Question

6.5.4.1 Consider the following Boolean function. It is required to implement it using 8x1 MUX and
4x1 MUX
F(x, y, z) = x' y' z' + x' y z' + x y' z + x y z

Answer 1

Solution:

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

Now we have to implement the above Boolean function uisng 8×1 MUX and 4×1 MUX.

Answer 2

Given Boolean Function:

$$F(x,y,z)=x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}{z}^{\mathrm{\prime }}+x^{\mathrm{\prime }}y{z}^{\mathrm{\prime }}+x{y}^{\mathrm{\prime }}z+xyz$$

Step 1: Simplify the Function (Optional)

First, check if the function can be simplified:

$$F(x,y,z)=x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}{z}^{\mathrm{\prime }}+x^{\mathrm{\prime }}y{z}^{\mathrm{\prime }}+x{y}^{\mathrm{\prime }}z+xyz$$

Factor out common terms:

$$=z^{\mathrm{\prime }}(x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}+x^{\mathrm{\prime }}y)+z(x{y}^{\mathrm{\prime }}+xy)$$$$=z^{\mathrm{\prime }}(x^{\mathrm{\prime }}(y^{\mathrm{\prime }}+y))+z(x(y^{\mathrm{\prime }}+y))$$

Since $y^{\mathrm{\prime }}+y=1$:

$$F(x,y,z)=z^{\mathrm{\prime }}{x}^{\mathrm{\prime }}+zx$$

This simplifies to:

$$F(x,y,z)=x\oplus z$$

 

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Option 1: Implement Using an 8×1 MUX

An 8×1 MUX has 3 select lines (for 8 inputs) and can directly implement any 3-variable function.

1. Truth Table for $F(x,y,z)$:

   $x$	$y$	$z$	$F$
   0	0	0	1	($x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}{z}^{\mathrm{\prime }}$)
   0	0	1	0
   0	1	0	1	($x^{\mathrm{\prime }}y{z}^{\mathrm{\prime }}$)
   0	1	1	0
   1	0	0	0
   1	0	1	1	($x{y}^{\mathrm{\prime }}z$)
   1	1	0	0
   1	1	1	1	($xyz$)

2. MUX Configuration:

• Select Lines: $S_2=x$, $S_1=y$, $S_0=z$.

• Input Lines (D0–D7):

  $$D0=1,D1=0,D2=1,D3=0,D4=0,D5=1,D6=0,D7=1$$

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Option 2: Implement Using a 4×1 MUX

A 4×1 MUX has 2 select lines, so we must reduce the function to 2 variables by fixing one variable as a select line.

1. Choose $x$ and $y$ as Select Lines ($S_1=x$, $S_0=y$):

• The remaining variable $z$ determines the MUX inputs.

• 
  

2. Express $F$ in terms of $z$ for each $(x,y)$ combination:

• $x=0,y=0$: $F=z^{\mathrm{\prime }}$

• $x=0,y=1$: $F=z^{\mathrm{\prime }}$

• $x=1,y=0$: $F=z$

• $x=1,y=1$: $F=z$

3. MUX Configuration:

• $D0$ and $D1$: Connect to $z^{\mathrm{\prime }}$ (NOT $z$).

• $D2$ and $D3$: Connect directly to $z$.

• Input Lines (D0–D3):

  $$D0=z^{\mathrm{\prime }},D1=z^{\mathrm{\prime }},D2=z,D3=z$$

• Logic for Inputs:

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Answer:

• 8×1 MUX: Directly implement using the truth table (inputs = 1, 0, 1, 0, 0, 1, 0, 1).

• 4×1 MUX: Use $x,y$ as select lines and $z\mathrm{/}{z}^{\mathrm{\prime }}$ as inputs.