逻辑与计算机设计基础课后答案及分析和实验
《逻辑与计算机设计基础》(原书第五版)课后习题答案-chapter02_solutions-5th
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《逻辑与计算机设计基础》(原书第五版)课后习题答案-chapter04_solutions-5th
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逻辑与计算机设计基础第4版第6章答案
CHAPTER 6© 2008 Pearson Education, Inc.6-1.*6-2.6-3.6-4.*6-5.a) F = (A + B) C Db) G = (A + B) (C + D)a) 3-input NAND gate 6 inputs 16 inputs 16 inputsThe longest path is from input C or D.0.073 ns + 0.073 ns + 0.048 ns + 0.073 ns = 0.267 nsa)b)c)4.03.02.01.06.05.08.07.09.0 ns6-6.6-7. 6-8.+If the rejection time for inertial delays is greater than the propagation delay, then an output change can occur before it can be predicted whether or not it is to occur due to the rejection time.For example, with a delay of 2 ns and a rejection time of 3 ns, for a 2.5 ns pulse, the initialedge will have already appeared at the output before the 3 ns has elapsed at which whether to reject or not is to be determined.a) The propagation delay is t pd = max(t PHL = 0.05, t PLH = 0.10) = 0.10 ns.Assuming that the gate is an inverter, for a positive output pulse, the following actually occurs:If the input pulse is narrower than 0.05 ns, no output pulse occurs so the rejection time is 0.05 ns. The resulting model predicts the following results, which differ from the actual delay behavior, but models the rejection behavior: :0.10 ns0.05 ns0.10 ns0.10 ns6-9. 6-10.*b) For a negative output pulse, the following actually occurs:The model predicts the following results, which differs from the actual delay behavior andthe actual rejection behavior:Overall, the model is inaccurate for both cases a and b, and provides a faulty rejectionmodel for case b. Using an average of t PHL and t PLH for t pd would improve the delayaccuracy of the model for circuit applications, but the rejection model still fails.0.10 ns0.05 ns0.15 ns0.10 ns0.10 nsa)There is a setup time violation at 28 ns. There is an inputb)There is a setup time violation just before 24 ns, There is an inputc)There is a setup time violation at 28ns.d)There is a hold time violation at 16ns and a setup time violation at 24ns.combination violation around 24 ns.combination violation around 24 ns.a) The longest direct path delay is from input X through the two XOR gates to the output Y.t delay = t pdXOR + t pdXOR = 0.20 + 0.20 = 0.40 nsb) The longest path from an external input to a positive clock edge is from input X through the XOR gate and the inverter to the B Flip-flop.t delay = t pdXOR + t pd INV + t sFF = 0.20 + 0.05 + 0.1 = 0.35 nsc) The longest path delay from the positive clock edge is from Flip-flop A through the two XOR gates to the output Y.t delay = t pdFF + 2 t pdXOR = 0.40 + 2(0.20) = 0.80 nsd) The longest path delay from positive clock edge to positive clock edge is from clock on Flip-flop A through t delay-clock edge to clock edge = t pdFF + t pdXOR + t pdINV + t sFF = 0.40 + 0.20 + 0.05 + 0.10 = 0.75 nse) The maximum frequency is 1/t delay- clock