20082ee102_1_HW-7_SOL

20082ee102_1_HW-7_SOL - UCLA Department of Electrical...

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UCLA Department of Electrical Engineering EE102: SYSTEMS & SIGNALS Spring 2008 Homework # 7 Solutions 1. # 4.13 Note that T = 2 π , so ω 0 = 1. Then, F n = 1 2 π Z π 0 1 · e - jnt dt = 1 2 π ± e - jπn - 1 - jn ² = 1 2 π ³ ( - 1) n - 1 - ´ = ( - j if n is odd. 0 if n is even. And F 0 = 1 2 π Z π 0 dt = 1 2 Since f ( t ) is real, f ( t ) = F 0 + X n =1 [2 a n cos( nt ) - 2 b n sin( nt )] Where a n = Re { F n } and Im { F n } Therefore, f ( t ) = 1 2 + X n =1 , n odd 2 sin( nt ) = 1 2 + 2 π X k =1 sin[(2 k - 1) t ] 2 k - 1 Fourier coefficients of the output y ( t ): Y n = H ( jnω 0 ) F n = ( + 1) 2 F n Hence Y n = 0 n = 0 1 π ( + 1) 2 n is odd. 0 otherwise 1
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Therefore, y ( t ) = X k = -∞ 1 π [1 + j (2 k - 1)] 2 e j (2 k - 1) t 2. Appex. # 12 (a) Note T = 2, so ω 0 = π . Then, G n = 1 2 Z 2 0 f ( t ) e - jnω 0 t dt = 1 2 Z 2 0 δ ( t - 1) e - jnπt dt = 1 2 e - jnπt (b) Y n = H ( jnω 0 ) G n = 1 1 + 20 jnπ G n = e - jnπt 2(1 + 20 jnπ ) (c) MSE = X -∞ | Y n | 2 - | Y 0 | 2 = 1 4 X n = -∞ , n 6 =0 1 1 + 400 n 2 π 2 < 1 4 X n = -∞ , n 6 =0 1 400 n 2 π 2 = 2 × 1 4 × 400 × π 2 X n =1 1 n 2 = 2 × 1 4 × 400 × π 2 · π 2 6 = 2 . 08 × 10 - 4 < 2 . 5 × 10 - 4 2
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3. Appex: # 65 (a) Note T = 4, so ω 0 = π 2 . Then, F
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This note was uploaded on 04/13/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

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20082ee102_1_HW-7_SOL - UCLA Department of Electrical...

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