. 1.1.1 一数字信号的波形如图1.1.1所示，试问该波形所代表的二进制数是什么？ N=20 (2) LSTTL驱动基本TTL门 解：0101 1010 1.2.1 试按表1.2.1所列的数字集成电路的分类依据，指出下列器件属于何种集成度器件：(1) 微处理器；(2) IC计算器；(3) IC加法器；(4) 逻辑门；(5) 4兆位存储器IC。 解：(1) 微处理器属于超大规模；(2) IC计算器属于大规模；(3) IC加法器属于中规模；(4) 逻辑门属于小规模；(5) 4兆位存储器IC属于 IOL(max)?8mA IIL(max)?1.6mA NOL?8mA?5 1.6mA IOH(max)?0.4mA IIH(max)?0.04mANOH?0.4mA?10 0.04mA甚大规模。 1.3.1 将下列十进制数转换为二进制数、八进制数、十六进制数和8421BCD码（要求转换误差不大于2-4）： (1) 43 (2) 127 (3) 254.25 (4) 2.718 解：(1) 43D=101011B=53O=2BH； 43的BCD编码为0100 0011BCD。 (2) 127D=1111111B=177O=7FH； 127的BCD编码为0001 0010 0111BCD。 (3) 254.25D=11111110.01B=376.2O=FE.4H； 0010 0101 0100.0010 0101BCD。 (4) 2.718D=10.1011 0111B=2.56O=2.B7H； 0010.0111 0001 1000BCD。 1.3.3 将下列每一二进制数转换为十六进制码： (1) 101001B (2) 11.01101B 解：(1) 101001B=29H (2) 11.01101B=3.68H 1.3.4 将下列十进制转换为十六进制数： (1) 500D (2) 59D (3) 0.34D (4) 1002.45D 解：(1) 500D=1F4H (2) 59D=3BH (3) 0.34D=0.570AH (4) 1002.45D=3EA.7333H 1.3.5 将下列十六进制数转换为二进制数： (1) 23F.45H (2) A040.51H 解：(1) 23F.45H=10 0011 1111.0100 0101B (2) A040.51H=1010 0000 0100 0000.0101 0001B 1.3.6 将下列十六进制数转换为十进制数： (1) 103.2H (2) A45D.0BCH 解：(1) 103.2H=259.125D (2) A45D.0BCH=41024.046D 2.4.3 解：(1) LSTTL驱动同类门 IOL(max)?8mA IIL(max)?0.4mA NmAOL?80.4mA?20 IOH(max)?0.4mA IIH(max)?0.02mA N0.4mAOH?0.02mA?20 .
N=5 2.4.5 解： __________________________ L?AB?BC?D?E?AB?BC?D?E 2.6.3 解： B=0时，传输门开通，L=A； B=1时，传输门关闭，A相当于经过3个反相器到达输出L，L=AA B L 0 0 0 0 1 1 1 0 1 1 1 0 所以，L?AB?AB?A?B 2.7.1 解： . 0 0 0 0 1 _____0 0 1 1 × 0 1 0 1 × 1 0 0 0 1 0 1 1 1 0 LA?A?A?B1?B2?A?B1?B2 C，BC_____?BC __________ LB?LA?A?B1?B2?A?B1?B2? (a) ____________________3.1.2 用逻辑代数证明下列不等式 D，ED?DE ________________________________________________________________A?AB?A?B A?BC?(A?B)(A?C)，得 由交换律 BC?DE，ABC?DE?A(BC?DE) _______________________________________________________________________ A?AB?(A?A)(A?B)?A?B __________ (b) ABC?ABC?ABC?AB?AC AF?GF，EAF?GF?E(AF?GF)?EF(A?G)__________________________________ABC?ABC?ABC?A(BC?BC?BC)?A(C?BC) L?A(BC?DE)?EF(A?G)?A(BC?DE)?EF(A?G)A(C?B)?AB?AC?__________________________________ 2.7.2 解： (c) A?ABC?ACD??C?D?E?A?CD?E A?ABC?ACD??C?D?E?A?ACD?(C?D)E?A?CD?CDE?A?CD?E3.1.3 用代数法化简下列等式 ______________________ (a) AB(BC?A) AB?B?A?AB?(A?B)?AB?AB?A?B L?A?B=A⊙B _________AB(BC?A)?ABC?AB?AB (b) 2.9.11 解： (A?B)(AB) (A?B)(AB)?AB _______ (c) ABC(B?C) _______ 当没有车辆行驶时，道路的状态设为0，有车辆行驶时，道路的状态为1；通道允许行驶时的状态设为1，不允许行驶时的状态设为0。 设A表示通道A有无车辆的状态，B1、B2表示通道B1、B2有无车辆的情况，LA表示通道A的允许行驶状态，LB表示通道B的允许行驶状态。由此列出真值表。 A .
