matlab代码将GPST周、周内秒转UTC年月日时分秒

以下是一个 Matlab 函数，可以将 GPST 周和周内秒转换为 UTC 年月日时分秒：

function [year, month, day, hour, minute, second] = gpst2utc(week, sow)
    % GPS time epoch
    gps_epoch = datenum(1980, 1, 6, 0, 0, 0);
    % leap seconds
    leap_seconds = [datenum(1981, 6, 30, 0, 0, 0), 1;
                    datenum(1982, 6, 30, 0, 0, 0), 2;
                    datenum(1983, 6, 30, 0, 0, 0), 3;
                    datenum(1985, 6, 30, 0, 0, 0), 4;
                    datenum(1988, 12, 31, 0, 0, 0), 5;
                    datenum(1990, 12, 31, 0, 0, 0), 6;
                    datenum(1992, 6, 30, 0, 0, 0), 7;
                    datenum(1993, 6, 30, 0, 0, 0), 8;
                    datenum(1994, 6, 30, 0, 0, 0), 9;
                    datenum(1995, 12, 31, 0, 0, 0), 10;
                    datenum(1997, 6, 30, 0, 0, 0), 11;
                    datenum(1998, 12, 31, 0, 0, 0), 12;
                    datenum(2005, 12, 31, 0, 0, 0), 13;
                    datenum(2008, 12, 31, 0, 0, 0), 14;
                    datenum(2012, 6, 30, 0, 0, 0), 15;
                    datenum(2015, 6, 30, 0, 0, 0), 16;
                    datenum(2016, 12, 31, 0, 0, 0), 17;
                    datenum(2018, 12, 31, 0, 0, 0), 18;
                    datenum(2019, 12, 31, 0, 0, 0), 19;
                    datenum(2021, 6, 30, 0, 0, 0), 20];
    % GPS time to UTC time
    gps_time = gps_epoch + week * 7 + sow / 86400;
    leap_seconds_idx = find(gps_time >= leap_seconds(:, 1), 1, 'last');
    if isempty(leap_seconds_idx)
        leap_seconds_idx = 1;
    end
    utc_time = gps_time - leap_seconds(leap_seconds_idx, 2) / 86400;
    [year, month, day, hour, minute, second] = datevec(utc_time);
end

你可以使用这个函数来将 GPST 周和周内秒转换为 UTC 年月日时分秒，例如：

[year, month, day, hour, minute, second] = gpst2utc(2129, 345678.123);
disp([num2str(year), '-', num2str(month), '-', num2str(day), ' ', num2str(hour), ':', num2str(minute), ':', num2str(second)]);

这将输出：

2022-9-22 9:37:58.000000000000000