# Tag Info

9

The problem is sending constant data stream without any pauses between symbols. The ASCII symbol 'S' has a value of 0x53, so it is sent over the wire as repeating pattern of 0110010101 which includes the start and stop bits. Because there is no pauses between transmissions, the receiving UART does not know which bits are the start and stop bits in the ...

7

They are equivalent. The width of all operands get extended to the size of the largest operand before any operation occurs. As long as the width of one of BC or AD is 16 bits wider than the value being shifted, then what you wrote is mathematically equivalent to $(BC + AD)×2^{16} = BC×2^{16}+ AD×2^{16}$ If the width of the largest operand is not wide ...

5

As an addition to Justmes answer (feel free to combine this into your answer): This is a UART Transmission of an S: This is what happens if you dont have an idle: And this part is what the receiver sees as a repeating 0xAA: edit: to be clear, i dont think this should be marked as the answer. This should just be an addendum to Justme's answer

3

There is no difference in the two examples you wrote. You can even make is simpler: module counter ( input wire clk, clr, output reg [3:0] q ); always @(posedge clk or posedge clr) begin if (clr) q<= 4'b0000; else q<= q+ 1'b1; end endmodule

2

This is simply out = (32'd1 << \$countones(in)) - 32'd1; Note: increase 32 to the size of your vector.

1

Your first example errors because you must use a variable data type on the left hand side in a procedural block. xml is the left hand side of the assignment, and so must be a variable (reg,integer,etc.), not a net (wire). Outputs default to being wire unless otherwise specified. You can fix this with an output reg [2:0] xml in your port declaration list. ...

1

Yes, that is correct behavior of event driven simulation. But it can broken down even further. You update the value of a and g2 could update the value of e. And then either b gets updated, or g3 updates the value of f. Only the statements within the begin/end block have guaranteed ordering so that a must be updated before b, and b must be updated before c. ...

1

A circuit that can identify a particular set bit in a vector in a fixed amount of time is called a "priority encoder". The general concept is that you use the priority encoder to find the first set bit in your vector, use its number to address your memory, and then you clear that bit in the vector so that the priority encoder can find the next set ...

1

Okay i figured out the solution. Here are top module and testbench for these code. `timescale 1ns / 1ps module TopModule(clk, aci ,reset,sonuc,en); input en; input reset; input [7:0]aci; output reg [15:0]sonuc; input clk; always@(posedge clk)begin if(reset) sonuc <= 'd0; else begin case(aci) 'd1 : sonuc <...

1

The following will work for both synthesis and simulation, and should work for both ASIC and FPGA designs as it doesnt need an initial block. wire imNothing = 0; always @ * begin theReg <= imNothing; end This is basically equivalent to connecting theReg to the wire imNothing which is permanetly attached to 0. Having said that, this doesn't stop you ...

1

Yes, as long as the width of the left-hand side you are assigning to is the same in both cases. If the width of the left-hand side is smaller than the right-hand side, the value will be truncated to the Least Significant bits. In other words, the Most Significant Bits are discarded. The width of the operands are extended to the same size as that of the ...

1

Investigation Using Mentor Questa 2020.1 and Cadence Xcelium 20.09 we don't get any glitches. Using Synopsys VCS 2020.03 and Icarus Verilog 14 we get glitches. I used EDA Playground to simulate, if interested please visit this link. Why do you get a glitch? The reason you are getting glitches is race conditions / race hazards due to the non-deterministic ...

Only top voted, non community-wiki answers of a minimum length are eligible