HW5_Solution

HW5_Solution - ECE 201A Homework 5 solution 1. A= jkr e J...

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ECE 201A Homework 5 solution 1. 'cos (' ) ' 4 jkr jkr AeJ r e d V r ξ μ π = ∫∫∫ G G ˆ ) s i n ' )(' ) 2 zm L Jr aI k z x y δδ ⎡⎤ ⎛⎞ =+ ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ G '' ' ' dV dx dy dz = 222 ''' 'cos x xy yz z x z r r xyz ++ == 2 2 ˆ sin ' ( ') ( ') ' ' ' 42 L xx yy zz jkr jk r L eL A a I k z x y e dx dy dz r μδ δ ∞∞ −∞ −∞ G 2 ' 2 ˆ sin ' ' L z jkr jk z r L Aa I kz e d z r G cos z r θ = 2 22 cos ' 2 ˆ ' L LL jk z jk z jkr jk z L ee e I e d z rj +− + = G () 2 1c o s ' o s ' 2 1 ˆ ' L jkr jk jk jk z jk z L e I e e e e d z θθ =− G o s ' o s ' 1 ˆ 4 2 1 cos 1 cos jk jk jk z jk z jkr e e e I j k j k πθ −− ⎧⎫ ⎪⎪ ⎨⎬ ⎩⎭ G
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() 1 cos 1 cos 1 cos 1 cos 2 2 2 2 22 1 ˆ 4 2 1 cos 1 cos LL L L L L jk jk jk jk jk jk jk jk jkr zm ee e e e e e e e Aa I r j jk jk θθ μ πθ θ + + −− ⎧⎫ ⎡⎤ ⎪⎪ ⎢⎥ =− ⎨⎬ +− ⎣⎦ ⎩⎭ G cos cos cos cos 2 2 1 ˆ
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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HW5_Solution - ECE 201A Homework 5 solution 1. A= jkr e J...

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