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GATE IT 2005 | Question: 7

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42 votes
42 votes

Which of the following expressions is equivalent to $(A \oplus B) \oplus C$

  1. $(A + B + C) (\bar A +\bar B +\bar C)$
  2. $(A + B + C) (\bar A +\bar B + C)$
  3. $ABC + \bar A (B \oplus C) + \bar B(A \oplus C)$
  4. None of these
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4 Answers

Best answer
47 votes
47 votes
Correct answer is C

$(A\oplus B)\oplus C$

At $C = 0$ ,$(A\oplus B)\oplus C = (A\oplus B)$ ----(I) [ as $0\oplus x = 0.x' +0.'x = x$]

At $C = 0$, $ABC + A'(B\oplus C)+ B'(A\oplus C)$

$= 0+A'(B\oplus 0)+B'(A\oplus 0) = A'B+AB' = A\oplus B$ -----(II)

At $C= 1$, $(A\oplus B)\oplus C =(A\odot B)$ --- (III) [as $1 \oplus x= 1.x'+1'.x = x'$]

At $C = 1$, $ABC + A'(B\oplus C)+ B'(A\oplus C)$

$=AB+A'(B\oplus 1)+B'(A\oplus 1) = AB +A'B' =(A\odot B)$ --(IV)

from eq (I), (II), (III) and (IV) it is clear
$(A\oplus B)\oplus C = ABC + A'(B\oplus C)+ B'(A \oplus C)$
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45 votes
45 votes

Consider truth table of three variable XOR function

A B C A⊕B⊕C
0 0 0 0
0 0 1 1 (m1)
0 1 0 1(m2)
0 1 1 0
1 0 0 1 (m4)
1 0 1 0
1 1 0 0
1 1 1 1 (m7)

Our XOR function is also known as Odd function because it outputs 1 whenever we have odd number of 1's in the input.

The minterms corresponding to the A⊕B⊕C is m1,m2,m4,m7 as we can see from truth table.

So, f(A⊕B⊕C)= ∑ m(1,2,4,7) = (A'B'C+ A'BC' + AB'C' + ABC) (Equivalent expression expressed as Sum of products)

So clearly A⊕B⊕C will be an SOP expression having 3 literals in each minterm.

Options (a) and (b) are ruled out as they cannot generate a 3 literal minterm(such as ABC) which can justify that they correspond to A⊕B⊕C.

Now if we expand option (c)

ABC + A'(B⊕C)+B'(A⊕C)

ABC + A'(B'C+BC') + B'(A'C + AC')

ABC (m7) + A'BC' (m2) + A'B'C (m1) + AB'C' (m4)

So above terms correspond exactly to our previously derived result of A⊕B⊕C.

Hence Option (c) is answer.

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5 votes
5 votes
Very Simple way to solve this question... just check the minterms.

As we know that for 3 XOR variable function (A⊕B)⊕C,  we have minterms Σ(1,2,4,7). i.e., odd number of 1s in minterms.

Now if you check with options, Option A and B will give minterm m5( i.e., AB'C). So both wrong.

check with option C, you'll get your desired minterms.

Answer: Option C

PS: Try to solve with the help of minterms (either question asked for XOR or XNOR)
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3 votes
3 votes 
(A ⊕ B) ⊕ C

By Solving, We get
= (A ⊕ B)′ C + (A ⊕ B) C′

The above expression can be written as:
= (A ⊙ B) C + (A ⊕ B) C′

Now,

= (AB + A′B′) C + (A ⊕ B) C′

= ABC + A′B′C + AB′C′ + A′BC′	     [As: X + X = X]				     
			         [So, A′B′C + A′B′C = A′B′C]
= ABC + (A′B′C + A′B′C) + AB′C′ + A′BC′

This can be written as:

= ABC + A′ (B ⊕ C) + B′ (A ⊕ C)
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