GATE EC 2016 Set 3 — Question 53
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Digital Circuits → Combinational Logic → K-Map Minimization
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Question
Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sum-of-product (SOP) expression for the function isFor :
For :
| RS \ PQ | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 0 | 0 | 0 | 0 |
| 01 | 1 | 0 | 0 | 1 |
| 11 | 1 | 0 | 0 | 1 |
| 10 | 0 | 0 | 0 | 0 |
| RS \ PQ | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 0 | 1 | 1 | 0 |
| 01 | 0 | 0 | 0 | 0 |
| 11 | 0 | 0 | 0 | 0 |
| 10 | 0 | 1 | 1 | 0 |
Correct answer
(B) QSX + QSX
Solution
map: the 1s are in columns (i.e. ) and rows (i.e. ). They form one quad: . With this gives . map: the 1s are in columns () and rows (), again one quad (the rows wrap around): . With this gives .So .
- (A) is logically equal to but uses four 4-literal terms, so it is not minimal.
- (C) swaps the literals, so it has 1s in the wrong cells.
- (D) ignores , so it covers 16 cells instead of 8.
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