电子科大数字电路,期末试题101102半期考试试卷-答案
电子科技大学期末数字电子技术考试题a卷-参考答案

电子科技大学二零零九至二零一零学年第 二 学期期 末 考试数字逻辑设计及应用 课程考试题 A 卷(120分钟)考试形式:闭卷 考试日期2010年7月12日课程成绩构成:平时 20 分, 期中 20 分, 实验 0 分, 期末 60 分一、To fill your answers in the blanks (1’×25)1. If [X]10= - 110, then [X]two's-complement =[ ]2,[X]one's-complement =[ ]2. (Assumed the number system is 8-bit long) 2. Performing the following number system conversions: A. [10101100]2=[ 0 ]2421B. [1625]10=[01001 ]excess-3C. [ 1010011 ]GRAY =[10011000 ]8421BCD3. If ∑=C B A F ,,)6,3,2,1(, then F D ∑=C B A ,,( 1,4,5,6 )=C B A ,,∏(0,2,3,7 ).4. If the parameters of 74LS-series are defined as follows: V OL max = 0.5 V , V OH min = 2.7 V , V IL max = 0.8 V , V IH min = 2.0 V , then the low-state DC noise margin is 0.3V ,the high-state DC noise margin is 0.7V .5. Assigning 0 to Low and 1 to High is called positive logic. A CMOS XOR gate in positive logic is called XNOR gate in negative logic.6. A sequential circuit whose output depends on the state alone is called a Moore machine.7. To design a "001010" serial sequence generator by shift registers, the shift register should need 4 bit as least.8. If we use the simplest state assignment method for 130 sates, then we need at least8state variables.9. One state transition equation is Q*=JQ'+K'Q. If we use D flip-flop to complete the equation, the D input terminal of D flip-flop should be have the function D= JQ'+K'Q.10.Which state in Fig. 1 is ambiguous D11.A CMOS circuit is shown as Fig. 2, its logic function z= A’B’+ABFig. 1 Fig. 212.If number [A]two's-complement =01101010 and [B]one's-complement =1001, calculate [A-B]two's-complement and indicate whether or not overflow occurs.(Assumed the number system is 8-bit long)[A-B]two's-complement = 01110000, overflow no13. If a RAM’s capacity is 16K words × 8 bits, the address inputs should be 14bits; We need 8chips of 8K ⨯8 bits RAM to form a 16 K ⨯ 32 bits ROM..14. Which is the XOR gate of the following circuit A .15.There are 2n-n invalid states in an n-bit ring counter state diagram.16.An unused CMOS NOR input should be tied to logic Low level or 0 .17.The function of a DAC is translating the Digital inputs to the same value of analogoutputs.二、Complete the following truth table of taking a vote by A,B,C, when more than two of A,B,C approve a resolution, the resolution is passed; at the same time, the resolution can’t go through if A don’t agree.For A,B,C, assume 1 is indicated approval, 0 is indicated opposition. For the F,A B C F三、The circuit to the below realizes a combinational function F of four variables. Fill in the Karnaugh map of the logic function F realized by the multiplexer-based circuit. (6’)四、(A) Minimize the logic function