Electronics and Communication Engineering - Digital Electronics - Discussion

I think the answer is 6 then how the answer is 4? Anyone, please explain.

As far as time is concerned u won't get that much time to solve this question here is the shortcut.

What is nand====(xy)'==x'+y'.

Sso here is shortcut==(x'+y')(w+z)=(xy)'(w'z')'==1nand for (xy)' + 1 nand for (w'z')' + 1 nand to convert z' to z +1 nand to convert w' to w as we r not alloweded to use not in the circuit.

Hope it helped you guys.

F=(x'+y').(Z+W)
F=(xy)'.(Z+W)
F=(xy)'Z + (xy)'W

NAND GATE 1:
INPUT 1 = x
INPUT 2 = y
OUTPUT = (xy)'
Take this output as K => [K = (xy)']

NAND GATE 2:
INPUT 1 = Z
INPUT 2 = K
OUTPUT = [(xy)'z]'
Take this output as L => L= [(xy)'z]'

NAND GATE 3:
INPUT 1 = W
INPUT 2 = K
OUTPUT = [(xy)'W]'
Take this output as M => M = [(xy)'W]'

NAND GATE 4:
INPUT 1 = L
INPUT 2 = M
OUTPUT = [ ((xy)'W)' . ((xy)'Z)' ]'

Using De Morgan's Law
(A.B)' = A' + B'
(A + B)' = (A)' . (B)'

OUTPUT = { [(xy)'W]' }' + { [(xy)'Z]' }'
OUTPUT = (xy)'W + (xy)'Z


This output is equal to F = (xy)'Z + (xy)'W


I Hope, this solution will give clarity

Can someone explain the same problem with circuit concept?

5 two input Nand gates.

For implementing I think we should not alter the given Boolean equation and according to the procedure it requires 6 NAND gates.

F=(x'+y')(w+z)=(xy)'(w+z) now 1st take a NAND gate with the input X,Y that will produce (xy)'=k.

Let assume then take another two NAND gate which inputs are K ,W and K,Z then take another NAND gate with the input of those outputs that will produce the desire output.

(x' + y') (z + w) --- overall, AND function - to make NAND -> DOUBLE COMPLEMENT
{ [ (x' + y') (z + w) ]' }' -- then simplify terms in parenthesis: FACTOR OUT Negation (remember De Morgans).

{ [ (x y)' (z' w')' ]' }'

(xy)' - 1 NAND
z' - 1 NAND (complementation: SHORT two inputs)
w' - 1 NAND (complementation: SHORT two inputs)
(z'w')' - 1 NAND
[ (x y)' (z' w')' ]' - 1 NAND for the overall gate (carries all operations)
{ [ (x y)' (z' w')' ]' }' - LAST NAND for complement, since original term has AND overall function.

Overall, 6 NAND gates.
Well, not including INVERTERS, that would be 4.

(x'+y')(z+w)--->xor gate. 4 nand gates are required.

Use DOT conversion rule, so we can convert F = (x' + y') (z + w) = xy + z'w'.

Now implement this using two input NAND gates.

Note: Use NAND as inverter to get z' and w'

The answer should be 5.