Question

2:
implement truth table in table 1 using only nand gates. (hint: use the circuit in figure 2.23a in the book 1 as a starting point.) (points: 5)
egr 21204 spring 2026 assignment 2
table 1: truth table for three - way light control (p2)
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$$\begin{tabular}{ccc||c} $x_1$ & $x_2$ & $x_3$ & $f$ \\\\ \\hline 0 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 1 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 1 & 1 & 0 \\\\ 1 & 0 & 0 & 1 \\\\ 1 & 0 & 1 & 0 \\\\ 1 & 1 & 0 & 0 \\\\ 1 & 1 & 1 & 1 \\\\ \\hline \\end{tabular}$$

3:

Explanation:

Step1: Write sum-of-minterms for $f$

From the truth table, identify rows where $f=1$: $m_1, m_2, m_4, m_7$.
$f(x_1,x_2,x_3) = \bar{x_1}\bar{x_2}x_3 + \bar{x_1}x_2\bar{x_3} + x_1\bar{x_2}\bar{x_3} + x_1x_2x_3$

Step2: Convert to NAND-only form

Use De Morgan's law: $A+B = \overline{\overline{A}\cdot\overline{B}}$, and $A = \overline{\overline{A}}$. First rewrite $f$ as:

$$ f = \overline{\overline{\bar{x_1}\bar{x_2}x_3 + \bar{x_1}x_2\bar{x_3} + x_1\bar{x_2}\bar{x_3} + x_1x_2x_3}} $$

Apply De Morgan's to the inner negation:

$$ f = \overline{\overline{\bar{x_1}\bar{x_2}x_3} \cdot \overline{\bar{x_1}x_2\bar{x_3}} \cdot \overline{x_1\bar{x_2}\bar{x_3}} \cdot \overline{x_1x_2x_3}} $$

Step3: Express each term as NAND

Each $\overline{ABC} = \text{NAND}(A,B,C)$, and single variables: $\bar{x} = \text{NAND}(x,x)$. Substitute:

• $\overline{\bar{x_1}\bar{x_2}x_3} = \text{NAND}(\text{NAND}(x_1,x_1), \text{NAND}(x_2,x_2), x_3)$
• $\overline{\bar{x_1}x_2\bar{x_3}} = \text{NAND}(\text{NAND}(x_1,x_1), x_2, \text{NAND}(x_3,x_3))$
• $\overline{x_1\bar{x_2}\bar{x_3}} = \text{NAND}(x_1, \text{NAND}(x_2,x_2), \text{NAND}(x_3,x_3))$
• $\overline{x_1x_2x_3} = \text{NAND}(x_1,x_2,x_3)$

Step4: Final NAND circuit

The final circuit is a 4-input NAND gate, where each input is the output of a 3-input NAND gate (with inverted variables implemented as 2-input NAND gates on single variables).

Answer:

The function implemented with only NAND gates is constructed as:

1. First, create inverted inputs: $\bar{x_1} = \text{NAND}(x_1,x_1)$, $\bar{x_2} = \text{NAND}(x_2,x_2)$, $\bar{x_3} = \text{NAND}(x_3,x_3)$
2. Build 3-input NAND gates for each minterm's negation:

• $G_1 = \text{NAND}(\bar{x_1},\bar{x_2},x_3)$
• $G_2 = \text{NAND}(\bar{x_1},x_2,\bar{x_3})$
• $G_3 = \text{NAND}(x_1,\bar{x_2},\bar{x_3})$
• $G_4 = \text{NAND}(x_1,x_2,x_3)$

1. Connect $G_1, G_2, G_3, G_4$ to a 4-input NAND gate: $f = \text{NAND}(G_1,G_2,G_3,G_4)$

The logical expression in NAND form is:

$$ f = \overline{\overline{\overline{x_1}\cdot\overline{x_2}\cdot x_3} \cdot \overline{\overline{x_1}\cdot x_2\cdot \overline{x_3}} \cdot \overline{x_1\cdot \overline{x_2}\cdot \overline{x_3}} \cdot \overline{x_1\cdot x_2\cdot x_3}} $$