expressionF = A·B + AC’ +B’·C+BC’+B’D+BD’+ADE(H+G) (5’)F = A·B + AC’ +B’·C+BC’+B’D+BD’ = A·(B ’C )’ +B’·C+BC’+B’D+BD’= A +B’·C+BC’+B’D+BD’+C ’D (或= A +B’·C+BC’+B’D+BD’+CD ’)= A +B’·C+BD’+C ’D (或= A + BC’+B’D +CD ’)(B) To find the minimum sum of product for F and use NAND-NAND gates to realize it (6’)),,,(Z Y X W F Π(1,3,4,6,9,11,12,14)------3分 F= X ’Z ’+XZ -----2分 =( X ’Z ’+XZ)’’=(( X ’Z ’)’(XZ)’)’ ------1分五、Realize the logic function using one chip of 74LS139 and two NAND gates.(8’)∑=)6,2(),,(C B A F ∑=)3,2,0(),,(E D C GF(A,B,C)=C’∑(1,3) ---- 3分 G(C,D,E)=C’∑(0,2,3) ----3分-六、Design a self-correcting modulo-6 counter with D flip-flops. Write out the excitation equations and output equation. Q2Q1Q0 denote the present states, Q2*Q1*Q0* denote the next states, Z denote the output. The state transition/output table is as following.(10’)Q2Q1Q0Q2*Q1*Q0*Z000 100 0100 110 0110 111 0111 011 0011 001 0001 000 1激励方程式:D2=Q0’(2分,错-2分)D1=Q2 (2分,错-2分)D0=Q1 (2分,错-2分)修改自启动:D2=Q0 +Q2Q1’(1分,错-1分)D1=Q2+Q1Q0’(1分,错-1分)D0=Q1+Q2Q0 (1分,错-1分)输出方程式:Z=Q1’Q0 (1分,错-1分)得分七、Construct a minimal state/output table for a moore sequential machine, that will detect the input sequences: x=101. If x=101 is detected, then Z=1.The input sequences DO NOT overlap one another. The states are denoted with S0~S3.(10’)For example:X:0 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 ……Z:0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 ……state/output table八、Please write out the state/output table and the transition/output table and theexcitation/output table of this state machine.(states Q2 Q1=00~11, use the state name A~D )(10’)Transition/output table State/output table Excitation/output table(4分) (3分) (3分)评分标准:转移/输出表正确,得4分;每错一处扣0.5分,扣完4分为止;由转移/输出表得到状态/输出表正确,得3分;每错一处扣0.5分,扣完3分为止;激励/输出表正确,得3分;每错一处扣0.5分,扣完3分为止。
电子科技大学2012年数电半期考试题参考解答

电子科技大学二零壹壹年至二零一贰学年第二学期“数字逻辑设计及应用”课程考试题(半期)(120分钟) 考试日期 2012年4月22日I. To fill the answers in the “( )” (2’ X 20=40) 1. [42.25 ]10 = ( 2A.4 )16 = ( 52.2 )8 .2. The binary two ’s complement is (1011), then its corresponding 8-bit two ’s complement is ( 11111011 ), and 8-bit one ’s complement is ( 11111010 ), and 8-bit signed-magnitude is ( 10000101 ).3. The 8421-BCD code is (10011000)8421-BCD ,then its corresponding decimal number is ( 98 ).4. The binary number code is (10101011)2, then its corresponding Gray code is ( 11111110 ).5. If F = ∏ABC (1,3,5),then its dual expression is =D F ∑ABC ( 2,4,6 ), and the complement expression of the function F is F ’=∑ABC ( 1,3,5)。
