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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. $X\bar{Y} + X\bar{Z} + \bar{X}Y\bar{Z}$

in Digital Logic by Active (2.2k points)
edited by | 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' .

2 Answers

0 votes
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.
0 votes

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?
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