Trending ▼   ResFinder  

ISC Class Notes 2020 : Computer Science

3 pages, 17 questions, 0 questions with responses, 0 total responses,    1    0
biz1in
   		+Fave Message
ProfileTimelineUploads
 Home > biz1in >

Formatting page ...

Subject :- Computer Science Class :XII Lesson No. :- 2 Topic :- Boolean Laws i) Indempotence Laws : (A + A ) = A (A.A) = A II) Elementary Laws : (A+1) = (1+A) =1 (A.1) = (1.A) =A (A+0) = (0+A) =A (A.0) = (0.A)= 0 III) Associative Laws: (A+B)+C = A + (B+C) (AB). C = A. (BC) IV) Commutative Laws : (A + B) = (B+A) (AB) = (BA) V) Distributive Laws : A + (BC) = (A+B)(A+C) A.(B+C) = AB +AC VI) Complementary Laws : (A + A ) = 1 (A . A ) = 0 VII) De Morgan s Laws/theorems : (A + B) = A . B (A.B) = A +B VIII) Involution Laws : (A ) = A IX) Absorption Laws : A + (AB) = A A . (A +B) = A Lesson No. :- 2 (Continued .. ) Proving Of Laws :1. State the Distributive laws. Prove any one using other Boolean laws. A + BC = (A +B ) (A+C) A . ( B+C) = AB + AC RTP :- A + BC = (A +B ) (A+C) R.H.S = (A +B) (A+C) = A.A + AC + AB + BC = A + AC + AB + BC [A.A = A, Indempotence Law] = A(1 + C) + AB + BC = A.1 + AB +BC [1 + A = 1, Elementary Law] = A +AB +BC = A(1+B) + BC = A.1 + BC = A + BC = L.H.S (Proved). 2. State and prove the De Morgan s laws using a truth table. (A + B ) + A +B (AB) = A . B A 0 0 1 1 B A B A + B AB (A+B) A .B (AB) A +B 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 1 1 0 0 0 1 1 1 0 0 1 1 0 0 0 0 3. State and Prove Absorption Laws using other Boolean laws. A + AB = A A. (A +B) = A RTP :- A + AB =A L.H.S = A + AB = A(1 + B) = A.1 = A = R.H.S RTP :- A. (A + B) = A L.H.S = A. (A +B) = A.A + AB = A + AB = A ( 1 + B) = A.1 = A = R.H.S Lesson No. :- 2 (Continued .. ) 1. Simplify :- X + (X + Y) = X + Y R.H.S = X + (X Y ) = (X + X ) .(X +Y ) = 1 ( X + Y ) = X + Y = R.H.S 2. Using Boolean laws prove that [ p. (p + q) ] + q = 1 L.H.S = [p.(p + q)] + q = [p.p + p.q] + q ( Distributive Law) =[pq] + q ( a.a = 0 ) =p +q +q (Demorgan s Law) =p +1 (Complementary law) =1 (Properties of 1) Exercise :1. 2. 3. 4. 5. 6. 7. 8. 9. Minimise F = AB + (AC) + AB C (AB +C). Simplify F= A B + AB C + A Simplify F(a,b,c) = [(a + b) (b +c)] + (a + c) . Simplify X Y Z + X Y Z+X YZ+X YZ +XY Z + XY Z using Boolean laws. Simplify (ab + x + y + z) (ab + x y z ) using Boolean laws. Simplify AB + A BC + (AC) + BC Simplify using Boolean laws [ (CD) + A] + A + CD +AB Minimise F = P Q + PQ R + P using Boolean laws. Write the expressions for (1),(2),(3)and (4) , reduce the final expression (4) using Boolean laws. Mention the laws used. x y 10. 11. 12. (1) y z (2) x z (3) Minimise F = ABC + ABC + A B C + AB C + A BC using Boolean laws. State Absorption Laws and prove any one of them using Boolean laws. Prove that (A+B)(A + C) = (A+B+C)(A+B+C )(A +B+C)(A +B +C) using Boolean laws. Mention the laws used. (4)

Formatting page ...

Formatting page ...

 

  Print intermediate debugging step

Show debugging info


 

 


© 2010 - 2025 ResPaper. Terms of ServiceContact Us Advertise with us

 

biz1in chat