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ps2sol_6243_2003

# ps2sol_6243_2003 - Massachusetts Institute of Technology...

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+ Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Problem Set 2 Solutions 1 Problem 2.1 Consider the feedback system with external input r = r ( t ) , a causal linear time invariant forward loop system G with input u = u ( t ) , output v = v ( t ) , a ) 1 / 2 e t , where ¯ and impulse response g ( t ) = 0 . 1 ( t ) + ( t + ¯ a 0 is a parameter, and a memoryless nonlinear feedback loop u ( t ) = r ( t ) + π ( v ( t )) , where π ( y ) = sin( y ) . It is customary to require well-posedness of such feedback models, r ± u ± G ± v π ( y ) Figure 2.1: Feedback setup for Problem 2.1 which will usually mean existence and uniqueness of solutions v = v ( t ) , u = u ( t ) of system equations t v ( t ) = 0 . 1 u ( t ) + h ( t δ ) u ( δ ) dδ, u ( t ) = r ( t ) + π ( v ( t )) 0 on the time interval t [0 , ) for every bounded input signal r = r ( t ) . 1 Version of October 8, 2003

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2 (a) Show how Theorem 3.1 from the lecture notes can be used to prove well-posedness in the case when ¯ a > 0 . In terms of the new signal variable y ( t ) = v ( t ) 0 . 1 π ( v ( t )) 0 . 1 r ( t ) system equations can be re-written as t y ( t ) = h ( t δ )[ r ( δ ) + ( y ( δ ) + 0 . 1 r ( δ ))] dδ, 0 where t ( t + a ) 1 / 2 e , t 0 h ( t ) = 0 , otherwise, and : R ∞� R is the function which maps z R into π ( q ), with q being the solution of q 0 . 1 π ( q ) = z. Since π is continuously differentiable, and its derivative ranges in [ 1 , 1], is con- tinuously differentiable as well, and its derivative ranges between 1 / 1 . 1 and 1 / 0 . 9. For every constant T [0 , ), the equation for y ( t ) with t T can be re-written as t y ( t ) = y ( T ) + a T ( y ( δ ) , δ, t ) dδ, T where a T y, δ, t ) = h ( t δ )[ r ( δ ) + ( y ( δ ) + 0 . 1 r ( δ ))] + h T ( t ) , T h T ( t ) = h ˙ ( t δ )[ r ( δ ) + ( y ( δ ) + 0 . 1 r ( δ ))] dδ. 0 When parameter a takes a positive value, function a = a T satisfies conditions of ¯ Theorem 3.1 with X = R n , ¯ a ) being a x 0 = y ( T ), r = 1, and t 0 = T , with K = K function of ¯ a = 0, and M = M T = M 0 ( a )(1 + max y ( t ) ) . t [0 ,T ] | | Hence a solution y = y ( · ) defined on an interval t [0 , T ] can be extended in a unique way to the interval t [0 , T + ], where T + T = min { 1 /M T , 1 / (2 K ) } , and max y ( t ) M T ( T + T ) + max y ( t ) ) .
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