Question

Design a circuit with three inputs (A, B, C) and two outputs (F1, F2). The first output F1 is logic 1 if the number of l’s in the binary number is less than the number of O's, otherwise F1 is logic 0. The second output F2 is 1 if the binary input is 2, 4, 5, 6,7 otherwise the second output F2 is logic 0. a. Derive the truth-table for F1 and F2 as a function of the 3 inputs. b. Simplify the output functions using K-map c. Draw the circuit diagram for F1 and F2 respectively. d. Convert the logic diagram for F2 to all NAND gate implementation.

Answer 1

Solution (a):-  

Truth table for F1 , F2 can be derived from the condition given in the problem.

A	B	C	F1	F2
0	0	0	0	0
0	0	1	1	0
0	1	0	1	1
0	1	1	0	0
1	0	0	1	1
1	0	1	0	1
1	1	0	0	1
1	1	1	0	1

Solution (b):-

Simplification using 3 input K-map

BU A. 00 01 10 0 FIE A'B'C + A'BC HAB'C'

БЕ A 00 О) 11 10 о F2 = A+Be/ 1)

Solution (c):-

Circuit diagram for F1 and F2

A B C A'B'C A Bic AB'C' Fi Valore A BC

A A F2 Bc

Solution (d):-
For NAND gate implementation F2 can be rewritten as F2 =(A'(BC')')'

A- B -F2 cs