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Boolean Algebra

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tintincute

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Hi i'm not sure if I'm doing it right here.
I have an equation which is like this and I have to change it with NOR Gates only.


X1 = A * B + C

is my solution correct here?
my other colleagues came up with a different answer. which confused me.

thanks
 

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Hi i'm not sure if I'm doing it right here.
I have an equation which is like this and I have to change it with NOR Gates only.


X1 = A * B + C

is my solution correct here?
my other colleagues came up with a different answer. which confused me.

thanks

Note: Thumbnail shows a different equation: X1'=A*B+C


Hi there,


You've added a bit of confusion here because you have stated
apparently the same equation two different ways.

X1=A*B+C
and
X1'=A*B+C

Now if you interpret both of these equations to be the same with
just a change of variable:

X1=A*B+C
and
X2=X1'=A*B+C

Then your equation is the correct one.

If, on the other hand, we interpret X1' to be equal to the
invert of X1, then their equation is the correct one.

The only difference between the two equations is that one
is inverted from the other.

There is another difference in the actual implementation however,
and that is that your equation seems to yield the simpler hardware
implementation, even if you have to add an additional inverter
on the output (if X1'=not(X1) really).

If you clarify the equations i can give you a simpler more direct answer.
 
Last edited:
oopsss sorry about the confusion. my bad.

ok, the equation is

X1 = A * B + C

i have to change this to an equation which only contains OR gate or the (+) sign. and if i check it with a "comparator" or any equality, the output should show that these two equations are the same.

thanks

ps. i attached the wave form picture. in the picture, you can see the output
 

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Last edited:
Hi again,


You've introduced another change, where now you are saying
that you need to use "OR" gates when before you were saying
that you need to use "NOR" gates. There's a big difference
in the implementation of the final circuit, so you need to
clarify this next.

In the mean time, let me just state that the following three
equations are the same, except the last one allows for the
simplest implementation when we are forced to use only
NOR gates for the solution...

Code:
X1 = A*B+C



X2 = (A+C)*(B+C)

     _____________
     _____   _____
X3 = (A+C) + (B+C)
Also, your colleagues solution is inverted from these solutions, so if
we call yours X4 then we have to call theirs not(X4), and theirs is
not equal to X1 above which was the main point i think.


As to your other idea about comparing the two waveforms, well
im afraid that that only works when you use binary weighted
frequencies as the input. For example, A=40kHz, B=20kHz, and
C=10kHz, and they should probably be sync'ed at 0 degrees.
Using arbitrary waveforms to test a logic network does not always
work (although you may get lucky) because there is the chance
that not every logic combination occurs on the inputs. Using a
binary weighted set of frequencies that are sync'ed at 0 degrees
makes sure this requirement is met.

If, on the other hand, your application has repeated pattern
waveforms as input, then other simplifications may also be
possible, and you should be able to use those patterns as
test inputs, as long as it is certain that they will never change.

One last note...

If you look at the original solution and your solution, you will find
that the number of inverts over each letter cancel to no inversions
(ie A=A'', B=B'', etc.).
If you look at your colleagues solution they have one more
inversion (ie A''' which equals A' which does not equal A).
 
Last edited:
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