ECE450_HW09_SOLN

ECE450_HW09_SOLN - ECE 450 Spring 2009 Homework Set 9...

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ECE 450: Spring 2009 Homework Set 9 Solutions Due on Friday, Apr 10, 2009 ()()() 222 22 1 2452 2 mnl 2 f GHz m n c abc µε ⎛⎞ ⎛⎞ ⎛⎞ =+ + + ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ The lowest 5 frequencies are: TE 101 = 4.47GHz TE 011 = 5.39 TE 102 = 5.66GHz TE 012 or TM 110 = 6.4GHz TE 111 or TM 111 = 6.71GHz 22 2 2 1 50 2 11 Thus, 50 1000 5 2000 50 1000 26 10400 50 fM H z abd ad bd ab ⎛ ⎞ ⎛⎞ ⎛⎞ + = + + ⎜ ⎟ ⎜⎟ ⎜⎟ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎛⎞ ⎛⎞ += = > += ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ = ⎛⎞ ⎛ + ⎜⎟ ⎝⎠ ⎝ 2 2 1 1000 30 12000 1 Thus, 1600, 10000 and 400 Thus a = 2.5cm, b = 1cm, d = 5cm db d d ⎞⎛⎞ = ⎠⎝⎠ ⎛⎞ == = ⎝⎠

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ECE 450: Spring 2009 Homework Set 9 Solutions Due on Friday, Apr 10, 2009 The maximum rate of increase of a function is simply the magnitude of its gradient vector. (a) 11 (2 ,2 ,2 ); At (3,4,12), (6,8,24), 26 xyz ∇Φ = ∇Φ = ∇Φ = 1 (b) 21 (1,2,2); At (3,4,12), (1,2,2), 3 ∇Φ = ∇Φ = ∇Φ = 1 (c)For this question, we must find the dot product of the gradient vector of Φ 1 with the direction of maximum increase of Φ 2 . 222 (1,2,2) 1 6 16 48 (3,4,12) (3,4,12) (6,8,24) (1,2,2) 23.33 33 122 u Du φφ ++ =∇ = = = =
ECE 450: Spring 2009 Homework Set 9 Solutions Due on Friday, Apr 10, 2009 222 00 22 2 2 2 0 2 2 2 (, ,) 44 ( ) 4( ) 4 ( ) ( ) 11 2 4 2 4 QQ Vxyz xyz xx yz xy zz xx y zz x x xxyz Q xyz z zz xxyz z Q πε =− ++ ∆++ −+ + + −∆ + + ⎡⎤ ⎢⎥ + = ++−∆ + +

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ECE450_HW09_SOLN - ECE 450 Spring 2009 Homework Set 9...

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