problem5_sol

problem5_sol - EE221A Problem Set 5 Solutions - Fall 2011...

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EE221A Problem Set 5 Solutions - Fall 2011 Problem 1. Dynamical systems, time invariance. i) To show that this is a dynamical system we have to identify all the ingredients: First we need a differential equation of the form ˙ x = f ( x,u,t ) : Let x ( t ) = y ( t ) (so h ( x,u,t ) = f ( x,u,t ) ) and differentiate (using Liebniz) the given integral equation to get d dt x ( t ) = - x ( t ) + u ( t ) this a linear time invariant dynamical system by inspection (it’s of the form ˙ x ( t ) = Ax ( t ) = Bu ( t ) ) but we can show the axioms. First let’s call the system D = ( U , Σ , Y ,s,r ) . The time domain is T = R . The input space U is as specified in the problem; the state space Σ and output space Y are identical and are R . The state transition function is s ( t,t 0 ,x 0 ,u ) = x ( t ) = e - ( t - t 0 ) x 0 + ˆ t t 0 e - ( t - τ ) u ( τ ) and the readout function is r ( t,x ( t ) ,u ( t )) = y ( t ) = x ( t ) Now to show the axioms. The state transition axiom is easy to prove, since u ( · ) only enters the state transition function within the integral where it is only evaluated on [ t 0 ,t 1 ] (where t 0 and t 1 are the limits of the integral). For the semi group axiom, let s ( t 1 ,t 0 ,x 0 ,u ) = x ( t 1 ) be as defined above. Then plug this into s ( t 2 ,t 1 ,s ( t 1 ,t 0 ,x 0 ,u ) ,u ) = e - ( t 2 - t 1 ) ± e - ( t 1 - t 0 ) x 0 + ˆ t 1 t 0 e - ( t 1 - τ ) u ( τ ) ² + ˆ t 2 t 1 e - ( t 2 - τ ) u ( τ ) = e - ( t 2 - t 0 ) x 0 + ˆ t 1 t 0 e - ( t 2 - τ ) u ( τ ) + ˆ t 2 t 1 e - ( t 2 - τ ) u ( τ ) = e - ( t 2 - t 0 ) x 0 + ˆ t 2 t 0 e - ( t 2 - τ ) u ( τ ) = s ( t 2 ,t 0 ,x 0 ,u ) , for all t 0 t 1 t 2 , as required. ii) To show that this d.s. is time invariant, we need to show that the space of inputs is closed under the time
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This note was uploaded on 12/09/2011 for the course EE 221A taught by Professor Clairetomlin during the Fall '10 term at Berkeley.

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problem5_sol - EE221A Problem Set 5 Solutions - Fall 2011...

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