电子科技大学2012 -2013学年第 二 学期期 末 考试 A 卷
课程名称：_数字逻辑设计及应用__ 考试形式： 闭卷 考试日期： 20 13 年 07 月 05 日 考试时长：_120___分钟
课程成绩构成：平时 30 %， 期中 30 %， 实验 0 %， 期末 40 % 本试卷试题由___七__部分构成，共__7___页。
题号 得分 得 分 一 二 三 四 五 六 七 八 九 十 合计 I. Fill out your answers in the blanks (3’ X 10=30’) 1. If a 74x138 binary decoder has 110 on its inputs CBA, the active LOW output Y5 should be ( 1 or high ). 2. If the next state of the unused states are marked as “don’t-cares” when designing a finite state machine, this approach is called minimal ( cost ) approach.
3.The RCO_L of 4-bit counter 74x169 is ( 0 or low ) when counting to 0000 in decreasing order.
4. To design a \001010\
( 4 ) bit at least.
5. One state transition equation is Q*=JQ’+K’Q. If we use T flip-flop with enable to complete the equation，the enable input of T flip-flop should have the function EN=( JQ’+KQ ).
6. A 4-bit Binary counter can have ( 16 ) normal states at most, 4-bit Johnson counter with no self-correction can have ( 8 ) normal states, 4-bit linear feedback shift-register (LFSR) counter with self-correction can have ( 16 ) normal states.
7. If we use a ROM, whose capacity is 16 × 4 bits, to construct a 4-bit binary code to gray code
converter, when the address inputs are 1001, ( 1101 ) will be the output.
8. When the input is 10000000 of an 8 bit DAC, the corresponding output voltage is 2V. The output voltage is ( 3.98 ) V when the input is 11111111. 得 分
II. Please select the only one correct answer in the following questions.(2’ X 5=10’)
1. If a 74x85 magnitude comparator has ALTBIN=1, AGTBIN=0, AEQBIN=0, A3A2A1A0=1101, B3B2B1B0=0111 on its inputs, the outputs are (
D ).
A) ALTBOUT=0, AEQBOUT=0, AGTBOUT=0 B) ALTBOUT=1, AEQBOUT=0, AGTBOUT=0 C) ALTBOUT=1, AEQBOUT=0, AGTBOUT=1
D) ALTBOUT=0, AEQBOUT=0, AGTBOUT=1
2. As shown in Figure 1, what would the outputs of the 4-bit adder 74x283 be ( B ) when A3A2A1A0=0100, B3B2B1B0=1110 and S/A=1. A) C4=1, S3S2S1S0=0010 D) C4=0, S3S2S1S0=1110
B) C4=0, S3S2S1S0=0110
C) C4=0, S3S2S1S0=1010
Figure 1
3. Which of the following statements is INCORRECT? ( A )
A) A D latch is edge triggered and it will follow the input as long as the control input C is active
low.
B) A D flip flop is edge triggered and its output will not change until the edge of the controlling
CLK signal.
C) An S-R latch may go into metastable state if both S and R are changing from 11 to 00
simultaneously.
D) The pulse applying to any input of an S -R latch must meet the minimum pulse width requirement. 4. The capacity of a memory that has 13 bits address bus and can store 8 bits at each address is ( B ). A) 8192 B) 65536 C) 104 D) 256 5. Which state in Figure 2 is NOT ambiguous ( C ).
A) A B) B C) C and D D) C
WXAW+YX’Z’ZBX’+YYZZ’1Figure 2
DC
得 分
III. Analyze the sequential-circuit as shown in Figure 3, D Flip-Flop with asynchronous preset
and clear inputs. [15’]
1.Write out the excitation equations, transition equations and output equation. [5’] 2.Assume the initial state Q2Q1=00, complete the timing diagram for Q2 ,Q1 and Z. [10’]
Figure 3
参考答案：
激励方程: D1=Q2/，D2= Q1
转移方程：Q1 *= D1=Q2/，Q2 *=D2= Q1 输出方程：Z= (CLK+Q2)/
参考评分标准：
1. 5个方程正确得5分；每错一个扣1分，扣完5分为止；
2. Q1、Q2、Z的波形边沿判断正确，得3分，错一个，扣1分，扣完3分为止；每个上升沿和下降沿各
0.5分，错1处扣0.5分，扣完7分为止。 得 分
IV. Analyze the sequential-circuit as shown below, which contains two 74x163 4-bit binary
counter. [15’]
1. Write out the logic expression LD_L for U1 and CLR_L for U2.[4’]
2. Assume the initial state is 310, write out the state sequence for the circuit. [8’] 3. Describe the modulus for the circuit. [3’]
The function table for 74x163 CLR_L 0 1 1 1 1 1 1 1 1 1 Inputs LD_L ENT X X 0 X 1 0 1 X 1 1 1 1 1 1 1 1 1 1 1 1 ENP X X X 0 1 1 1 1 1 1 Current state QD QC QB QA X X X X X X X X X X X X X X X X 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 …………. 1 1 1 1 Next state QD* QC* QB* QA* 0 0 0 0 D C B A QD QC QB QA QD QC QB QA 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 ………….. 0 0 0 0 Outputs RCO 0 0 0 0 0 0 0 0 0 1 参考评分标准：
1. LD_L=Q3/,CLR_L=(Q5Q4Q3)/
[4’]
2. 状态序列：十六进制数表示：03,…08,13,…18,23,…28,33,…38,03,…08
或十进制数表示：3,…8,19,…24,35,…40,51,…56,3,…8[8’]
错1处扣1分，扣完为止。 3. m=24 得 分
[3’]
V. Design a sequence signal generator with self-correcting to generate a serial output sequence
of 101100, using a 74x194 and a 74x151.[15’] 1. List the transition table .[4’]
2. Write out the canonical sum of feedback
function LIN.[[4’]
3. Draw the circuit diagram.[7’]
参考评分标准：
1. 转移表正确4分，错1行扣0.5分。 2. 反馈函数正确LIN=D0=∑m(0,2,4,5)
The function table for 74x194 Iutputs S1 S0 Next state QA* QB* QC* QD* Function 0 0 0 1 1 0 1 1 QA QB QC QD RIN QA QB QC QB QC QD LIN A B C D Hold Shift right Shift left Load [4’]
3. 电路图正确7分，错1处扣0.5分，扣完为止。
《数字逻辑设计及应用》期末考试题0A参考解答
