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A logic network has two data inputs $A$ and $B$, and two control inputs $C_0$ and $C_1$. It implements the function $F$ according to the following table.
$${\begin{array}{|cc|c|}\hline
\textbf{$C_1$}&    \textbf{$C_0$}&  \textbf{F}\\\hline
0&0&\text{$\overline{A + B}$} \\ 0&1& \text{A + B} \\     1&0& \text{$A \oplus B$ } \\ \hline
\end{array}}$$Implement the circuit using one $4$ to $1$ Multiplexer, one $2-$input Exclusive OR gate, one $2-$input AND gate, one $2-$input OR gate and one Inverter.
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Best answer

This is the implementation asked in question

  • $C_0 = 0 , C_1 = 0$  line $00$ will be selected and $F$ will give $(A+B)'$
  • $C_0 = 0 , C_1 = 1$  line $01$ will be selected and $F$ will give $(A\oplus B)$
  • $C_0 = 1 , C_1 = 0$  line $10$ will be selected and $F$ will give $(A+B)$
  • $C_0 = 1 , C_1 = 1$ line $11$ will be selected and $F$ will give $(A+B)'.(A+B) =0$
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DOUBT: why input is taken as (A+B)’.(A+B)  for c0=1,c1=1?

just to make zero?

CORRECTION:in the question output is EXOR  for c0=1,c1=0 ? if  wrong then pls explain?

 

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edited by

In question "using one 2 input AND gate" is mention, so for that purpose (A+B)(A+B)’

is used.

–1
–1

@Pranavpurkar

for the correction part
no there is no problem in solution in the question the sequence is C1 then C0
and in solution the sequence is C0 then C1 

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