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:
Convert the following logic circuit below to its corresponding logic function without any simplification. 2. Assume...
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