Γe jβ 0 z Be jβz 0 < z < t t < z The associated magnetic fields are found using Maxwell s equation H = 1. e jβ 0 z = ˆx β 0

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1 ECE 6310 Spring 01 Assignment 3 Solutions Balanis 5.10 The plane waves in the three regions are given by E i ŷe 0 e jβ 0 z E r ŷe 0 Γe jβ 0 z z < 0 E a ŷe 0 Ae jβz E b ŷe 0 Be jβz 0 < z < t E t ŷe 0 Te jβ 0 z t < z The associated magnetic fields are found using Maxwell s equation H 1 jωµ E From which we find the incident magnetic field is H i 1 jωµ 0 jβ 0ẑ ( ) ŷe 0 e jβ 0 z ˆx β 0 E 0 e jβ z 0 ˆx E 0 e jβ 0 z ωµ 0 The remaining magnetic fields are found similarly H i ˆx E 0 e jβ 0 z H a ˆx E 0 η Ae jβz H t ˆx E 0 Te jβ 0 z H r ˆx E 0 Γe jβ 0 z H b ˆx E 0 jβz Be η z < 0 0 < z < t Applying the boundary condition that the tangential component of the electric and magnetic fields must be continuous at the two interfaces. r a b ( z 0) + ( z 0) ( z 0) + ( z 0) r a b ( z 0) + H x ( z 0) H x ( z 0) + H x ( z 0) b t ( z t) + ( z t) ( z t) b t ( z t) + H x ( z t) H x ( z t) i H x i a H x a

2 We arrive at the four equations for our four unknowns: Γ, T, A and B. 1+ Γ A + B 1 + Γ A η + B η Ae jβt + Be jβt Te jβ 0t A η e jβt + B η e jβt T e jβ 0t Solving these equations we find the solution for the reflection coefficient, either in terms of trigonometric or exponential functions. (I used Mathematica.) η ( )sinβt Γ η ( + )sinβt jη cosβt η + η + ( ) ( )( η ) e jβt e jβt ( ) e jβt ( η ) e jβt Similarly we find the transmission coefficient, with the transmitted wave referred to the first interface jηη T 0 η ( + )sinβt jη cosβt e jβ0t or to the second interface. 4η ( η + ) e jβt η ( ) e jβt e jβ0t T E t y z t i z 0 ( ) ( ) E 0 Te jβ0t E 0 E 0 e jβ 0 0 Te jβ 0t jηη 0 η ( + )sinβt jη cosβt 4η η + ( ) e jβt ( η ) e jβt Balanis 5.31 (a) The reflection coefficient, which is valid for complex impedance media is Γ η η 1 η + η 1 Re( η i ) 0 where the square root evaluated for the impedance must have a positive real part. Since the permeability is the same (that of free space) in both mediums, the reflection coefficient reduces to

3 Γ ε ε 1 Re ε i ε 1 + ε ( ) 0 Since this expression (and the one above) is a ratio of material properties, it applies to both absolute or relative values. The complex permittivity of the second medium, the sea water, is + j j σ ωε 0 Plugging in free space relative permittivity for medium one, and the above complex relative permittivity for medium two we find Γ 1 j 1+ j σ ωε 0 σ ωε 0 which allows us to find the magnitude of the reflected electric field. (Note that standard branch used in computational systems gives the correct branch of the squared root, satisfying the condition mentioned above.) The text asks for intensity which would imply the power density, but since that is asked for later, I assume it is field magnitude that is meant here. E r Γ E i mv/m (b) The standing wave ratio is SWR E + + E E + E 1+ Γ 1 Γ 189 (c) The incident and reflected power densities are obtained from the time averaged Poynting vector. The impedance is that of air. i S av r S av Ei W/m E r Γ i S av W/m

4 (d) The transmission coefficient is T 1+ Γ ε 1 ε 1 + ε σ 1+ j ωε 0 which yields the transmitted electric field magnitude E t T E i mv/m (e) From (4-31) we have S av E Re η e α z Thus for the transmitted wave just below the surface t S av Et Re ε r ( ) Et σ Re j η 0 ωε W/m (f) Since the sea water is a good conductor as specified in Table 4-1 σ ωε the expressions in that table may be used. The penetration depth (the depth at which the electric field magnitude attenuates to 1/e of its surface value) is δ ωµ 0 σ m (g) The wavelength in the sea water is λ π ωµ 0 σ πδ so that the depth at 00 wavelengths is

