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+24 votes

Given the function $F = P' +QR$, where $F$ is a function in three Boolean variables $P, Q$ and $R$ and $P'=!P$, consider the following statements.

$(S1) F = \sum(4, 5, 6)$

$(S2) F = \sum(0, 1, 2, 3, 7)$

$(S3) F = \Pi (4, 5, 6)$

$(S4) F = \Pi (0, 1, 2, 3, 7)$

Which of the following is true?

  1. (S1)-False, (S2)-True, (S3)-True, (S4)-False
  2. (S1)-True, (S2)-False, (S3)-False, (S4)-True
  3. (S1)-False, (S2)-False, (S3)-True, (S4)-True
  4. (S1)-True, (S2)-True, (S3)-False, (S4)-False
asked in Digital Logic by Veteran (97.7k points)
edited by | 2.3k views
This question are concept building
what does ' !P ' mean?

P ′ = ! P  means  ??

6 Answers

+24 votes
Best answer
$F=P'+QR$,  draw the Kmap for this

We can find the minterm $\sum (0,1,2,3,7)$ and maxterm $\Pi (4,5,6) $

So, option A is correct: $(S1)$-False, $(S2)$-True, $(S3)$-True, $(S4)$-False
answered by Active (5k points)
edited by
+8 votes

$F = P{}' + QR$

for SOP we have :

$F = P{}.1.1 + 1.QR'=P{}'(Q+Q{}')(R+R{}') + (P+P{}')QR$

$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR+P{}'QR$

$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR$

$F=\sum(0, 1, 2, 3, 7)$ (considering barred terms as 0 and unbarred as 1 and converting them to binary and then to decimal).

now for POS we have :

$F = P{}' + QR = (P{}'+Q)(P{}'+R) = (P{}'+Q+0)(P{}'+R+0)$

$(P{}'+Q+R.R{}')(P{}'+R+Q.Q{}') $



$F=\prod (4, 5, 6)$(considering barred terms as 1 and unbarred as 0 and converting them to binary and then to decimal).

answered by Active (2.3k points)
edited by
But K-map is better time saving approach rt?
may be but that also kind of varies from individual to individual don't you think ?
Yes. Sure :)
But you needn't solve for both min terms and max terms rt? Solve one and take complement for other.
i was thinking of doing that, but just did it to show the general method to solve it both ways( POS ans SOP). Feel free to edit.
+8 votes

Answer = A

answered by Boss (30.5k points)
+1 vote

S2 and S3 are true

answered by Boss (34.2k points)
0 votes
Option A is correct

P'QR + P'QR' +P'Q'R + P'Q'R' + PQR = ⋿(0,1,2,3,7) = ∏(4,5,6)
answered by Boss (10.6k points)
0 votes
option A is correct.


answered by Active (1.3k points)

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