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NPP Number System, Boolean Algebra and Logic Circuits 181

Problem 3.35 àíZ 3.35
Obtain the expression for Y and Simplify: Y hoVw ì`§OH$ kmV H$aHo$ gab H$amo…

A
Y
B

Solution: hc:

The output of first NAND Gate is: NAND :
AB.

The output of upper NAND Gate is: NAND :
  A AB..  
 

The output of lower NAND Gate is: NAND :
  B AB..  
 
The output of last NAND Gate is: NAND :

Y =   A A . B .  .   .B A B .  
   
Now applying De Morgan’s Theorems:

Y = ( A A B +.. ) ( B AB.. )

( A.A B . ) ( A.B+ ) B .

Y = AB. (A B+ )
Y = ( A + B . ) B
) (A +

Y = A A + . A B + . BA + . BB.
Y = A B +. BA.