Answer: \((156)_{10} = (10011100)_2\)

Verification: \[\begin{aligned} &128 + 16 + 8 + 4 \\ &= 156 \checkmark \end{aligned}\]

Reading remainders from bottom to top

Answer: \((237)_{10} = (355)_8\)

Verification: \[\begin{aligned} &3 \times 8^2 + 5 \times 8^1 + 5 \times 8^0 \\ &= 192 + 40 + 5 \\ &= 237 \checkmark \end{aligned}\]

Answer: \((2890)_{10} = (B4A)_{16}\)

Verification: \[\begin{aligned} &11 \times 256 + 4 \times 16 + 10 \\ &= 2816 + 64 + 10 \\ &= 2890 \checkmark \end{aligned}\]

Step 1: Octal to Binary
(each octal digit \(\to\) 3 bits)

Result: \((111010101)_2\)

Step 2: Binary to Hexadecimal
(group 4 bits from right)

Answer: \((725)_8 = (1D5)_{16}\)

Step-by-step:

• Position 0: \(1 + 0 = 0\) (no carry)

• Position 1: \(1 + 1 = 10\) (write 0, carry 1)

• Position 2: \(0 + 1 + 1_{\text{carry}} = 10\) (write 0, carry 1)

• Position 3: \(1 + 1 + 1_{\text{carry}} = 11\) (write 1, carry 1)

Answer: \((1011)_2 + (1110)_2 = (11001)_2\)

Verification: \(11 + 14 = 25~\checkmark\)

Answer:

\((10110101)_2 + (11011011)_2 = (110010000)_2\)

Verification:

• \((10110101)_2 = 181_{10}\)

• \((11011011)_2 = 219_{10}\)

• Sum = \(400_{10} = (110010000)_2~\checkmark\)

Step 2: Add minuend and 1’s complement

Step 3: Add end-around carry to result \[0011 + 1_{\text{carry}} = 0100\] Answer: \((1010)_2 - (0110)_2 = (0100)_2 = 4_{10}\)

Verification: \(10 - 6 = 4~ \checkmark\)

Step 2: Add minuend and 2’s complement

Step 3: Discard the carry (indicates positive result) Answer: \[\begin{aligned} (10110)_2 - (01101)_2 & = (01001)_2 \\ &= 9_{10} \end{aligned}\]

Verification: \(22 - 13 = 9~\checkmark\)

Step 1:
Find 2’s complement of subtrahend (1010)

• 1’s complement: 0101

• 2’s complement: \(0101 + 1 = 0110\)

Step 2: Add minuend and 2’s complement

Step 3: No carry! Result is negative. Find 2’s complement of result:

• 1’s complement of 1100: 0011

• 2’s complement: \(0011 + 1 = 0100\)

Answer: \[\begin{aligned} (0110)_2 - (1010)_2 & = -(0100)_2 \\ &= -4_{10} \end{aligned}\]

Verification: \(6 - 10 = -4\) \(\checkmark\)

Step 2: Add

Step 3: No end-around carry! Result is negative.

Take 1’s complement of result: \(\overline{1001} = 0110\)

Answer: \[\begin{aligned} (0011)_2 - (1001)_2 &= -(0110)_2 \\ &= -6_{10} \end{aligned}\]

Verification: \(3 - 9 = -6\) \(\checkmark\)

• Answer: Function outputs 1 for rows 1, 3, 6, and 7

• In minterm notation: \(F(A,B,C) = \sum m(1, 3, 6, 7)\)

\[\begin{aligned} F &= A\overline{B}C + ABC + A\overline{B}\,\overline{C} \\ &= A\overline{B}(C + \overline{C}) + ABC \quad \text{(Factor } A\overline{B}\text{)} \\ &= A\overline{B}(1) + ABC \quad \text{(Complement law)} \\ &= A\overline{B} + ABC \\ &= A(\overline{B} + BC) \quad \text{(Factor } A\text{)} \\ &= A(\overline{B} + C) \quad \text{(Absorption: } \overline{B} + BC = \overline{B} + C\text{)} \end{aligned}\]

• Answer:

  \[F = A(\overline{B} + C) = A\overline{B} + AC\]

• Gate Reduction: Original requires 7 gates, simplified requires 4 gates!

