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A $3$ line to $8$ line Decoder is used to implement a $3$-variable Boolean function as shown in figure.

The simplified form of output $Y$ is.

1. $\bar{X}Y + \bar{Y}Z + XY\bar{Z}$
2. $\bar{X}Z +\bar{Y}Z + XYZ$
3. $X\bar{Y} + X\bar{Z} +\bar{X}YZ$
4. $X\bar{Y} + X\bar{Z} + \bar{X}Y\bar{Z}$

edited | 378 views
+2
here ans is c if we are not applying bubble operation

simply considering when x'y'z' stands for 0 and so on and if we apply bubble operation we will get as ans output of c with complemented inputs
0
Her​e for a particular input combination, decoder shud give  exactly one output as '0'  rest all as '1'.

For eg -   for input  ZYX  =  0 0 0

output pin D0  will be  '0'   and other o/p pins will be '1' .

all bubble take ahead of OR gate that becomes NAND gate....

inputs to NAND gate are(1,3,5,6) which gives same output...using k-map...ans is C...
by Junior (739 points)
0
input to NAND gate is 1,3,5,6

output of NAND gate is 1,3,5,6

can you explain how both are same?
0
How it is giving the same output ?? can you explain a bit more.

HERE its written like this(Z is MSB)

ZYX ....

NOW TAKE 0's outside and it will be NAND GATE now it says when enteries are either (1,3,5,6) we get 1

so fill 1 at those places and it will turn out to be Z'Y'X+Z'YX+ZY'X+ZYX' THAT MATCHES WITH c

JUST MAKE THE OPTION C IN CANONICAL FORM

by Active (2.1k points)
0
But, doesn't a NAND gate mean "Atleast 1 among (1,3,5,6) should be 0" so that it can output 1.. So how to draw a K-Map for it?? Can you plz post a diagram too? Thanx in davance :)
0
I agree with Tushar can u elaborate on this?

+1 vote