I've got a problem understanding formal writing down of complements on inputs.

During my laboratory exercise I was working with a device that had D flip flop embedded in it. The goal was to investigate the working way of asynchronous inputs s and r. Hence, it was not mandatory to plug in the D input or clock input.

Pressed s button and released r button gave Q = 0 result.

Released s button and pressed r button gave Q = 1 result.

So we were dealing with complemented (active-low) async inputs s and r.

Released both s and r inputs gave Q = 0 output (so Reset was dominating over Set).

Pressed both s and r inputs gave Q = previous output value.

That being said, I am having trouble organizing the truth table for it.

Which truth table (characteristic table) presents the above described situation?

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The left table is a 1:1 translation of what has been above said, and I would be willing to use that table. But what about right table? In my opinion, the right table is exactly the same as left table, but it just explicity says that the inputs are complemented.

In the case of the left table, it is implicitly said that the inputs are complemented.

Regardless: I believe both truth tables are correct, aren't they? The only difference is that the left table implicitly says that inputs are complemented, and the right table says its explicitly.

Essentially: are those truth tables equivalent, or not? If not, then which truth-table is coresponding to my situation?

Also, adding timing chart. Not sure if it's correct as it is, or perhaps it will be correct only if I change the notation from s and r to ~s and ~r.

enter image description here

  • \$\begingroup\$ Is "pressed" defined as a logic 1 or a logic 0? \$\endgroup\$ – Hearth Oct 26 '19 at 15:32
  • \$\begingroup\$ @Hearth I suppose pressing sets the logical value to 1. But because inputs are complemented (somewhere inside the device), then the "real" logical value sent is 0. \$\endgroup\$ – weno Oct 26 '19 at 15:34
  • \$\begingroup\$ "I suppose pressing sets the logical value to 1" - suppose nothing. Does a pressed button produce logic 1 or 0 in the logic (positive or negative) of the flip flop? en.wiktionary.org/wiki/negative_logic \$\endgroup\$ – Bruce Abbott Oct 26 '19 at 15:51
  • \$\begingroup\$ I don't want to guess. What I know for sure is that both inputs were complemented and pressing/releasing buttons led to above described Q states. Isn't that enough information? \$\endgroup\$ – weno Oct 26 '19 at 16:06

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