Logic Gates: Important Solved Problems

Applied Physics:

UNIT II: Logic Gates

IMPORTANT SOLVED PROBLEMS

1. Prove that A+ B+ AB= A + B.

Solution

Hence, Proved

2. Simplify the following Boolean expressions:

Solution

3. Prove the following Boolean identities:

Solution

4. Simplfy the following Boolean expression:

Y= (AB+C) (AB+D)

Solution

Y = (AB+C) (AB+ D) = AB AB + ABD + ABC + CD

But A • A= A and B • B=B

Therefore, Y = AB+ ABD + ABC + CD

(or) Y = AB(1+D+C) + CD

But, (1+D+C) = 1

∴ Y = AB(1) + CD = AB + CD

5. Using De Morgan's theorem simplify:

Solution

∴ Y= 1

6. Simplify the following logic expressions using Boolean algebra.

 F = AB + A(B+C) + B(B+C)

Solution

Applying the distributive law to the second and third terms in the given expression, the expression becomes

F = AB+ AB+ AC + B•B + BC

= AB+ AB + AC + B + BC   [ because B•B=B]

= AB+ AC+B+BC       [ because AB + AB = AB]

= AB+AC+B (1+C)           [ because AB+ AB = AB]

= AB+ AC+ B•1              [ because B • 1=B]

= AB+ AC+ B            [ because (1+A)=1]

= B(A+1) + AC = B+AC            [ because (1+A)=1]

F= B+AC

7. Implement the following Boolean equation using only NAND gates.

Y = AB + CDE + F

Solution

Step 1: Realisation using basic gates.

Step 2: Replace

AND → NAND

OR → Bubbled‒OR

NOT → NAND Inverter.

Step 3: Draw the logic circuit using only NAND gates.

8. Reduce the function using K‒map F(A,B) = Σ m(0,2,3)

Solution

9. Reduce the function Y(A,B,C) = Σ(0,1,6,7)

Solution

10. Simplify the SOP function F (A,B,C) = Σ (1,2,3,7)

Solution

11. Reduce the given function F(A,B,C,D) = E (1,2,3,5,8,10,11,12)

Solution

12. Simplify the POS function F(A,B,C) = Π M(0,1,5,7)

Solution

13. Simplify the POS function F(A,B,C,D)= Σ m (3,5,7,8,10,11,12,13) or F (A,B,C,D) = Π M (0,1,2,4,6,9,14,15)

Solution

14. Minimize the following 4‒variable Boolean expression in SOP form using K‒map.

F (A,B,C,D) = Σ m (0,1,4,5,6,10,13) + d (2,3)

Solution

Note: In above example don't care X in cell order 2 is treated as "1" and don't care X in Cell order 3 is treated as "0"

15. Simplify the following Boolean function for minimal SOP & POS form using K‒map

F (A, B, C, D) = Σ (0,1,4,5,11,13,15)

Solution

For SOP

For POS

The given minterm function can be written in terms of Maxterm as follows.

F (A, B, C, D) = Σ (0,1,4,5,11,13,15) = Π M (2,3,6,7,8,9,10,12,14)

ASSIGNMENT PROBLEMS

1. Find the decimal equivalent of the binary number (11111)₂. (Ans: (31)₁₀)

2. Explain the following decimal number in the binary form.

a) 25.5

b) 10.625

c) 0.6875.

(Ans: (a) (11001.1)₂ (b) (1010.101)₂ (c) (0.1011)₂)

3. Convert each of the following decimal numbers to BCD.

(a) 53

(b) 89

(c) 170.

(Ans:

(a) 5          3    

   0101  0011

(b) 8  9

1000   1001

(c) 1    7      0 

0001 0111 0000

4. Convert each of the following BCD codes to decimal.

(a) 0001 0111 0000

(b) 1000 1001

(c) 0101 0011.

 (Ans: (a) 170 (b) 89 (c) 53)

5. Simplify

6. Simplify the expression, Y = +C+B +BC. (Ans: y=)

7. If A+B=C, show that A+C=B

8. Find one the minimal expression for the switching function given below using the Karnaugh Map.

F (A, B, C, D) = Σ(1, 3, 6, 7, 9, 13, 14, 15).

(Ans: D + AD + BC.)

9. Reduce the following function using K map technique.

F (A, B, C, D) = Σm (0, 1, 4, 8, 9, 10)

(Ans:  +A+)

10. Reduce the following function using K‒map Technique.

 f(A, B, C, D) = π M (0, 2, 3, 8, 9, 12, 13, 15)

(Ans: (++D) (A+B+) (+C) (A+B+D))