# A8_SOL - Signals and Systems Fall 2013 Solution to Homework...

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Signals and Systems Fall 2013 Solution to Homework Assignment #8 1. (a) (E 0.5){ y ( k )} 0, y (0) 0.   1 0.5 0 y k y k Taking the z-transform, we have:           0 0.5 0 0.5 0 z Y z y Y z zY z Y z   0 ZI Y z It then follows that:     0 ZI y k y k (b) (E 2 3E 2){ y ( k )} u s ( k ), y (0) 2, y (1) 0.                       2 1 2 2 0 1 3 0 2 3 2 1 3 0 S S z Y z y z y z Y z y Y z U z z z Y z U z z y z z y     2 2 2 2 2 2 1 3 2 3 2 1 3 2 3 2 2 6 2 1 2 1 ZIR ZSR ZIR ZSR z z z z Y z z z z z z z z z z z z z z z Using partial-fraction expansion, we have:    2 2 2 1 2 1 1 ZSR z z z z Y z A B C z z z z z Multiply   ZSR Y z by 2 z and set 2 z , we have: 2 2 2 1 2 1 A A Multiply   ZSR Y z by 2 1 z and set 1 z , we have: 1 1 1 1 2 C C     Then let z   , we have: 0 1 A B B     2 2 1 1 ZSR z z z Y z z z z Taking the inverse z-transform yields:         2 k ZS s s s y k u k u k ku k

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   2 2 6 2 1 2 1 ZIR z z z z Y z A B z z z z Multiply   ZIR Y z by 2 z and set 2 z : 2 2 2 6 2 2 2 2 1 A A   Multiply   ZIR Y z by 1 z and set 1 z : 2 2 1 6 1 1 4 1 2 B B     2 4 2 1 ZIR z z Y z z z Taking the inverse z-transform yields:         2 2 4 k ZI s s y k u k u k   Finally, the complete response is given as:   2 2 2 4 3 2 k k ZS ZI s s s s s k s s y k y k y k u k u k ku k u k u k ku k u k   (c) (E 2 2E 2){ y ( k )} (2) k u s ( k ), y (0) y (1) 0.             2 1 0 1 2 0 2 1 2 z z Y z y z y z Y z y Y z z It can be seen that we only have the zero-state response.
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• Fall '09
• MALIKELBULUK
• SIGNALS, Harshad number, Impulse response, ylabel

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