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Is it me not understanding this table and diagram or something? If in a Moore machine, the output only depends on the current state Then why does the table for states F and H say the output is independent of the input? At F, if the input is a 0 the state changes to I and output would be a 1 At F, if the input is a 1 the state stays the same at F and the output would be a 0

The state table does not reflect this?

The state table should show multiple outputs...

enter image description here

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  • \$\begingroup\$ Is it because im stupid or something? \$\endgroup\$ – user220808 Apr 30 at 12:41
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    \$\begingroup\$ You're confusing "next state" with "output". The combination of current state and inputs determine the next state, but the output doesn't change until it moves to that next state. The output depends ONLY on the current state. \$\endgroup\$ – Finbarr Apr 30 at 12:49
  • \$\begingroup\$ So what youre saying is if i am just landed on F, the current output is 0. I set an input of 1, the output is still 0, i set an input of 0 at t1, the output is still 0 at t1+deltat until t2 where the state is I and the output is 1? \$\endgroup\$ – user220808 Apr 30 at 13:09
  • \$\begingroup\$ Very ambiguous to me \$\endgroup\$ – user220808 Apr 30 at 13:15
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    \$\begingroup\$ @Oldfart, they aren't totally independent. The first one can be inferred from the second one. \$\endgroup\$ – The Photon Apr 30 at 16:10
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the output only depends on the current state Then why does the table for states F and H say the output is independent of the input?

Because if it depends only on the current state, then it can't also depend on the inputs.

The inputs determine whether you get to a certain state. But after you're in that state the output only depends on the state. Assuming a typical clocked design, the output doesn't change instantly if the inputs change before the next clock edge. And it only changes at the clock edge if the changed inputs lead to a new state.

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  • \$\begingroup\$ A Moore machine doesn't have to have to be clocked. \$\endgroup\$ – Finbarr Apr 30 at 16:14
  • \$\begingroup\$ @Finbarr, edited. \$\endgroup\$ – The Photon Apr 30 at 16:16
  • \$\begingroup\$ Have already figured this out, clearly trying to get free browny points. Should be negged \$\endgroup\$ – user220808 Apr 30 at 17:13
  • \$\begingroup\$ @LuckyBlueCherry, somebody has to post an actual answer or the question will be bumped to the front page indefinitely. If Finbarr wants to post their comment as an answer, and expresses it equally or more clearly, I will delete my answer. \$\endgroup\$ – The Photon Apr 30 at 17:15

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