1..Question:Use Quine-McCluskey method to generate the set of prime implicants and obtain all minimal sum-of-products expression for the following switching functions: f(x1,x2,x3,x4,x5) = m(0,2,4,6,7,8,10,11,12,13,14,16,18,19,29,30)No of 1s Minter ms Check PI I Stage Check PI II Stage Check PI III Stage Check PI

0

0

1

2 4 8 16

0 , 2(2) 0 , 4(4) 0 , 8(8) 0 , 16(16)

0,2,4,6(2,4) 0,2,8,10(2,8) 0,2,16,18(2,16) 0,4,8,12(4,8)

b

0,2,4,6,8,10, 12,14(2,4,8) All others are equivalent

a

2

6 10 12 18

2 , 6(4) 2 ,10(8) 2 ,18(16) 4 , 6(2) 4,12(8) 8 ,10(2) 8 ,12(4) 16 ,18(2)

2,6,10,14(4,8) 4,6,12,14(2,8) 8,10,12,14(2,4)

3

7 11 13 14 19

6 ,7(1) 6 ,14(8) 10 ,11(1) 10 ,14(4) 12 ,13(1) 12 ,14(2) 18 ,19(1)

h g f e

4

29 30

13 ,29(16) 14 ,30(16)

d c

Table 1.1

Prime implicant chart:

N 0 2 4 6 7 8 10 11 12 13 14 16 18 19 29 30 o *a 0,2,4,6,8,10,12,14(2,4,8 *b *c *d *e f *g *h) 0,16,2,8(16,2) 14,30(16) 13,29(16) 18,19(1) 12,13(1) 10,11(1) 6,7(1)

Table 1.2

After completing the table for all prime implicants of the function, a search is made for the column having a single check. In this example we see that there is only one check in the column corresponding to minterms 4, 7,8,11,16,19,29 and 30. We, therefore, say that the P.I.s which cover these minterms are essential P.I.s since this P.I.s are the only one covering the corresponding minterms. The asterisk is placed in the left on the row corresponding to essential P.I.s.We also place checks in the bottom row in columns to the minterms which are covered by the essential P.I,s.The essential P.I.s for this example are a,b,c,d,e,g,h. Here P.I.s f is redundant because minterms 12 and 13 are already covered by essential P.I.s. The minimal sum of product expression is therefore given by, f(x1,x2,x3,x4,x5) = a+b+c+d+e+g+h An easier method for writing the value of the P.I.s is to write down the binary equivalent of any minterms in the prime implicant and then crossing out the variables which have Binary weight corresponding to the numbers given in the parenthesis of the P.I., since these variables are eliminated in the process of finding P.I.s. We have for the above example:

16

8

4 2

1P.I.

x1 x2 x3 x4 x5 a b c d e g h 0,2,4,6,8,10,12,14(2,4,8) 0,16,2,18(16,2) 14,30(16) 13,29(16) 18,19(1) 10,11(1) 6,7(1) 0 1 0 0 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 1 1 1

x1'x5 ' x2' x3' x5' x2 x3 x4 x5' x2 x3 x4' x5 x1 x2' x3' x4 x1' x2 x3'x4 x1' x2' x3 x4

Therefore minimal sum of product expression for the switching function is given as f(x1,x2,x3,x4,x5)= x1'x5 '+ x2' x3' x5' + x2 x3 x4 x5' + x2 x3 x4' x5 + x1 x2' x3' x4 + x1' x2 x3'x4 + x1' x2' x3 x4

2..Question:a. Use Quine-McCluskey method to generate the set of prime implicants and obtain all minimal sum-of-products expression for the following switching functions: f(x1,x2,x3,x4,x5) = m(1,3,6,7,8,12,14,16,18,22,23,24,28,30,31)No of 1s1

Minter ms1 8 16

Check PI

I Stage

Check PIj i h

II Stage

Check PIe

1 , 3(2) 8 ,12(4) 8, 24(16) 16,18(2) 16,24(8)

8,12,24,28(4,16)

2

3 6 12 18 24

3

7 14 22 28

3,7(4) 6,7(1) 6,14(8) 6,22(16) 12,14(2) 12,28(16) 18,22(4) 24,28(4)

g f

6,7,22,23(1,16) 6,14,22,23(8,16) 12,14,28,30(2,16)

d c b

22,23,30,31(1,8) 7,23(16) 14,30(16) 22,23(1) 22,30(8) 28,30(2)

a

4

23 30

23,31 (8) 30,31(1)

5

31

Table 2.1

Prime implicant chart:

N o *a b c d *e f g h i *j

1 3 6 7 8 12 14 16 18 22 23 24 28 30 3122,23,30,31(1,8) 12,14,28,30(2,16) 6,14,22,30(8,16) 6,7,22,23(1,16) 8,12,24,28(4,16) 18,22(4) 3,7(4) 16,24(8) 16,18(2) 1,3(2)

Table 2.2

After completing the table for all prime implicants of the function, a search is made for the column having a single check. In this example we see that there is only one check in the column corresponding to minterms 1,8 and 31. We, therefore, say that the P.I.s which cover these minterms are essential P.I.s since this P.I.s are the only one covering the corresponding minterms. The asterisk is placed in the left on the row corresponding to essential P.I.s.We also place checks in the bottom row in columns to the minterms which are covered by the essential P.I,s.The essential P.I.s for this example are a, e, j. The minimal sum of product expression is therefore given by, f(x1,x2,x3,x4,x5) = a+e+j. Now the reduced P.I. chart is made for the remaining minterms and P.I.s. P.I. b * c * d * f g h * i 6 12,14,28,30(2,16) 6,14,22,30(8,16) 6,7,22,23(1,16) 18,22(4) 3,7(4) 16,24(8) 16,18(2) 7 14

