BD?ACD?ABD
(e)
L(A,B,C,D)??m(3,4,5,6,7,8,9,10,12,13,14,15)
(f)
AB?BC?BD?ACD?ACD?ABCD
L(A,B,C,D)??m(0,1,2,5,6,7,8,913,14)
(g)
BC?CD?ABC?ACD?BCD
L(A,B,C,D)??m(0,1,4,6,9,13)??d(2,3,5,7,11,15)
(h)
A?D
L(A,B,C,D)??m(0,13,14,15)??d(1,2,3,9,10,11)
AB?AD?AC
3.3.4 试分析图题3.3.4所示逻辑电路的功能。
S?A?B?C
_____________________________________C?(A?B)CAB?AB?(A?B)C
全加器
3.3.6 分析图题3.3.6所示逻辑电路的功能。
S0?A0?B0 C0?A0B0
S1?A1?B1?C0 C1?A1B1?(A1?B1)C0
二位加法电路
3.4.3 试用2输入与非门和反相器设计一个4位的奇偶校验器，即当4位数中有奇数个1时输出为0，否则输出为1。
L?A?B?C?D
_______________________L?AB?AB?AB?AB
3.4.7 某雷达站有3部雷达A、B、C，其中A和B功率消耗相等，C的功率是A的功率的两倍。这些雷达由两台发电机X和Y供电，发电机X的最大输出功率等于雷达A的功率消耗，发电机Y的最大输出功率是X的3倍。要求设计一个逻辑电路，能够根据各雷达的启动和关闭信号，以最节约电能的方式启、停发电机。
A B 0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 X Y 0 0 1 0 1 0 0 1 0 1 0 1 0 1 1 1
X?ABC?ABC?ABC
Y?m1?m3?m5?m6?m7?AB?C
4.1.1 解：
D7?I3I2I1I0?I3I2I1I0?I3I2I1I0______________________________________________________________________?I3I2I1I0?I3I2I1I0?I3I2I1I0_______________________________________________________________________________________________________________________________________________________________________________________________________?I3?I2?I1?I0?I3?I2?I1?I0?I3?I2?I1?I0D6?I3I2I1I0?I3I2I1I0?I3I2I1I0____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________D5?I3?I2?I1?I0?I3?I2?I1?I0
?I3?I2?I1?I0?I3?I2?I1?I0?I3?I2?I1?I0____________________________________________________________________________________________________________________________________________________________________________________________________D4?I3?I2?I1?I0?I3?I2?I1?I0?I3?I2?I1?I0__________________________________________________________________________________________________________________________________
D3?I3?I2?I1?I0?I3?I2?I1?I0______________________________________________________________________________________________________________________________
D2?I3?I2?I1?I0?I3?I2?I1?I0D1?D5 D0?D7
4.1.2 解：
P?B9B8B7B6B5B4B3B2B1B0