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The finite state machine described by the following state diagram with A as starting state, where an arc label is and x stands for 1-bit input and y stands for 2-bit output

1. outputs the sum of the present and the previous bits of the input
2. outputs 01 whenever the input sequence contains 11
3. outputs 00 whenever the input sequence contains 10
4. none of the above
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This might help ....

option (B) and (C)  are clearly wrong . it says input 11 then o/p 01 and i/p 10 then o/p 00 but here at single bit o/p is 2 bit sequence

now for option (a) we can trace out .. suppose string is 0111

at A---0---> A----1---> B--1-->C---1-->C

O/P            00            01       10         10

we can see here at (A,0)---> (A,00) which sum of 0+0=00, (previous i/p bit + present i/p bit)

(A,1)--->(B,01) which is sum of 0+1= 1=01,

(B.1)--->(C,10) which is sum of 1+1= (previous i/p bit + present i/p bit)=10 ,

(C,1)---> (C,10) which is sum of 1+1=10

selected
Then why output of 101 in not 10???

And as per statement A it says sum of present and "previous bits" (not bit).

So according to me answer should be D)
it says sum of last two bits, not the sum whole string. So A should be correct.