113_1_example_midterm_sols

# 113_1_example_midterm_sols - MIDTERM SOLUTIONS 1 We can...

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Unformatted text preview: MIDTERM SOLUTIONS 1. We can find the input–output relation of the system by iteration as follows: y (0) = x (0) y (1) = 1 2 x (0) + x (2) y (2) = 1 4 x (0) + 1 2 x (2) + x (4) . . . y ( n ) = 1 2 ¶ n £ x (0) + 2 x 2 (1) + ... + 2 n x 2 ( n ) / Then the general form of y ( n ) is y ( n ) = 1 2 ¶ n n X i =0 2 i x (2 i ) , n ≥ (a) Let y 1 ( n ) and y 2 ( n ) denote the output sequences that correspond to an input sequences x 1 ( n ) and x 2 ( n ), respectively. Then y 1 ( n ) and y 2 ( n ) can be expressed as y 1 ( n ) = S [ x 1 ( n )] = 1 2 ¶ n n X i =0 2 i x 1 (2 i ) , n ≥ y 2 ( n ) = S [ x 2 ( n )] = 1 2 ¶ n n X i =0 2 i x 2 (2 i ) , n ≥ Now, the output y ( n ) that corresponds to the linear combination ax 1 ( n ) + bx 2 ( n ) is y ( n ) = S [ ax 1 ( n ) + bx 2 ( n )] = 1 2 ¶ n n X i =0 2 i ( ax 1 (2 i ) + bx 2 (2 i )) , n ≥ whereas ay 1 ( n ) + by 2 ( n ) = a 1 2 ¶ n n X i =0 2 i x 1 (2 i ) + b 1 2 ¶ n n X i =0 2 i x 2 (2 i ) , n ≥ clearly, S [ ax 1 ( n ) + bx 2 ( n )] = ay 1 ( n ) + by 2 ( n ) Therefore, the system is linear . (b) The system response to the input x ( n- k ) is given by y k ( n ) = S [ x ( n- k )] = 1 2 ¶ n n X i = k 2 i x (2 i- k ) Note that the lower limit of the sum has changed from 0 to k 2 for even k or k +1 2 for odd k...
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## This note was uploaded on 02/09/2011 for the course EE 113 taught by Professor Walker during the Spring '08 term at UCLA.

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113_1_example_midterm_sols - MIDTERM SOLUTIONS 1 We can...

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