I found two different sources that explain inertial delay in Verilog HDL in two different ways.
1) The first one says that any input signal shorter than the specified delay will be ignored.
2) The second one says that, given a change at one of the inputs, the output signal will be evaluated at the scheduled time using the values of the input signal at that time.
For example: Consider a delayed buffer
assign #4 out = in;
initial
begin
in = 0;
#5 in = 1;
#1 in = 0;
#1 in = 1;
end
If we monitor the signals, this would result in
0 5 6 7 8 9 10 11
| | | | | | | |
__ _________________
in _________| |__|
_____
out1) XXXXXXX____________________|
___________
out2) XXXXXXX______________|
out1) ignored the input (5,6) pulse in "up" and the (6,7) pulse in "down" state for being too short in time, only 1 time unit. But then "in" stayed up long enough (7,11) and thus out changed at 11.
out2) scheduled an evaluation at time step 9, because the input changed at time 5. Similarly at times 10 and 11 for "in" having changed at times 6 and 7 respectively. So, at times 9, 10 and 11, "out" takes the current value of "in" at these times, which is always "up" in this case.
Which evaluation is the right one?
The easiest way to find out - test it. Let's add a simple line:
$monitor("%g out = %b", $time, out);
to monitor out signal and run some simulator (e.g. Riviera). The result will be:
0 out = x
4 out = 0
11 out = 1
So your first approach is the correct one.
I would suggest you to look at the following links to get a good idea of how to implement inertial delay and transport delay in Verilog:
pdf_1
pdf_2
Related
I've designed a ROM for coefficients and an up-down counter to read these coefficients one by one but there are two cases for the starting point where a specific number of coefficients for type1 and another set of coefficients for type 2 ... so for example for type 1 I want to start from address zero and for type 2 start from address 30 ... I remember that someone told me it is possible using some # or something but I don't remember what is the actual way to do this
this for my counter code
module UDcounter(input clk,rst,up,GItype,
output reg [5:0]addr);
always #(posedge clk,posedge rst)
if (rst)
addr<=6'b0;
else
begin
if (GItype) //assume 1 is a long GI type
begin
// addr=6'b000000;
if (up)
addr=addr+1;
else addr=addr-1;
end
else //for short GI
begin
//addr=6'b100000;
if (up)
addr=addr+1;
else addr=addr-1;
end
end
endmodule
the error here is that every clock cycle it start addressing from addr=0 for example and the output address is always 1 (for the +1) line
So what I understood from your question is that you want to design a ROM which will store coefficients.
Going by your question I assume that you have two types of coefficient viz type a & type b stored in the ROM, say the starting address for type a is 0 and for type b is 30. To go about accessing the ROM you would want two counters viz addr_ptr_a and addr_ptr_b which will act as address pointers, lets assume that the ROM has about 60 address locations then addr_ptr_a will count from 0 to 29 and addr_ptr_b will count from 30 to 60.
The GItype signal can be used to determine which counter to enable.
I am assuming a sequential read operation, for a random read operation you would need a separate logic to generate the read address.
While reading the syntax of Verilog, I came across the four logic values: 0 1 x z.
After searching the web, seeking to find the difference between x and z, I found only that x is unknown value and z is high impedance (tristate). I think that I understand the definition of x but didn't quite understood the one of z - what does it mean "high impedance (tristate)"?
I would like to see an example for each logic value out of the two: x z
Z means the signal is in a high-impedance state also called tri-state. Another signal connected to it can change the value: a 0 will pull it low, a 1 will pull it high.
To understand impedance (and thus high impedance) you should have some understanding of resistance, voltage and current and their relations as defined by Ohms law.
I can't give you an example of 'X' or 'Z', just as I can't give you an example of '1' or '0'. These are just definitions of signal states. In fact in Verilog there are more then four states. There are seven strengths.
(See this webpage).
Here is a principle diagram of how a chip output port makes a zero, one or Z. In reality the switches are MOSFETs.
Tri-state signals are no longer used inside chips or inside FPGA's. They are only used outside for connecting signals together.
x, as you had already found describes an unknown state. By default verilog simulation starts with all variables initialized to this value. One of the task of the designer is to provide correct reset sequences to bring the model into a known state, without 'x', i.e.
always #(posedge clk)
if (rst)
q <= 0;
In the above example initial value of q which was x is replaced by a known value of 0.
