Reformat the date and time, and also show nanoseconds using mathlab
显示 更早的评论
I made a program about the moving object distance calculation,I have problem in finding time to receive and send the difference to be precise in order of nano, how to find the mathlab retrieve data with nanosecond.thx
采纳的回答
Walter Roberson
2011-3-21
You need to tell us more about the data representation.
Are these value serial date numbers? If so then
>> eps(now)
ans =
1.16415321826935e-10
>> ans * 24*60*60
ans =
1.00582838058472e-05
which tells you that you cannot get higher resolution than 1000 nanoseconds using that format.
If you are trying to time events using tic and toc, then you will find that those clocks are only defined down to milliseconds (1000 Hz). To go beyond that requires system-dependent methods.
4 个评论
Paskah Nainggolan
2011-3-21
Walter Roberson
2011-3-21
How are the signals being sent and how are they being received? Is there a clock that is synchronized between the two of them? Is this a theoretical question based upon the distance and the medium and the path length and the temperature, or is this a practical question of how a receiver can figure out how long a signal was in transit?
If this is the practical matter of a receiver figuring out how long a signal was in transit, then I am not certain whether it is possible with a single source, but I would suggest that you study the way that it is done for GPS.
Paskah Nainggolan
2011-3-22
Walter Roberson
2011-3-31
Mathworks does not provide any means to time serial device events to nanoseconds. If you had the Data Acquisition Toolbox and were receiving the data with an appropriate card, you might be able to get access to higher resolution time stamps; I do not know how accurate they can go.
Recall that a nanosecond only about 2 clock cycles on a 2 GHz CPU. There is no way that an interpreted language such as MATLAB running on a multiprocessing system is going to be able to reliably interrupt and access hardware clock cycle counters within 2 instructions. For nanosecond measurements, you need a real-time device or additional hardware support.
更多回答（2 个）
James Tursa
2011-3-22
The "now" function is only good to about 0.015 sec accuracy. You should not use it for any timing that you need more precisely than that. e.g., run this code (Ctrl-C to stop it):
n = now;
while( true )
m = now;
if( m ~= n )
disp((m-n)*86400);
n = m;
end
end
You will have to get more precise timing some other way. e.g., if you have an OpenMP capable compiler you could get MTIMESX from the FEX and use its omp_get_wtime functionality. e.g.,
n = mtimesx('omp_get_wtime');
while( true )
m = mtimesx('omp_get_wtime');
if( m ~= n )
disp(m-n);
n = m;
end
end
On my system the above returns differences on the order of 1.5e-5 seconds (I presume this is all system dependent), so maybe something like this will work for you. MTIMESX (which admittedly is overkill for your situation) can be found here:
3 个评论
James Tursa
2011-3-22
Note that eps(now)*86400, which gives a result in the 1e-5 range, is misleading. The "now" function is *not* that accurate as the above test code will show you.
Paskah Nainggolan
2011-3-31
Walter Roberson
2011-3-31
You could multiply the difference by 10^9, but as discussed above that number not be meaningful to within 15000 nanoseconds.
Paskah Nainggolan
2011-3-31
0 个投票
1 个评论
Walter Roberson
2011-3-31
That is a new question on a different topic; please start a new thread.
类别
在 帮助中心 和 File Exchange 中查找有关 Data Acquisition Toolbox Supported Hardware 的更多信息
2011-3-21
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
