# A4_SOL - Signals and Systems Fall 2013 Solution to Homework...

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Signals and Systems Fall 2013 Solution to Homework Assignment #4 1. x ( k ) y ( k ) 0.45 E 1 1.05 E 1 1.4 E 1 2 y 1 ( k ) y 2 ( k ) y 3 ( k ) (a) From the diagram, we have: 2 1 ( 1) ( ) y k y k + = 2 1 { ( )} ( ) E y k y k = (1) 1 1 2 ( 1) 1.4 ( ) 0.45 ( ) ( ) y k y k y k x k + = + 1 2 ( 1.4){ ( )} 0.45 ( ) ( ) E y k y k x k = + (2) Substituting (1) into (2) yields 2 2 ( 1.4)( ){ ( )} 0.45 ( ) ( ) E E y k y k x k = + 2 2 ( 1.4 0.45){ ( )} ( ) E E y k x k + = (3) 3 2 ( ) 2 ( ) ( ) y k y k y k = + (4) 3 3 ( 1) 1.05 ( ) y k y k + = 3 ( 1.05){ ( )} 0 E y k = (5) 3 ( ) (1.05) k y k m . Note that m depends on the initial condition. Observe that 3 ( ) y k does not depend on the input ( ) x k . Furthermore, we have 2 3 ( ) ( ) 2 ( ) y k y k y k = + . As such, the operational transfer function can be obtained by using just the component 2 ( ) y k in the output. In particular, substituting (6) into (3) yields 2 ( 1.4 0.45){ ( )} ( ) E E y k x k + = Finally, we obtain the operational transfer function of the system as: 2 1 ( ) 1.4 0.45 H E E E = +

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