Section 11.12. Find the values, if any, of the Boolean variable x that satisfies these equations.a) x l = 0b) x + x = 0c ) x 1 = x d ) x x = 1Solution:a) x = 0b) x = 0c) x = 0,1d) None6. Use a table to express the values of each of these Boolean functions.a) F(x, y, z) = zb) F(x, y, z) = xy + yzc ) F(x, y , z) = xyz + (xyz)d ) F(x, y, z) = y(xz + xz)Solution:a)xyzz

0001

0010

0101

0110

1001

1010

1101

1110

b)xyzxyxyyzxy+yz

00011000

00111011

01010101

01110101

10001000

10101011

11000000

11100000

c)xyzyxyzxyzxyz xyz + xyz

00010010

00110011

01000011

01100011

10010010

10111011

11000010

11100100

d)xyzxyzxzxzxz+xzy( xz+xz)

0001110111

0011100000

0101010110

0111000000

1000110000

1010101011

1100010000

1110001010

10. How many different Boolean functions are there of degree 7?Solution:There are 2n different n- tuples of 0s and 1s.A Boolean function is an assignment of 0 or 1 to each of these 2n different n-tuples.Therefore, there are different Boolean functions.So for n=7 it will be = 2128

24. Simplify these expressions.a) x 0 b) x 1c) x x d) x xSolution:We know that XOR for Boolean algebra as below:0 0 = 00 1 = 11 0 = 11 1 = 0So from above table we can give the answer of given problem as below:a) x 0 = xb) x 1 = xc) x x = 0d) x x = 1

28. Find the duals of these Boolean expressions.a) x + y b) x yc) xyz + x y z d) xz + x . 0 + x . 1Solution:The dual of a Boolean expression is obtained by interchanging Boolean sums and Boolean products and interchanging 0s and 1s.a) x . yb) x + yc) (x + y + z) . (x + y + z)d) (x + z) . (x + 1) . (x + 0)

Section 11.22. Find the sum-of-products expansions of these Boolean functions.a) F(x,y)= x + yb) F(x, y) = x yc) F(x,y) = 1d) F(x,y)= ySolution:a) F(x,y)= x + y= x.1 + y.1Identity Law= x. (y + y) + y.(x + x)Unit Property= xy + xy + xy + xyDistributive Law= xy + xy + xyIdempotent Lawb) F(x,y) = x yc) F(x,y) = 1= 1.1= (x + x) . (y + y)Unit Property= xy + xy + xy + xyDistributive Lawd) F(x,y)= y= y.1Identity Law= y.(x + x)Unit Property= xy + xyDistributive Law

4. Find the sum-of-products expansions of the Boolean function F(x, y, z) that equals 1 if and only ifa) x = 0b) xy = 0c) x + y = 0 d) xyz = 0

Solution:Sum of products for F(x,y,z) = 1= (x+).(y+).(z+)= xyz + xy + xz + x + yz + y + z + Eqn 1Nowa) x = 0 so put x = 0 in Eqn 1F(x,y,z) = yz + y + z +

b) xy = 0 so put xy = 0 in Eqn 1F(x,y,z) = z + x + yz + y + z +

c) x + y = 0 i.e. x = 0 and y = 0 ; so put in Eqn 1F(x,y,z) = z +

d) xy = 0 so put xyz = 0 in Eqn 1F(x,y,z) = xy + xz + x + yz + y + z +

8. Find a Boolean product of Boolean sums of literals that has the value 0 if and only if either x= y = 1 and z = 0, x = z = 0 and y = 1, or x = y = z = 0. [Hint: Take the Boolean product of the Boolean sums found in parts (a), (b), and (c) in Exercise 7.]Solution:F(x, y, z) = + z). + z). (x + y + z)

14. Show thata) = x | x. b) xy = (x | y) | (x | y).c) x + y = (x | x) | (y | y).

Solution: We can prove that all using the truth table.a) xx | x

011

100

b)xyxyx | y(x | y) | (x | y)

00010

01010

10010

11101

c)xyx + yx | xy | y(x | x) | (y | y)

000110

011101

101011

111001

Section 11.32,4,6,10,162. Find the output of the given circuit.

Solution: 4. Find the output of the given circuit.

Solution: () . ()

6. Construct circuits from inverters, AND gates, and OR gates to produce these outputs.a) + y b) c) xyz + d) ()

Solution:a)

b)c)d)

10. Construct a circuit for a half subtractor using AND gates, OR gates, and inverters. A half subtractor has two bits as input and produces as output a difference bit and a borrow.Solution: Output, a dierence bit d = x - y and a borrow b, can be represented by the following truth table:xyd=x-yb

0000

0111

1010

1100

From the above table we can define dierence bit as d = x y = . The borrow is b =

So the logic gates is as follows

There may be many other design also possible like

16. Use NOR gates to construct circuits for the outputs given in Exercise 15a) b) x + y c) xy d) x ySolution:a)

b) =

c) =

d)