Best代写-最专业靠谱代写IT | CS | 留学生作业 | 编程代写Java | Python |C/C++ | PHP | Matlab | Assignment Project Homework代写

电子电路代写 | EL 6463: Advanced Hardware Design

电子电路代写 | EL 6463: Advanced Hardware Design



● Read and complete all questions carefully.

● This exam is to be completed individually. Collaboration/plagiarism will not be tolerated.

● Please sign the code of conduct (located at the bottom of this exam) and include it in your submission.

● Clarification questions can be posted on Slack in #finalexam, but do not put any specifics about your own answers in the channel.

● During the exam period, you will not be allowed to receive assistance from the teaching staff regarding the use of the tools or implementation/simulation.

Overview: In this examination you will design, implement, and simulate digital hardware with sequential and combinational logic to implement a pseudo-random number generator algorithm. This algorithm will be implemented via hardware known as a linear feedback shift register (LFSR). These have applications in generating simple randomness in gaming, cryptography, audio and signal processing.

An example of a 16-bit LFSR is shown below:

Image courtesy: CC0

The depicted LFSR has XOR “taps” at positions [16, 14, 13, 11] (or zero-indexed: 15, 13, 12, 10).

In each clock cycle, the rightmost tap (which is known as the output bit) is XOR’d sequentially with the next right-most taps and the result is “shifted in” to the left-most bit (the most significant bit (MSB)).

In this final exam you will implement an LFSR with the following specifications (essential features).

Note: this is the same base specification as what was used in the mid-term exam.

1. It will be 32 bits long (from position 0 (MSB) to 31 (LSB)), with taps at positions as determined:

a. Each of the unique adjacent 2-digit numbers in your N number modulo 32.

b. The 32 bit (position 31, the LSB).

c. For example, if your N number is 01123497, you will have taps at positions

i. [01, 11, 12, 23, (34 % 32 =) 2, (49 % 32 =) 17, (97 % 32 =) 1] (“1” value is not unique) ii. = [1, 2, 11, 12, 17, 23, 31] (31 comes from point 1b) – remember these are zero-indexed

2. Top I/O:

a. clk: input

b. rstn: active-low reset signal

c. data_out: 8 bit output

d. get_random: input 1 bit control signal

e. 1-bit input “load_seed”

f. 8-bit input “data_in”.

3. Unless otherwise specified, the LFSR will update (shift-in the XOR result) upon every clock cycle. The bits shift into position 0 (the MSB).

4. If “output” is requested by setting “get_random” to 1, the least significant 8 bits will be output to “data_out” in the next clock cycle, the next 8 in the next clock cycle, the next 8 in the next cycle, and the most significant 8 bits will be output in the final (fourth) cycle. No LFSR shifts will occur during this output sequence. At all other times the “data_out” will be all zeros.

a. NOTE:

i. If the get_random signal is held high for longer than 4 cycles then the LFSR would presumably immediately begin exporting the contents again, and so it wouldn’t perform any shifts, however there might be one clock cycle where the internals could shift. This part of the spec is upto to you.

ii. Similarly if get_random is high for less than 4 cycles, it should continue with the output until it is completed.

5. Reset signal “rstn” will reset the LFSR contents to 0x02468ACD

a. (Reset signal “rstn” should always take precedence over every other signal)

6. If “load_seed” is set to 1, the lower 8 bits will be loaded into the LFSR from “data_in” in the first clock cycle, the next 8 in the next, the next 8 in the next, and the upper 8 bits will be loaded in the final (fourth) cycle. No LFSR shifts will occur during loading. After loading, shifts will resume as normal.