A digital circuit which compares two numbers A2A1A0 and B2B1B0 is shown in figure

Question

How it is that EXNOR is a odd function?

Answer 1

$\overline{(A_2\oplus B_2)\oplus (A_1\oplus B_1)\oplus(A_0\oplus B_0)} = (A_2\oplus B_2)\odot (A_1\oplus B_1)\oplus(A_0\oplus B_0)= (A_2\oplus B_2)\oplus (A_1\oplus B_1)\odot(A_0\oplus B_0)$

a) $((0\oplus 1)\oplus (1\oplus 1))\odot(0\oplus1) = 1\odot 1 = 1$

b) $((0\oplus 1)\oplus (1\oplus 0))\odot(0\oplus1) = 0\odot 1 = 0$

c) $((1\oplus 1)\oplus (0\oplus 1))\odot(1\oplus 0) = 1\odot 1 = 1$

d)$((1\oplus 0)\oplus (0\oplus 1))\odot(1\oplus 1) = 0\odot 0 = 1$

I think they mean to ask invalid combination and did typo :D

Another way,
Xor gives $1$ when having odd no's of $1's$ in input, otherwise $0$. so take the complement of that result of xor, will get $1$ for option a,c,d and $0$ for option b.

Comments

$\overline{A \oplus B \oplus C } = A \odot B \oplus C$

Please explain this step. I am not able to remove the complement.

Also, I thought about the problem the way you described in alternate method. But, the solution approaches the problem by assuming that EXNOR will output 1 only if either all inputs are 1 or 0. By this approach, option (b.) looks right too. Please explain.

---

Only problem is that they write the options for xor and design the circuit with xnor 

---

Brilliant explanation Sir! Cleared my doubts. :)

Answer 2

Firstly EXNOR gate gives Output as 1 When all the inputs are 0's or all the inputs are 1's

Now go from bottom to Top 3 inputs must be 0's or 1's

If you observe in option 2  when u do xor with corresponding bits all are 1's

but in option 1 A2 violates this condition.