7. Implement the F from the following truth table using 4 input multiplexers and any necessary logic gates. A B C D F 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1
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From the truth table, we can see that F is 0 for all inputs except when A=1, B=1, C=0, and D=1. In that case, F is 1. To implement this behavior using 4-input multiplexers, we can use the following steps: Show more…
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Truth table for system of four NAND gates as shown in Fig. $20.10$ is $$ \begin{aligned} &\text { (A) }\\ &\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (B) }\\ &\begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (C) }\\ &\begin{array}{|l|l|l|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { D) }\\ &\begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{array} \end{aligned} $$
B/ Write a Boolean expression for this truth table, then simplify that expression as much as possible, and draw a logic gate circuit equivalent to that simplified expression: AB C Output 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1
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