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I've only one simple problem about JK flip-flops. The last columns in the following table (Flip-flop inputs). How were they genereted from the other parts on the left? Does this has a relationship with the equation: J Q!+K! Q?

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  • \$\begingroup\$ Could you add scheme(s) for the table, please? \$\endgroup\$ – Roman Apr 26 '17 at 8:17
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At first: The table presents a machine which has one input (=X) an four states. The number of the current state is the binary number AB. A and B are the outputs of the two JK ffs which are used as the memory devices. A and B are called "state variables" in the automate theory.

The machine has a four output- three input combination circuit (=gates) which generates the right inputs for the JK ffs for the wanted state transitions. The inputs of that combination circuit are A, B and X. The right hand columns show the outputs of that combination circuit. The purpose is to show what values should be applied to the J/K inputs in order to progress through the states. The Boolean equations for the outputs Ja, Ka, Jb and Kb consist A, B and X as the variables. Your own version J Q!+K! Q is nonsense.

I write one of the four Boolean equations: Ja=B. You must find the rest by yourself.

The two columns under Next State are the outputs, and they are derived from the columns under Present State and Flip-Flop Inputs, the actual change taking place when the active edge of the clock pulse occurs. Note: X is not the clock, it's the input of the machine.

Here's the truth table for a J-K flip flop from Wikipedia for comparison. The table in your question simply has two such devices. Qnext is either A or B under Next State and Q is either A or B under Present State:

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