习题 2 — 布尔函数与组合电路
2-1 Boolean Function Analysis¶
For the Boolean functions \(E\) and \(F\), as given in the following truth table:
| X | Y | Z | E | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 |
(a) List the minterms and maxterms of each function. - minterms:\(m_1,m_2,m_4,m_6\) - maxterms:\(M_0,M_3,M_5,M_7\) (b) List the minterms of \(\overline{E}\) and \(\overline{F}\). - minterms:\(m_0,m_3,m_5,m_7\) - maxterms:\(M_1,M_2,M_4,M_6\) (c) List the minterms of \(E + F\) and \(E \cdot F\). - E+F: \(m_0,m_1,m_2,m_4,m_6,m_7\) - EF: \(m_2,m_4\) (d) Express \(E\) and \(F\) in sum-of-minterms algebraic form. - \(E(X, Y, Z) = \overline{X}\overline{Y}Z + \overline{X}Y\overline{Z} + X\overline{Y}\overline{Z} + XY\overline{Z}\) - \(F(X, Y, Z) = \overline{X}\overline{Y}\overline{Z} + \overline{X}Y\overline{Z} + X\overline{Y}\overline{Z} + XYZ\) (e)* Simplify \(E\) and \(F\) to expressions with a minimum of literals. - \(E=\overline{XY}Z+Y\overline{Z}+X\overline{Z}\) - \(F = \overline{Z} + XY\)
2-2 Prime Implicants¶
*Find all the prime implicants for the following Boolean functions, and determine which are essential:
(a) \(F(W, X, Y, Z) = \sum m (0, 2, 5, 7, 8, 10, 12, 13, 14, 15)\) prime implicants: - \(\overline{XZ}\) - \(XZ\) - \(WX\) - \(WY\) Essential Prime Implicants: - \(\overline{XZ}\) - \(XZ\) - \(WX\)
2-3 Optimization with Don't-Cares¶
*Optimize the following Boolean functions \(F\) together with the don't-care conditions \(d\). Find all prime implicants and essential prime implicants, and apply the selection rule.
(b) \(F(W, X, Y, Z) = \sum m (0, 2, 4, 5, 8, 14, 15)\),
- Prime Implicants: \(\overline{X}\overline{Z}, \overline{W}\overline{Y}Z\), \(XZ, WX, \overline{W}X\overline{Y}, \overline{Y}\overline{Z}\).
- Simplified Expression: \(F = \overline{X}\overline{Z} + XZ + WX\)
- Essential Prime Implicants: \(\overline{X}\overline{Z}, XZ, WX\).
3-1 Hierarchical Components and Shannon's Expansion¶
A hierarchical component with the function \(H = XY + XZ\) is to be used along with inverters to implement the following equation:
- \(G = ABC + ABD + \overline{A}\overline{B}C + \overline{A}\overline{B}D\)
The overall circuit can be obtained by using Shannon’s expansion theorem: \(F = \overline{X} F_0(X) + X F_1(X)\) where \(F_0(X)\) is \(F\) evaluated with variable \(X = 0\) and \(F_1(X)\) is \(F\) evaluated with variable \(X = 1\). This expansion \(F\) can be implemented with function \(H\) by letting \(Y = F_0\) and \(Z = F_1\).
Requirements:
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The process can be repeated until all \(F_i\)’s are single literals or constants.
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For \(G\), use \(X = A\) to find \(G_0\) and \(G_1\) and then use \(X = B\) for \(G_0\) and \(G_1\).
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Draw the top-level diagram for \(G\) using \(H\) as a hierarchical component.
my provement:
3-2 Decoder Design¶
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Design a 4–to–16-line decoder using two 3–to–8-line decoders and 16 2-input AND gates.
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Design a 4–to–16-line decoder with enable using five 2–to–4-line decoders with enable.
3-3 Combinational Circuit Design¶
Design a combinational circuit that compares two 4-bit unsigned numbers \(A\) and \(B\) to see whether \(B\) is greater than \(A\). The circuit has one output \(X\), so that \(X = 1\) if \(A < B\) and \(X = 0\) if \(A \geq B\).