deriving excitation equation and table.

Thread Starter

circuitwizz31241

Joined Jul 27, 2017
3
This is for Asynchronous state machine design.

So so far ive got the equation to be:
Y=A'BC+AB+C'Y

Converting this to an excitation table however not sure to use 5 variables as theres a Z on the schematic or not, ive shown the Kmap i think i need to use however with there only being three terms it would only be three 1's on the k map?
---------A B C Y Z
A'BC= 0 1 1 0 0
AB= 1 1 0 0 0
C'Y= 0 0 0 1 0

Any clearance would be much appreciated,Thanks.
 

Attachments

WBahn

Joined Mar 31, 2012
26,398
You are basically claiming that there is no difference between AB and ABC'Y'Z'. But there's a HUGE difference -- AB is True any time both A and B are True, regardless of the value of any other variable, while you are requiring that all of the other variables be False.

Also, you have Y on both sides of your equation. Are you sure that's what you want?

If there's no Z on the schematic, then where is Z coming from?
 

Thread Starter

circuitwizz31241

Joined Jul 27, 2017
3
You are basically claiming that there is no difference between AB and ABC'Y'Z'. But there's a HUGE difference -- AB is True any time both A and B are True, regardless of the value of any other variable, while you are requiring that all of the other variables be False.

Also, you have Y on both sides of your equation. Are you sure that's what you want?

If there's no Z on the schematic, then where is Z coming from?
Hey thanks for looking at this for me.

I think the equation should have Y on both sides and also theres a Z on the bottom right of the schematic shown in the second diagram so i think i need to include this also? Do you know whether the k-map ive set up is the correct way of going abouts it?

Thanks very much.
 

WBahn

Joined Mar 31, 2012
26,398
You have two outputs, so treat them as two different circuits.

Y has feedback, which makes it technically a feedback sequential circuit. That complicates the analysis, though this particular circuit may sidestep those issues.

You don't need a five-variable K-map because Z is not an input. Your outputs depend only on A, B, C and Y (due to the feedback). But the feedback means that you may have state information that you need to track, meaning that the present output is not just a function of the present inputs, but also the history of the inputs.
 

Thread Starter

circuitwizz31241

Joined Jul 27, 2017
3
You have two outputs, so treat them as two different circuits.

Y has feedback, which makes it technically a feedback sequential circuit. That complicates the analysis, though this particular circuit may sidestep those issues.

You don't need a five-variable K-map because Z is not an input. Your outputs depend only on A, B, C and Y (due to the feedback). But the feedback means that you may have state information that you need to track, meaning that the present output is not just a function of the present inputs, but also the history of the inputs.
Hey thanks for that makes a little more sense now treating them as two seperate circuits, With that in mind i have remade the K-maps as shown in the attachments, does this seem more correct?
 

Attachments

WBahn

Joined Mar 31, 2012
26,398
Hey thanks for that makes a little more sense now treating them as two seperate circuits, With that in mind i have remade the K-maps as shown in the attachments, does this seem more correct?
Z is not an input. It is an output. The labels on a K-map are the inputs -- the values in the cells are the value of the output for the given values of the inputs for that row and column.

A four input K-map is usually draw as a 4x4 table with two variables on each side. This makes it easier to identify groupings.

I can't tell what you are trying to do in our two tables at the bottom in which you have Y as the fourth column on one and Z as the fourth column on the other.

What are Y and Z when A, B, and C are all LO?

Can you even tell for sure?

Consider A,B,C = LO, Y = LO, and Z = HI. Is this stable?
Consider A,B,C = LO, Y = HI, and Z = LO. Is this stable?

If both are stable, then you have a latching circuit that maintains state.
 
Top