\circ) - 377 \cos(60^\circ)}{251.33 \cos(35.26^\circ) + 377 \cos(60^\circ)} = \frac{205.4 - 188.5}{205.4 + 188.5} = \frac{16.9}{393.9} = 0.042 Γ ⊥ = 251.33 c o s ( 35.2 6 ∘ ) + 377 c o s ( 6 0 ∘ ) 251.33 c o s ( 35.2 6 ∘ ) − 377 c o s ( 6 0 ∘ ) = 205.4 + 188.5 205.4 − 188.5 = 393.9 16.9 = 0.042 Given that Γ = − 0.42 \Gamma = -0.42 Γ = − 0.42 Reflected electric field: E ⃗ < e m > r = Γ < / e m > ⊥ E < e m > 0 y ^ e − j ( 4.3 x − 2.5 z ) = − 0.42 E < / e m > 0 y ^ e − j ( 4.3 x − 2.5 z ) \vec{E}<em>r = \Gamma</em>{\perp} E<em>0 \hat{y} e^{-j(4.3x - 2.5z)} = -0.42 E</em>0 \hat{y} e^{-j(4.3x - 2.5z)} E < e m > r = Γ < / e m > ⊥ E < e m > 0 y ^ e − j ( 4.3 x − 2.5 z ) = − 0.42 E < / e m > 0 y ^ e − j ( 4.3 x − 2.5 z ) Frequency: f = c λ = c k 2 π f = \frac{c}{\lambda} = \frac{ck}{2\pi} f = λ c = 2 π c k Since k = ω c k = \frac{\omega}{c} k = c ω ω = c k \omega = ck ω = c k f = c k 2 π f = \frac{ck}{2\pi} f = 2 π c k f = 3 × 10 8 ⋅ 5 2 π = 238.7 MHz f = \frac{3 \times 10^8 \cdot 5}{2\pi} = 238.7 \text{ MHz} f = 2 π 3 × 1 0 8 ⋅ 5 = 238.7 MHz 2013 Test 2 Q2 A plane wave is incident at 30 degrees from medium 1 to medium 2. Medium 1 has parameters μ < e m > 0 \mu<em>0 μ < e m > 0 ϵ < / e m > 0 ϵ r \epsilon</em>0 \epsilon_r ϵ < / e m > 0 ϵ r Medium 2 has parameters μ < e m > 0 \mu<em>0 μ < e m > 0 ϵ < / e m > 0 \epsilon</em>0 ϵ < / e m > 0 Polarization is not explicitly stated but inferred from the problem. Brewster angle occurs when Γ ∣ ∣ = 0 \Gamma_{||} = 0 Γ ∣∣ = 0 tan ( θ < e m > B ) = ϵ < / e m > 2 ϵ 1 \tan(\theta<em>B) = \sqrt{\frac{\epsilon</em>2}{\epsilon_1}} tan ( θ < e m > B ) = ϵ 1 ϵ < / e m > 2 Given θ < e m > i = 30 ∘ \theta<em>i = 30^\circ θ < e m > i = 3 0 ∘ 0 = Γ < / e m > ∣ ∣ = η < e m > 2 cos ( θ < / e m > t ) − η < e m > 1 cos ( θ < / e m > i ) η < e m > 2 cos ( θ < / e m > t ) + η < e m > 1 cos ( θ < / e m > i ) 0 = \Gamma</em>{||} = \frac{\frac{\eta<em>2}{\cos(\theta</em>t)} - \frac{\eta<em>1}{\cos(\theta</em>i)}}{\frac{\eta<em>2}{\cos(\theta</em>t)} + \frac{\eta<em>1}{\cos(\theta</em>i)}} 0 = Γ < / e m > ∣∣ = c o s ( θ < / e m > t ) η < e m > 2 + c o s ( θ < / e m > i ) η < e m > 1 c o s ( θ < / e m > t ) η < e m > 2 − c o s ( θ < / e m > i ) η < e m > 1 ϵ r 1 = 3 \epsilon_{r1} = 3 ϵ r 1 = 3 2011 Test 1 Q2 Given that tan ( θ < e m > i ) = 4 3 \tan(\theta<em>i) = \frac{4}{3} tan ( θ < e