Subjects digital logic design

Boolean Bcd Addition

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Boolean Bcd Addition


1. **Write the SPOS Boolean logic expression for** $f(A,B,C,D) = \Sigma m(0,2,5,7,8,10,13,15)$. To find the Sum of Products (SOP) form, list minterms corresponding to the decimal indices: $$m_0 = \overline{A}\overline{B}\overline{C}\overline{D}, m_2 = \overline{A}\overline{B}C\overline{D}, m_5 = \overline{A}B\overline{C}D, m_7 = \overline{A}BCD, m_8 = A\overline{B}\overline{C}\overline{D}, m_{10} = A\overline{B}C\overline{D}, m_{13} = AB\overline{C}D, m_{15} = ABCD$$ Hence, $$f = \overline{A}\overline{B}\overline{C}\overline{D} + \overline{A}\overline{B}C\overline{D} + \overline{A}B\overline{C}D + \overline{A}BCD + A\overline{B}\overline{C}\overline{D} + A\overline{B}C\overline{D} + AB\overline{C}D + ABCD$$ 2. **Simplify the given logic circuit output X.** From the problem description, the logic gates invert A, B, C, then use AND, OR gates combining the signals. Assuming the expression from description: $$X = [A\overline{B}(C + BD) + A + \overline{B} + \overline{C}]C$$ Simplify inside the bracket first: - Note that $A + \overline{B} + \overline{C}$ covers many cases, so combined with $A\overline{B}(C+BD)$ it simplifies. Rewrite: $$X = [A + \overline{B} + \overline{C}]C$$ Because $A + \overline{B} + \overline{C}$ dominates the other term. Distribute $C$: $$X = AC + \overline{B}C + \overline{C}C$$ Note $\overline{C}C=0$, So, $$X = AC + \overline{B}C = C(A + \overline{B})$$ This is the simplified Boolean expression. 3. **Perform BCD Addition of A and B:** Given: $$A = 0101\ 0110\ 0111 \, (BCD \ digits: 5, 6, 7)$$ $$B = 0011\ 1000\ 1001 \, (BCD \ digits: 3, 8, 9)$$ Add digit by digit (right to left): - 7 + 9 = 16 decimal (invalid BCD, over 9), So add 6 (0110) to correct: 7 (0111) + 9 (1001) = 16 (10000 binary, 5 bits too large), after adding 0110 (6): 10000 + 0110 = 10110 (22 decimal), take right 4 bits (0110) for digit value 6 and carry 1 to next digit. - Next digit: 6 + 8 + carry 1 = 15 decimal 15 > 9, add 6: 15 + 6 = 21 decimal Represented in BCD as carry 1 and digit 5 - Next digit: 5 + 3 + carry 1 = 9 decimal (valid, no correction needed) Final BCD result: $$\boxed{1001\ 0101\ 0110}$$ Which is decimal 9 5 6. --- **Summary of answers:** - SPOS expression for $f$ as above. - Simplified $X = C(A + \overline{B})$ - BCD addition result = 1001 0101 0110