Crowdly
Add to Chrome
ECE 564 (001) Fall 2025 ASIC and FPGA Design with Verilog
Looking for ECE 564 (001) Fall 2025 ASIC and FPGA Design with Verilog test answers and solutions? Browse our comprehensive collection of verified answers for ECE 564 (001) Fall 2025 ASIC and FPGA Design with Verilog at moodle-courses2527.wolfware.ncsu.edu.
Get instant access to accurate answers and detailed explanations for your course questions. Our community-driven platform helps students succeed!
What is wrong with the following code fragment intended for synthesis?
reg[3:0] A;initialA =’4’b000 1;always@(posedge clock)A <= C;Consider the following code (implemented exactly as described):
reg A, B, C;always@(posedge clock) A <= B;always@(posedge clock)
B <= (A | C) & B;always@(posedge clock)
C <= A;The minimum and maximum delays between each set of successive gates are marked as #(min: typical: max) ns and are marked on the output node of the driving gate. (Remember the timing equations are ≤ and ≥ constraints). You also need the following:
- T_setup = #(1 : 1 : 2) ns.
- T_hold = #(1: 1 :2) ns.
- T_skew = #(1: 1 : 1) ns for the clock.
- T_clock-Q = #(2 : 3 : 5) ns.
- T_logic = #(1 : 2 : 3) ns for each and every logic gate from any input to any output.
Is there potential for a hold violation? T/F
In general, the designer can choose whether the clock skew is positive or negative between any pair of flip-flops.
In the following test fixture, what is the timing diagram being specified?
initialbegindec = 0; run = 0; ck = 1;#10 dec =1; run = 1;#10 run = 0;#20 $finish;endalways #5 ck = ~ck;What is wrong with the following code fragment intended for synthesis?
wire foo;always@(*)foo = A | B;Consider the following code (implemented exactly as described):
reg A, B, C;always@(posedge clock) A <= B;always@(posedge clock)
B <= (A | C) & B;always@(posedge clock)
C <= A;The minimum and maximum delays between each set of successive gates are marked as #(min: typical: max) ns and are marked on the output node of the driving gate. (Remember the timing equations are ≤ and ≥ constraints). You also need the following:
- T_setup = #(1 : 1 : 2) ns.
- T_hold = #(1: 1 :2) ns.
- T_skew = #(1: 1 : 1) ns for the clock.
- T_clock-Q = #(2 : 3 : 5) ns.
- T_logic = #(1 : 2 : 3) ns for each and every logic gate from any input to any output.
What is the fastest possible clock period that satisfies setup constraints across all variations? Give the answer in ns without the units
Consider the following code. It is the only code in the module.
always@(posedge clock)begin A <= C & D; endalways@(*)E = A+B;If it is synthesized with the following constraints
Create_clock -period 8 -waveform {0 4} –name clockset_clock_skew -uncertainty 1.0 clockset_input_delay 1.0 -clock clock all_inputs() – clockset_output_delay 2.0 -clock clock all_outputs() – clockFlip flop tck-Q delay is 1 ns, setup time is 1 ns, and hold time 1 ns, all logic gate delays vary between 1 and 2 ns.What is the maximum allowed delay for the logic A+B? Give your answer in ns without the units.
Consider the 3-level clock tree shown. If the 6-sigma variation in clock buffer delay is +/- 20 ps, and the post-routed delay of each buffer-to-buffer or buffer-flip-flop wire can vary by +/- 5 ps from the pre-layout delay estimate, what is the expected clock skew? Assume you can just add the 6-sigma variation to get the final variation.Give your answer as a number (in ps). For example, if the answer is 25 ps, just enter 25
The details of the logic design have a strong impact on achievable clock speed
Consider the following design, with these delaysInverter = FO-4 Flip-flop t_cp-Q = 2 FO4 Flip-flop t_su / t_h = FO4 Clock skew = FO4Clock has 25% duty cycleWhat is the clock cycle in a latch based design, assuming cycle stealing? Give the answer in units of FO4. For example, if its 10 FO4 just answer 10. For the purposes of this question, only solve the problem for setup violations, since I have only gave you the max logic delays and not the mins to permit a hold calculation. Give the answer to one decimal point of precision.
Note this question is quite difficult because of the feedback and answer given by straightforward application of the formula in the notes is incorrect.