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I'm following the course From Nand to tetris. In chapter 3, the authors introduce the notion of Data Flip-flop (DFF), which computes the output(t) = input(t-1) where t is the time.

My question is that why is output(t) = input(t-1) but not input(t) ? I think that to have output(t), we should supply an input(t). So in this case, how can output(t) be evaluated without having an input ? (i.e. input(t))

Many thanks for your enlightments!

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  • \$\begingroup\$ The input must be stable before the clock. Thus t-1. \$\endgroup\$
    – Kartman
    Commented Aug 22 at 14:32
  • \$\begingroup\$ @Kartman: Hi, so what is the role of input(t) ?? \$\endgroup\$ Commented Aug 22 at 15:45
  • \$\begingroup\$ After the clock, the input has no effect. Think of it like taking a picture - the camera captures the image when triggered. The output of the f/f represents the image that was captured and the image was the state of the input prior to the clock transition. A real f/f has three main parameters - setup time, hold time and clock to output time. The setup time the time the input must be stable before the clock edge. The hold time is the time the input must remain stable after the clock. The clock to output time is the time it takes for the captured input state to propagate to the output. \$\endgroup\$
    – Kartman
    Commented Aug 22 at 22:58
  • \$\begingroup\$ Hello, thanks! However I'm still very confused. can you please explain me a little bit about output(t-1) ? Here can I understand that output(t) = input(t-1) = output(t-1), as the goal of the FF is to "remember" the previous output, and I really don't see how output(t-1) enters into the picture. Many thanks! \$\endgroup\$
    – VDT-QHH
    Commented Aug 23 at 0:05
  • \$\begingroup\$ if output(t) = input(t-1), then how can output(t-1) = input(t-1)? That would mean output(t)=output(t-1) \$\endgroup\$
    – MrGerber
    Commented Aug 23 at 7:40

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