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3 votes
3 votes

f implements,

A) (ABC)' + A'BC' + ABC

B) A + B + C

C) A ⊕ B ⊕ C

D) AB + BC + CA

4 Answers

2 votes
2 votes

plz refer here: https://gateoverflow.in/2751/gate1996_2-22

According to the  given $4\times 1$ multiplexer diagram, the output equations is as follows:

$f=\bar{A} \bar{B} I_0+\bar{A} B I_1+A \bar{B} I_2+A B I_3$
$$
\begin{aligned}
& \Rightarrow \bar{A} \bar B C+\bar{A} B C+A \bar{B} C+A B C \\
& \Rightarrow \bar{B} C(A+\bar{A})+B C(A+\bar{A})[\because A+\bar{A}=1] \\
& \Rightarrow \quad \bar{B} C+B C \\
& \Rightarrow \quad C(B+\bar{B}) \quad[B+\bar{B}=1] \\
& \text { f } \Rightarrow C \\
&
\end{aligned}
$$

edited by
2 votes
2 votes
Ans is "C". Not option C.

the Equation made from MuX = A'B'C + A'BC + AB'C + ABC

By solving this equation using K Map it will give ans as C

A'B'C + A'BC + AB'C + ABC = C.
0 votes
0 votes
f=A'B'C+A'BC+AB'C+ABC

=1

OPTION 1.   (ABC)'+A'BC'+ABC

                    =(A'+B'+C')+A'BC'+ABC

                     =A'(1+BC')+B'+C'+ABC

                      =A'+B'+C'+ABC

                     =(ABC)'+ABC

                      =1

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