# A5_SOL - Signals and Systems Fall 2013 Solution to Homework...

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Signals and Systems Fall 2013 Solution to Homework Assignment #5 1. (a) y ( k ) 25 . 0 1 E { x ( k )} [Ans: h (0) 0; h ( k ) (0.25) k 1 , k 1,2,3,  .] From the input-output equation, the transfer function can be expressed as: 1 ( ) 0.25 H E E ( 1) 0.25 ( ) ( ) y k y k x k The diagram for the system is plotted below: Impulse response: 2 (0) 0 (1) 0.25(0) 1 1 (2) (0.25)(1) 0 0.25 (3) (0.25)(0.25) (0.25) y y y y Therefore, the impulse response is: 1 ( ) (0.25) ( 1) k s h k u k (b) y ( k ) ) 25 . 0 ( 1 E E { x ( k )} The transfer function is given as: 1 ( ) 0.25 H E E E ( 2) 0.25 ( 1) ( ) y k y k x k The diagram for the system is plotted below: E 1 x ( k ) y ( k ) 0.25 E 1 x ( k ) y ( k ) 0.25 E 1

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Impulse response: (0) 0 (1) 0 (2) 1 (3) 0.25 y y y y The impulse response is: 2 ( ) (0.25) ( 2) k s h k u k (c) y ( k ) 1 2 1 2 E E { x ( k )} From the input-output equation, the transfer function is: 2 1 ( ) 2 1 H E E E ( 2) 2 ( 1) ( ) ( ) y k y k y k x k   The diagram for the system is shown below: Impulse response: (0) 0 (1) 0 (2) (0) 1 (3) 2 (2) 2 (4) 2 (3) (2) 3 (5) 2 (4) (3) 4 y y y x y y y y y y y y           The impulse response is: 2 ( ) ( 1)( 1) ( 1) k s h k k u k . E 1 x ( k ) y ( k ) -2 E 1 -1
E 1 x ( k ) y ( k ) E 1 E 1 x ( k ) y ( k ) 2. (a) h ( k ) 2 , 0 2 , 1 k k The impulse response can be written as: ( ) ( 2) h k k It then follows that ( ) ( 2) y k x k 2 1 ( ) { ( )} y k x k E Hence, the operational transfer function can be expressed as: 2 1 ( ) H E E (The simulation diagram is depicted on the right) (b) h ( k ) 0 , 0 0 , 1 k k The above impulse response can be interpreted as follows: ( ) ( ) ( ) k s n h k u k n  Equivalently, ( ) (

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