practice_problemsSolution - EECS 20N Final Exam Practice...

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EECS 20N Final Exam Practice Problems SOLUTIONS, Spring 2006 1 (a) s ( n + 1) = As ( n ) + Bx ( n ) s (0) = s 1 (0) s 2 (0) Zero-Input State Response x ( n ) = 0 , n s ( n + 1) = As ( n ); A = 1 1 0 1 s (1) = As (0) = 1 1 0 1 ‚• s 1 (0) s 2 (0) = s 1 (0) + s 2 (0) s 2 (0) s (2) = As (1) = 1 1 0 1 ‚• s 1 (0) + s 2 (0) s 2 (0) = s 1 (0) + 2 s 2 (0) s 2 (0) s ( n ) = s 1 (0) + ns 2 (0) s 2 (0) (b) h ( n ) = c T A n - 1 b, n 1; h (0) = d = 0 A 0 = 1 0 0 1 A 1 = 1 1 0 1 A 2 = 1 2 0 1 A n = 1 n 0 1 , n 0 h (1) = c T A 0 b = £ 1 0 / 1 0 0 1 ‚• 0 1 = £ 1 0 / 0 1 = [0] h (2) = c T A 1 b = £ 1 0 / 1 1 0 1 ‚• 0 1 = £ 1 1 / 0 1 = [1] h ( n ) = c T A n - 1 b = £ 1 0 / 1 n - 1 0 1 ‚• 0 1 = £ 1 n - 1 / 0 1 = [ n - 1] 1
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(c) s (0) = 1 1 ; x ( n ) = δ ( n - 1) = 1 , n = 1 0 , otherwise n 0 , y ( n ) = c T A n s (0) + n X k =0 h ( n - k ) x ( k ) y (0) = c T A 0 s (0) + h (0) x (0) = c T A 0 s (0) = £ 1 0 / 1 0 0 1 ‚• 1 1 = £ 1 0 / 1 1 = [1] y (1) = c T A 1 s (0) + 1 X k =0 h (1 - k ) x ( k ) = c T A 1 s (0) + h (0) = £ 1 0 / 1 1 0 1 ‚• 1 1 + [0] = £ 1 1 / 1 1 + [0] = [2] y ( n ) = c T A n s (0) + n X k =0 h (1 - k ) x ( k ) = c T A n s (0) + h ( n - 1) = £ 1 0 / 1 n 0 1 ‚• 1 1 + h [ n - 1] = £ 1 n / 1 1 + h [ n - 1] = [1 + n ] + [ n - 2] = [2 n - 1] 2 y ( n ) - y ( n - 1) = x ( n ) - 2 x ( n - 1) y ( n ) = x ( n ) - 2 x ( n - 1) + y ( n - 1) 2
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(a) s ( n ) = y ( n - 1) x ( n - 1) s ( n + 1) = y ( n ) x ( n ) = A y ( n - 1) x ( n - 1) + Bx ( n ) = 1 - 2 0 0 | {z } A y ( n - 1) x ( n - 1) + 1 1 | {z } B x ( n ) y ( n ) = C y ( n - 1) x ( n - 1) + dx ( n ) = £ 1 - 2 / | {z } C y ( n - 1) x ( n - 1) + 1 |{z} d x ( n ) Zero State Impulse Response s (0) = 0 0 ; x ( n ) = δ ( n ) y (0) = d = 1 s (0) = 0 0 T y (1) = Cs (1) s (1) = B = 1 1 T y (2) = Cs (2) s (2) = As (1) y (3) = Cs (3) s (3) = A 2 s (1) . . . . . . y ( n ) = Cs ( n ) s ( n ) = A n - 1 s (1) A n = 1 - 2 0 0 A n 1 1 = - 1 0 h (0) = 1 h (1) = £ 1 - 2 / 1 1 = - 1 h (2) = £ 1 - 2 / - 1 0 = - 1 . . . h ( n ) = - 1 3
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0 1 -1 1 2 3 n h(n) (b) y ( n ) = x ( n ) - 2 x ( n - 1) + y ( n - 1) + + z -1 z -1 -2 y(n) x(n) (c) s ( n ) = - 2 x ( n - 1) + y ( n - 1) s ( n + 1) = - 2 x ( n ) + y ( n ) = - x ( n ) + s ( n ) A = 1 ,B = - 1 ,C = 1 ,d = 1 Zero-State Impulse Response s (0) = 0 y (0) = 1 s (1) = - 1 y (1) = - 1 s (2) = - 1 y (2) = - 1 . . . . . . + -2 + Z -1 X(n) Y(n) S(n) (d) Same as (a) 4
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(e) Frequency response x ( n ) = e jωn , y ( n ) = H ( ω ) e jωn H ( ω ) e jωn = e jωn - 2 e ( n - 1) + H ( ω ) e ( n - 1) H ( ω )(1 - e - ) = 1 - 2 e - H ( ω ) = 1 - 2 e - 1 - e - Using impulse response: h ( n ) = δ ( n ) - u ( n - 1) δ ( n ) DTFT -→ 1 u ( n ) DTFT -→ 1 1 - e - u ( n - 1) DTFT -→ e - 1 - e - H ( ω
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This note was uploaded on 04/01/2008 for the course EECS 20n taught by Professor Babakayazifar during the Spring '08 term at University of California, Berkeley.

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practice_problemsSolution - EECS 20N Final Exam Practice...

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