| 摘要 |
<p>PROBLEM TO BE SOLVED: To obtain a time synchronizing device having high reliability for time information. SOLUTION: This device finds E=5m+TOVER (33). When 36 is NO, it directly proceeds to 38, and when it is YES, it sets LOOP=p-1 (37) and proceeds to 38. It decides whether or not m=0 in 38, and in the case of NO, it waits until the second digit of a reference calendar 11 becomes 5(m-1)+TOVER (40). When 38 is YES, it decides whether or not E=0 (39), and in the case of YES, correction is finished. In the case of NO, it waits until the second digit of the reference calendar becomes 0 (41). When 40 or 41 is YES, it decides whether a system calendar is faster or slower than the reference calendar (42). When it is faster, it outputs a clock delay signal (44) and when it is slower, it outputs a clock increase signal (43). It waits for 50 seconds (45), defines COUNT= COUNT+1 (46), sets START=0 (47) and returns to 33.</p> |
