Complete the following:
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Simplify the following equation using Boolean Identities:
~x~yz + ~xy~z + x~y~z + x~yz + xy~z + xyz
Construct truth tables for the following:
xyz+ x~(yz) + ~(xyz)
(x + y)(x + z)(~x + z)
Use the truth tables from the previous question to create equations in SOP and POS format.
The Correct Answer and Explanation is :
Simplification of Boolean Expression:
The given Boolean expression is:
[
\sim x \sim y z + \sim x y \sim z + x \sim y \sim z + x \sim y z + x y \sim z + x y z
]
Now, we will simplify it step-by-step using Boolean identities.
- Group terms based on common factors:
- ( \sim x \sim y z + \sim x y \sim z )
- ( x \sim y \sim z + x \sim y z )
- ( x y \sim z + x y z )
- Simplify each group:
- ( \sim x \sim y z + \sim x y \sim z = \sim x (\sim y z + y \sim z) )
- Using distributive law, simplify ( \sim y z + y \sim z ) as it represents an XOR operation: ( \sim x (y \oplus z) )
- ( x \sim y \sim z + x \sim y z = x \sim y (\sim z + z) = x \sim y \cdot 1 = x \sim y )
- By the complement law ( \sim z + z = 1 )
- ( x y \sim z + x y z = x y (\sim z + z) = x y \cdot 1 = x y )
- Again using the complement law ( \sim z + z = 1 )
- Final expression:
Combining all the simplified terms:
[
\sim x (y \oplus z) + x \sim y + x y
]
This is the simplified Boolean expression.
Truth Table Construction:
Let’s construct the truth tables for the following Boolean expressions:
1. ( xyz + x \sim (yz) + \sim (xyz) )
| x | y | z | yz | x(yz) | (\sim(xyz)) | Expression ( xyz + x \sim (yz) + \sim (xyz) ) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 | 1 |
2. ( (x + y)(x + z)(\sim x + z) )
| x | y | z | (x + y) | (x + z) | ((\sim x + z)) | Expression ( (x + y)(x + z)(\sim x + z) ) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
SOP and POS Equations from Truth Tables:
From the first truth table for ( xyz + x \sim (yz) + \sim (xyz) ):
- SOP (Sum of Products): From the rows where the output is 1 (rows 1, 2, 3, 4, 5, 6, 7, and 8), the SOP equation is:
[
xyz + x \sim yz + \sim xyz
]
- POS (Product of Sums): We can take the rows where the output is 0 (no rows have 0 in the table), so there is no POS expression for this.
From the second truth table for ( (x + y)(x + z)(\sim x + z) ):
- SOP (Sum of Products): From the rows where the output is 1 (rows 4, 6, 7, and 8), the SOP equation is:
[
(x + y)(x + z)(\sim x + z)
]
- POS (Product of Sums): From the rows where the output is 0 (rows 1, 2, 3, and 5), the POS equation is:
[
(x + y + z)(x + y + \sim z)(x + \sim y + z)
]
Explanation:
- SOP: The Sum of Products (SOP) format consists of AND operations followed by OR operations. For each row in the truth table where the output is 1, we write the corresponding product (AND of the variables). Finally, all the terms are OR’d together to form the SOP equation.
- POS: The Product of Sums (POS) format is the reverse of SOP. For each row in the truth table where the output is 0, we write the corresponding sum (OR of the variables). Finally, all the sums are ANDed together to form the POS equation.
In both the cases, simplifying the Boolean expressions helps in reducing the complexity of the logic circuits that could be built based on these expressions.