sequential circuits, MMExer5.6

Georacer

Joined Nov 25, 2009
5,181
Welcome to the wonderful chapter of sequential circuits. They're my favourite.

Off the bat, I can see that, if you copied your question correctly, your FF schematic isn't in agreement with it.

A(t+1) corresponds to the D input of A. It should accept the function Da=x'y+xB, but instead you feed it x'y+xA.
Similarly for Db=x'A+xB you feed it xA+xB.

Take a better look.
 

Thread Starter

PG1995

Joined Apr 15, 2011
816
Hi

I copied the question statement correctly.

PS: Yes, you are right. My schematic isn't correct. But see the schematic in the manual is also wrong. Isn't it? Perhaps, I'm relying too much on the manual! :)

I have corrected the schematic. Now please help me. I need to understand all this quickly.
 

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Georacer

Joined Nov 25, 2009
5,181
The solution's circuit has nothing to do with the question. Even the output is in the wrong pin. Maybe it's from an older version.

Your solution is correct. Took me a while to revise everything.
However, you shouldn't write the output on the transition arrows. The output isn't related to the transition, as this is a Moore machine, not a Mealy one. The output is related only to the current state.
You could do a double circle around the states that have a HIGH output to notify them.
 

Thread Starter

PG1995

Joined Apr 15, 2011
816
Thank you very much. It's very kind of you.

I don't exactly get this:
You could do a double circle around the states that have a HIGH output to notify them.
For instance, which one of the states in the attachment could have a double circle? "10"?

Best wishes
PG
 

Thread Starter

PG1995

Joined Apr 15, 2011
816
Hi

Please have a look on the attachment. Hopefully, I have everything correct this time.

I have one question, though. What do those functions A(t+1) and B(t+1) tell us? I mean, do they have anything to do with the circuit I have drawn in part (a)? The functions A(t+1) and B(t+1) don't say anything about inputs J and K, so I don't think we can replace the circuit in part (a) with any of these functions. Please let me know. Thanks.

Best wishes
PG
 

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Georacer

Joined Nov 25, 2009
5,181
Those functions were referring to the D-FF, where the input D is by definition the same as the state of the FF in the next cycle, hence the A(t+1). You can use them for JK-FFs too, but you will have to extract the input functions for the J and K inputs with a K-map in order to build the circuit.

Let me have a look at your attachment and I 'll get back to you.
 

Georacer

Joined Nov 25, 2009
5,181
Okay, everything is correct.

Make a habit of producing the inverted inputs using a NOT gate right on the edge of your schematic where the input comes in. That way you will always have the x' available and you will use only one NOT gate, regardless of the times you will use the signal.

I also recommend this free software, developed by the University of Deusto. http://sourceforge.net/projects/boole-deusto/
It automatically produces minimized expressions and FSM checks, so you can verify your results on your own.
 

Thread Starter

PG1995

Joined Apr 15, 2011
816
Thank you very much, GeoRacer.

I understand that it would have taken you considerable time to check everything.

I believe everything is correct in the attachment except one thing. Please help me with the "Q", and what does the circuit represent? Thanks a lot.

Best wishes
PG
 

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Georacer

Joined Nov 25, 2009
5,181
xy+x'y' is actually equivalent to (x XOR y)'.

Do a 2 by 2 K-map, and fill it with the 1s of (x XOR y)'. After you extract the SoP you will see that it equals xy+x'y'.

I won't try to answer the "what it represents" question. It could mean anything and we know better than try to guess the professors' intentions.
 
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Zazoo

Joined Jul 27, 2011
114
xy+x'y' = x XNOR y

As for what the circuit represents it functions as a serial/sequential multi-bit adder. e.g. If x and y are attached to the serial-out of two shift registers and S is connected to the serial-in of a third shift register then after a full shift (n clock cycles) the output shift reister will hold the sum of the two input registers with the final carry in the D-FF.

Edit: As Georacer said, this may not be the intended purpose but it's one purpose that jumps out at me. Check with you instructor.
 
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Thread Starter

PG1995

Joined Apr 15, 2011
816
Hi again, :)

Just did another problem. Once again, the manual isn't in agreement with me. When you have time, just give it a look though I'm quite confident that I'm correct! Thank you.

Best regards
PG
 

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Thread Starter

PG1995

Joined Apr 15, 2011
816
The solution manual is mostly incorrect. As you can see even the inputs aren't correct. But still it provides me a guideline that how to proceed. Thank you very much.

Best wishes
PG
 

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Thread Starter

PG1995

Joined Apr 15, 2011
816
I had a signals and systems book like that. It is really frustrating because you end up second guessing yourself and spending extra time re-doing problems that were correct to begin with.
Exactly.

Yes, it gets on my nerves too when they do that. It makes you want to slap somebody.
Well, if I didn't have you and good people like Zazoo to help me, then at least I would have slapped myself many times!:p

Best wishes
PG
 
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