Get total number of leap seconds relative to gps time

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user86753
user86753 2022-10-13
编辑： David Hill 2022-10-13
I have a gps time stamp from a gps receiver, I'd like to convert that to GMT/UTC date and time.
GpsTime = 1667089162.11771
DayToSecs = 24*60*60;
TimeZero = datetime(1970,1,1, 'Format', 'd-MMM-y HH:mm:ss.SSS Z');
Tvec = GpsTime/DayToSecs + TdmsTimeZero;
Do I need to also adjust for leap seconds? If so, How do I return the total number of leap seconds from 'leapseconds' instead of the timetable?

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回答（1 个）

David Hill
David Hill 2022-10-13
编辑：David Hill 2022-10-13
Ran in:
As long as the gps collection date is after 1 January 2017, then just subtract 18 seconds (37-19) see
https://hpiers.obspm.fr/eop-pc/index.php?index=TAI-UTC_tab&lang=en
Although your gps time seems to be in the future (not sure why).
GpsTime = 1667089162.11771;
utc=datetime(1980, 1, 6)+seconds(GpsTime-18);
utc=datetime(utc,'Format','yyyy-MM-dd HH:mm:ss.SSS')
utc = datetime
2032-11-03 00:19:04.117

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