UNIT 06. DIGITAL LOGIC AND DESIGN - Lab Activities - Computer Science For Class X

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Computer Science For Class X
Unit 06: Digital Logic And Design

LAB ACTIVITIES

1. Draw a logic circuit of 10^th Law of Boolean Algebra.
Ans: Logic Circuit Of 10^th Law of Boolean Algebra:
(Hint: 10^th Law of Boolean Algebra is same as 10^th Rule of Boolean Algebra given in Text Book on page number 115)

10^th Law of Boolean Algebra
A + A . B = A

Logic Circuit:

2. Design a logic circuit from Boolean expression Q = (A.B) + (A.B). (A + B)
Ans: Logic Circuit From Boolean Expression Q = (A.B) + (A.B). (A + B)

3. Derive the Boolean expression from the given circuit and make a truth table and simply that Boolean expression

Ans: Solution:

• Three Inputs A, B and C.
• Above circuit consist of AND, OR, and NOT gates.
• A, B and C gates are connected with AND gate.
• The output of the first AND gate is ABC.
• B and C are input for NOT gate.
• The output of the NOT gate are B and C
• The expression B NOT and C NOT are converted into OR gate.
• The output of OR gate are B + C
• A gate and B NOT + C NOT are connected with AND gate.
• Finally, ABC and A(B + C) are connected with OR gate.
• Therefore, the final output for Boolean expression for above circuit is ABC + A(B + C).

TRUTH TABLE

A	B	C	ABC	B'	C'	A(B+C)	ABC+A(B+C)
1	1	1	1	0	0	0	1
1	1	0	0	0	1	1	1
1	0	1	0	1	0	1	1
0	1	1	0	0	0	0	0
0	0	0	0	1	1	0	0
0	0	1	0	1	0	0	0
0	1	0	0	0	1	0	0
1	0	0	0	1	1	1	1

OR

Simplify Boolean Expression Using Boolean Algebra

ABC + A(B + C)
⇒ ABC + AB + AC (Open parenthesis or bracket)
⇒ A(BC + B + C (Take common variable A)
⇒ A{BC + (BC) }  (By De Morgan's LaW)
(As B and C is the NOR  gate of B and C, equivalently, the complement of BC)
⇒ BC + (BC) = 1 (According to 6^th rule of Boolean Algebra, A.A = 1)
⇒ Therefore  A(1)
⇒ The final Answer is A