State registers

Discussion in 'Homework Help' started by ibcoding, Apr 29, 2013.

  1. ibcoding

    Thread Starter New Member

    Apr 13, 2013
    7
    0
    If I have six state registers, I have 2^n values or 64 states. What if I have 3 inputs? Do I simply multiply it by 3? Also, with 6 registers, the total transitions can be calculated using 2^m * 2^n. So that's, what, 512 total transitions. And by that logic I should have 8 transitions from each state, right?
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Multiply WHAT by 3? The number of states?

    You then state that the total transistions can be calculated using a formula that has an 'm' in it and never mention what 'm' is? You also don't say what 'n' is, but the values you use make that clear.

    Get in the habit of clearly stating the problem, your information, and your questions. Many times just doing that will reveal the answer to you. If it doesn't, is certainly makes it much easier for someone else to help you.

    Reading between the lines, however, you are correct. Your state transition table will have 8 entries for every possible state because, in a given state, there are 2^3 possible input combinations of the three inputs. Since there are 64 states, that means that there would be 256 rows in the table.

    That's a big table. So you want to really look for "don't care" conditions to shrink it down.
     
  3. ibcoding

    Thread Starter New Member

    Apr 13, 2013
    7
    0
    Thanks for the response. m = input and n = flip/flops. So, 256 rows and 256 columns for 512 total transitions. Got it. Just wanted to see if I was on the right track.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Where are you getting 256 columns from?

    Please describe the organization of the table you are talking about. We are not mind readers.
     
  5. ibcoding

    Thread Starter New Member

    Apr 13, 2013
    7
    0
    Well I didn't want to post the question verbatim, but since I already submitted the assignment, the question reads as such:

    You have a sequential circuit with three inputs and six state registers. How many states does it have? How many transitions from each state does it have? How many total transitions does it have?

    My answers: 64, 8, 512. According to my book, the total transitions is 2^m * 2^n, as I specified earlier.
     
  6. ibcoding

    Thread Starter New Member

    Apr 13, 2013
    7
    0
    I didn't mean columns. That was an error. I meant 512 transitions.
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Okay, you answered correctly (as you already know). You had also said 256 rows. That should be 512 rows. An alternative is to have 64 rows (one per state) and 8 columns (1 per input combination) not counting the column with the state ID in it.
     
Loading...