6. The range of 8-bit two ’s complement is (-128 ~ 127), and the range of 8-bit unsigned binary number is (0 ~ 255).7. If there are 2012 different states, we need at least ( 11 ) bits binary code to represent them.8. For the two ’s complement addition and subtraction operation, if [ A ] two’s -complement =11011011, and [ B] two’s -complement =10011111 , calculate [-A-B ] two’s -complement , [A-B ] two’s -complement , and indicate whether or not overflow occurs.[-A-B ] two’s -complement = [ 10000110 ], overflow: [ yes ] [A-B ] two’s -complement = [ 00111100 ], overflow: [ no ]9. The maximum LOW-state output current I OLmax for an HC-series CMOS gate driving CMOS inputs is 0.02mA, the maximum HIGH-state output current I OHmax is -0.02mA, and the maximum input current I Imax for an HC-series CMOS input in any states is A μ1±, the DC fanout at HIGH-state is ( 20 ).10. The unused CMOS NAND gate inputs should be tied to logic ( 1 ).11. The following logic diagram Fig.1 implements a function of 3-variable with a 74x138. The logicfunction can be expressed as F (A,B,C) =∏A,B,C ( 2, 3,4,5,7 ).Fig.112. The CMOS circuit is shown in Fig.2. Write the function of the circuit. ( F=(AB+C+D)’ )Fig.2II. There is only one correct answer in the following questions.(3’ X 10=30)1. What is the correct 2’s -complement representation of the decimal number -325?( A ) A) 1010111011 B) 1101000101 C) 1011010011 D) 10101001102. A 20-to-1 multiplexer need ( B ) selection control inputs at least.A) 4B) 5C) 6D) 203. In the 8-radix number system, the result of operation 721/20 is: ( B )A) 36.05B) 35.04C) 35.05D) 36.044. What is the duality logic function of the logic function: F = ∑ABC (0,3,5,7)( C )A),,(1,2,4,6)A B C ∑ B),,(0,2,4,7)A B C ∑ C),,(0,2,4,7)A B C ∏D),,(1,2,4,6)A B C ∏5. The inputs waveform A,B,C and output waveform F of a combinational circuit are shown as Fig.3. The canonical product-of-sums expression of this circuit is ( D )A)(),,2,3,5,7A B C∑B)(),,0,2,4,6A B C∑C) ,,(1,2,4,7)A B C ∏ D),,(0,3,5,6)A B C ∏Fig.36. For each of the following logic expressions, ( B ) is the hazard-free circuit.A) F=A’·B + A·C + B’·C B ) F=A’·B + A·C + B·C C) F=(A+B)·(B’+C)·(C+D) D) F=(A+B’)·(B+C)·(C’+D) 7. For the logic function )''()''(),,,(C B D C AB D C B A F '++=, the corresponding minimal sum is ( A ).A) A’+B+C’D’ B ) (A’+B+C’)(A’+B+D’) C) A’+B+B’C’D’ D ) A’+B+AC’D’8. The INVERTER and AND-OR-INVERTER circuits are shown as Fig.4 (a), (b) respectively, which conclusion below is correct? ( C )A) The delay between input and output of (a) circuit is much less than (b) circuit. B) The delay between input and output of (a) circuit is much greater than (b) circuit. C) The delay between input and output of (a) circuit is about same as (b) circuit. D) The delay relationship between circuit (a) and (b) is uncertainty.Fig.4 (a)Fig.4 (b)9. The circuit shown in Fig.5 realize a logic functin F