The difference between 'x' and 'z' is that 'z' is a known state of high impedance, meaning actually disconnected. As such, it could be driven to any other value with some other driver. It is used to express tri-state buses or some other logic.
wire bus;
assign bus = en1 ? value1 : 1'bz;
...
assign bus = en2 ? value2 : 1'bz;
In the above example the bus is driven by 2 different drivers. If 'en1' or 'en2' is high, the bus is driven with a real 'value1' or 'value2'. Otherwise its state is 'z'.
verilog has truth tables for every operator for all the values. You can check how they are used. i.e. for '&'
& 0 1 x z
0 0 0 0 0
1 0 1 x x
x 0 x x x
z 0 x x x
you can find for every other gate as well. Note that there are no 'z' in the result, just 'x's.
In system verilog X is treated like unconnected wire and Z is Weak HIGH.
Suppose a situation where you have wire connecting 2 modules m1 and m2.
If you are driving Z onto that wire from m1 then you can pull down this wire by assigning it to zero by m2.
As I figured out :
"tristate" or "high impedance" In transistors occures when you have "nothing" in the output.
that may occur, for example :
In a situation that you have an nMOS transistor let's call that T1:
the gate value of T1 is for example 0
so T1 would not conduct and there is no conduction path between your supply (probably 0 ) and the drain(output)
-that may occur a "Z" or tristate
--
It may occur for PMOS transistors with value -> 1 too.
I'm trying to realize a counting module. My basic setup:
FPGA (Digilent's Arty with Xilinx Artix-35T) with two BNC cables attached to IO ports connected to a signal generator and via USB/UART to the PC for reading out. My signal generator produces with, say, 1 Hz some TTL signal.
I now want to count the amount of events in channel 1, in channel 2 and the coincidences of channels 1 and 2. While the basic principle works, I see channels 1 and 2 separate, even though they have the same input (via BNC-T connector). Also, sometimes one of the output channels jumps - in either direction, see figure.
The violet channel ("Channel 1") has a different slope than green ("Channel 2"). Also the coincidences make here two little lossy jumps.
My sequential counting code looks like
reg [15:0] coinciInt [(numCoincidences -1):0]; // internally store events
always #(posedge clk or posedge reset) // every time the clock rises...
begin
signalDelay <= signal; // delayed signal for not counting the same event twice
if(reset) // reset
begin
for(i=0;i<numCoincidences;i=i+1)
coinciInt[i] <= 16'b0;
end
else // No reset
begin
for(i=1;i<numCoincidences;i=i+1) // loop through all coincidence possibilities:
begin
if( ((signal & i) == i) && ((signalDelay & i) != i) ) // only if signal give coincidence, but did not give before, it's a coincidence
begin // "(signal & i) == i" means that "signal" is checked if bitmask of "i" is contained:
// ((0011 & 0010) == 0010) is true, since 0011 & 0010 = 0010 == 0010
coinciInt[i] <= coinciInt[i] + 1'b1; // the i-th coincidence triggered, store it
end
end
end
end // end of always
assign coinci = coinciInt; // the output variable is called coinci, so assign to this one
Please note that all events are in the register coinci - coincidences as well as 'single events'. Ideally, coinci[1] should store events of channel 1, coinci[2] these of channel 2 and coinci[3] coincidences between 1 and 2, since channels are labelled by 1,2,4,8,...,2^n and coincidences by the respective sum. coinci[0] is used for some kind of checksum, but that's off-topic now.
Are there any ideas for the missing counts? For the different slopes?
Thank you very much
Edit 1
#Brian Magnuson pointed to the meta stability issue. Using multi-buffered inputs solved the issue of diverging channels. That works nicely. Although I don't fully understand the reason for this, I also did not see any jumps in the coincidence channel so far. You probably save me a lot of time, thanks!
I would suspect a meta-stability problem. Your incoming pulses on ch1/ch2 are probably not synchronized with the system clock you are using. See here.
Because of this you are probably sometimes catching the counter updates 'mid-stride' so to speak which will cause unexpected behavior.
To fix this you can flop the inputs twice (called a dual-rank synchronizer) before feeding them into the rest of your logic. Usually multi-bit synchronization requires a bit more careful handling but in your case each bit can be treated independently.
I have the following 8-bit array:
0
4
0
0
5
0
2
0
How do I make it to the following in a single cycle (without iterating the element one by one)?
4
5
2
0
0
0
0
0
I know how to do it in software (MATLAB), but I'm not sure how to do it with combinational logic.
