A combinational circuit is to be designed that counts the number of occurrences of 1 bits in a 4 bit input, however, an input 1111 is an invalid input for the circuit and output in such a case will be 00. One valid input for such circuit may be 1110 having the output 11; another valid input may be 1010 with the output 10. Draw the truth table for the circuit. Use the Karnaugh's map to design the circuit and draw it using AND, OR and NOT gates
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Combinatorial circuit is in the diagram enclosed.
Truth table and Karnaugh map are first drawn. Then the circuit using the logic gates AND, OR, NOT.
The circuit will be simpler with use of exclusive or gates.
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')
Truth table and Karnaugh map are first drawn. Then the circuit using the logic gates AND, OR, NOT.
The circuit will be simpler with use of exclusive or gates.
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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