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113_1_midterm_A_sol

# 113_1_midterm_A_sol - EE113 Digital Signal Processing...

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EE113: Digital Signal Processing Prof. Mihaela van der Schaar Midterm Exam Solution(Version A) Spring 2009 1. (a) The modes are the roots of the equatoin λ 2 - 5 λ + 6 = 0, which are λ 1 = 2 and λ 2 = 3. (b) The general homogeneous solution is C 1 2 n + C 2 3 n . The initial conditions are h (0) = 0 and h (1) = 2. We have C 1 + C 2 = 0 2 C 1 + 3 C 2 = 2 and the solutions are C 1 = - 2 and C 2 = 2. The impulse response is h ( n ) = [ - 2(2) n + 2(3) n ] u ( n ). (c) The step response can be computed by u ( n ) * h ( n ) = X k = -∞ u ( n - k )[ - 2(2) k + 2(3) k ] u ( k ) = n X k =0 [ - 2(2) k + 2(3) k ] , n 0 0 , n < 0 = 3 n +1 - 2(2) n +1 + 1 , n 0 0 , n < 0 We can also write it as [3 n +1 - 2(2) n +1 + 1] u ( n ). (d) We can get the step response without using convolution. Let the step response be w ( n ), then we first solve the particular solution, which is w p ( n ) = Ku ( n ). Substituting w p ( n ) = Ku ( n ) into the difference equation, we get Ku ( n ) - 5 Ku ( n - 1) + 6 Ku ( n - 2) = 2 u ( n - 1) For n 2, we have K = 1 and hence w p ( n ) = u ( n ). The homogeneous solution is w h ( n ) = A 1 (2) n + A 2 (3) n , hence the complete step response is w ( n ) = w p ( n )+ w h ( n )) = A 1 (2) n + A 2 (3) n

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113_1_midterm_A_sol - EE113 Digital Signal Processing...

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