Combinational Circuit

Question

A digital circuit shown below has two 3-bit input A₂A₁A₀ and B₂B₁B₀

To obtain output Y = 1, the number of possible cases are ________.

Comments

8 ??

Answer 1

It is a 3 input EXOR GATE which produces an output 1 if all the inputs are 1 or exactly one input is 1.
We have 6 inputs - so possible combinations $=2^6 = 64$ as each of them can be either 0 or 1.
We get output 1 if all 3 inputs to the final EXOR gate is 1 or exactly one of them is 1.

• Lets take all 1's. This happens when $A_i \neq B_i$. So, 2 cases for each of the 3 EXOR gates - giving $2 \times 2 \times 2 = 8$ possible cases.
• Now, when we consider a single gate, it produces a 1 for 2 possible inputs out of 4, meaning 2 possibilities for 0. So, for say output 100, we have $2\times 2 \times 2 = 8$ possible inputs and same for outputs 010 and 001. Thus we get $8+8+8 = 24$ possible inputs.

Thus we get a 1 output for $8+24 = 32$ possible inputs.

Comments

sir it's 3 input xor gate so output 0 only when all inputs are 0 or any two inputs are 1's.

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Yes, thanks. Let me correct.

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Y = $A_0  xor   B_0   xor   A_1   xor   B_1   xor   A_1   xor  B_2 $

Now XOR of a binary string of 6 bit is 1 if it contains odd no of 1's.

So, Ans =  $6C_1 + 6C_3 + 6C_5 $

             = 6 + 20 + 6 = 32.

@Arjun sir , Can we do like this..?

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@vijaycs you are right.