# Sequential Network Problem-Mealy model

Discussion in 'Homework Help' started by laguna92651, Mar 29, 2008.

1. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
I'm constructing a transition table for a synchronous sequential network. It is 2 JK flip flops. I've developed the transition expressions okay, but am not sure of the format to use for the Transition Table. The network reflects a Mealy machine. It has 2 states and 2 inputs.

I only have 1 example of a Mealy and 1 example of a Moore machine, so I don't have much reference to figure out the methodology or format. Sorry for all the dashes, it was the only way I could get the table to line up.

This is my current Transition Table, I'm not sure if this correct.

Present State---------- Next State-------- Output
(Q1Q2) -------------------(Q+1Q+2) -------------z
-------------------------------Inputs (xy) ---------Input(x)
00 01 10 11 ----------0 1
---------------------------------------------------------------------------

00 -------------------------01 00 00 00 ----------0 1
01 -------------------------00 01 00 01 ----------0 1
10 -------------------------01 10 10 10 ----------1 0
11 -------------------------00 10 10 10 ----------0 0

One area I'm confused on is the output, there are 2 inputs, x&y. It is my understanding there should be one output column for every combination of input variables (4 in this case). The z output expression doesn't have a y in it, z = Q1Q'2 + x'Q'1. I assumed I don't need to include y in the output.

Second question, when I evaluate a Next State expression do I treat it as an OR?

As an example for x=1 y=0, Q1Q2 = 10

Q+1 = yQ1 + xQ1Q'2 + xy'Q1 + y'Q1Q'2 + xQ1Q2

Evaluates to: 01 + 111 + 111 +111 + 110
Is this like a Or statement, if any one of the terms evaluates to 1, Q+1 = 1?

Thanks

Feb 22, 2007
52
0
To clear you thoughts the diference between a Moore and Mealy machine is as simple as this: in a Moore machine the output depends on the current state (say the value in the flip-flops) but in a Mealy machine the output depends on the current state AND the current inputs.

Now the table. I used to draw the transition tables like this:

...PS........I........NS.....O
.Q1 | Q2 | X | Y | Q1+ | Q2+ | Z
---------------------------------
.0. | 0. | 0 | 0 | ... | ... | ...
..........| 0 | 1 | ... | ... | ...
..........| 1 | 0 | ... | ... | ...
..........| 1 | 1 | ... | ... | ...
.0. | 1. | 0 | 0 | ... | ... | ...
...........| 0 | 1 | ... | ... | ...
..........| 1 | 0 | ... | ... | ...
..........| 1 | 1 | ... | ... | ...
.1. | 0. | 0 | 0 | ... | ... | ...
..........| 0 | 1 | ... | ... | ...
..........| 1 | 0 | ... | ... | ...
..........| 1 | 1 | ... | ... | ...
.1. | 1. | 0 | 0 | ... | ... | ...
..........| 0 | 1 | ... | ... | ...
..........| 1 | 0 | ... | ... | ...
..........| 1 | 1 | ... | ... | ...
Where:
PS - present state
I - inputs
NS - next state
O - outputs

About the output question. There's no need for a y.

I didin't understand quit well your last question but I'll try to explain what I undestand of it.

Imagine you have this equations:

Q1+ = Q1 x + Q2
Q2+ = Q1 y + Q2' x
z = Q1 Q2 y

The next state if (x = y = 1, Q1 = Q2 = 0) will be:
(using boolean algebra)
Q1+ = 0 * 1 + 0 = 0
Q2+ = 0 * 1 + 1 * 1 = 0
z = 0 * 1 * 1 = 0

Hope helped...

EDIT: you were right its hard to make the table to line up.

3. ### razzler Member

Feb 9, 2008
16
0
use k maps.it's easier and you get the equations immediately.for the output, you only get a "1" if it satisfies your conditions example if x and y are 0, you get z=1 or when the present state is in this condition, the output is high regardless of the inputs

4. ### razzler Member

Feb 9, 2008
16
0
oh i just noticed, your table is lacking the flip-flop input columns. if you have this, it is easier to get the equations needed

5. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
On the second answer, why does;
Q2+ = 0*1 + 1*1 = 0 wouldn't it equal 1?

If Q1+ evaluated to;

Q1+ = 1*1*1 + 1*1 + 1*1*1 what would Q1+ equal?