Electrical – How to determine the flip-flops input in truth table

You can work it out by going through the steps.
We know if T is at 1 then Q will toggle, if it's at 0 Q will stay the same.

Say we start off with Q0, Q1 and Q2 at 0, and the multiplexer input is set to 0 (so basically ignore the Qbar paths)

At the first clock pulse, since T0 is at 1, then Q0 will toggle from 0 to 1.
So we now have 1 0 0
At the second clock, since T1 is at 1, Q1 will toggle from 0 to 1. Q0 will toggle back to 0.
Now we have 0 1 0
At the third clock, Q1 will remain at 1 since T1 is at 0. Q0 will toggle back to 1.
Now we have 1 1 0
At the fourth clock, since the and gate now has both inputs (Q0 and Q1) at 1, then T2 will be at 1 therefore Q2 will toggle from 0 to 1. T1 is at one so Q1 will toggle back to 0, as will Q0.
Now we have 0 0 1
At the fifth clock, T2 = 0 so Q2 will remain at 1, T1 = 0 so Q1 stays at 0, and Q0 always toggles so it will change to 1.
Now we have 1 0 1
At the sixth clock, T2 = 0 so Q2 stays at 1, T1 = 1 so Q1 toggles to 1 and Q0 toggles to 0.
Now we have 0 1 1
At the seventh clock, T2 = 0 so Q2 stays at 1, T1 = 0 so Q1 stays at 1, and Q0 toggles to 1.
Now we have 1 1 1
At the eighth clock, both inputs to the and gate are high so T2 = 1, so Q2 toggles to 0. T1 is at 1 so Q1 toggles to 0, and Q0 toggles to 0.
Now we have 0 0 0 which is where we started.
If you change the multiplexer input to 1 and select the Qbar path, you can see how it would count down (just complement all the above results, e.g. 0 0 0 becomes 1 1 1, 1 0 0 becomes 0 1 1 and so on)

This looks like a homework question, which I won't answer. But here are some pointers:

• You cannot represent more states than what you have available -> that means that you will have to go to the next higher state (i.e. 2 FF's for 4 states) or use only one FF per state. (i.e. FF #1 correlates directly to A).

• option #1 ( 2 FF's and 4 states) is more economical but you have to make sure that the unused state does n't not get activated and then locks out.

  • you might draw this as a 4th state "D" with loops back to itself.
  • what is generally considered safe design is that you always have explicit transitions AWAY from the unused state in case it gets activated.

• option #2 uses more FF's but cannot have any hidden states.

  • it is inherently safer.
  • it is known as a "one hot" design and thermometer codes are examples of this.

Your choice of states "A = 00" etc. will make the design simpler or more complicated. SO may want to go with what you decribe or you may want to go with state C = "10". You should look at all possibilities.

The first SM, only uses 0 or 1 as an input because it only has one input variable. They should have used a variable for clarity anyways. You'll notice in the table that it is marked as "x" but not in the diagram.