((A exor B) .c) wholebar

answer which i got from a web site is A bar B bar +AB +C bar

can any one please explain how this comes..

 

Use De Morgan's Laws.

 

((A exor B) .c) wholebar

answer which i got from a web site is A bar B bar +AB +C bar

can any one please explain how this comes..

I would recommend adopting a cleaner, more common way of writing your expressions.

F = ((A XOR B)C)'

F = (A')(B') + AB + C'

First, expand the XOR into its full expression using only AND, OR, NOT.

The apply DeMorgan's.

An alternative is to consider the logic expression for the XNOR gate.

Finally, show your work for as far as you are able to get it. We will help you see the next step, we won't just show you how to do it.

 

thanks WBahn for the explanation. now i was able to solve it.

 

- >> (a`b+ab`)`+c`

(a`b)`.(ab`)`+c`

(a`+b)(a+b`)+c`
that gives my answer --> ab+a`b`+c`

 

- >> (a`b+ab`)`+c`

(a`b)`.(ab`)`+c`

(a`+b)(a+b`)+c`
that gives my answer --> ab+a`b`+c`

Yep. That's the most fundamental way to approach it.

Another way to come at it at a somewhat higher level would be:

F = [ (a XOR b) c ]'

F = (a XOR b)' + c'

F = (a XNOR b) + c'

The XNOR gate is also sometimes called the equality gate (for two inputs only), so (a XNOR b) is true if both a and b are T or if both a and b are F.

F = (ab + a'b') + c'

F = ab + a'b' + c'