ECE
soln7

# soln7 - ECE320 Solution Notes 7 Cornell University Spring...

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ECE320 Solution Notes 7 Spring 2006 Cornell University T.L.Fine 1. Formally describe the prefix decoder shown in Figure 1 by specifying the components Σ , I , O , f, g . start 0 1 2 σ σ σ [0]/0 [1]/ ε [1]/ ε [0]/10 [0]/110 [1]/111 Figure 1: Prefix Decoder Σ = { σ 0 , σ 1 , σ 2 } , I = { 0 , 1 } = B , O = { , 0 , 10 , 110 , 111 } or = I * . f : Σ × I → Σ , f ( σ 0 , 0) = σ 0 , f ( σ 0 , 1) = σ 1 , f ( σ 1 , 0) = σ 0 , f ( σ 1 , 1) = σ 2 , f ( σ 2 , 0) = σ 0 , f ( σ 2 , 1) = σ 0 . g : Σ × I → O , g ( σ 0 , 0) = 0 , g ( σ 0 , 1) = , g ( σ 1 , 0) = 10 , g ( σ 1 , 1) = , g ( σ 2 , 0) = 110 , g ( σ 2 , 1) = 111 . 2. Design a door access electronic combination lock as an FSA . The lock characteristics are a keypad containing keys for the five numerals 1 , . . . , 5 and a doorknob. You need to enter the following three digits 3, 2, 5, immediately after the lock has been reset, to ready the lock to be unlocked. Turning the doorknob either opens the door (if the combination is correct) or resets the lock to its locked position (if the combination is in error). Releasing the doorknob also resets the lock to its locked position. Your FSA should include the three states lock reset, activated doorknob, lock open . The final state can be taken to be lock reset , although this is irrelevant. Your inputs should include the two commands

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• Spring '06
• FINE
• Combination lock, doorknob

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