CDA 103
Tutorial Sheet-4
Group 1
1. For the given Boolean function, convert the following truth table into SOP form:
A B C Output
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
2. Convert the following truth table into POS forms:
A B C Output
0 0 0 0
0 0 1 ‘1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0
3. Given the Boolean function F (A, B, C) = AB + AC + BC, convert it into the POS form.
4. Express the Boolean function F (A, B, C) = AB + AC + BC in POS form.
5. For the Boolean function F (A, B, C) = (A + B)(C + D), derive its POS form.
6. Convert the following Boolean function into SOP form:
F (A, B, C) = (A + B)(C + D)(E + F )
7. Convert the following Boolean function into SOP form:
F (A, B, C) = (A + B + C)(D + E + F )
1
Tutorial Sheet-4
Group 1
1. For the given Boolean function, convert the following truth table into SOP form:
A B C Output
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
2. Convert the following truth table into POS forms:
A B C Output
0 0 0 0
0 0 1 ‘1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0
3. Given the Boolean function F (A, B, C) = AB + AC + BC, convert it into the POS form.
4. Express the Boolean function F (A, B, C) = AB + AC + BC in POS form.
5. For the Boolean function F (A, B, C) = (A + B)(C + D), derive its POS form.
6. Convert the following Boolean function into SOP form:
F (A, B, C) = (A + B)(C + D)(E + F )
7. Convert the following Boolean function into SOP form:
F (A, B, C) = (A + B + C)(D + E + F )
1