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Q.Derive a Canonical POS expression for a Boolean function F, represented by the following truth table: X Y Z F(X,Y,Z) 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 0
CBSECBSE Class XII Board 2019Subjective· 1mImportance★★★★★est
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Start your 14-day free trial to unlock the full solution →Canonical POS is formed by multiplying (ANDing) a Maxterm for every row where F = 0. Here F = 0 for rows XYZ = 001, 011, 110, 111, giving F(X,Y,Z) = (X+Y+Z′)(X+Y′+Z′)(X′+Y′+Z)(X′+Y′+Z′) = π(1,3,6,7).
Reconstructing the truth table
| X | Y | Z | F |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
Rule for Canonical POS: for every row where F = 0, write a Maxterm — an OR of the three variables — where a variable that is 0 in that row is written as-is, and a variable that is 1 in that row is written complemented. The Canonical POS is the AND (product) of all these maxterms.
Rows where F = 0:
- X=0, Y=0, Z=1 → M₁ = (X + Y + Z′) …
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