% initialise temporary vectors
TempType = zeros(maxType,1);
TempStart = zeros(maxType,1);
TempStop = zeros(maxType,1);
index = 1;
% remove zero elements from the middle
for j = 1:maxType
if (PreType(j) > 0 && PreStart(j) > 0 && PreStop(j) > 0)
TempType(index) = PreType(j);
TempStart(index) = PreStart(j);
TempStop(index) = PreStop(j);
index = index + 1;
end
end
I think any simplified sorting algorithm can do the job. For example, here is a modified bubble sort solution implemented in a single cycle:
module MoveZeros;
parameter W1 = 8;
parameter W2 = 10;
integer i, j;
logic [W1-1:0] array[W2-1:0] = {0,4,0,0,5,0,2,0,0,1};
logic [W1-1:0] temp;
always_comb begin
for (i=W2-1 ; i >=0 ; i=i-1)
for (j=W2-1 ; j >= 0 ; j=j-1)
begin
if (array[j]==0 && array[j-1] != 0) begin
temp = array[j];
array[j] = array [j-1];
array[j-1] = temp;
end
end
end
endmodule
output:
# array = '{4, 5, 2, 1, 0, 0, 0, 0, 0, 0}
Working example on edaplayground. Depending on your cycle time and the width of your input array (W2), you may want to break this algorithm into multiple cycles.
Synthesis tools unroll loops, therefore, the synthesized circuit will have O(W2^2) comparators and multiplexers, which can explode. Hence for bigger arrays, a multi-cycle solution is the way to go.
This is not an answer, which would take several hours of work, but SO's comments are not up to this sort of question. You should ask on comp.arch.fpga, if it's still alive.
Start by finding a datasheet for one of the old asynchronous fall-through FIFOs; these will include a circuit diagram. You don't really want to do anything like this, because the stage-to-stage handshaking is hairy, and you can't apply all 8 values simultaneously, but it'll give you ideas for a more synchronous implementation. Adapting a fall-through FIFO to do what you want is trivial - just ignore zero inputs.
If you can go up to 8 clock cycles, a more synchronous implementation is easy, with relatively limited hardware.
One cycle doesn't look too difficult, but will use more hardware. How sure are you that you must do it in one cycle? How much hardware can you use? If you've got a free PLL/DLL I'd be inclined to use that to get an 8x clock.
EDIT
Actually, with the benefit of more than 2 minutes thought, this seems pretty easy, even in one cycle.
Say you've got 8 registers with your 8 inputs (I0-I7), and 8 output registers (Q0-Q7). Each output register has associated logic which selects an input register for source data. The Q0 selector finds the lowest-numbered I register which contains non-zero data. The Q1 selector finds the next highest I register which contains non-zero data, and so on. Each selector drives a mux which loads the corresponding output register. Q0 requires an 8-1 mux (eight 8-bit inputs from I0-I7, one 8-bit output which goes to the input of Q0). Q1 requires a 7-1 mux (the inputs can only be I1-I7), and so on, until Q7, which doesn't require a mux at all (it can only be driven by I7).
The only smarts are in the selectors which find the source data for each output register. The Q7 selector is trivial; Q7 can only select I7, and only if all of I0-I7 contain non-zero data. Q6 is a bit more complicated, and so on.
If you can't see how to code a selector, ask specifically about that one in a new question, to avoid all the comments.
i have just started to learn some verilog and came across a problem that would normally be solved using a for loop in other languages.
s2 s1 s0 m
0 0 0 u
0 0 1 v
0 1 0 w
0 1 1 x
1 0 0 y
1 0 1 z
This is the truth table of the problem where s2,s1,s0 are switches, (u,v,w,x,y,z) are inputs and if certain switches are set, the output(m) will be set as one of the inputs. However, this is only 1 bit wide and we can just assign each m to be one of the following inputs provided that s0,s1,s2 matches. If it were 3 bit wide input and output ex:(u0,u1,u2),(m0,m1,m2) but each of the variables still follows the same logic with one another(ex: u0,u1,u2 will all be the outputs m0,m1,m2 provided that all switches are 0) how can i shorten my code. Am i allowed to do:
assign m[3:0] = (~s0&~s1&~s2&u[3:0])
No, that is not a good solution. Verilog will extend your select signals to be the same width as m before performing the & operation, and the default behavior is to add zeros on the left to extend a signal. It might work if you declared the select signals to be signed but I think you can do better.
Furthermore, the example you provide only deals with the first row of your truth table. For the other seven combinations of s0, s1, and s2 you would be assigning all zeros to m. If you want to do this in a single combinational assignment then you would need 8 logical expressions like the one you have, all ORed together to get the final result.
A much cleaner way of doing this is with an if-then-else or case statement inside an always block.
Try a case statement:
always #* begin
case ({s2,s1,s0})
3'b000 : m = u;
3'b001 : m = v;
3'b010 : m = w;
3'b011 : m = x;
3'b100 : m = y;
3'b101 : m = z;
default: m = 0; /* <-- catch all */
endcase
end
Note: Use default if the truth table is incomplete. Otherwise a latch will be inferred.
Working example: http://www.edaplayground.com/s/6/187