5 d 0λ 63. m (h) Using the phase velocity from the table (since this is a steady state harmonic problem and and not a pulsed transient problem where the group velocity should be used) the time for phase front traversal to the 100 meter depth is t d v p d µ 0σ ω 31.6 µs (i) And finally, the relative phase velocity in the sea water compared to air v p 1 ω v 0 v 0 µ 0 σ Balanis 5.45 (a) We need to solve equation (5-80) Δf 4 Γ m f 0 π cos 1 η L ( ) / ( η L + ) 1/ N to get the magnitude of the maximum reflection coefficient in the allowable bandwidth, Γm. Γ m η L η L + cos N π 4 Δf f 0 We need to be careful, since the zero subscript in this section does not refer to free space properties, but to the infinite medium in front of the layered media, which in this problem is not free space. I will use the subscript air to refer to free space values. The ratio of the backside impedance to the front side impedance is η L µ air / ε air µ air / ε air The desired magnitude is then Γ m cos π 4 Δf f

6 (b) The junction reflection coefficients in our two layer problem are given by η Γ n N L η C N n N L N! η L + η L + ( N n)!n! ( n)!n! The three junction reflection coefficients are then Γ Γ 1 1 Γ (c) Using equation (5-47a) Γ n η n+1 η n η n+1 + η n we can sequentially solve for the impedances η n+1 η n 1+ Γ n 1 Γ n with the results η 1 η air η air 1+ Γ 0 1 Γ 0 η air 1 1+ Γ 0 1 Γ Ω η η 1 1+ Γ 1 1 Γ Ω The impedance is given in terms of the permeability and permittivity as η µ air ε air η air Solving for the relative permittivity

7 η air η For our two layers we have The thickness is given by a quarter wavelength in the medium d λ c 0 / n f 1 4 f c 0 For our two layers we have d mm d 33.7 mm (d) The upper and lower frequencies in terms of the fractional bandwidth are f lower f 0 Δf f f upper f 0 + Δf f Δf f 0 Δf f GHz.5 GHz (e) From equation (5-76), the input reflection coefficient is approximately Γ in η ( f ) e jnθ L cos N θ θ π η L + f f 0 For our two layer structure we have a cosine squared dependence on frequency Γ in 1 π + 1 cos f f 0

8 G in êg max f êf 0 Balanis 6.9 From Maxwell s curl E equation we have three equations. E jωµh E z y z jωµh x E x z E z x jωµh y x E x y jωµh z From Maxwell s curl H equation we have another three equations. H jωεe H z y H y z jωεe x H x z H z x jωε H y x H x y jωεe z We will use just the first two equations in each of these sets. Replace derivatives with respect to z by the propagation constant.

9 z jβ z and set the z-component of the electric field to zero (since we are looking for TE fields) we arrive at the four equations jβ z jωµh x jβ z E x jωµh y H z y + jβ z H y jωεe x jβ z H x H z x jωε I will use the coma notation for derivatives, for example, H z,y H z y makes things more compact. Treat these derivatives are just constants with regard to solving the equations. β z ωµh x β z E x ωµh y H z,y + jβ z H y jωεe x jβ z H x H z,x jωε Solve these four equations for the four unknowns: Ex, Ey, Hx, Hy. (I used Mathematica.) E x j j H x j H y j ωµ β β H j β z,y z β β ηh z,y z ωµ β β H j β z,x z β β ηh z,x z β z β β z H z,x β z β β z H z,y

10 Balanis 6.34 The aperture electric field and resulting effective surface currents are E a ˆxE 0 cos π b / x b / a / z a / b / x b / ˆn E J MS a a / z a / 0 elsewhere J S 0 In the aperture the magnetic surface current evaluates to J MS ˆn E a ŷ ˆxE 0 cos π ẑe 0 cos π From this we can calculate the L integrals. sinθ L φ 0 e jβ J MSz S r cosψ d s Remember that the the L and N integrals are part of a far-field approximation. If you need exact results in all regions, use the exact integrals for the magnetic and electric potential. The argument of the exponential is given by the dot product of the source position vector and the observation point unit vector. r cosψ r ˆr x ˆx + z ẑ ( ) ( sinθ cosφ ˆx + sinθ cosφŷ + cosθẑ) x sinθ cosφ + z cosθ Plugging things in we have for the Lθ integral b/ a/ E 0 sinθ cos π e jβ x sinθ cosφ + z ( cosθ) d z d x b/ a/ The two integrals can be split into independent integrals.

11 b/ jβ sinθ cosφ E 0 sinθ e x d x cos π e jβ cosθ z d z b/ a/ a/ The first integral, compliments of Mathematica b/ e jβ sinθ cosφ x d x b/ sin b β sinθ cosφ β sinθ cosφ and the second integral, compliments of Mathematica cos π aπ cos a e jβ cosθ z β cosθ d z π a β cos θ a/ a/ The Lθ integral becomes 8πaE 0 β sin b β sinθ cosφ cos a β cosθ cosφ π a β cos θ ( ) The non-zero field components in terms of Lθ are, according to equation (6-1) E φ H θ jβr jβe 4πr jβe jβr 4πr η Collecting and simplifying we have for the far-field components E φ ηh θ j a sin b β sinθ cosφ r e jβr cos a E β cosθ 0 cosφ π a β cos θ E r E θ H r H φ 0 ( )

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