Question

Convert the following logic circuit below to its corresponding logic function without any

simplification.

2. Assume that B consists of three bits: B2, B1, and B0. Determine the logic functions that evaluate

to 1 (“true”) if and only if the following situations are true. (Reduce each function to the fewest

terms and gates using Boolean algebra rules or DeMorgan’s Theorem, as needed.)

a. B contains only one 0.

b. B contains only one 1.

c. B contains an odd number of 1’s.

d. B is an unsigned number less than 5.

3. Construct logic circuits for each of the functions from problem 2.

Answer 1

2)  B is an unsigned number less than 5 is 1 and 3

Truth Table:

S.NO	B2	B1	Bo	F
0	0	0	0	0
1	0	0	1	1
2	0	1	0	1
3	0	1	1	1
4	1	0	0	1
5	1	0	1	1
6	1	1	0	1
7	1	1	1	1

F (B2, B1, Bo) = m (1, 2, 3, 4, 5, 6, 7)

The minterms of F = B2'B1'Bo+B2'B1Bo'+B2'B1Bo+B2B1'Bo'+B2B1'Bo+B2B1Bo'+B2B1Bo

F = B2'B1'Bo+B2'B1Bo'+B2'B1Bo+B2B1'Bo'+B2B1'Bo+B2B1Bo'+B2B1Bo

F = B2'B1'Bo+B2'B1(Bo'+Bo)+B2B1'(Bo'+Bo)+B2B1(Bo'+Bo)   { By Distributive law PQ+PR = P(Q+R) }

F = B2'B1'Bo+B2'B1(1)+B2B1'(1)+B2B1(1)   { We know that P+P'= 1 }

F = B2'(B1'Bo+B1)+B2(B1'+B1)  { By Distributive law PQ+PR = P(Q+R) }

F = B2'(B1'+B1)(Bo+B1)+B2(B1'+B1)  { By Distributive law P+QR = (P+Q)(P+R) }

F = B2'(1)(Bo+B1)+B2(1)  { We know that P+P'= 1 }

F = (B2'Bo+B2'B1)+B2  { By Distributive law P(Q+R) = PQ+PR }

F = B2'Bo+(B2'B1+B2)

F = B2'Bo+(B2'+B2)(B1+B2)  { By Distributive law P+QR = (P+Q)(P+R) }

F = B2'Bo+(1)(B1+B2)  { We know that P+P'= 1 }

F = B2'Bo+B1+B2

F = B2'Bo+B2+B1

F = (B2'+B2)(Bo+B2)+B1  { By Distributive law P+QR = (P+Q)(P+R) }

F = (1)(Bo+B2)+B1  { We know that P+P'= 1 }

F = Bo + B1 + B2

The Simplified SOP of F (B2, B1, Bo) = Bo + B1 + B2

Using K-map:

F (B2, B1, Bo) = m (1, 2, 3, 4, 5, 6, 7)

Above Function in K-map & Simplified K-map as follows

The Simplified SOP of F (B2, B1, Bo) = Bo + B1 + B2

3) Circuit: