EE 20N - Spring 2000 - Varaiya - Midterm 1 Solutions

EE 20N - Spring 2000 - Varaiya - Midterm 1 Solutions - are...

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EECS20, Spring 2002 – Solutions to Midterm 1 1. 15 points (a) Not well-formed : The feedback composition has more than one £xed point solution when the initial state of B is 1 and of C is x . (b) Well-formed : The feedback composition has a unique non-stuttering in- put for all reachable states. (c) Not well-formed : The output of C is not a subset of the inputs of A . 2. 20 points (a) (7 points) State transition diagram for A is shown below: a b c d {1}/0 {0}/0 {reset}/0 {0,reset}/0 {reset}/0 {0}/0 {1}/0 {1}/1 {reset}/0 {1}/0 {0}/0 (b) (5 points) The following machine is bisimilar to A : initialState = a Inputs = { 0,1,reset,absent } Outputs = { 0,1,absent } States = { a’,c’,d’ } 1
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a' c' d' {reset,0}/0 {0,reset}/0 {1}/0 {1}/1 {reset}/0 {1}/0 {0}/0 (c) (3 points) { (a,a’),(b,a’),(c,c’),(d,d’) } (d) (5 points) a {0,reset,1}/0 {1}/1 3. 20 points (a) Yes, the machine B can be de£ned via a £nite state machine model. The set of Inputs , Outputs and States , and initialState
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Unformatted text preview: are the same as in A ; only the update function changes appropriately. (Note: It is possible to concoct examples where B stays in its initial state for all possible inputs; i.e., all other states are unreachable. Nevertheless, B still satises the properties of a FSM.) (b) No. Consider the following counterexample: 2 a {1}/0 b c {1}/0 {1}/0 In the arc-reversed machine, there are two arcs emanating from state b for the same input. 4. 10 points In order for the system to be well-formed, it must have a unique non-stuttering xed point. Using the input-output relationship y ( n ) = cs ( n ) + dx ( n ) and the feedback law x ( n ) = ky ( n ) , we obtain y ( n ) = cs ( n ) + dk y ( n ) y ( n ) = cs ( n ) 1-dk . This leads to the requirement that 1-dk 6 = 0 , or equivalently, dk 6 = 1 . 3...
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EE 20N - Spring 2000 - Varaiya - Midterm 1 Solutions - are...

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