m > i ) = 3 4 θ < / e m > i = 53.13 ∘ \theta</em>i = 53.13^\circ θ < / e m > i = 53.1 3 ∘ Therefore, sin ( θ < e m > i ) = 4 5 \sin(\theta<em>i) = \frac{4}{5} sin ( θ < e m > i ) = 5 4 cos ( θ < / e m > i ) = 3 5 \cos(\theta</em>i) = \frac{3}{5} cos ( θ < / e m > i ) = 5 3 The wave vector k ⃗ = x ^ k < e m > x + z ^ k < / e m > z = x ^ k sin ( θ < e m > i ) + z ^ k cos ( θ < / e m > i ) \vec{k} = \hat{x} k<em>x + \hat{z} k</em>z = \hat{x} k \sin(\theta<em>i) + \hat{z} k \cos(\theta</em>i) k = x ^ k < e m > x + z ^ k < / e m > z = x ^ k sin ( θ < e m > i ) + z ^ k cos ( θ < / e m > i ) H ⃗ = 1 η y ^ × E ⃗ = 1 η ( − z ^ cos ( θ < e m > i ) + x ^ sin ( θ < / e m > i ) ) E < e m > 0 e − j ( k < / e m > x x + k z z ) \vec{H} = \frac{1}{\eta} \hat{y} \times \vec{E} = \frac{1}{\eta} (-\hat{z} \cos(\theta<em>i) + \hat{x} \sin(\theta</em>i)) E<em>0 e^{-j(k</em>x x + k_z z)} H = η 1 y ^ × E = η 1 ( − z ^ cos ( θ < e m > i ) + x ^ sin ( θ < / e m > i )) E < e m > 0 e − j ( k < / e m > xx + k z z ) sin ( θ < e m > t ) = μ < / e m > 1 ϵ < e m > 1 μ < / e m > 2 ϵ < e m > 2 sin ( θ < / e m > i ) \sin(\theta<em>t) = \sqrt{\frac{\mu</em>1 \epsilon<em>1}{\mu</em>2 \epsilon<em>2}} \sin(\theta</em>i) sin ( θ < e m > t ) = μ < / e m > 2 ϵ < e m > 2 μ < / e m > 1 ϵ < e m > 1 sin ( θ < / e m > i ) θ t = 11.537 ∘ \theta_t = 11.537^\circ θ t = 11.53 7 ∘ Reflection coefficient: Γ = η < e m > 2 cos ( θ < / e m > i ) − η < e m > 1 cos ( θ < / e m > t ) η < e m > 2 cos ( θ < / e m > i ) + η < e m > 1 cos ( θ < / e m > t ) = 0.355 \Gamma = \frac{\eta<em>2 \cos(\theta</em>i) - \eta<em>1 \cos(\theta</em>t)}{\eta<em>2 \cos(\theta</em>i) + \eta<em>1 \cos(\theta</em>t)} = 0.355 Γ = η < e m > 2 c o s ( θ < / e m > i ) + η < e m > 1 c o s ( θ < / e m > t ) η < e m > 2 c o s ( θ < / e m > i ) − η < e m > 1 c o s ( θ < / e m > t ) = 0.355 η = 120 π \eta = 120\pi η = 120 π Incident electric field: E ⃗ i = 12 ( 0.6 x ^ + 0.8 z ^ ) e − j ( 4 x + 3 z ) V/m \vec{E}_i = 12 (0.6 \hat{x} + 0.8 \hat{z}) e^{-j(4x + 3z)} \text{ V/m} E i = 12 ( 0.6 x ^ + 0.8 z ^ ) e − j ( 4 x + 3 z ) V/m Reflected electric field: E ⃗ < e m > r = Γ E < / e m > i = 0.42 ⋅ 12 ( 0.6 x ^ − 0.8 z ^ ) e − j ( 4 x − 3 z ) V/m \vec{E}<em>r = \Gamma E</em>i = 0.42 \cdot 12 (0.6 \hat{x} - 0.8 \hat{z}) e^{-j(4x - 3z)} \text{ V/m} E < e m > r = Γ E < / e m > i = 0.42 ⋅ 12 ( 