a)
b) A= x2+(x1*x0) y B=x0+(x1*x2) c) A= ((x2)'(x1*x0)')' y B ((x0)'(x2*x1)')' |
a | n | l | Ea | En | El |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 0 | 0 |
* Ea = an'l'+an'l+anl' = an'(l'+l)+anl'= an'+anl' =an(n'+l')
* En = anl'
* El = an'l
a)
A | B | C | D | S |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
b) S = Σ(m6,m10,m12,m14)
S = A'BCD' + AB'CD' + ABC'D' + ABCD'
d)
a | b | c | d | A | B | C | D |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | X | X | X | X |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | X | X | X | X |
1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | X | X | X | X |
1 | 1 | 1 | 0 | X | X | X | X |
1 | 1 | 1 | 1 | X | X | X | X |
1) Se necesitaran un total de 4 entradas, las cuáles serán las monedas (m1 y m0), ya que son 3 tipos de monedas, o ninguna; y los productos (p1 y p0), ya que son 4 productos diferentes. Las salidas serán 3, donde las monedas de cambio (c1 y c0) serán también 3 tipos de moneda, o ninguna; y la salida (S), si entrega o no el producto.
2)
m1 | m0 | p1 | p0 | S | c1 | c0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 1 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 0 |
3)
4)
a)
T | V | H | P | M |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
b)
c) M = ( H + V + T)’’ = H’ x V’ x T’
Al implementar Morgan, la ecuación queda como :
F = ((AE')'(BC)'(BD')'(CD')')'
a)
b)
c)
B)
C)
D | C | B | A | F |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
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