SI反射仿真问题

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VIH overshoot: VIH,max = VDDQ + 1.2V for a pulse width ≤ 3ns, and the pulse width cannot
be greater than one-third of the cycle rate. VIL undershoot: VIL,min = –1.2V for a pulse
- z* a. ~; H1 {) r  h  Owidth ≤3ns, and the pulse width cannot be greater than one-third of the cycle rate.

$ J0 T8 a/ `0 V$ w8 b- }大家是如何理解这段话的，比如供电电压为3.3V，是不是只要过冲电压小于4.5V，下冲电压大于-1.2V就可以了，我不大理解for a pulse width ≤ 3ns, and the pulse width cannot be greater than one-third of the cycle rate ，这句话是说的啥意思呢？？？