0.6 x ^ − 0.8 z ^ ) e − j ( 4 x − 3 z ) V/m Reflected magnetic field: H ⃗ < e m > r = − Γ E < / e m > 0 η ( z ^ cos ( θ < e m > i ) + x ^ sin ( θ < / e m > i ) ) e − j ( 4 x − 3 z ) = 1.33 mA/m \vec{H}<em>r = \frac{-\Gamma E</em>0}{\eta} (\hat{z} \cos(\theta<em>i) + \hat{x} \sin(\theta</em>i)) e^{-j(4x - 3z)} = 1.33 \text{ mA/m} H < e m > r = η − Γ E < / e m > 0 ( z ^ cos ( θ < e m > i ) + x ^ sin ( θ < / e m > i )) e − j ( 4 x − 3 z ) = 1.33 mA/m 2019 Exam Q5 Problem Setup A uniform plane wave with H ⃗ = y ^ H < e m > 0 e − j ( 10 x + 10 3 z ) A/m \vec{H} = \hat{y} H<em>0 e^{-j(10x + 10\sqrt{3}z)} \text{ A/m} H = y ^ H < e m > 0 e − j ( 10 x + 10 3 z ) A/m μ < / e m > 1 = μ < e m > 0 \mu</em>1 = \mu<em>0 μ < / e m > 1 = μ < e m > 0 ϵ < / e m > 1 = ϵ < e m > 0 \epsilon</em>1 = \epsilon<em>0 ϵ < / e m > 1 = ϵ < e m > 0 σ < / e m > 1 = 0 \sigma</em>1 = 0 σ < / e m > 1 = 0 μ < e m > 2 = μ < / e m > 0 \mu<em>2 = \mu</em>0 μ < e m > 2 = μ < / e m > 0 ϵ < e m > 2 = 9 ϵ < / e m > 0 \epsilon<em>2 = 9\epsilon</em>0 ϵ < e m > 2 = 9 ϵ < / e m > 0 σ 2 = 0 \sigma_2 = 0 σ 2 = 0 Task Determine the angle of incidence, θ i \theta_i θ i Determine the phase constant, k 1 k_1 k 1 Determine the intrinsic impedance, η 2 \eta_2 η 2 Determine the average power density in the incident wave in medium 1. Solution Angle of incidence: tan ( θ < e m > i ) = k < / e m > x k < e m > z = 10 10 3 = 1 3 ⇒ θ < / e m > i = 30 ∘ \tan(\theta<em>i) = \frac{k</em>x}{k<em>z} = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \theta</em>i = 30^\circ tan ( θ < e m > i ) = k < e m > z k < / e m > x = 10 3 10 = 3 1 ⇒ θ < / e m > i = 3 0 ∘ Phase constant: k < e m > 1 = k < / e m > x 2 + k z 2 = 10 2 + ( 10 3 ) 2 = 100 + 300 = 400 = 20 rad/m k<em>1 = \sqrt{k</em>x^2 + k_z^2} = \sqrt{10^2 + (10\sqrt{3})^2} = \sqrt{100 + 300} = \sqrt{400} = 20 \text{ rad/m} k < e m > 1 = k < / e m > x 2 + k z 2 = 1 0 2 + ( 10 3 ) 2 = 100 + 300 = 400 = 20 rad/m Intrinsic impedance: η < e m > 2 = μ < / e m > 2 ϵ < e m > 2 = μ < / e m > 0 9 ϵ < e m > 0 = 1 3 μ < / e m > 0 ϵ < e m > 0 = 1 3 η < / e m > 0 = 1 3 ⋅ 120 π = 40 π ≈ 125.67 Ω \eta<em>2 = \sqrt{\frac{\mu</em>2}{\epsilon<em>2}} = \sqrt{\frac{\mu</em>0}{9\epsilon<em>0}} = \frac{1}{3} \sqrt{\frac{\mu</em>0}{\epsilon<em>0}} = \frac{1}{3} \eta</em>0 = \frac{1}{3} \cdot 120\pi = 40\pi \approx 125.67 \Omega η < e m > 2 = ϵ < e m > 2 μ < / e m > 2 = 9 ϵ < e m > 0 μ < / e m > 0 = 3 1 ϵ < e m > 0 μ < / e m > 0 = 3 1 η < / e m > 0 = 3 1 ⋅ 120 π = 40 π ≈ 125.67Ω Average power density: S ⃗ < e m > a v = 1 2 Re [ E ⃗ × H ⃗ ∗ ] = 1 2 Re [ ( η H ⃗ ) × H ⃗ ∗ ] = 1 2 η ∣ H < / e m > 0 ∣ 2 k ^ \vec{S}<em>{av} = \frac{1}{2} \text{Re} [\vec{E} \times \vec{H}^* ] = \frac{1}{2} \text{Re} [(\eta \vec{H}) \times \vec{H}^* ] = \frac{1}{2} \eta |H</em>0|^2 \hat{k} S < e m > a v = 2 1 Re [ E × H ∗ ] = 2 1 Re [( η H ) × H ∗ ] = 2 1 η ∣ H < / e m > 0 ∣ 2 k ^ S ⃗ a v = 1 2 ( 120 π ) ( 1 ) 2 ( 0.5 x ^ + 0.5 3 z ^ ) = 60 π ( 0.5 x ^ + 0.5 3 z ^ ) = ( 30 π x ^ + 30 3 π z ^ ) W/m 2 \vec{S}_{av} = \frac{1}{2} (120\pi) (1)^2 (0.5 \hat{x} + 0.5\sqrt{3} \hat{z}) = 60\pi (0.5 \hat{x} + 0.5\sqrt{3} \hat{z}) = (30\pi \hat{x} + 30\sqrt{3}\pi \hat{z}) \text{ W/m}^2 S a v = 2 1 ( 120 π ) ( 1 ) 2 ( 0.5 x ^ + 0.5 3 z ^ ) = 60 π ( 0.5 x ^ + 0.5 3 z ^ ) = ( 30 π x ^ + 30 3 π z ^ ) W/m 2 2013 Exam Q2 A plane wave is incident at 50 degrees. Medium 1: μ < e m > 0 \mu<em>0 μ < e m > 0 ϵ < / e m > 0 ϵ r \epsilon</em>0 \epsilon_r ϵ < / e m > 0 ϵ r Medium 2: μ < e m > 0 \mu<em>0 μ < e m > 0 ϵ < / e m > 0 \epsilon</em>0 ϵ < / e m > 0 Good conductor: \frac{\sigma}{\omega \epsilon} >> 1 If σ ω ϵ = 200 \frac{\sigma}{\omega \epsilon} = 200 ω ϵ σ = 200 Calculate α \alpha α η \eta η α = π f μ σ \alpha = \sqrt{\pi f \mu \sigma} α = π f μ σ η = j ω μ σ + j ω ϵ ≈ j ω μ σ = ( 1 + j ) π f μ σ \eta = \sqrt{\frac{j \omega \mu}{\sigma + j \omega \epsilon}} \approx \sqrt{\frac{j \omega \mu}{\sigma}} = (1+j) \sqrt{\frac{\pi f \mu}{\sigma}} η = σ + jω ϵ jω μ ≈ σ jω μ = ( 1 + j ) σ π f μ Problem Setup - Electric Field in Medium 2: The general form for the electric field in medium 2 is:E < e m > t ( x , z ) = E < / e m > 0 e − α z e − j ( β < e m > x x + β < / e m > z z ) y ^ \mathbf{E}<em>{t} (x, z) = E</em>{0} e^{-\alpha z} e^{-j(\beta<em>{x} x + \beta</em>{z} z)} \hat{y} E < e m > t ( x , z ) = E < / e m > 0 e − α z e − j ( β < e m > x x + β < / e m > z z ) y ^
For a good condutor β ≈ α = π f μ σ \beta \approx \alpha = \sqrt{ \pi f \mu \sigma} β ≈ α = π f μ σ
Given σ = 2.15 ⋅ ω ⋅ ϵ 0 ⋅ ( 1 + j ) \sigma = 2.15 \cdot \omega \cdot \epsilon_0 \cdot (1 + j) σ = 2.15 ⋅ ω ⋅ ϵ 0 ⋅ ( 1 + j )
It uses generic values and results to illustrate that α ≈ β \alpha \approx \beta α ≈ β