5555555555
19. Use Boolean algebra to find the most simplified SOP expression for F = ABD + CD + ACD + ABC + ABCD.
(a) F = ABD + ABC + CD
(b) F = CD + AD
(c) F = BC + AB
(d) F = AC + AD
| Subject Name : Electronics Engineering |
| Exam Name : IIT GATE, UPSC ESE, RRB, SSC, DMRC, NMRC, BSNL, DRDO, ISRO, BARC, NIELIT |
| Posts Name : Assistant Engineer, Management Trainee, Junior Engineer, Technical Assistant |
| Electronics & Communication Engineering Books |
| GATE 2023 Total Info | ENGG DIPLOMA | UGC NET Total Info |
| IES 2023 Total Info | PSUs 2022 Total Info | CSIR UGC NET Total Info |
| JAM 2023 Total Info | M TECH 2023 Total Info | RAILWAY 2022 Total Info |
Related Posts
555555555513. Mapping the SOP expression , we get ________. (a) (A) (b) (B) (c) (C) (d) (D)- 555555555547. The Boolean expression C + CD is equal to ________. (a) C (b) D (c) C + D (d) 1
- 555555555564. The Boolean expression for a 3-input OR gate is ________. (a) X = A + B (b) X = A + B + C (c) X = ABC (d) X = A + BC
- Boolean Algebra » Exercise - 1 6. Solution of A.0 is : (a) 1 (b) A (c) 0 (d) A or 1
- 555555555513. Simplify the expression using DeMorgan's theorems. (a) (b) (c) (d)