about input variable W, X, Y . Then, the Fis:( A )A) F=,,,(0,1,3,7,9,13,14)w x y z ∑B) F=,,,(0,2,5,7,9,13,14)w x y z ∑C) F=,,,(0,1,3,7,8,12,15)w x y z ∑D) F=,,,(1,2,5,7,9,12,15)w x y z ∑Fig.510. Which of the following statements are NOT correct about logic function? ( D ) A) There are multi-expressions of a logic function ’s minimal sum. B) The canonical sum of a circuit is a sum of minterms.C) Any logic function can be expressed using a sum of minterms or a product of maxterms. D) A sum of prime implicants must be the logic function ’s minimal sum. III. Combinational Circuit Analysis And Design: [30’]1.Write the truth table and the logic function performed by the CMOS circuit in Figure 6. (7’)Fig.6Solution :Z=S ’A+SB评分标准:真值表正确 4 分, 错一个扣0.5分;表达式正确 3分。
数电期末试卷及答案(1)

数电期末试卷及答案(1)数电期末试卷及答案第一部分:选择题1. 下列哪段代码可以实现2个二进制数的加法?A. y <= x1 + x2;B. y <= x1 xor x2;C. y <= x1 or x2;D. y <= x1 and x2;答案:A2. 如图,若P=0,Q=1,则S2S1S0的值为(MSI)S0 S1 S20 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1A. 010B. 001C. 110D. 100答案:A3. 下列哪个逻辑门的输出始终是0?A. 与门B. 或门C. 非门D. 异或门答案:C4. 如图,若编号相同的控制输入端和定时输入端连接,输出端Q1与标志组Z1连接,D1为1,则经计时电路处理后,输出端Q3的值为 [image: 图片]A. 1B. 0C. 不确定D. 未能实现功能答案:B5. 数字电路门电路中,进行加法运算的电路是A. 选择器B. 与门C. 或门D. 全加器答案:D第二部分:填空题6. 数据位错误率BER是信道传输误码率的一种度量方式,其公式为____________。
答案:BER=已传输的数据包中错误数据包的数量÷已传输的数据包的总数量7. 在卡诺图化简时,最小项指的是只有1个取值为1的项,最大项指的是只有1个取值为0的项。
请将下面的卡诺图进行化简,给出化简后的表达式:[image: 图片]答案:B'D+ACD8. SR型触发器的名称是基于其2个输入端S和R的名称组合而来的,其中S和R代表什么?答案:S代表Set(置位),R代表Reset(清零)9. 四位二进制数1011的十进制数值为___________。
答案:1110. 计数器的输出数值加1的功能被称为“向上计数”,输出数值减1的功能被称为“向下计数”,向上计数的计数器称为___________,向下计数的计数器称为___________。
数字电子技术期末考试题及答案(经典)

xxx~xxx学年第x学期《数字电子技术》期末复习题第一部分题目一、判断题(每题2分,共30分。
描述正确的在题号前的括号中打“√”,错误的打“×”)【】1、二进制有0 ~ 9十个数码,进位关系为逢十进一。
【】2、(325)8 >(225)10【】3、十进制数整数转换为二进制数的方法是采用“除2取余法”。
【】4、在二进制与十六进制的转换中,有下列关系:(100111010001)2=(9D1)16【】5、8421 BCD码是唯一能表示十进制数的编码。
【】6、十进制数85的8421 BCD码是101101。
【】7、格雷码为无权码,8421 BCD为有权码。
【】8、数字电路中用“1”和“0”分别表示两种状态,二者无大小之分。
【】9、逻辑变量的取值,1比0大。
【】10、在逻辑代数中,逻辑变量和函数均只有0和1两个取值,且不表示数量的大小。
【】11、逻辑运算1+1=1【】12、逻辑运算A+1+0=A【】13、因为逻辑表达式A+B+AB=A+B成立,所以AB=0成立。
【】14、在时间和幅度上均不连续的信号是数字信号,所以语音信号是数字信号。
【】15、逻辑函数的运算次序为:先算括号内,后算括号外;先求与,再求或,最后求非。
【】16、AB A C BC AB A C++=+【】17、逻辑函数表达式的化简结果是唯一的。
【】18、逻辑真值表、逻辑表达式、逻辑图均是逻辑关系的描述方法。
【】19、n个变量组成的最小项总数是2n个。
【】20、逻辑函数的化简方法主要有代数化简法和卡诺图化简法。
【】21、逻辑函数化简过程中的无关项一律按取值为0处理。
【】22、数字电路中晶体管工作在开关状态,即不是工作在饱和区,就是工作在截止区。
【】23、TTL或非门的多余输入端可以接高电平。
【】24、某一门电路有三个输入端A、B、C,当输入A、B、C不全为“1”时,输出Y为“0”,输入A、B、C全为高电平“1”时,输出Y为“1”,此门电路是或门电路。
电子科大数字电路期末试题半期测验

