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Changing 3 input or gates into 2 input xnor gates

  • Thread starter lalosfire
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  • #1
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Realize (a'b'c'+a'bc+ab'c+abc') using only 2-input x-nor gates.

I've taken a shot at this, however I'm only able to but it into more or gates.
a'(b'c'+bc)+a(b'c+bc')
a'[(b'+c)(b+c')]+a[(b'+c')(b+c)]
a'(b'c+bc')+a(b'c'+bc)
That's where I got, however it's really no simpler and I doubt is even close to being on the right track.
 

Answers and Replies

  • #2
CWatters
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It's not far off.

Your first equation is...

a'(b'c'+bc)+a(b'c+bc') ................................(1)

Then you seem to have missed that the first part on the left is in the right form for an XNOR.

An XNOR performs the function (x'y'+xy) so perhaps leave the bit a'(b'c'+bc) alone. That would make your second equation....

a'(b'c'+bc) + a[(b'+c')(b+c)] .......................(2)

Now over to you to have another go. Leave the left hand side alone.
 
  • #3
CWatters
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Aside:

I've found it helpful to remember DeMorgans rule as "split/join the bar and change the sign" also remembering that adding two inverters in series has no effect.

For example to manipulate x.y you can do the following steps..

x.y

Invert twice..

= (x.y)''

then split the lower bar and change the . to +

= (x'+y')'
 

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