Exercici portes lògiques 3 (resolt)
Per a les 24 = 16 combinacions que hi ha dels dígits xyzt, troba la funció s i implementa'n el circuit lògic
| a | b | s |
| 0 | 0 | x |
| 0 | 1 | y |
| 1 | 0 | z |
| 1 | 1 | t |
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
\(s = 0\)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
\(s = a · b\)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
\(s = a · \overline{b}\)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
\(s = a · \overline{b} + a · b\)
\(s = a · (b + \overline{b})\)
Com que \((x + \overline{x}) = 1\)
\(s = a\)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
\(s = \overline{a} · b\)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
\(s = \overline{a} · b + a · b\)
\(s = b · (a + \overline{a})\)
Com que \((x + \overline{x}) = 1\)
\(s = b\)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
\(s = \overline{a} · b + a · \overline{b}\)
\(s = a \oplus b\)
(XOR)
| a | b | s |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
\(s = \overline{a} · b + a · \overline{b} + a · b\)
\(s = a · (b + \overline{b}) + \overline{a} · b \)
Com que:
\((x + \overline{x}) = 1\)
i
\(x + \overline{x} · y = x + y\)
(Llei de redundància)
\(s = a + b\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
\(s = \overline{a} · \overline{b}\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
\(s = \overline{a} · \overline{b} +a · b\)
\(s = \overline{a \oplus b}\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
\(s = \overline{a} · \overline{b} + a · \overline{b}\)
\(s = \overline{b}\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
\(s = \overline{a} · \overline{b} + a · \overline{b} + a · b\)
\(s = \overline{b} + a · b\)
Amb la llei de la redundància
\(s = a + \overline{b}\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
\(s = \overline{a} · \overline{b} + \overline{a} · b\)
\(s = \overline{a} · (\overline{b} + b)\)
Com:
\(x + \overline{x} = 1\)
\(s = \overline{a}\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
\(s = \overline{a} · \overline{b}+ \overline{a} · b + a · b\)
\(s = \overline{a} · (\overline{b}+ b) + a · b\)
\(s = \overline{a} + a · b\)
Llei de la redundància:
\(s = \overline{a} + b\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
\(s = \overline{a} · \overline{b} + \overline{a} · b + a · b\)
\(s = \overline{a} · (\overline{b} + b) + a · b\)
\(s = \overline{a} + a · b\)
Llei de la redundància
\(s = \overline{a} + \overline{b}\)
| a | b | s |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
\(s = 1\)
En resum:
| x | y | z | t | s |
| 0 | 0 | 0 | 0 | \(0\) |
| 0 | 0 | 0 | 1 | \(a · b\) |
| 0 | 0 | 1 | 0 | \(a ·\overline{b}\) |
| 0 | 0 | 1 | 1 | \(a\) |
| 0 | 1 | 0 | 0 | \(\overline{a} · b\) |
| 0 | 1 | 0 | 1 | \( b\) |
| 0 | 1 | 1 | 0 | \(a \oplus b\) |
| 0 | 1 | 1 | 1 | \(a + b\) |
| 1 | 0 | 0 | 0 | \(\overline{a} · \overline{b}\) |
| 1 | 0 | 0 | 1 | \(a \odot b\) |
| 1 | 0 | 1 | 0 | \(\overline{b}\) |
| 1 | 0 | 1 | 1 | \(a + \overline{b}\) |
| 1 | 1 | 0 | 0 | \(\overline{a}\) |
| 1 | 1 | 0 | 1 | \( \overline{a} + b\) |
| 1 | 1 | 1 | 0 | \(\overline{a} + \overline{b}\) |
| 1 | 1 | 1 | 1 | \(1\) |