I'm learning about sequential synchronous circuits, and I was trying to do an exercise, but I'm not going as I expected.

The exercise is about

  • Elaborate the excitation equation of flip-flops;
  • Write the output equation;
  • Say if it is a Moore or Mealy Machine;
  • Elaborate transition table and outputs;
  • Draw the state diagram;

enter image description here

So, my biggest problem is the first question "Elaborate the excitation equation of flip-flops", once I done the first question I think I can handle this exercise Something is getting confusing in my head.

1 - Should I consider "A-B-C" as inputs or outputs? Taking into account a Mealy machine the Output is a function of current state and inputs, if I consider "A-B-C" as my inputs what would be my output for example?

2 - I tried excitation equation of each flip-flop (considering "A-B-C" as inputs)

Q0 = B

enter image description here

Q1 = C


Q3 = (C'+ B')  ^ ( B ^ A )             -> " ^ = Xor "

As you can see i'm a little confused, if sameone can help me I would be grateful. Thanks everyone!

  • 1
    \$\begingroup\$ Why do you think they are inputs? Imagine Qo is driving '0', and you drive A to '1'. What will happen? \$\endgroup\$
    – Mitu Raj
    Commented Jun 1, 2021 at 18:37
  • 1
    \$\begingroup\$ On a D flip-flop, D is the input and Q is the output. On any flip-flop, Q is the output. \$\endgroup\$
    – Aaron
    Commented Jun 1, 2021 at 21:48

1 Answer 1


This state machine doesn't actually have an input (apart from the clock signal). It just generates a sequence of output values on signals A, B and C once you apply a clock.

If you look more closely, you'll see that A, B and C are each connected to the output of one of the flipflops. Since a node (wire) can only be driven by exactly one (regular) output, this means that A/B/C are themselves outputs and not inputs. If they were inputs, the signal put in would collide with the output signal of the flipflops.

To analyze it, you could start with 0 stored in each flipflop and then compute the resulting values at the inputs of the flipflops. If you repeat this with the new values that you get out, you'll be able to determine the sequence of output values that this circuit produces.

  • \$\begingroup\$ Thanks for your helping me, but one more question, if the circuit has no inputs, we can't consider it as a Mealy machine right? Because a Mealy machine, the Output is a function of current state and INPUTS. \$\endgroup\$
    – Jeep nN
    Commented Jun 2, 2021 at 16:52
  • 1
    \$\begingroup\$ Yes, this is a Moore machine. However, Moore machines are a subset of the Mealy machines - every Moore machine can be considered as a Mealy machine that ignores its inputs in the output calculation, but that's just a technicality. \$\endgroup\$ Commented Jun 2, 2021 at 17:02

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