How to define time unit and time precision

system-verilog
Akshay Patil picture Akshay Patil · Dec 22, 2015 · Viewed 16.9k times · Source 
//`timescale 10ps/1fs
module time_presion();
    timeunit 100ps/10ps; //If We change this to 100ns/10ps it works fine
    parameter p=11.49;
    int a;
    initial begin
        $monitor("%t ,My values Changes %d",$time,a);
        #p a = 10;
        #p a = 30;
        #p a = 40;
        //#100us;
        #p a = 50;
        #1 $finish(1);
    end
endmodule

When I run this code, I get this error: 

file: time_prcision.sv
    timeunit 100ps/10ps;
           |
ncvlog: *E,TUSERR (time_prcision.sv,4|11): timeunit is smaller than the specified time precision [IEEE Std 1800-2009].
    module worklib.time_presion:sv
        errors: 1, warnings: 0

If I change the timeunit to 100ns/10ps, then the code runs properly. What is wrong in the above code?

Answer

sharvil111 picture sharvil111 · Dec 22, 2015

From SystemVerilog LRM 1800-2012, section 3.14.2.2:

The time unit and precision can be declared by the timeunit and timeprecision keywords, respectively, and set to a time literal (see 5.8).

The line timeunit 100ps/10ps; defines the time unit in current module,program, package or interface, locally.

If specified, the timeunit and timeprecision declarations shall precede any other items in the current time scope.

The time unit tells that when you give #1 delay (for example) the unit of that delay. If we select 100ps as time unit, then providing #1 delay shall result in 100ps delay.

The time precision tells the smallest delay you can configure in a given time unit. Precision represent how many decimal points of precision to use relative to the time units. For example: 

timescale 100ps/10ps shall have a #1 delay of 100ps 
while you can give #0.1 as the smallest delay i.e. of 10ps.

timescale 1ns/1ps shall have `#1` as 1ns and `#0.001` 
as 1ps as the smallest delay.

In your code, timescale 10ps/1fs shall represent a #1 of 10ps delay and #0.0001 is the smallest measurable delay. Now, coming to the error:

timeunit is smaller than the specified time precision

Intuitively, it can be said that time unit must never be smaller than time precision. This is stated in the above error.

A timescale of 100ps/10ps shall round-off every delay after one decimal place. Providing 11.49 shall be rounded off to 11.5 and multiplied by time precision and then displayed.

In short, using timescale 1ns/1ps, delays are interpreted to be in nanoseconds and any fractions to be rounded-off to the nearest picosecond. I used $realtime in display statement and the output is as follows. The display 115 is due to default time scaling (time*timeunit/timeprecision) in timeformat.

              0 ,My values Changes           0
             115 ,My values Changes          10
             230 ,My values Changes          30
             345 ,My values Changes          40
             460 ,My values Changes          50

For more information, refer to timeunit, difference between time unit and time precision and time scale tutorial links.

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