awk: outputting the last ten minutes of a log file

OK, I've got this log file that has the date/time stamp in field one ($1) and I just wanted to extract the last ten minutes worth of data. Using the gawk functions mktime() and strftime() I wrote the following solution:

Getting the last ten minutes (600 seconds) of a log file:

awk -F, '{ tenago = strftime("%H:%M:%S", mktime(strftime("%Y %m %d %H %M %S", systime() - 600))); if ( substr($1,14) > tenago ) print $0 }' logfile.txt

Getting Hours:Minutes:Seconds timestamp for ten minutes ago was accomplished using:

tenago = strftime("%H:%M:%S", mktime(strftime("%Y %m %d %H %M %S", systime() - 600)))

It makes sense if you read it from the inside out. The systime() function, because it hasn't been supplied any parameters returns the current datetime minus (-) 600 seconds. It's formatted as a string using: "%Y %m %d %H %M %S" and passed to the mktime() function where it's converted to a string again... I think you may be able to simplify this - let me know if you can! :)

EDIT: Here we go, I knew there was a way to simplify the above command! Here's my latest version:

awk -F, '{ tenago = strftime("%H:%M:%S", systime() - 600); if ( substr($1,14) > tenago ) print $0 }'