# SOL05 - Signals and Systems Fall 2012 Solution to Homework...

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Signals and Systems Fall 2012 Solution to Homework Assignment #5 1. (a) (E 2 + 3E + 2){ y ( k )} = x ( k ) The characteristic polynomial is 2 ( ) 3 2 ( 1)( 2) D E E E E E = + + = + + It can be seen that we have two roots 1, 2 , which are not inside the unit circle. The system is therefore unstable. (b) (10E 2 + 3E + 2){ y ( k )} = (E 2){ x ( k )} We have 2 ( ) 10 3 2 D E E E = + + The two conjugate roots are: 3 9 80 3 71 0.15 0.4218 20 20 20 j j ± = ± = ± It is not hard to verify that these roots are inside the unit circle. Hence, the system is stable. (c) (E 2 + 1.2E + 0.2){ y ( k )} = x ( k ) We have: 2 ( ) 1.2 0.2 ( 1)( 0.2) D E E E E E = + + = + + We therefore have two roots 1 and 0.2 . Since the root -1 is on the unit circle, the system is marginally stable. It is not BIBO stable. (d) (E 3 + 3E 2 + 3E + 1){ y ( k )} = (E 2 0.1E){ x ( k )} We have: 3 2 3 ( ) 3 3 1 ( 1) D E E E E E = + + + = + Obviously, we have a repeated root on the unit circle. Therefore, the system is not BIBO stable

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2. (a) Given the operation transfer function H 1 (E) = 5 . 0 E E . We have: 1 1 ( ) 1 0.5(1 ) H E E = 1 1 0.5 { ( )} ( ) ( ) 0.5 ( 1) ( ) y k x k E y k y k x k Λ Ξ × = Μ Ο Ν Π = The corresponding simulation diagram is plotted below (Matlab codes to generate impulse and step responses are provided in the next page): (b) Given the operation transfer function H 2 (E) = 4 2 3 4 0625 . 0 125 . 0 25 . 0 5 . 0 E E E E E + + + + .
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