邏輯閘
AND閘
A
|
B
|
A˙B
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
1
|
1
|
OR閘
A
|
B
|
A+B
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
NOT閘
X
|
F=X’
|
0
|
1
|
1
|
0
|
XOR閘
X
|
Y
|
F(X,Y)=X’Y+XY’
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
EX: X+0=X
X+1=1
X˙0=0
X˙1=X
X+X’=1
X˙X=0
分配律
(a).
X+(Y˙Z)=(X+Y)˙(X+Z)
X
|
Y
|
Z
|
Y˙Z
|
X+( Y˙Z)
|
(X+Y)
|
(X+Z)
|
(X+Y)˙(X+Z)
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
(b). X˙(Y+Z)=(X˙Y)+(X˙Z)
X
|
Y
|
Z
|
Y+Z
|
X˙ ( Y+Z)
|
(X˙Y)
|
(X˙Z)
|
(X˙Y) + (X˙Z)
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
狄摩根定理
(a). (X+Y )’=X’˙Y’
X
|
Y
|
(X+Y)
|
(X+Y) ‘
|
X’
|
Y’
|
X’˙Y’
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
0
|
1
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
’
(b). (X˙Y )’=X’ +
Y’
X
|
Y
|
(X+Y)
|
(X+Y) ‘
|
X’
|
Y’
|
X’˙Y’
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
吸收定理:
(a) X+XY=X
(b) X˙(X+Y)=X
(c) X+X’Y=X+Y
(d) X˙(X’+Y’)=X˙Y
沒有留言:
張貼留言