# Digital-Grandtest1-Q3

A digital circuit shown below has 2 3-bit inputs. To obtain Y = 1, the number of possible cases are?

reshown Jun 20

+1 vote

Solution:

XOR gates are used. The output of XOR gate is 1, when inputs have odd number of ones.

i.e  For Y=1, we  have 4 possible cases: 001,010,100,111

Now for each case: For example consider case 1(001):

->A0&oplus;B0=1 & A1&oplus;B1=0 & A2&oplus;B2=1

-> For A0&oplus;B0=1 : two possiblities - A0,B0= 0,1 or 1,0

A1&oplus;B1=0 : two possiblities - A1,B1= 0,1 or 1,0

A2&oplus;B2=1 : two possiblities - A2,B2= 0,1 or 1,0

Hence total no. of combinations=2*2*2 = 8

Hence for each case: 8 combinations.

Hence total no. of possible cases= 4 * 8 = 32.
answered Jun 21 by (14,390 points)
Here all gate are xor xor true only when

Odd number of 1 in input. For three variable

111,100,010,001

Or for 2 input

01,10.  True and 00,11.  False

For 111

We have 2*2*2=8 options

For 100

We have 2*2*2=8 options

For 010

We have 8 options

For 001 we have 2*2*2=8 options

So total number of options 8+8+8+8=32

answered Jun 20 by (17,270 points)
+1 vote

 $l$ $m$ $n$ $Y$ 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1

There are 4 cases for which we are getting Y = 1

Lets take one of that case

we can get 0 at L by 00,11 so 2 values at m by 2 and at n by 2 (01,10)

so for 1 case : 2X2X2 = 8

And there are 4 such cases so 4X8 = 32

answered Jun 20 by (6,560 points)