Electronics and Communication Engineering - Digital Electronics - Discussion

Discussion Forum : Digital Electronics - Section 1 (Q.No. 31)
31.
Minimum number of 2-input NAND gates required to implement the function F = (x + y) (Z + W) is
3
4
5
6
Answer: Option
Explanation:

F = (x + y) (z + w) = xy.(z + w)

= xyz + xyw

= minimum no. of 2 input NAND gate.

Discussion:
14 comments Page 2 of 2.

Apoorva said:   6 years ago
Can someone explain the same problem with circuit concept?

Harshita Sharma said:   5 years ago
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
(6)

Vishnu said:   4 years ago
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.
(3)

Vraj said:   2 months ago
I think the answer is 6 then how the answer is 4? Anyone, please explain.


Post your comments here:

Your comments will be displayed after verification.