Electrical – J K Flip Flop and Boolean Algebra

Ignore what they are calling it and draw a truth table for what they describe. There appears three inputs: "the current state" which you can arbitrarily call Q(n), N, and M. There is one output which we can arbitrarily call Q(n+1). The names are not totally arbitrary, because Q(n+1) is by definition a one-time-step delayed version of Q(n) (i.e. Q(n) is the previous output).

  Inputs           Outputs
 Q(n)    M    N    Q(n+1) 
  0      0    0      ?
  0      0    1      ?
  0      1    0      ?
  0      1    1      ?
  1      0    0      ?
  1      0    1      ?
  1      1    0      ?
  1      1    1      ?

You should be able to fill in the question marks from the narrative description trivially. This is a truth table.

I believe an excitation table is just a rephrasing of the truth table where the valid entries are include 0, 1, and Q(n) and it's complement /Q(n) (so there are only two "real" inputs). And removing "redundant" rows (i.e. if a variable is a don't-care under some conditions, then express that collapsed row with an "x" for that variable).

Inputs       Outputs
M    N       Q(n+1)
0    0       Q(n), /Q(n), 0, or 1?
0    1       Q(n), /Q(n), 0, or 1?
1    0       Q(n), /Q(n), 0, or 1?
1    1       Q(n), /Q(n), 0, or 1?

Can you reduce this further once you put answers in to the question marks?

The first one is not master-slave; you would have to cascade two copies of that circuit (with the clock inverted between them) to get the same functionality as the second circuit, which is indeed a master-slave (edge-triggered) flip-flop.