edge to clock edge. For this circuit, t delay-clock edge to clock edgeis 0.75 ns, so the maximum frequency is 1/0.75 ns = 1.33 GHz.the XOR gate and inverter to clock on Flip-flop B.into its environment. Calculation of this frequency cannot be performed in this case since data for paths through the environment is not provided.Comment: The clock frequency may need to be lower due to other delay paths that pass outside of the circuit6-11.6-12.a) The longest direct path delay is from input X through the four XOR gates to the output Y.t delay = 4 t pdXOR = 4(0.20) = 0.80 nsb) The longest path from an external input to a positive clock edge is from input X through three XOR gates and the inverter to the clock of the second B Flip-flop.t delay = 3 t pdXOR + t pd INV + t sFF = 3(0.20) + 0.5 + 0.1 = 0.75 nsc) The longest path delay from the positive clock edge is from the first Flip-flop A through the four XOR gates to the output Y.t delay = t pdFF + 4 t pdXORR = 0.40 + 4(0.20) = 1.2 nsd) The longest path delay from positive clock edge to positive clock edge is from the first Flip-flop A through three XOR gates and one inverter to the clock of the second Flip-flop B.t delay-clock edge to clock edge = t pdFF + 3 t pdXOR + t pdINV + t sFF = 0.40+ 3(0.20) + 0.5 + 0.1 = 1.15 nse) The maximum frequency is 1/t delay-clock edge to clock edge. For this circuit, the delay is 1.15 nsso the maximum frequency is 1/1.15 ns = 870 MHz.Comment: The clock frequency may need to be lower due to other delay paths that pass outside of the circuit into its environment. Calculation of this frequency cannot be performed in this case since data for paths through the environment is not provided.AA6-13.*(Errata: Change "32 X 8" to "64 X 8" ROM)6-14. 6-15.6-16.IN OUT IN OUT IN OUT IN OUT 0000000000 00000100000001 01101000000011 00101100000100 1000 0000010000 00010100010001 01111000010011 00111100010100 1001 0000100000 00100100100001 10001000100011 01001100100101 0000 0000110000 00110100110001 10011000110011 01011100110101 0001 0001000000 01000101000010 00001001000011 01101101000101 0010 0001010000 01010101010010 00011001010011 01111101010101 0011 0001100000 01100101100010 00101001100011 10001101100101 0100 0001110000 01110101110010 00111001110011 10011101110101 0101 0010000000 10000110000010 01001010000100 00001110000101 0110 0010010000 10010110010010 01011010010100 00011110010101 0111 0010100001 00000110100010 01101010100100 00101110100101 1000 0010110001 00010110110010 01111010110100 00111110110101 1001 0011000001 00100111000010 10001011000100 01001111000110 0000 0011010001 00110111010010 10011011010100 01011111010110 0001 0011100001 01000111100011 00001011100100 01101111100110 0010 0011110001 01010111110011 00011011110100 01111111110110 0011a)16 + 16 + 1 = 33 address bits and 16 + 1 = 17 output bits, 8G × 17b)8 + 8 + 1 + 1 = 18 address bits and 8 + 1 = 9 output bitsc) 4 × 4 = 16 address bits and 14 output bits are needed, 64K × 14XYZXYZXYZXYZA B C D1111111111111111B = XY + XY + YZC = YZ C = YZ + Z A = XY + XY + YZBy using A instead of A and YZ instead of Y in D, YZ can be shared by all four functions. Further, since A is the complement of B, terms XY and XY can be shared between A and B. Thus, only four product terms YZ, XY, XY, and Z are required.An inversion must be programmed for A.6-17.6-18.6-19.