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au10_312ps4_sol

au10_312ps4_sol - ECE312 Autumn 2010 Problem Set 4...

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Unformatted text preview: ECE312 Autumn 2010: Problem Set 4 Solutions Problem 1 : Let η = radicalbig μ /ǫ = 377 Ω. (a) ˜ E = ˆ yje jπz ; by comparing with ˜ E = ˜ E e − j k · r , we can see that the the vector k = − π ˆ z (i) The field propagates in the direction ˆ k = − ˆ z (ii) The magnetic field ˜ H = 1 η ˆ k × ˜ E = 1 η ( − ˆ z ) × ˆ yje jπz = ˆ x 1 η je jπz (iii) In free space k = ω/c , therefore f = ck/ (2 π ) = 150 MHz. Also λ = 2 π/k = 2 m. (iv) Since this field has only a single component (i.e. ˆ y ), the polarization is linear . (b) ˜ E = ( j ˆ y − ˆ z ) e − j 1000 x (i) ˆ k = ˆ x (ii) The magnetic field ˜ H = 1 η ˆ k × ˜ E = 1 η ˆ x × ( j ˆ y − ˆ z ) e − j 1000 x = 1 η (ˆ y + j ˆ z ) e − j 1000 x (iii) k = 1000, so f = 1000 c/ (2 π ) = 47 . 75 GHz, and λ = 2 π/ 1000 = 6 . 28 mm. (iv) Here the field has two components (ˆ y and ˆ z ) of equal amplitude, 90 ◦ out of phase, so the polarization is circular. To classify the handedness, look in the time domain: E ( t ) = − ˆ y sin( ωt − 1000 x ) − ˆ z cos( ωt − 1000 x ). Fix a point in space ( x = 0), we can see the field rotate from − ˆ z to − ˆ y as time goes from ωt = 0 to ωt = π/ 2. By the right hand rule, curling with right-hand fingers in the direction of field rotation results in the right-hand thumb pointing in the ( − ˆ z ) × (...
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au10_312ps4_sol - ECE312 Autumn 2010 Problem Set 4...

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