B1 B2 LA LB ABC(B?C)?(A?B?C)(B?C)?AB?BC?AC?BC?C?AB?C_____ (d) A?ABC?ABC?CB?CB A?ABC?ABC?CB?CB?A?C _________________________________(e) AB?AB?AB?AB . _____________________________________3.1.4 将下列各式转换成与 – 或形式 ________ AB?AB?AB?AB?A?A?0 ____________________________________________________________________________ _______________________________(a) A?B?C?D(1)当A?B__________________ __________ (f) (A?B)?(A?B)?(AB)?(AB) ?0，C?D?1时，真值为1。于是 ____________________________________________________________________________ AB=01，CD=00或CD=11时，真值为1； AB=10，CD=00或CD=11时，真值为1。 (A?B)?(A?B)?(AB)?(AB)?(A?B)?(A?B)?(AB)?(AB) ?(AB?BA?B)(AB?AB)?B(AB?AB)?AB则有四个最小项不为0，即 (g) ABCD、ABCD、ABCD、ABCD (A?B?C)(A?B?C) (A?B?C)(A?B?C)?A?B (h) ABC?ABC?ABC?A?BC ABC?ABC?ABC?A?BC?A?ABC?BC?A?BC?BC?A?C _____________________________ (i) AB?(A?B) ___________________AB?__________(A?B)?_____AB?(A?B)?(A?B)(A?B)?A?B (j) B?ABC?AC?AB B?ABC?AC?AB?B?ABC?AC?B?AC?AC (k) ABCD?ABD?BCD?ABCD?BC ABCD?ABD?BCD?ABCD?BC?ABC?ABD?B(CD?C)?ABC?ABD?B(C?D)?ABC?ABD?BC?BD?B(AC?AD?C?D)?B(A?C?A?D)?AB?BC?BD__________________________________________________(l) AC?ABC?BC?ABC __________________________________________________AC?ABC?BC?ABC?(AC?ABC)?(B?C)?(A?B?C)?(ABC?ABC)(A?B?C)?BC(A?B?C)?ABC?BC?BC__________________________________________________ (m) AB?ABC?A(B?AB) ____________________________________________________________________ ________________________________AB?ABC?A(B?AB)?A(B?BC)?AB?AB ________________________________________________________________?A(B?C)?A?A?B?C?A?0. ________?B?1__________ (2)当A，C?D?0时，真值为1。 AB=00，CD=10或CD=01时，真值为1； AB=11，CD=10或CD=01时，真值为1。 则有四个最小项不为0，即ABCD、ABCD、ABCD、ABCD__________________ A?B?C?D??m(1,2,4,7,8,11,13,14) __________________________________________________________________ (b) A?B?C?D?C?D?A?D __________________________________________________________________ A?B?C?D?C?D?A?D?(A?B)(C?D)?(C?D)(A?D)?(C?D)(A?B?D)?AC?AD?BC?BD?CD?D ?AC?BC?D_______________________________________________________AC?_____BD?_____BC?_____ (c) AB _______________________________________________________AC?_____BD?_____BC?_____AB?_____AC?_____BD?_____BC?_____AB?(A?C)(B?D)?(B?C)(A?B) ?AB?BC?AD?CD?AB?AC?B?BC?B?AD?CD?AC3.1.7 利用与非门实现下列函数 (a) L=AB+AC _______________________ L?AB?AC L?__________D(A?C___ (b) ) ____________________?C_________ L?D(A)?DAC ____________________ (c) L?(A?B)(C?D) __________________________________________ L?(A?B)(C?D)?ABCD . 3.2.2 用卡诺图法化简下列各式 ______________________ (a) AC?ABC?BC?ABC _________________________________________ AC?ABC?BC?ABC?AC?BC?BC?ABC?AC?C?ABC?C?ABC?C___________ (b) ABCD?ABCD?AB?AD?ABC ABCD?ABCD?AB?AD?ABC?AB?ABCD?AD ?A(B?BCD)?AD?AB?ACD?AD?AB?A(D?DC)?AB?AD?AC__________________ (c) (AB?BD)C?BD(AC)?D(A?B) __________________(AB?BD)C?BD(AC)?D(A?B) ?ABC?BCD?BD(A?C)?DAB ?ABC?BCD?ABD?BCD?ABD?ABC?BCD?AB?BCD__________ (d) ABCD?D(BCD)?(A?C)BD?A(B?C) __________ABCD?D(BCD)?(A?C)BD?A(B?C) ?ABCD?BCD?ABD?BCD?ABC ?m11?m1?m9?m12?m14?m6?m14?m4?m5??m(1,4,5,6,9,11,12,14) BD?ACD?ABD (e) L(A,B,C,D)??m(3,4,5,6,7,8,9,10,12,13,14,15). AB?BC?BD?ACD?ACD?ABCD (f) L(A,B,C,D)??m(0,1,2,5,6,7,8,913,14) BC?CD?ABC?ACD?BCD (g) L(A,B,C,D)??m(0,1,4,6,9,13)??d(2,3,5,7,11,15) A?D (h) 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。 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 1 0 1 1 1 S?A?B?C C?(A?B)CAB?AB?(A?B)C _____________________________________X?ABC?ABC?ABC 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 C X Y .
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____________________________________________________________________________________________________________________________________________________________________________________________________________?I3?I2?I1?I0?I3?I2?I1?I0?I3?I2?I1?I0________________________________________________________________________________________________________________________________D5?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?I0 D1?D5 D0?D7 4.1.2 解：
数字电子技术 第五版 康华光习题解答
