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Number of minterms at the output of 8 input XOR gate is ________
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This can be done using some observation :

Number of minterms that are present at the output of a 2 input XOR gate  =  2   [i.e. for A ⊕ B , we have 2 minterms]

Number of minterms that are present at the output of a 3 input XOR gate  =  4

=  22 [i.e. for A ⊕ B ⊕ C , we have 4 minterms]

Number of minterms that are present at the output of a 4 input XOR gate  =  23

In this way ,

Number of minterms that are present at the output of an 8 input XOR gate  =  27

= 128

answered by Veteran (100k points)
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For XOR gate, if the no. of 1s in the input is even then it results in 0 otherwise 1.

eg- 11010100 will result in 0 whereas 11010101 will give 1

So, possible inputs from 8 bits are 28 =256

Minterms are those which result in 1 or those combinations which give 1 OR Minterms = Total possible inputs(28) - Combinations of inputs which give 0.

Inputs which result in 0 value have 0,2,4,6,8 number of 1s = 8C0 + 8C2 + 8C4 + 8C6 + 8C8 = 128

Minterms = 256-128 =128

answered by Active (1.9k points)

+1 vote