113_1_SOLUTION_MIDTERM-2011-VERSION-B

# 113_1_SOLUTION_MIDTERM-2011-VERSION-B - EE113 Digital...

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Unformatted text preview: EE113: Digital Signal Processing Spring 2011 Solution to Midterm Exam (Version B) 1. The difference equation of a relaxed system is: y ( n ) − 1 6 y ( n − 1) − 1 6 y ( n − 2) = x ( n ) (a) Find a closed form for the impulse response h ( n ) (i.e., the zero state system output when the input is an impulse). Solution : By setting x ( n ) = δ ( n ), we have h ( n ) − 1 6 h ( n − 1) − 1 6 h ( n − 2) = δ ( n ) = braceleftbigg 1 , n = 0 , n negationslash = 0 . Hence, for n ≥ 1, h ( n ) can be found by solving homogeneous difference equation h ( n ) − 1 6 h ( n − 1) − 1 6 h ( n − 2) = 0 . (1) The characteristic polynomial for (1) can be solved by parenleftbigg λ − 1 2 parenrightbiggparenleftbigg λ + 1 3 parenrightbigg = 0 , thus, the modes are λ 1 = 1 2 , λ 2 = − 1 3 . Hence, h ( n ) = C 1 parenleftbigg 1 2 parenrightbigg n + C 2 parenleftbigg − 1 3 parenrightbigg n , n ≥ . Since the system is relaxed, y ( − 1) = h ( − 1) = 0 and h (0) = δ (0) = 1, which gives C 1 = 3 5 , C 2 = 2 5 . Therefore, h ( n ) = bracketleftbigg 3 5 parenleftbigg 1 2 parenrightbigg n + 2 5 parenleftbigg − 1 3 parenrightbigg n bracketrightbigg u ( n ) . (b) If the input to the system is x ( n ) = u ( n − 2), what is the output? Solution : The output y ( n ) of the system can be expressed as the convolu- tion of x ( n ) and h ( n ), i.e., y ( n ) = x ( n ) ∗ h ( n ) = u ( n − 2) ∗ h ( n ) . Hence, using h ( n ) in part (a) gives us i) n ≤ 1, y ( n ) = 0 ii) n ≥ 2, y ( n ) = ∞ summationdisplay k =-∞ x ( k ) h ( n − k ) = ∞ summationdisplay k =-∞ u ( k − 2) bracketleftBigg 3 5 parenleftbigg 1 2 parenrightbigg n- k + 2 5 parenleftbigg − 1 3 parenrightbigg n- k bracketrightBigg u ( n − k ) = n summationdisplay k =2 bracketleftBigg 3 5 parenleftbigg 1 2 parenrightbigg n- k + 2 5 parenleftbigg − 1 3 parenrightbigg n- k bracketrightBigg = n- 2 summationdisplay k =0 bracketleftBigg 3 5 parenleftbigg 1 2 parenrightbigg k + 2 5 parenleftbigg − 1 3 parenrightbigg k bracketrightBigg = 3 5 ....
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## This note was uploaded on 10/03/2011 for the course EE 113 taught by Professor Walker during the Spring '08 term at UCLA.

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113_1_SOLUTION_MIDTERM-2011-VERSION-B - EE113 Digital...

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