Chap3StudentSolutions

Chap3StudentSolutions - M J Roberts Chapter 3 Mathematical...

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M. J. Roberts - 8/16/04 Solutions 3-1 Chapter 3 - Mathematical Description and Analysis of Systems Selected Solutions 1. Show that a system with excitation, x t ( ) , and response, y t ( ) , described by y u x t t ( ) = ( ) ( ) is non-linear, time invariant, stable and non-invertible. Homogeneity: Let x g 1 t t ( ) = ( ) . Then y u g 1 t t ( ) = ( ) ( ) . Let x g 2 t K t ( ) = ( ) . Then y u g y u g 2 1 t K t K t K t ( ) = ( ) ( ) ( ) = ( ) ( ) . Not homogeneous Additivity: Let x g 1 t t ( ) = ( ) . Then y u g 1 t t ( ) = ( ) ( ) . Let x h 2 t t ( ) = ( ) . Then y u h 2 t t ( ) = ( ) ( ) . Let x g h 3 t t t ( ) = ( ) + ( ) . Then y u g h y y u g u h 3 1 2 t t t t t t t ( ) = ( ) + ( ) ( ) ( ) + ( ) = ( ) ( ) + ( ) ( ) Not additive Since it is not homogeneous and not additive, it is not linear. It is also not incrementally linear because incremental changes in the excitation do not produce proportional incremental changes in the response. It is statically non-linear because it is non-linear without memory (lack of memory proven below). Time Invariance: Let x g 1 t t ( ) = ( ) . Then y u g 1 t t ( ) = ( ) ( ) . Let x g 2 0 t t t ( ) = ( ) . Then y u g y 2 0 1 0 t t t t t ( ) = ( ) ( ) = ( ) . Time Invariant Stability: The unit step function can only have the values, zero or one, therefore any bounded (or unbounded) excitation produces a bounded response. Stable Causality: The response at any time, t t = 0 , depends only on the excitation at time, t t = 0 and not on any future values.
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M. J. Roberts - 8/16/04 Solutions 3-2 Causal Memory: The response at any time, t t = 0 , depends only on the excitation at time, t t = 0 and not on any past values. System has no memory. Invertibility: There are many value of the excitation that all cause a response of zero and there are many values of the excitation that all cause a response of one. Therefore the system is not invertible. 2. Show that a system with excitation, x t ( ) , and response, y t ( ) , described by y x x t t t ( ) = ( ) ( ) 5 3 is linear but not causal and not invertible. Causality: At time, t = 0, y x x 0 5 3 ( ) = ( ) ( ) . Therefore the response at time, t = 0, depends on the excitation at a later time, t = 3. Not Causal Memory: At time, t = 0, y x x 0 5 3 ( ) = ( ) ( ) . Therefore the response at time, t = 0, depends on the excitation at a previous time, t = − 5. System has memory. Invertibility: A counterexample will demonstrate that the system is not invertible. Let the excitation be a constant, K . Then the response is y t K K ( ) = = 0. This is the response, no matter what K is. Therefore when the response is a constant zero, the excitation cannot be determined. Not Invertible. 3. Show that a system with excitation, x t ( ) , and response, y t ( ) , described by y x t t ( ) = 2 is linear, time variant and non-causal. Time Invariance: Let x g 1 t t ( ) = ( ) . Then y g 1 2 t t ( ) = . Let x g 2 0 t t t ( ) = ( ) . Then y g y g 2 0 1 0 0 2 2 t t t t t t t ( ) =  ≠ ( ) = . Time Variant Causality:
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M. J. Roberts - 8/16/04 Solutions 3-3 At time, t = − 2, y x ( ) = ( ) 2 1 . Therefore the response at time, t = − 2, depends on the excitation at a later time, t = − 1.
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