A logic circuit consists of two $2 \times 4$ decoders as shown below.

The output of decoder are given below

$D_{0} = 1 w h e n A_{0} = 0, A_{1} = 0$

$D_{1} = 1 w h e n A_{0} = 1, A_{1} = 0$

$D_{2} = 1 w h e n A_{0} = 0, A_{1} = 1$

$D_{3} = 1 w h e n A_{0} = 1, A_{1} = 1$

The value of f(a, b, c) will be _________.
1

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Solution

The correct option is A 1
In the given circuit

$D_{0} = {¯ ¯¯ ¯ A}_{1} . {¯ ¯¯ ¯ A}_{0}, D_{1} = {¯ ¯¯ ¯ A}_{1} A_{0}, D_{2} = A_{1} {¯ ¯¯ ¯ A}_{0}, D_{3} = A_{1} A_{0}$

For first decoder,

$A_{0} = a; A_{1} = b$

$D_{2} = b . ¯ ¯ ¯ a a n d D_{3} = a b$

For second decoder,

$A_{1} = D_{2} D_{3} = b . ¯ ¯ ¯ a, a . b = 0$

$A_{0} = c$

and $D_{0} = {¯ ¯¯ ¯ A}_{1} . {¯ ¯¯ ¯ A}_{0} + {¯ ¯¯ ¯ A}_{1} A_{0} = {¯ ¯¯ ¯ A}_{1} = 1$

The value of f(x, y, z) = 1

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