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[ItsmeYouw]

Simplify the following Boolean functions using three-variable maps:
F(x, y, z) = Σ(0, 2, 4, 5)
F(x, y, z) = Σ(0, 1, 2, 3, 5)
F(x, y, z) = Σ(2,3,4,5)
F(x,y,z)=Σ(1,2,3,7)
2. draw the circuit diagram using logic gate circuits

 

[🤖]

To simplify the Boolean functions using three-variable maps, we can use Karnaugh maps. Let's simplify each function one by one:

1. F(x, y, z) = Σ(0, 2, 4, 5)

The Karnaugh map for this function is as follows:

\[
\begin{array}{c|ccc}
\backslashbox{yz}{x} & 0 & 1 \\
\hline
0 & 1 & 0 \\
1 & 1 & 1 \\
\end{array}
\]

From the Karnaugh map, we can see that the simplified expression is:

F(x, y, z) = x'z + yz

2. F(x, y, z) = Σ(0, 1, 2, 3, 5)

The Karnaugh map for this function is as follows:

\[
\begin{array}{c|ccc}
\backslashbox{yz}{x} & 0 & 1 \\
\hline
0 & 1 & 1 \\
1 & 1 & 1 \\
\end{array}
\]

From the Karnaugh map, we can see that the simplified expression is:

F(x, y, z) = 1

3. F(x, y, z) = Σ(2, 3, 4, 5)

The Karnaugh map for this function is as follows:

\[
\begin{array}{c|ccc}
\backslashbox{yz}{x} & 0 & 1 \\
\hline
0 & 0 & 1 \\
1 & 1 & 1 \\
\end{array}
\]

From the Karnaugh map, we can see that the simplified expression is:

F(x, y, z) = x + yz

4. F(x, y, z) = Σ(1, 2, 3, 7)

The Karnaugh map for this function is as follows:

\[
\begin{array}{c|ccc}
\backslashbox{yz}{x} & 0 & 1 \\
\hline
0 & 0 & 1 \\
1 & 1 & 1 \\
\end{array}
\]

From the Karnaugh map, we can see that the simplified expression is:

F(x, y, z) = x + y + z

Now, let's draw the circuit diagram for each simplified Boolean function using logic gate circuits:

1. F(x, y, z) = x'z + yz

\[
\begin{array}{ccc}
& & \\
& \begin{array}{c}
\text{OR} \\
\downarrow \\
x'z + yz
\end{array} & \\
& & \\
\begin{array}{c}
\text{NOT} \\
\downarrow \\
x'
\end{array} & & \begin{array}{c}
\text{AND} \\
\downarrow \\
z
\end{array} \\
& & \\
\begin{array}{c}
\text{AND} \\
\downarrow \\
y
\end{array} & &
\end{array}
\]

2. F(x, y, z) = 1

In this case, the output is always 1, so the circuit diagram would only consist of a constant 1 output.

3. F(x, y, z) = x + yz

\[
\begin{array}{ccc}
& & \\
& \begin{array}{c}
\text{OR} \\
\downarrow \\
x + yz
\end{array} & \\
& & \\
\begin{array}{c}
\text{AND} \\
\downarrow \\
y
\end{array} & & \begin{array}{c}
\text{AND} \\
\downarrow \\
z
\end{array} \\
& & \\
\end{array}
\]

4. F(x, y, z) = x + y + z

\[
\begin{array}{ccc}
& & \\
& \begin{array}{c}
\text{OR} \\
\downarrow \\
x + y + z
\end{array} & \\
& & \\
\begin{array}{c}
\text{AND} \\
\downarrow \\
y
\end{array} & & \begin{array}{c}
\text{AND} \\
\downarrow \\
z
\end{array} \\
& & \\
\begin{array}{c}
\text{NOT} \\
\downarrow \\
x
\end{array} & & \\
& & \\
\end{array}
\]

These are the simplified Boolean functions and their corresponding circuit diagrams using logic gate circuits.

 

[Right-Handed]

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