• Consensus Theorem:

  \[XY + \overline{X}Z + YZ = XY + \overline{X}Z\]

• The term \(BC\) is the consensus term and can be eliminated.

  \[\begin{aligned} F &= AB + \overline{A}C + BC \\ &= AB + \overline{A}C \quad \text{(Apply consensus theorem)} \end{aligned}\]

• Verification by expansion: \[\begin{aligned} BC &= BC(A + \overline{A}) \\ &= ABC + \overline{A}BC \\ &= ABC + \overline{A}BC \subseteq AB + \overline{A}C \end{aligned}\]

• Answer: \(F = AB + \overline{A}C\)

\[\begin{aligned} F &= \overline{A}BC + A\overline{B}C + AB\overline{C} + ABC \\ &= \overline{A}BC + ABC + A\overline{B}C + AB\overline{C} \quad \text{(Rearrange)} \\ &= BC(\overline{A} + A) + A\overline{B}C + AB\overline{C} \quad \text{(Factor)} \\ &= BC + A\overline{B}C + AB\overline{C} \quad \text{(Complement law)} \\ &= BC + AC(\overline{B} + B) \quad \text{(Factor } AC\text{, note: } AB\overline{C} = A\overline{C}B\text{)} \\ &= BC + AC \end{aligned}\]

• Alternative approach: \[\begin{aligned} F & = C(\overline{A}B + A\overline{B} + AB) + AB\overline{C}\\ & = C(A + B) + AB\overline{C}\\ &= AC + BC + AB\overline{C} \\ &= AC + BC \end{aligned}\]

• Answer:

  \(F = BC + AC = C(A + B)\)

Canonical SOP \[\begin{aligned} F &= m_0 + m_2 + m_5 + m_7 \\ &= \overline{A}\,\overline{B}\,\overline{C} + \overline{A}B\overline{C} + A\overline{B}C + ABC \end{aligned}\]

Answer: Both forms represent the same function! Canonical POS

Missing minterms: 1, 3, 4, 6 \(\to\) Use maxterms: \(M_1, M_3, M_4, M_6\) \[\begin{aligned} F &= \prod M(1, 3, 4, 6) \\ &= (A+B+\overline{C})(A+\overline{B}+\overline{C})\\ &(\overline{A}+B+C)(\overline{A}+\overline{B}+C) \end{aligned}\]

Method 1: Direct from SOP \[\begin{aligned} F &= A\overline{B} + \overline{A}B \\ &= \overline{\overline{A\overline{B}} \cdot \overline{\overline{A}B}} \end{aligned}\] Implementation:

• NOT gates: \(\overline{A}\), \(\overline{B}\) (2 NAND gates with tied inputs)

• Term 1: NAND(A, \(\overline{B}\)) \(\to\) \(\overline{A\overline{B}}\)

• Term 2: NAND(\(\overline{A}\), B) \(\to\) \(\overline{\overline{A}B}\)

• Final: NAND of above two outputs \(\to\) F

Method 2: Efficient 4-NAND implementation

• Gate 1: NAND(A, B) \(\to\) \(X = \overline{AB}\)

• Gate 2: NAND(A, X) \(\to\) \(Y = \overline{A\overline{AB}} = \overline{A} + B\)

• Gate 3: NAND(B, X) \(\to\) \(Z = \overline{B\overline{AB}} = A + \overline{B}\)

• Gate 4: NAND(Y, Z) \(\to\) \(F = A \oplus B\)

Step 1: Simplify first if possible \[\begin{aligned} F &= ABC + \overline{A}BC + A\overline{B}C \\ &= BC(A + \overline{A}) + A\overline{B}C \\ &= BC + A\overline{B}C \\ &= C(B + A\overline{B}) \\ &= C(A + B) \end{aligned}\] Step 2: Express in NAND form \[F = C(A + B) = \overline{\overline{C(A+B)}} = \overline{\overline{C} + \overline{(A+B)}}\]

Step 3: NAND implementation

• Gate 1: NAND(A, B) with both inverted = OR

• Gate 2: NAND(C, output of gate 1) = Final output

Answer: 3 NAND gates (2 for OR, 1 for AND)

Analysis: This is a multilevel function

Step 1: Identify levels

• Level 1: \(AB\)

• Level 2: \((AB)C\)

• Level 3: \((AB)C + D\)

• Level 4: \(((AB)C + D)E\)

Step 2: Convert each level to NAND

• \(AB\) \(\to\) NAND(A,B) followed by NOT \(\to\) 2 NAND gates

• \((AB)C\) \(\to\) NAND of previous with C, then NOT \(\to\) 2 NAND gates

• \((AB)C + D\) \(\to\) Requires OR, implement with NAND

• Final AND with E \(\to\) NAND then NOT

Step 1: Write maxterms \[\begin{aligned} M_0 &= A + B + C \\ M_2 &= A + \overline{B} + C \\ M_4 &= \overline{A} + B + C \\ M_6 &= \overline{A} + \overline{B} + C \end{aligned}\]

Step 2: Product of sums \[\begin{aligned} F &= (A+B+C)(A+\overline{B}+C)\\ &~~(\overline{A}+B+C)(\overline{A}+\overline{B}+C) \end{aligned}\] Step 3: Apply De Morgan’s for NOR-NOR \[F = \overline{\overline{(A+B+C)} + \overline{(A+\overline{B}+C)} + \overline{(\overline{A}+B+C)} + \overline{(\overline{A}+\overline{B}+C)}}\]

Step 4: Implementation

• Level 1: 4 NOR gates (one for each maxterm)

• Level 2: 1 four-input NOR gate for final output