ASSIGNMENT (Boolean Algebra)

Q1.		Design NAND-to-NAND circuit for the following expressions
	a.	XYZ’ + X’Y’Z
	b.	A’B.(A’B’C’ + B’C)
	c.	A’B’C’ + A’B’C + A’BC + ABC’ + ABC
	d.	(X + Y’).Z
	e.	Y’ Z + Z X’
	f.	(A+B) (B+D)
	g.	3 input AND gate
	h.	3 input OR gate
Q2.		Design NOR-to-NOR circuit for the following expressions
	a.	X(Y’ + Z’) + XY’
	b.	A’B.(A’B’C’ + B’C)
	c.	X.Y’ + Z
	d.	(X + Y) (Y + Z) (X+Z)
	e.	F(A,B,C) = (A+B’) (B+C)
	f.	F(X,Y,Z) = (X’+Y) (Y’+Z)
	g.	(A’+B’+C’) (A+B’+C’) (A+B+C’)
Q3.		Obtain a simplified form of a Boolean Expression using Karnaugh Map
	a.	F(W,X,Y,Z)=S(2,3,6,10,11,14)
	b.	F(A,B,C,D)=S(0,1,3,45,6,7,9,10,11,13,15)
	c.	F(A,B,C,D)=P(0,1,3,45,6,7,9,10,11,13,15)
	d.	F(X,Y,Z)=P(3,4,5,6,7)
	e.	F(A,B,C,D)=S(0,1,2,5,6,8,9,10,13,15)
	f.	F(W,X,Y,Z)=S(2,3,6,10,11,14) Find expression in POS form
	g.	F(X,Y,Z)=S (3,4,5,6,7)
	h.	F(U,V,W,X)=S(7,9,10,11,12,13,14,15)
	i.	F(U,V,W,X)=P(0,2,3,7,8,10,11)
	j.	F(U,V,W,X)=P(3,4,5,6,7,9,11,12,13,14,15) Draw circuit diagrams for the simplified Boolean Expression

Q4. Find the minimal SOP expressions for F1 and F2 using K-Map for the given truth table for a function F(X,Y,Z), X Y Z F1 F2 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 1 1 0 1 1 0 0 1 1 1 1 1 1 Q5. A combination circuit has 4 inputs and single output. Output is 1 if All inputs are 1 Even number of inputs are 1 All inputs are 0 Make the truth table · Write Canonical SOP and POS forms of expressions ·Simplify using K-Map (Both SOP and POS expressions) ·Draw logic circuit diagram for simplified expression Q6. A truth table having 4 input variables ( A, B, C, D) has an output 1 when ABCD=0001 ABCD=0110 ABCD=1110 Express in SOP and POS Q7. Evaluate logic circuit diagram For i) a=1, b=1, c=0 ii) a=0, b=0, c=0 iii) a=1, b=1, c=1 Q8. Evaluate Boolean expression (w.x.y’ + (x’. y)’) for w=1, x=1, y=0, z=1 Q9. Simplify using algebric method AC+A’BC XZ’ + X’Y’Z + X’YZ’ + Y’Z (A+B)(A+B’) abc+a’bc+ab’c+abc’+ab’c’+a’bc’a’b’c’ AB+AB’+A’C+(AC)’ XY ( X’YZ’+XY’Z’) Q10. Verify the following : A + A’B = A+B (A+B)’ . (A’ + B’) = 0 AB + AB’ = A X’Y+Z = (X’+Y’+Z).(X’+Y+ Z).(X+Y+Z) (X+Y)(X+Z) = X+YZ x’y’z’+x’y’z+x’yz+x’yz’+xy’z’+xy’z = x’+y’ a.(a+b)=a (a’+b’).(a’+b).(a+b’) = a’b’ Q11. Write dual of following: 1. (x+y’) 3. a+1.bc 2. xy+xy’+x’y 4. a.a’=0 Q12. Write complement of following: 1. x’yz’ + x’y’z 3. xy(y+z) 2. ab’ +c’d’ 4. x(y’z+yz) Q13. State and verify Absorption Law using truth table. Q14. State and verify Involution Law using truth table. Q15. State and verify Idempotent Law using truth table. Q16. State and verify Distributive Law using truth table. Q17. State and verify DeMorgan’s Law using truth table.