HW7Solution

HW7Solution - ECE 201A Homework 7 solution 1. (a) Y...

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ECE 201A Homework 7 solution 1. (a) X Z Y Ground plane Waveguide a b σ =∞ 0 z = z y b , ii EH GG We can find an equivalent problem by placing a perfectly conducting plane that covers the waveguide aperture as shown below. In the original problem the tangential electric field is zero everywhere on this plane except on the rectangular waveguide aperture. To keep the tangential electric field unchanged on the aperture we need to introduce a magnetic surface current whose value is he same as the tangential electric field, i.e., ( ) 1 ,,0 s Mn E x y G and ( ) 2 s E x y =− × G . This choice makes sure that tangential electric field jumps from zero to () nE x y × G G on the perfect conductor covering the aperture so that an observer in space does not see a change on the tangential electric field at 0 z = . The sign of the magnetic surface current is different on both sides of the aperture because we pick a single normal shown in the figure. Obviously normal to the surface for 0 z points in the other direction. In other words 1 nn = and 2 =− . In this formulation Exy G is unknown and we are trying to formulate a solution for ( ) G . 0 z = z y b , 0 z = z y b , 2 s M G 1 s M G n G The equivalent problem shown above can be solved on both sides of the perfect conductor at 0 z = . In other words we should solve the following two problems.
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σ =∞ For 0 z z y b 1 s M G n G 0 z = z y b , ii EH GG 2 s M G n G For 0 z For 0 z we need to find the electric vector potential F G . In this case we can eliminate the perfectly conducting plane using image theory. This requires doubling the magnetic surface current. F G is found by solving () () () 22 Fr kFr Mr ε ∇+ = G G This non homogenous partial differential equation can be solved using the Green's function techniques as () ( ) ( ) '; ' ' Mr Grr dV = ∫∫∫ G G In this case the radiation is into free space and () ' ;' 4' jk r r e Grr rr ε π −− = G G so ' 11 '' jk r r e dV ε π = ∫∫∫ G G Once , Fxz G is found fields are found in source free regions using 1 E F ε =− ∇× and E jH ωμ ∇× =− . The second equation (Faraday's law) can be arranged to yield 1 1 HE F F F F k F jj j j ωεμ = = ∇×∇× = ∇∇ −∇ = + G G G G G So 1 E F and 2 1 1 HF k F j =∇ + G i
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For 0 z we can use the principle of superposition and break the problem into the superposition of two problems. The first problem is the reflection of the incident field from a short circuit at the end of the waveguide. The solution to his part is ir EEE =+ GGG and HHH . The second problem is the radiation of the magnetic surface current into a shorted rectangular waveguide. The solution to this part can again be written as 22 1 E F ε =− ∇× GG and () 2 2 1 HF k F j ωεμ =∇ + G i where () () ( ) () ( ) 2 1 2 '; ' ' ' ' Fr MrGrrd V V == ∫∫∫ ∫∫∫ G G G G G G In this case 2 ;' Grr is the appropriate Green’s function that describes fields generated by unit a magnetic point current source within a shorted rectangular waveguide. This can be found using the well known techniques. Then the overall solution for 0 z is 2 1 F =+− × G and 2 1 Fk F j =++ ∇+ G G G
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HW7Solution - ECE 201A Homework 7 solution 1. (a) Y...

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