Topic: Digital Electronics
In given Combinational logic, X is given by
(The diagram shows a 4:1 MUX with inputs $I_0=0, I_1=1, I_2=1, I_3=0$ selected by A, B outputting $F_o$, connected to a second 4:1 MUX with same inputs selected by $F_o, C$.)
- A. $X=A\overline{B}\overline{C}+\overline{A}B\overline{C}+\overline{A}\overline{B}C+ABC$
- B. $X=\overline{A}BC+A\overline{B}C+AB\overline{C}+\overline{A}\overline{B}\overline{C}$
- C. $X=AB+BC+AC$
- D. $X=\overline{A}\overline{B}+\overline{B}\overline{C}+\overline{A}\overline{C}$
Answer: A. $X=A\overline{B}\overline{C}+\overline{A}B\overline{C}+\overline{A}\overline{B}C+ABC$
Explanation: For the first 4:1 MUX, $F_o = \overline{A}B(1) + A\overline{B}(1) = A \oplus B$.
For the second 4:1 MUX, $X = \overline{F_o}C(1) + F_o\overline{C}(1) = F_o \oplus C = A \oplus B \oplus C$.
Expanding the XOR of 3 variables gives $X = A\overline{B}\overline{C} + \overline{A}B\overline{C} + \overline{A}\overline{B}C + ABC$.