*XY ZXYZ X YZ XY Z XYZ111111111X YZ 11AB CDEF11A = XYB = X + YZC = XY + X Y + ZD = YZE = 0F = ZImplementation of A, D, and E requires only two terms, XY and YZ. Straightforward implementation of B, C, and F requires four terms, XY , XYZ, XYZ, and Z. By implementing B, C, and F, only three additional termsX, X Y , and Z are required. So we form the solution using five product terms: XY , YZ, X Y, X, and Z. The solution is described by the equations given with the six K-maps.ABC D A BC D A BC D A BC D W X Y Z The values given in the four K-maps come from Table 3-1 on page 99.100110001100001111110000110011001111100d d d d dd d d d d dd d d d ddd d d d d dd W = A B + BC DX = BC D + BC + BD Y = CD + C D Z = D In this case, shared terms are limited. One such term B C D is generated in W.Assume 3-input OR gates.ABC DABC D ABCD ABC D W 100110001100001111110000110011001111100d d d d dd d d d ddd d d d ddd d d d d dd W = A + BC + BDX = BC D + BC + BD Y= CD + C D Z = D Each of the equations above is implemented using one 3-input OR gate. Four gates are used.6-20.6-21.X Y ZXYZX YZX Y Z11111111A B CD1A = XZ + YZ + X YZ B = XY + YZ + X YC = A + XYD = XY + Z1111111111Figure 6-23 uses 3-input OR gates.A, B, and D each require three or fewer product terms so can be implemented with 3-input OR gates.C requires four terms so cannot be implemented with a 3-input OR gate. But because the first PAL device outputcan used as an input to implement other functions it can be assigned to A and A can then be used to implement C using just two inputs of a 3-input OR gate.A ABCDABCD F111111111111111111GFigure 6-23 uses 3-input OR gates.Straightforward implementation of F requires five prime implicants and of G requires four prime implicants, but only 3 inputs are available on the PAL OR gates. So sum-of-products that can be factored from F and G or both and implemented by the other PAL cells are needed. A single sum of products that will work is H = ABC + BCD + BCD. The terms of H are shown with dotted lines on the K-maps. Using H:F = H + CD + ABG =H + AB There are other possible functions for H and corresponding results for F and H.。
数字电路与逻辑设计习题答案
数字电路与逻辑设计习题答案数字电路与逻辑设计习题答案数字电路与逻辑设计是计算机科学与工程领域中的重要基础课程，它涉及到数字信号的处理和转换，以及逻辑门电路的设计和分析。
学习这门课程时，习题是巩固知识和提高能力的重要途径。
下面将给出一些常见的数字电路与逻辑设计习题的答案，希望对大家的学习有所帮助。
1. 设计一个4位二进制加法器电路，实现两个4位二进制数的相加。
答案：一个4位二进制加法器电路可以由四个全加器电路组成。
每个全加器电路有三个输入：两个被加数和一个进位输入，以及两个输出：和位和进位输出。
将四个全加器电路按位级联，将进位输出连接到下一位的进位输入，最后一个全加器的进位输出作为最高位的进位输出。
和位的输出即为两个4位二进制数的和。
2. 给定一个3输入的逻辑电路，输出为1的条件是至少有两个输入为1。
请设计一个电路，实现这个逻辑功能。
答案：可以使用与门和或门组合的方式来实现这个逻辑功能。
首先，将三个输入分别与一个与门的三个输入相连，将输出连接到一个或门的输入。
然后，将三个输入分别与一个或门的三个输入相连，将输出连接到与门的输入。
