SUMITA ARORA Class 11 CS ASSIGNMENT CHAPTER-3 TYPE-B PART-2

Chapter-3 Type-B Solution for Assignment Question No 16-30

16. Find the dual of Boolean Expression \(\overline{A}D+\overline{C}D+\overline{A}B \)

Sol: Apply duality
= \( (\overline{A}+D).(\overline{C}+D).(\overline{A}+B) \)
= \( (\overline{A}\overline{C}+ \overline{A}D+D\overline{C}+DD) (\overline{A}+B) \) – Apply Distribution law
= \( (\overline{A}\overline{C}+ \overline{A}D+D\overline{C}+D) (\overline{A}+B) \) – Apply Distribution law
= \( (\overline{A}\overline{A}\overline{C}+ \overline{A}\overline{A}D+ \overline{A}D\overline{C}+\overline{A}D+B\overline{A}\overline{C}+B\overline{A}D+BD\overline{C}+ BD \) – Apply Idempotent law
= \( \overline{A}\overline{C}+ \overline{A}D+ \overline{A}D\overline{C}+ \overline{A}D+ B\overline{A}\overline{C}+ B\overline{A}D+BD\overline{C}+ BD \) – Apply Absorption law
= \( \overline{A}\overline{C}+ \overline{A}D+ \overline{A}D+ B\overline{A}\overline{C}+ B\overline{A}D+BD\overline{C}+ BD \) – Apply Idempotent law
= \( \overline{A}\overline{C}+ \overline{A}D+ B\overline{A}\overline{C}+ B\overline{A}D+BD\overline{C}+ BD \) – Apply Absorption law
= \( \overline{A}\overline{C}+ \overline{A}D+ B\overline{A}\overline{C}+ B\overline{A}D+BD \) – Apply Absorption law
= \( \overline{A}\overline{C}+ \overline{A}D+ B\overline{A}\overline{C}+BD \) – Apply Absorption law
= \( \overline{A}\overline{C}+ \overline{A}D+ BD \)

17. Find the dual of Boolean Expression \( B+\overline{A}C+\overline{B}A \)

Sol: Apply duality
= \( B(\overline{A}+C) (\overline{B}+A) \) – Apply Distribution law
= \( B(\overline{B}+A)\overline{A}+B(\overline{B}+A)C \) – Apply Distribution law
= \( B\overline{A}\overline{B}+B\overline{A}A+B(\overline{B}+A)C \) – Apply Complement law
= \( 0+B\overline{A}A+B(\overline{B}+A)C \) – Apply Identity law
= \( B\overline{A}A+B(\overline{B}+A)C \) – Apply Complement law
= \( 0+B(\overline{B}+A)C \) – Apply Identity law
= \( B(\overline{B}+A)C \) – Apply Distribution law
= \( BC\overline{B}+BCA \) – Apply Complement law
= \( 0+BCA \) – Apply Identity law
= \( BCA \)

18. Find the dual of Boolean Expression \( X\overline{Y}Z+\overline{X}\overline{Y}Z \)

Sol: Apply Duality
= \( (X+\overline{Y}+Z).(\overline{X}+\overline{Y}+Z) \) – Apply Distribution law
= \( (\overline{X}+\overline{Y}+Z)X+(\overline{X}+\overline{Y}+Z)\overline{Y}+(\overline{X}+\overline{Y}+Z)Z \) – Apply Distribution law
= \( X\overline{X}+X\overline{Y}+XZ+\overline{Y}\overline{X}+\overline{Y}\overline{Y}+\overline{Y}Z+Z\overline{X}+Z\overline{Y}+ZZ \) – Apply Idempotent & Complement law
= \( 0+X\overline{Y}+XZ+\overline{Y}\overline{X}+\overline{Y}+\overline{Y}Z+Z\overline{X}+Z\overline{Y}+Z \) – Apply Identity & Absorption law
= \( X\overline{Y}+XZ+\overline{Y}+\overline{Y}Z+Z\overline{X}+Z\overline{Y}+Z \) – Apply Absorption law
= \( X\overline{Y}+XZ+\overline{Y}+Z\overline{X}+Z\overline{Y}+Z \) – Apply Absorption law
= \( X\overline{Y}+XZ+\overline{Y}+Z\overline{X}+Z \) – Apply Absorption law
= \( X\overline{Y}+XZ+\overline{Y}+Z \) – Apply Absorption law
= \( XZ+\overline{Y}+Z \) – Apply Absorption law
= \( \overline{Y}+Z \)

