Help with digital technology task?
Hey,
I've been given a digital technology assignment that's certainly not too difficult. Since I'm really not good at the module, I have to ask you for help.
I gave the following table:
Flip-flop Q1 in the right column is a typo and should be Q0. Furthermore, the outputs of the state flip-flops are also the outputs.
Now I'm supposed to use that to create the state diagram in Moore notation and Mealy structure—which I've done—and in the next steps, create tables for the subsequent states, from which I can then determine the Boolean equation (for the later circuit diagram). And this is exactly where I'm going wrong. I haven't provided any information about state transition conditions, nor any information about inputs (which would be useful for Mealy anyway). Maybe I've simply overlooked something, and someone can help me.
Many thanks in advance
If you have any questions, just ask! 😀
(1 votes)
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What is shown is, in the end, a counter that counts up to 7, i.e. a sequencer.
Your state sequence table has a fairly simple structure, and you first have an index column on the left. You can number them from 0 to 7.
The next three adjacent columns are Q2, Q1 and Q0 for the states of the FFs.
Right next to these, you have three columns for the following states Q2*, Q1*, Q0*.
In these, each line enters what would occur in the next line.
From this, you will construct a KV diagram for Q2*, Q1* and Q0*. The fields in this one get only one 1 that concerns the respective Qx* FF.
I see. However, I would like to make a table with, for example, current status 000 and then an input (button) 0 and 1. I don’t come to a conclusion because I didn’t give any transitional logic, did I? If now, Z0 would have jumped to Z3 when entering 1 I would have 011 and then it would also be easy to understand for me.
Understand, yes, you do not have separate input variables, but the input variables of the succession state are the pre-condition itself. That’s what they call an unconditional transition. The counter is running on its own as soon as it hits. If you determine the equations for your FFs by means of KV diagram as above, logical equations result at the end 3 . In none of these will appear an input variable that is not in the tables.
At the end there are three flip-flops which are connected to each other. Then it depends on whether you want to use D-FFs or JK-FFs.
Can you post the complete task?
For Q0*, let the groups overlap with the 1en, can be that if you don’t do it out something else. It’s so desired.
Then you do something wrong. What you mean by the starting values is also creepy to me.
You get three equations in the end.
Q2* = ….
Q1* = ….
Q0* = …
Here you have the complete table:
Now a KV diagram for Q2*, Q1* and Q0* is available.
A KV diagram contains 3 variables, namely Q2, Q1, Q0. There you only enter a 1 in the fields where your Qx* variable has a 1 in the table.
Q2* has a 1. The equation for this from the KV diagram is
You have to create the other two equations independently.
Do you think you could look over a part of my solution again? I want to be sure. I have created a truth table from the table from above by website where the following initial values come out: Z0 = 0, Z1 = 1, Z2 = 1, Z3 = 0, Z4 = 1, Z5 = 1, Z6 = 0, Z7 = 1. For my DNF, I then took those with the output value 1: (Q2 & ~Q1 & Q0) OR (~Q2 & Q1 & ~Q0) OR (~Q2 & ~Q1 & Q0) OR (Q2 & ~Q1 & ~Q0) OR (Q2 & Q1 & Q0). If I minimize this now using KV diagram, only C comes out, which seems to me somewhat suspicious.
happy, great success continues!
You really helped me very well, be a little bit careful, because I had no touch points with digital technology as I said. I can definitely not thank you enough!!!
Okay.
Your state sequence is exactly described:
PON-> Z5 -> Z3 -> Z2 -> Z1 -> Z4 -> Z0 -> Z6 -> Z7 -> PON
The PON stands for Power On, so it is the basic state when switching on.
All states are still given the output variables. In addition, you have 3 FFs (Q2, Q1, Q0) and, as already suspected, there are D-flipflops because their inputs are called “D”.
The characteristic equation for a D-flipflop is: Qd=D, i.e. the output switches to 1, when the input D=1. So your previously “missed” input vectors are the “D”s of the flipflops, and because their Q is now =D, it’s a jacket like pants, which you represent in the table.
Your table would have the following form:
This is nothing more than a tabular list of
Condition number | Output of the FFs in this state | Output of the FFs of the state following this.
Now you ask for your “inputs” and these are as described above the output at the D-FF. If you put them in the table, it would be nothing more than a copy of the three columns 😉 That’s the nice thing about D-FFs because you can save them. Here to show:
As you can see, the D’s are just a copy of the Q’s.
Yes, here the link from the image of the task https://drive.google.com/file/d/1VZIfsKoNF0O2GN543AcIutPWiLQaN_TC/view?usp=drive_link and here the picture of my table I have designed (below point 5 of the task) https://drive.google.com/file/d/1VYhyrafyJ9OcC6aPWD_eyb7s72cKYFeh/view?usp=drive_link