这样，当至少有两个输入为1时，与门的输出为1，或门的输出也为1。
3. 给定一个4输入的逻辑电路，输出为1的条件是输入中有奇数个1。
请设计一个电路，实现这个逻辑功能。
答案：可以使用异或门实现这个逻辑功能。
首先，将四个输入两两分组，然后将每组的输出与另一组的输出进行异或操作。
最后，将四个异或门的输出连接到一个或门的输入。
这样，当输入中有奇数个1时，异或门的输出为1，或门的输出也为1。
4. 设计一个4位比较器电路，实现两个4位二进制数的大小比较。
答案：一个4位比较器电路可以由四个比较器组成。
每个比较器有两个输入：两个被比较的位，以及一个输出：比较结果。
将四个比较器电路按位级联，将每个比较器的输出连接到下一位比较器的输入。
最后一个比较器的输出即为两个4位二进制数的大小比较结果。
5. 给定一个3输入的逻辑电路，输出为1的条件是输入中的1的个数大于等于2。
逻辑与计算机设计基础答案 chapter02
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数字设计基础与应用（第2版）第1章习题解答
第1章 数字逻辑基础1－1 将下列二进制数转换为十进制数。
(1) 2(1101) (2) 2(10110110) (3) 2(0.1101) (4) 2(11011011.101) 解（1）3210210(1101)12120212(13)=⨯+⨯+⨯+⨯=（2）75421210(10110110)1212121212(182)=⨯+⨯+⨯+⨯+⨯= （3） 124210(0.1101)1212120.50.250.0625(0.8125)---=⨯+⨯+⨯=++=（4）76431013210(11011011.101)22222222 12864168210.50.125 (219.625)--=+++++++=+++++++= 1－2 将下列十进制数转换为二进制数和十六进制数(1) 10(39) (2) 10(0.625) (3) 10(0.24) (4) 10(237.375) 解（1）10216(39)(100111)(27)== (2) 10216(0.625)(0.101)(0.A)==（3）近似结果： 16210)3.0()00111101.0()24.0(D =≈ (4) 10216(237.375)(1110'1101.011)(0ED.6)== 1－3 将下列十六进制数转换为二进制数和十进制数(1) 16(6F.8) (2) 16(10A.C) (3) 16(0C.24) (4) 16(37.4) 解（1） 16210(6F.8)(1101111.1)(111.5)== (2) 16210(10A.C)(1'0000'1010.11)(266.75)== (3) 16210(0C.24)(1100.0010'01)(12.140625)== (4) 16210(37.4)(11'0111.01)(55.25)== 1－4 求出下列各数的8位二进制原码和补码(1) 10(39)- (2) 10(0.625) (3) 16(5B) (4) 2(0.10011)- 解（1）10(39)(1'0100111)(1'1011001)-==原码补码 (2) (0.1010000)(0.1010000)==10原码补码(0.625) (3) 16(5B)(01011011)(01011011)==原码补码(4) 2(0.10011)(1.1001100)(1.0110100)-==原码补码1－5 已知10X (92)=-，10Y (42)=，利用补码计算X ＋Y 和X －Y 的数值。
大学计算机基础教程课后习题答案
第1章计算机概述一、术语解释ENIAC: Electronic Numerical Integrator And ComputerALU: Arithmetic Logic UnitCU: Control UnitIC:Integrated CircuitULSI:Ultra Large Scale Integrated CircuitCPU: Central Processing UnitASCII: American Standard Code for Information InterchangeCAD:Computer-Aided DesignAI:Artificial Intelligence二、选择题1-5 ABACB 6-10 BBDDA 11-15 CADBD三、简答题1.计算机的特点是什么？（1）记忆能力强;（2）计算精度高与逻辑判断准确;（3）高速的处理能力;（4）能自动完成各种操作.2.计算机的发展经历了哪几代？各以什么器件为其主要特征？（1）电子管计算机(1946-1957)（2）晶体管计算机(1958-1964)（3）集成电路计算机(1965-1971)（4）大规模集成电路计算机(1972-现在)3.计算机有哪些方面的应用？请举例说明。
(1) 科学计算(2) 过程检测与控制(3) 信息管理(4) 计算机辅助系统(5) 人工智能4.什么是冯·诺依曼原理？（1）计算机应由运算器、控制器、存储器、输入设备和输出设备五个基本部件组成。
（2）内部以二进制表示指令和数据。
每条指令由操作码和地址码两部分组成。
操作码指出操作类型，地址码指出操作数的地址。
由一串指令组成程序。
（3）采用“存储程序”工作方式。
5.简述计算机的五大部件及其功能。
存储器不仅能存放数据，而且也能存放指令，形式上两者没有区别，但计算机应能区分数据还是指令；控制器应能自动执行指令；运算器应能进行加、减、乘、除四种基本算术运算，并且也能进行一些逻辑运算和附加运算；操作人员可以通过输入设备、输出设备和主机进行通信。
《逻辑与计算机设计基础》(原书第五版)课后习题答案-chapter03_solutions-5th
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第二章布置习题参考解
2-1 用真值表验证定律2-2 用代数化简来证明下列布尔议程的性质
a)
c)
2-3 用代数化简来证明下列布尔议程的性质
2-6 化简下列布尔表达式，使表达式中包含的因子最少
2-10
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c)
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(a)
用两个三态缓冲器和一个非门实现函数H=XY+XZ。
把两个缓冲器和两个非门互联实现异或门。
(b)
2-33
把三个三态缓冲器的输出连在一起，增加一些逻辑（门）实现函数 (a)
F=ABC+ABD+ABD。
假设C、D以及D是三态缓冲器的输入，A、B通过逻辑电路产生便能输入。
对于(a) 中设计的三态缓冲器的输出是不是没有冲突？如果不是，更改必 (b)
要的设计，使设计的电路没有冲突。
上述设计没有三态输出冲突。