**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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16 votes

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.

$${\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.

27 votes

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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