I am trying to write an assertion, the spec goes like:
if a is high in any cycle, then for the next 3 cycles, c should be assert if b is not asserted.
If anytime b is asserted, c should be deasserted in the next cycle.
I tried below but not sure how to add b in this scenario.
a |-> c[*3]
Should I just disable the assertion when b is asserted?
Thanks for help.
if a is high in any cycle, then for the next 3 cycles, c should be assert if b is not asserted
Try this sequence:
a |-> (c [*3] && !B [*3])
If anytime b is asserted, c should be deasserted in the next cycle.
Try adding this other sequence:
b |-> !c
It will check b at the clocking event you describe in the sequence or a defaulting clock, I believe.
Draw a waveform of the assertions of these sequences and check whether they provide what you need or not.
If you want to see the combined behavior of these two assertions, to check for conflicts for example, you can make a third sequence by anding them and checking the waveform.
I am not sure I am fully clear on the spec but goes something along the lines of :
a |-> ##1 (c && !b)[*3] or (b[->1] ##1 $fell(c))
Now this in itself does not make much sense as the second seq for b being asserted and negedge on c does not have an end point. If the spec is for all of this to happen within 3 cycles:
sequence option1;
(c && !b)[*3];
endsequence
sequence option2;
(b[->1] ##1 $fell(c));
endsequence
a |-> ##1 ((option1 or option2) throughout ##3 1);
Note a few things:
- the extra cycle as your spec said for the next 3 cycles. My interpretation of the spec could be wrong here
- I coded the two options in two separate sequence to emphasize the point. As long as either one holds, your prop will hold.
- This is a literal transcription of your spec (or my interpretation of it). If you actually work out all legal options you can probably simplify the property as a single sequence though it will likely impact readability wrt specification
Related
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 am trying to successively subtract a particular number to get the last digit of the number (without division). For example when q=54, we get q=4 after the loop. Same goes for q=205, output is q=5.
if(q>10)
while(q>10)
begin
q=q-10;
end
The iteration should converge logically. However, I am getting an error:
"[Synth 8-3380] loop condition does not converge after 2000 iterations"
I checked the post - Use of For loop in always block. It says that the number of iterations in a loop must be fixed.
Then I tried to implement this loop with fixed iterations as well like below (just for checking if this atleast synthesizes):
if(q>10)
while(loopco<9)
begin
q=q-10;
loopco=loopco-1;
end
But the above does not work too. Getting the same error "[Synth 8-3380] loop condition does not converge after 2000 iterations". Logically, it should be 10 iterations as I had declared the value of loopco=8.
Any suggestions on how to implement the above functionality in verilog will be helpful.
That code can not be synthesized. For synthesis the loop has to have a compile time known number of iterations. Thus it has to know how many subtractions to make. In this case it can't.
Never forget that for synthesis you are converting a language to hardware. In this case the tool needs to generate the code for N subtractions but the value of N is not known.
You are already stating that you are trying to avoid division. That suggest to me you know the generic division operator can not be synthesized. Trying to work around that using repeated subtract will not work. You should have been suspicious: If it was the easy it would have been done by now.
You could build it yourself if you know the upper limit of q (which you do from the number of bits):
wire [5:0] q;
reg [3:0] rem;
always #( * )
if (q<6'd10)
rem = q;
else if (q<6'd20)
rem = q - 6'd10;
else if (q<6'd30)
rem = q - 6'd20;
etc.
else
rem = q - 6'd60;
Just noticed this link which pops up next to your question which shows it has been asked in the past:
How to NOT use while() loops in verilog (for synthesis)?
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 am new to verilog, can anyone please explain me how does these statements execute.
always#(posedge clock) begin
A <= B ^ C;
D <= E & F;
G <= H | J;
K <= G ? ~&{A,D} : ^{A,D}
end
As far I can tell, the right side is first executed. Hence, the values for A, D, G, K are first calculated. While calculating value for K, depending on value of G the either first or second expression would execute. Can anyone please explain this operation. Please also tell how the last statement gets synthesized as this entire code is inside a always block and with positive edge clock. Thanks in advance.
A nonblocking assignment evaluates the RHS expression at the
beginning of a time step and schedules the LHS update to take place at the end of the time step.
In Verilog, there is a well defined event queue as shown below. For each and every timestamp, all the regions are evaluated. If there are any events to be executed in current timestamp, then they are triggered. Once all the events of current timestamp are triggered, then only simulation time moves forward.
Here, the RHS of ALL the expressions are evaluated at the beginning of the timestamp of posedge of clock. Hence, the values of B^C, E&F,H|J,G ? ~&{A,D} : ^{A,D} are evaluated and stored internally in the simulator.
Thereafter, the LHS are updated as the simulation progresses to NBA region of the same timestamp.
The values of G,AandDare not updated in active region. Hence, while calculating the value ofK, the previous values ofG,AandDare taken in the Active region. Then, all the veriables;G,A,DandK` are updated simulatanously.
I have made an example code over EDAPlayground. The waveforms might be helpful.
As far as last statement is concerned, I guess it will create a flop with mux (with Select=G and Inputs as nand(A,D) , xor(A,D) ) as input.
I would like to configure the guard of an edge to be:
(turn % 4) == me
where turn is a clock variable and me is an int representing a process.
Please give me an example of how to make a guard for the above predicate.
Thanks,
Kevin
My answer is not fully complete (so I won't mark it as Complete).
However, if you have a "clock x," and want the clock to wrap around from n --> 0, then you add this guard to an edge.
if (x == n) ? 0 : x