design a circuit that counts only 1 invalid input 1111,output is 00, valid input 1110 and output is 10 valid input 1010output is 11
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valid input 1110, output should be 11
1010 10.
see the diagram for truth table, Karnaugh map, then the Logic functions in terms of variables. Then the combinatorial circuit with gates AND, OR, NOT.
If exclusive OR gates could be used , then circuit will be simpler.
X = ABC' + AC'D + BC'D + A'CD + BCD' + AB'C
= A(B'C + C'B) + B (C'D + D'C) + D (A C' + A' C)
[ALSO, X' = A'B'C' + A'B' D' + A'C'D' + B'C'D' + ABCD
By taking the zeros in the table.]
Y = (A'B + A B') C'D' + (A'B' + AB) C'D + CD (A'B + A B')+ (A'B'+AB)CD'
= (A'B + A B') (C'D' + CD) + ( A'B' + AB) (C'D + C D')
valid input 1110, output should be 11
1010 10.
see the diagram for truth table, Karnaugh map, then the Logic functions in terms of variables. Then the combinatorial circuit with gates AND, OR, NOT.
If exclusive OR gates could be used , then circuit will be simpler.
X = ABC' + AC'D + BC'D + A'CD + BCD' + AB'C
= A(B'C + C'B) + B (C'D + D'C) + D (A C' + A' C)
[ALSO, X' = A'B'C' + A'B' D' + A'C'D' + B'C'D' + ABCD
By taking the zeros in the table.]
Y = (A'B + A B') C'D' + (A'B' + AB) C'D + CD (A'B + A B')+ (A'B'+AB)CD'
= (A'B + A B') (C'D' + CD) + ( A'B' + AB) (C'D + C D')
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