20101ee102_1_EE102Hw1Solution-1

# 20101ee102_1_EE102Hw1Solution-1 - EE102 HW 1 Solution: 1. 0...

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Unformatted text preview: EE102 HW 1 Solution: 1. 0 | x | > 0 f ( x) = 1 x = 0 The integral of a single point is zero. 2. ∞ π 1∞ F = ∫ e − (4+i 6π )t sin(12π t + )dt = ∫ e − (4+i 6π ) t [ei (12π t +π /5) − e − i (12π t +π /5) ]dt 0 5 2i 0 iπ /5 − iπ /5 e e eiπ /5 e − iπ /5 eiπ /5 (4 + 18π i ) − e −iπ /5 (4 − 6π i ) − = − = 2i (4 + 6π i − 12π i ) 2i (4 + 6π i + 12π i ) 2i (4 − 6π i ) 2i (4 + 18π i ) 2i (16 + 108π 2 + 48π i) ∫ 4 −1 f ( x) 2 dx = 0 = = 8sin(π / 5) + eiπ /5 (18π ) + e −iπ /5 (6π ) 8sin(π / 5) + 24π cos(π / 5) + i12π sin(π / 5) = 2(16 + 108π 2 + 48π i ) 2(16 + 108π 2 + 48π i ) (2sin(π / 5) + 6π cos(π / 5)) 2 + (3π sin(π / 5)) 2 2 sin(π / 5) + 6π cos(π / 5) + i3π sin(π / 5) →| F |= (8 + 54π 2 + 24π i ) (8 + 54π 2 ) 2 + (24π ) 2 3π sin(π / 5) 24π ) − tan −1 ( ) 2sin(π / 5) + 6π cos(π / 5) 8 + 54π 2 Some students used e − ( 4+16π )t instead of e − ( 4+i 6π )t to solve the problem which gives a different answer: R F = tan −1 ( ∞ π 1∞ F = ∫ e − (4+16π )t sin(12π t + )dt = ∫ e − (4+16π )t [ei (12π t +π /5) − e − i (12π t +π /5) ]dt 0 5 2i 0 iπ /5 − iπ /5 e e 2i sin(π / 5)(4 + 16π ) + 24iπ cos(π / 5) − = 2i (4 + 16π − 12π i ) 2i (4 + 16π + 12π i ) 2i ((4 + 16π ) 2 + (12π ) 2 ) sin(π / 5)(4 + 16π ) + 12π cos(π / 5) = ((4 + 16π ) 2 + (12π ) 2 ) sin(π / 5)(4 + 16π ) + 12π cos(π / 5) →| F |= , RF = 0 ((4 + 16π )2 + (12π ) 2 ) We will accept both solutions, since many people miss-read i6 to be 16. 3. i ) ∑ e 2π ik = k =0 ∞ N ∑1 = N + 1 k =0 N ∞ ∞ −1 1 1 1 1 1 1/ 2 ii ) ∑ ( )|k | e −2π ik = ∑ ( )|k | =∑ ( ) k + ∑ ( ) − k = + =3 1 − 1/ 2 1 − 1/ 2 k =−∞ 2 k =−∞ 2 k =0 2 k =−∞ 2 4. Ln(1 + x 4 ) x3 i)∫ dx = =0 −∞ 1 + x 4 4 −∞ ∞ ∞ ii ) ∫ cos( x)dx = [ sin( x) ] 0 = sin(∞) doesnt ' converge ∞ 0 ∞ 5. F= F= 1 1 1 8π v = →F= , R F = − tan −1 ( ) 22 2 22 22 z + 4 z + 5 −4v π + 8π vi + 5 (5 − 4v 2π 2 ) (5 − 4v π ) + 64π v 2 1 25 + 16v 4π 4 + 24v 2π 2 → Fmax v =0 = 1 5 ...
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## This note was uploaded on 09/03/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

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20101ee102_1_EE102Hw1Solution-1 - EE102 HW 1 Solution: 1. 0...

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