19. Design a logic circuit to realize the Boolean function \( f(x,y) = x.y+ \overline{x}.\overline{y} \)

Sol:

20. Draw the logic circuit for this boolean expression: \( y = \overline{A}\overline{B}\overline{C}D+A\overline{B}\overline{C}D+AB\overline{C}D+ABC\overline{D} \)

Sol: Reduce the equation:

= \( \overline{A}\overline{B}\overline{C}D + A\overline{B}\overline{C}D + AB\overline{C}D + ABC\overline{D} \) – Apply Distribution law (extract common terms)
= \( \overline{B}\overline{C}D (\overline{A}+A) + AB\overline{C}D + ABC\overline{D} \) – Apply Complement law
= \( \overline{B}\overline{C}D (1) + AB\overline{C}D + ABC\overline{D} \) – Apply Identity law
= \( \overline{B}\overline{C}D + AB\overline{C}D + ABC\overline{D} \) – Apply Distribution law
= \( \overline{C}D (\overline{B}+AB) + ABC\overline{D} \) – Apply Absorption law  \( (AB+\overline{A}=B+\overline{A}) \)
= \( \overline{C}D (\overline{B}+A) + ABC\overline{D} \) – Apply Distribution law
= \( \overline{C}D\overline{B}+ \overline{C}DA+ABC\overline{D} \)

Circuit:

 

21. Draw the AND-OR circuit for \( y = A\overline{B}\overline{C}\overline{D} + AB\overline{C}\overline{D} + ABCD \)

Sol: Reduce the expression:
= \( A\overline{B}\overline{C}\overline{D} + AB\overline{C}\overline{D} + ABCD \) – Apply Distributive law (AB+AC = AB+C)
= \( A\overline{C}\overline{D}(\overline{B} + B) + ABCD \) – (Taking common terms and boolean identity)
= \( A\overline{C}\overline{D} + ABCD \)

Circuit:

22. Given the Boolean function \(\overline{A}D+C\overline{D}+A\overline{B} \)
(i) Obtain the truth table of the function        (ii) Draw the logic circuit diagram

A	B	C	D	\(\overline{A} \)	\( \overline{C}\)	\(\overline{B}\)	\(\overline{A}D\)	\(\overline{C}D\)	\(A\overline{B}\)	\( \overline{A}D+\overline{C}D\)	\( \overline{A}D+\overline{C}D+A\overline{B} \)	Output
0	0	0	0	1	1	1	0	0	0	0	0	F
0	0	0	1	1	1	1	1	1	0	1	1	T
0	0	1	0	1	0	1	0	0	0	0	0	F
0	0	1	1	1	0	1	1	0	0	1	1	T
0	1	0	0	1	1	0	0	0	0	0	0	F
0	1	0	1	1	1	0	1	1	0	1	1	T
0	1	1	0	1	0	0	0	0	0	0	0	F
0	1	1	1	1	0	0	1	0	0	1	1	T
1	0	0	0	0	1	1	0	0	1	0	1	T
1	0	0	1	0	1	1	0	1	1	1	1	T
1	0	1	0	0	0	1	0	0	1	0	1	T
1	0	1	1	0	0	1	0	0	1	0	1	T
1	1	0	0	0	1	0	0	0	0	0	0	F
1	1	0	1	0	1	0	0	1	0	1	1	T
1	1	1	0	0	0	0	0	0	0	0	0	F
1	1	1	1	0	0	0	0	0	0	0	0	F

