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The minimum number of $2$-input $NAND$ gates required to implement the function $F = (x' + y')(z + w)$ is ______
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10 votes

4 two input NAND gates

F = (x' + y').z  +  (x' +y').w

  = (xy)'.z  + (xy)'.w

  = ( ( (xy)'.z )' . ( (xy)'.w )' )'

We can tie the inputs 'x' and 'y' to 1st NAND gate to implement the function (xy)' and the output of this gate can be connected to the inputs of the  2nd  and 3rd NAND gate.The other remaining input of the  2nd  and 3rd NAND gate will be 'w' and 'z'.The output of 2nd and 3rd NAND gate will than be connected to the inputs of the 4th NAND gate.

6 votes
6 votes

aa

for this type of problem first try to convert expression into nand logic.
f= (x' + y')(z+w)
=  (xy)'(z+w) ----- here  (x' + y') is now converted into NAND logic  (xy)'
now sppose A = (xy)'
f= A(z+w)
=Az+Aw ----- it is a sop expression requires 2 level and- or circuit or 2-level NAND-NAND circuit.
u can also convert this into NAND logic
f=((Az)'(Aw)')' ---- this expression shows require 3 NAND gate 1 for  (Az)' 2nd for  (Aw)' and 3rd for ((Az)'(Aw)')'
and before this 1 NAND gate also reqire for  A=(xy)'
so total no of NAND gates = 4 
---------  hope u ll get it in a proper way after this explanation....

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2 votes
2 votes
We can reduce this function to minimum one first

(x'+y')(z+w) = x'z +x'w + y'z + y'w. which in turn minimize to => y'z + wy' + zx'

so F = y'z + wy' + zx'

F' = (y'z)' .(wy')' .( zx')'

F = ((y'z)' .(wy')' .( zx')')'

so we can find that it needs 4 nand gate.
2 votes
2 votes
see point no1 for nand gates question should be in the form of sop/ssop(standard sum  of products)

so i consider (x'+y') as p

therfore p(z+w)=pz+pw  for this 3 nand gates required and

for the implementaion of p we require 1 nanad gate the ftotal are 4nand gates

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