113_1_SOLUTION_MIDTERM-2011-VERSION-A

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

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Unformatted text preview: EE113: Digital Signal Processing Spring 2011 Solution to Midterm Exam (Version A) 1. Consider the following two systems y 1 ( n ) = 1 3 n summationdisplay k = n- 2 x ( k ) , y 2 ( n ) = braceleftBigg x ( n 4 ) n = 4 k,k integer n negationslash = 4 k,k integer . (a) Determine whether the systems are linear, time-invariant, re- laxed, BIBO stable, and causal. Justify your answer to receive full credit. Solution : Properties y 1 ( n ) y 2 ( n ) Relaxed Yes No Linear Yes Yes Time-Invariant Yes No BIBO Stable Yes Yes Causal Yes No • System y 1 ( n ): · y 1 ( n ) = 1 3 ∑ n k = n- 2 x ( k ) = 1 3 [ x ( n − 2) + x ( n − 1) + x ( n )] can be considered as a relaxed constant coefficient differ- ence equation. It is relaxed, linear, time-invariant, BIBO stable, and causal. • System y 2 ( n ): · Not relaxed: Counter example: if the first non-zero sam- ple of x ( n ) is at n = − 1, then y 2 ( − 4) will be non-zero, which is before − 1. · Linear: Let two outputs be y a 2 ( n ) = braceleftBigg x a ( n 4 ) n = 4 k,k integer n negationslash = 4 k,k integer . , and y b 2 ( n ) = braceleftBigg x b ( n 4 ) n = 4 k,k integer n negationslash = 4 k,k integer . , for an input x a ( n ) and x b ( n ), respectively. The output y ( n ) for an input Ax a ( n ) + Bx b ( n ) with some constants A and B is given by i) if n = 4 k , k integer, y ( n ) = Ax a parenleftBig n 4 parenrightBig + Bx b parenleftBig n 4 parenrightBig = Ay a 2 ( n ) + By b 2 ( n ) . ii) if n negationslash = 4 k , k integer, y ( n ) = 0 . Based on i) and ii), y ( n ) for an input Ax a ( n )+ Bx b ( n ) is y ( n ) = Ay a 2 ( n ) + By b 2 ( n ) , which implies that system y 2 ( n ) is linear. · BIBO Stable: For a bounded sequence | x ( n ) | ≤ M < ∞ with a finite positive number M , | y 2 ( n ) | = braceleftBigg | x ( n 4 ) | ≤ M < ∞ n = 4 k,k integer < ∞ n negationslash = 4 k,k integer ....
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113_1_SOLUTION_MIDTERM-2011-VERSION-A - EE113 Digital...

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