电子科大数字电路期末试题半期测验————————————————————————————————作者:————————————————————————————————日期:电子科技大学二零零七至二零零八学年第二学期期中考试“数字逻辑设计及应用”课程考试题 期中卷(120分钟)考试形式:闭卷 考试日期 2008年4月26日课程成绩构成:平时 20 分, 期中 20 分, 实验 0 分, 期末60 分一 二 三 四 五 六 七 八 九 十 合计一、选择填空题(单选、每空2分,共30分)1-1.与十进制数 (0. 4375 )10 等值的二进制数表达是 ( A ) A. ( 0.0111 ) 2 B. ( 0.1001 ) 2 C. ( 0.0101 ) 2 D. ( 0.01101 ) 2 1-2. 与十六进制数(FD .A )16等值的八进制数是( A )8A. ( 375.5 )8B. ( 375.6 )8C. ( 275.5 )8D. ( 365.5)8 1-3.与二进制数(11010011) 2 对应的格雷码表达是 ( C ) GrayA. ( 11111010 ) GrayB. (00111010 ) GrayC. ( 10111010 )GrayD. (11111011 ) Gray 1-4.下列数字中与(34.42)8 相同 的是( B )A.(011010.100101)2B.(1C.88)16 C.(27.56)10D.(54.28)5 1-5.已知[A]补=(10010011),下列表达式中正确的是( C )A. [–A]反=(01101100)B. [A]反=(10010100)C. [-A]原=(01101101)D. [A]原=(00010011)1-6.一个十六路数据选择器,其选择控制输入端的数量为( A )A .4个 B. 6个 C. 8个 D. 3个1-7.四个逻辑相邻的最小项合并,可以消去( B )个因子。
《数字电路》期末模拟试题及答案 (2)(2020年整理).doc

数字电子电路 模拟试题-2一、填空题(共30分)1. 三极管有NPN 和PNP 两种类型,当它工作在放大状态时,发射结____,集电结______;NPN 型三极管的基区是______型半导体,集电区和发射区是______型半导体。
2. 把高电压作为逻辑1,低电平作为逻辑0的赋值方法称作_______逻辑赋值。
一种电路若在正逻辑赋值时为与非门,则在负逻辑赋值时为________。
3. 四位二进制编码器有____个输入端;____个输出端。
4. 将十进制数287转换成二进制数是________;十六进制数是_______。
5. 根据触发器功能的不同,可将触发器分成四种,分别是____触发器、____触发器、____触发器和____触发器。
6. 下图所示电路中,Y 1 =______; 7. Y 2 =______;Y 3 =______。
二、选择题(共 20分)1. 当晶体三极管____时处于饱和状态。
A. 发射结和集电结均处于反向偏置B.发射结正向偏置,集电结反向偏置 C. 发射结和集电结均处于正向偏置2. 在下列三个逻辑函数表达式中,____是最小项表达式。
A .B A B A )B ,A (Y += B.C B C B A BC A )C ,B ,A (Y ++=C. C AB ABC B C A C B A )D ,C ,B ,A (Y +++⋅⋅=3.用8421码表示的十进制数45,可以写成__________A .45 B. [101101]BCD C. [01000101]BCD D. [101101]2 4.采用OC 门主要解决了_____A .TTL 与非门不能相与的问题 B. TTL 与非门不能线与的问题 C. TTL 与非门不能相或的问题5.已知某触发的特性表如下(A 、B 为触发器的输入)其输出信号的逻辑表达式为___A . Q n+1 =A B. n n 1n Q A Q A Q +=+ C. n n 1n QB Q A Q +=+三、化简下列逻辑函数,写出最简与或表达式:(共20分) 1. BC A C B A C B B A Y 1+⋅++= 2. Y 2=Σm (0,1,8,9,10,11)3. Y 3见如下卡诺图1.四选一数据选择器的功能见下表,要实现Y (A ,B ,C )=Σm (1,4,6,7)功能,芯片应如何连接,画出电路连接图(需写出必要的解题步骤)(20分)2.触发器电路如下图所示,试根据CP 及输入波形画出输出端Q 1 、Q 2 的波形。
最新电子科大数字电路,期末试题-2006-1数电半期考试

电子科技大学二零零五至二零零六学年第二学期期中考试“数字逻辑设计及应用”课程考试题 期中卷(120分钟)考试形式:闭卷 考试日期 2006年4月22日课程成绩构成:平时 20 分, 期中 20 分, 实验 0 分, 期末60 分一、填空题(每空1分,共15分)1、( 323 )10 =( 101000011 ) 22、(0. 4375 )10 =( 0.0111 ) 23、(1101.0011) 2 = ( 13.1875 )104、(FD .A )16 = ( 11110000.1010 ) 2= ( 360.50 )85、( 4531 )10 = ( 0100 0101 0011 0001 ) 8421BCD 。
6、写出与下列十进制数对应的8-bit 原码(signed-magnitude),补码(two ’s-complement)和反码 (one ’s-complement)表达:7、已知二进制数 A = 10110100,对应的格雷码(GRAY CODE )表达为( 1110 1110 ) 8、与非逻辑门电路的未用输入端应接在( 高电平或某一个输入信号端 )上。
9、已知二进制数 A 的补码为:[A]补= 10110100,求 [-A]补=( 01001100 )二、填空题(每空3分,共30分)1、已知一个函数的积之和(与或式, The sum of productions )列表表达式为 F =∑ABC (1,4,5,6,7),问与其对应的最简积之和表达式为:F =( A + B ’C )。
2、对于按照逻辑式 F AC BC '=+ 实现的电路,存在静态( 1 )型冒险。
3、四变量逻辑函数F = ∑ABCD (2,4,5,7,9,14)的反函数 F ’=∏ABCD ( 2,4,5,7,9,14 )。
4、已知带符号的二进制数 X1 = +1110 ,X2 = -1011,求以下的表达,并要求字长为8位。
2020秋西安电子科技大学《数字逻辑电路》大作业期末试题及答案