16

18

From the above table the prime implicant b is dominated by c. The prime implicant g is dominated by d. The prime implicant h is dominated by i. Therefore b, g, h are cancelled out. Now we look for the column in which there is a single check to find out secondary essential P.I.s .It can be seen that c, d, f, i are secondary essential prime implicants and they cover rest of the terms. The minimal sum of product expression for the switching function is therefore given by f = a+c+d+e+f+i+j An easier method for writing the value of the P.I.s is to write down the binary equivalent of any minterms in the prime implicant and then crossing out the variables which have Binary weight corresponding to the numbers given in the parenthesis of the P.I., since these variables are eliminated in the process of finding P.I.s. We have for the above example:

16

8

4

2

1P.I.

x1 x2 x3 x4 x5 a c d e f i j 22,23,30,31(1,8) 6,14,22,30(8,16) 6,7,22,23(1,16) 8,12,24,28(4,16) 18,22 (4) 16,18(2) 1,3(2) 1 1 1 1 1

x1x3 x4 x3 x4 x5' x2' x3 x4 x2 x4' x5' x1 x2' x4 x5' x1x2' x3' x5' x1' x2' x3' x5

0 1 1 0

1 0 1 0 0 1 0 0 1

1 0 0 0 0 0

Therefore minimal sum of product expression for the switching function is given as f(x1,x2,x3,x4,x5)= x1x3 x4 + x3 x4 x5' +x2' x3 x4 +x2 x4' x5' + x1 x2' x4 x5' + x1x2' x3' x5' + x1' x2' x3' x5

3..Question:c. Use Quine-McCluskey method to generate the set of prime implicants and obtain all minimal sum-of-products expression for the following switching functions: f(x1,x2,x3,x4,x5) = m(4,8,10,16,21,27,28) + d.c.(1,5,23,25,30,31)

No of 1s1

Minter ms1 4 8 16 5 10

Check PI k

I Stage

Check PIj i h

1 , 5(4) 4,5(1) 8,10(2)

2

5,21(16)

g

3

21 25 28

21,23(2) 25,27(2) 28,30(2)

f e d

4

23 27 30

23,31 (8) 27,31(4) 30,31(1)

c b a

5

31

Table 3.1

Prime implicant chart:

No a b c *d *e f g *h *i *j *k30,31(1) 27,31(4) 23,31(8) 28,30(2) 25,27(2) 21,23(2) 5,21 (16) 8,10 (2) 4,5(1) 1,5(4) 16

1 4 5 8 10 16 21 23 25 27 28 30 31 Table 3.2After completing the table for all prime implicants of the function, a search is made for the column having a single check. In this example we see that there is only one check in the column corresponding to minterms 1, 4, 8,10,16,25 and 28. We, therefore, say that the P.I.s which cover these minterms are essential P.I.s since this P.I.s are the only one covering the corresponding minterms. The asterisk is placed in the left on the row corresponding to essential P.I.s.We also place checks in the bottom row in columns to the minterms which are covered by the essential P.I,s.The essential P.I.s for this example are d,e,h,i,j,k.. The minimal sum of product expression is therefore given by, f(x1,x2,x3,x4,x5) = d+e+h+i+j+k. Now the reduced P.I. chart is made for the remaining minterms and P.I.s.

P.I. a b * c * f g

21 30,31(1) 27,31(4) 23,31(8) 21,23(2) 5,21(16)

23

31

From the above table the prime implicant a is dominated by c. The prime implicant b is dominated by c. The prime implicant g is dominated by f. Therefore a, b, g is cancelled out. Now we look for the column in which there is a single check to find out secondary essential P.I.s .It can be seen that c, f, are secondary essential prime implicants and they cover rest of the terms. The minimal sum of product expression for the switching function is therefore given by f = c+d+e+f+h+i+j+k An easier method for writing the value of the P.I.s is to write down the binary equivalent of any minterms in the prime implicant and then crossing out the variables which have Binary weight corresponding to the numbers given in the parenthesis of the P.I., since these variables are eliminated in the process of finding P.I.s. We have for the above example:

16 8

4

2 1P.I.

x1 x2 x3 x4 x5 c d e f h i j k 23,31(8) 28,30 (2) 25,27(2) 21,31(8) 8,10(2) 4,5(1) 1,5(4) 16 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 0 1

x1 x 3 x 4 x 5 x1 x2 x3 x5' x1 x2 x3' x5 x1 x2' x3 x5 x1' x2 x3' x5' x1' x2' x3 x4' x1' x2' x4' x5 x1x2' x3' x4' x5'

0 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0

Therefore minimal sum of product expression for the switching function is given as f(x1,x2,x3,x4,x5)= x1 x3 x4 x5 + x1 x2 x3 x5' + x1 x2 x3' x5 + x1 x2' x3 x5 + x1' x2 x3' x5' + x1' x2' x3 x4' + x1' x2' x4' x5+ x1x2' x3' x4' x5'