数字电路代写（有偿）

Question

数字电路任务代写（有偿）

要求用英语书写，转为Word文档，完成任务中各项要求

Answer 1

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I'll provide a structured solution for the digital circuit assignment. Since the assignment requires personalized calculations based on student ID and birthday, I'll demonstrate the methodology using the example student ID (2279803) and birthday (23/01/2005) provided in the instructions. You'll need to replace these with your actual details.

1. Number System Conversions

a) 4 Least Significant Digits to Binary

Given student ID: 2279803 → Last 4 digits: 9803

Convert 9803 to binary:

9803 ÷ 2 = 4901 R1
4901 ÷ 2 = 2450 R1
2450 ÷ 2 = 1225 R0
1225 ÷ 2 = 612 R1
612 ÷ 2 = 306 R0
306 ÷ 2 = 153 R0
153 ÷ 2 = 76 R1
76 ÷ 2 = 38 R0
38 ÷ 2 = 19 R0
19 ÷ 2 = 9 R1
9 ÷ 2 = 4 R1
4 ÷ 2 = 2 R0
2 ÷ 2 = 1 R0
1 ÷ 2 = 0 R1

Reading remainders upwards: 10011001001011

Verification:

1×2^13 + 0×2^12 + 0×2^11 + 1×2^10 + 1×2^9 + 0×2^8 + 0×2^7 + 1×2^6 + 0×2^5 + 0×2^4 + 1×2^3 + 0×2^2 + 1×2^1 + 1×2^0
= 8192 + 512 + 256 + 64 + 8 + 2 + 1 = 9803

b) Conversions

i. Hexadecimal:
Group binary into 4s: 1001 1001 0010 1011 → 9 9 2 B → 0x992B

ii. Base 12:
Convert 9803 to base 12:

9803 ÷ 12 = 816 R11 (B)
816 ÷ 12 = 68 R0
68 ÷ 12 = 5 R8
5 ÷ 12 = 0 R5

Result: 580B₁₂

2. Fixed-Point Binary Conversion

Take 98.03:
Integer part (98):

98 ÷ 2 = 49 R0
49 ÷ 2 = 24 R1
24 ÷ 2 = 12 R0
12 ÷ 2 = 6 R0
6 ÷ 2 = 3 R0
3 ÷ 2 = 1 R1
1 ÷ 2 = 0 R1

→ 1100010

Fractional part (0.03):

0.03 × 2 = 0.06 → 0
0.06 × 2 = 0.12 → 0
0.12 × 2 = 0.24 → 0
0.24 × 2 = 0.48 → 0

→ 0.0000 (to 4 bits)

Combined: 1100010.0000

3. BCD Conversions

Full student ID: 2279803

i. 8421 BCD:

2: 0010
2: 0010
7: 0111
9: 1001
8: 1000
0: 0000
3: 0011

→ 0010 0010 0111 1001 1000 0000 0011

ii. 74-2-1 BCD:

2: 0100
2: 0100
7: 1010
9: 1110
8: 1101
0: 0000
3: 0101

→ 0100 0100 1010 1110 1101 0000 0101

4. Gray Code Generation

Last 2 digits: 03 → 03 + 20 = 23 (since <33)

Generate Gray codes from 23 to 38 (23+15):

Decimal Binary Gray
23      10111  11100
24      11000  10100
...     ...     ...

5. Error Detection

For the 8421 BCD (28 bits → 7×4 bits), add parity:
Example with even parity:

0010 (0) → 00100
0111 (1) → 01111
...

If one bit flips, parity check will detect and correct.

B) Combinational Logic System

Birthday: 23/01/2005 → 230105 → 23+01+05=29 → binary 11101 → LSB=1 → Use F1

1. Canonical Forms

F1 = Σ(2,3,5,7,8,10,12,13)

SOP:

F1 = A'B'CD' + A'B'CD + A'BC'D + A'BCD + AB'C'D' + AB'CD' + ABC'D' + ABC'D

POS:

F1 = Π(0,1,4,6,9,11,14,15)

2. Minimization

Using K-map:

CD\AB 00 01 11 10
00    0  0  1  1
01    0  1  1  0
11    1  1  0  0
10    1  0  1  1

Minimized SOP:

F1 = A'D + B'CD' + ABC' + A'B'C

3. NAND/NOR Implementations

NAND version:

F1 = NAND(NAND(A',D), NAND(B',C,D'), NAND(A,B,C'), NAND(A',B',C))

C) Sequential Logic System

Using odd number set: {0,6,2,4,5,1,7,3}

1. Asynchronous Counter

3-bit counter with output logic to map states to the sequence.

2. Synchronous Design

More reliable, uses clocked flip-flops with state transitions directly implementing the sequence.

3. Modified Sequence

Remove first/last 0 → {6,2,4,5,1,7,3} → duplicate middle → {6,2,4,5,5,1,7,3}

Design 7-state machine using 3 flip-flops with custom state transitions.

Final Notes

1. All solutions must be adapted to your specific student ID and birthday
2. Complete circuit diagrams would be included in the Word document
3. Timing analysis would be performed for sequential circuits
4. All minimization steps would be shown with proper K-maps
5. Estimated completion time: 12-15 hours for full solution

Would you like me to elaborate on any specific part of the solution?