学习中心/函授站_姓名学号西安电子科技大学网络与继续教育学院2020 学年下学期《数字逻辑电路》期末考试试题(综合大作业)题号一二三四总分题分30 10 30 30得分考试说明:1、大作业试题于2020 年10 月15 日公布:(1)毕业班学生于2020 年10 月15 日至2020 年11 月1 日在线上传大作业答卷;(2)非毕业班学生于2020 年10 月22 日至2020 年11 月8 日在线上传大作业答卷;(3)上传时一张图片对应一张A4 纸答题纸,要求拍照清晰、上传完整;2、考试必须独立完成,如发现抄袭、雷同均按零分计;3、答案须用《西安电子科技大学网络与继续教育学院标准答题纸》手写完成,要求字迹工整、卷面干净。
一、单项选择题(每小题2 分,共40 分)1、下列各数中与十进制数101 不相等的数是( D )。
A.(0100 0011 0100)余3BCD B.(141)8C.(1100101)2D.(66)162、八进制数(35)8的8421BCD 是( B )。
A.0011 1000B.0010 1001C.0011 0101D.0010 11003、为使与非门输出为1 则输入( A )。
A.只要有0 即可B.必须全为0C.必须全为1D.只要有1 即可4、函数F AC BC AB与其相等的表达式是( B )。
A.BC B.C+AB C.AC AB D.AB5、使函数F AB AC BC 等于 0 的输入 ABC 的组合是( C )。
A .ABC=111 B .ABC=110 C .ABC=100 D .ABC=0116、四变量的最小项ABCD 的逻辑相邻项是( A )。
A .ABCDB .ABCDC .ABCD D .ABCD 7、函数 F ABC B .C (A D )BC 的对偶式是( C )。
A .G (A B C )(B C )(AD B C )B .G A BC (B C )ADB CC .G A B C (B C )(AD B C )D .G A BC (B C )AD B C8、FA B C ADE BDE ABC 的最简式为( A )。
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电子科技大学二零零九年至二零一零学年第二学期“数字逻辑设计及应用”课程考试题(半期)(120分钟)考试日期2011年4月23日一二三四五六七八九十总分评卷教师I. To fill the answers in the “( )” (2’ X 19=38)1. [1776 ]8 = ( 3FE )16 = ( 1111111110 )2= ( 1000000001 ) Gray .2. (365)10 = ( 001101100101 )8421BCD=( 001111001011 ) 2421 BCD.3.Given an 12-bit binary number N. if the integer’s part is 9 bits and the fraction’s part is 3 bits ( N = a8 a7 a6 a5 a4 a3 a2 a1 a0 . a-1 a-2 a-3), then the maximum decimal number it can represent is ( 511.875 ); the smallest non-zero decimal number it can represent is ( 0.125 ).4. If X’s signed-magnitude representation X SM is(110101)2, then it’s 8-bit two’s complement representation X2’s COMP is( 11101011 ) , and (–X)’s 8-bit complement representation (–X) 2’s COMP is ( 00010101 )2 .5. If there are 2011 different states, we need at least ( 11 ) bits binary code to represent them.6.If a positive logic function expression is F=AC’+B’C(D+E),then the negative logic function expression F = ( (A+C’)(B’+(C+DE)) ).7. A particular Schmitt-trigger inverter has V ILmax = 0.7 V, V IHmin = 2.1 V, V T+= 1.7 V, and V T-= 1.3 V, V OLmax=0.3V, V OHmin=2.7V. Then the DC noise margin in the HIGH state is ( 0.6V ), the hysteresis is ( 0.4V ). 8.The unused CMOS NAND gate input in Fig. 1 should be tied to logic ( 1 ).Fig.1Circuit of problem I-89. If number [ A ] two’s-complement =11011001and [ B] two’s-complement=10011101 , calculate[-A-B ]two’s-complement, [-A+B ]two’s-complement and indicate whether or not overflow occurs.[-A-B ] two’s-complement=[ 10001010 ], overflow: [ yes ][-A+B ] two’s-complement=[ 11000100 ], overflow: [ no ].10.The following logic diagram Fig.2 implements a function of 3-variable with a 74138. The logic function can be expressed as F (A,B,C) = ∑A,B,C ( 0,1,2 ).Fig.2 Circuit of problem I-10II. There is only one correct answer in the following questions.(3’ X 9 = 27)1. Which of the following Boolean equations is NOT correct? ( B )A) A+0=A B) A1 = AC) D)2. Suppose A2’s COMP =(1011),B2’s COMP =(1010),C2’s COMP =(0010). In the following equations, the most unlikely to produce overflow is( C )。