23. Given the boolean function \( F=\overline{w}xy+w\overline{x}y+wxy \)
(i) Obtain the truth table of the function        (ii) Draw the logic circuit diagram

Sol: Truth Table

W	X	Y	Output
0	0	0	F
0	0	1	F
0	1	0	F
0	1	1	T
1	0	0	F
1	0	1	T
1	1	0	F
1	1	1	T

(ii) Reduce the expression
= \(overline{w}xy+w\overline{x}y+wxy \) – Apply distributive law (AB+AC = AB+C)
= \( xy(\overline{w}+w)+ w\overline{x}y \) – Apply complement law \((A+\overline{A}=1) (A1=A) \)
= \(xy+ w\overline{x}y \) – Apply Distributive law
= \(y(x+w\overline{x}) \) – Apply Absorption law \((\overline{A}B+A = B+A) \)
= \( y(w+x) \) – Apply Distributive law
= \( yw+yx \)

Circuit:

24. Given the Boolean function \((ZX+\overline{Y})(XY+\overline{Z}) \)
(i) Obtain the truth table of the function       (ii) Draw the logic circuit
Sol:

X	Y	Z	Output
0	0	0	T
0	0	1	F
0	1	0	F
0	1	1	F
1	0	0	T
1	0	1	F
1	1	0	F
1	1	1	T

Reduce expression
= \((ZX+\overline{Y})(XY+\overline{Z}) \) – Apply Distribution Law
= \( (XY+\overline{Z})ZX+(XY+\overline{Z})\overline{Y} \) – Apply Distribution
= \( ZXXY+ZX\overline{Z}+\overline{Y}XY+\overline{Y}\overline{Z} \) – Apply Idempotent law & Complement law
= \( ZXY+0+0+ \overline{Y}\overline{Z} \)  – Apply Identity law (A+0=A)
= \( ZXY+ \overline{Y}\overline{Z} \)

Circuit:

 

 

 

 

 

25. Given the boolean function \((A+\overline{B}+\overline{C})(\overline{A}\overline{B}+B\overline{C}) \)
(i) Obtain the truth table of the function       (ii) Draw the logic circuit

Sol:

A	B	C	Output
0	0	0	T
0	0	1	T
0	1	0	T
0	1	1	F
1	0	0	F
1	0	1	F
1	1	0	T
1	1	1	F

Reduce the expression:
= \((A+\overline{B}+\overline{C})(\overline{A}\overline{B}+B\overline{C}) \) – Apply Disbution Law
= \((\overline{A}\overline{B}+B\overline{C})A+(\overline{A}\overline{B}+B\overline{C})\overline{B}+(\overline{A}\overline{B}+B\overline{C})\overline{C}\) – Apply Distribution Law
= \(A\overline{A}\overline{B}+AB\overline{C}+\overline{B}\overline{A}\overline{B}+\overline{B}\overline{B}\overline{C}+\overline{C}\overline{A}\overline{B}+\overline{C}B\overline{C} \) –  Apply Complement, Idempotent
= \(0+AB\overline{C}+ \overline{B}\overline{A}+0+\overline{C}\overline{A}\overline{B}+\overline{C}B \) – Apply Identity Law
= \(AB\overline{C}+ \overline{B}\overline{A}+\overline{C}\overline{A}\overline{B}+\overline{C}B \) –  Apply Distribution law
= \(AB\overline{C}+\overline{B}\overline{A}+\overline{C}(\overline{A}\overline{B}+B) \) – Apply Distribution and Absorption law
= \(AB\overline{C}+\overline{B}\overline{A}+\overline{C}B \) – Apply Commutative law
= \(AB\overline{C}+B\overline{C}+ \overline{A}\overline{B} \) – Apply Absorption law
= \(\overline{A}\overline{B}+ B\overline{C} \)

See Part-III Here