Get total number of leap seconds relative to gps time

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?

Answer 1

[David Hill]

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