A)(A+B)2’s COMP B) (A-C) 2’s COMP C) (B+C) 2’s COMP D) (C-A)2’s COMP3. What is the possible radix of the number system in operation of 302/20=12.1. ( B )A) 7 B) 4 C) 5 D) 84. A 16-to-1 multiplexer need ( B ) control inputs.A) 3 B) 4 C) 6 D) 85.If the canonical sum for an n-input logic function is also a minimal sum, how many literals are in each product term of the sum? ( A )A) n B) 2n C) n-1 D) cannot be determined.6.A priority encoder 74LS148’s input is:I0-L, I1-L, I2-L, I3-L, I4-L, I5-L, I6-L, I7-L,output is Y2-L,Y1-L,Y0-L.The inputs and output are all active-low. When active-low enable input S_L=0,and I2-L=I4-L=I5-L =0, then Y2-L,Y1-L,Y0-L is ( B ).A) 110 B) 010 C) 001 D) 1017. There are two logic functions with same 4 input variables: F1 =∑ABCD(1,2,6,7,10,12) and F2 = ∏ABCD(3,5,8,9,13,14). The relationship between them is(C)。
A). Shannon’s expansion B). Equivalent C). Duality D). Complement8. How many inverters can be driven by the same inverter for the 74LS TTL families in Fig. 3.When the output of in hign and low levels are,, the input currents are, the maximum output current, the maximum output current= —0.4mA. ( D )A) 5 B) 10 C) 15 D) 20Fig. 3 Circuit of problem II-89. A CMOS gate circuit is shown as Fig 4. The function expression for the circuit is ( C ).A) (AB+BC)’ B) (AC+BC)’C)( AB+AC)’ D) AB+ACIII. Combinational Circuit Analysis And Design: [35]1. Minimize the logic function expression to realize the circuit with as less gates as possible. [5’]F = AB’(C + D) + BC’ + A’B’ + A’C + BCF = AB’(C+D) + B(C+C’)+A’B’+A’C (1’)Fig 4 circuit of problem II-9= AB’(C+D)+ B+A’B’+ A’C (1’)= A(C+D)+ B+A’+ A’C (2’)= C+D+A’+B (1’)2. Simplify the following logic function into the minimal sum expression and minimal product expression using Karnaugh map.F(A,B,C,D)= ∑m(0,4,7,11,15)+ ∑d(1,5,8,9,10,12,13,14) [5’]A B C F A F B F C 000000 001001 010010 011001 100100 101001 110010 111001(3’)the minimal sum expression :F = A+C’+BD (1’)the minimal product expression F =(A+B+ C’)(C’+ D) (1’) 3. There is a logic circuit and the inputs A,B and C are corresponding to theoutputs F A,F B and F C .The waveforms of inputs and outputs are shown in Fig.5.Please write out the truth table include all the inputs and outputs, and the logic expression of the outputs F A, F B and F C, and describe the logic function.[10’]ABCF AF BF C(6’)Fig. 5 Waveforms ofproblem III-3F A= AB’C’ (1’)F B= BC’ (1’)F C= C (1’)逻辑功能:该电路为优先权排序电路,C具有最高优先权,B次之,A的优先权最低。