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1) Draw the Truth-Table for the majority-vote problem (assuming 3 inputs). 2) No

ID: 3730169 • Letter: 1

Question

1) Draw the Truth-Table for the majority-vote problem (assuming 3 inputs). 2) Now, assuming our 8-to-1 multiplexer can be visualized like this: DO DI D2 D3 D4 DS D6 D7 We set the inputs (the D's) the way we want. The output Y will be the appropriate D depending upon the value of S2 SI SO S2 S1 So Use a box like this to represent a 8-to-1 multiplexer, and sketch a circuit which will implement the majority-vote logic from 1). 3) Consider the following 4-input function F(A, B, C, D)-(0, l, 3, 4, 8, 9, 15). Draw the truth table for this function, and then sketch a circuit which will implement this function, using an 8-to-1 multiplexer (use a box like the one above) plus a single NOT gate. Yes, it is possible (but you have to spot a pattern or two in the truth table)!

Explanation / Answer

1)

Let 3 inputs be A,B,C and output represented by Z.

According to majority problem, output Z will be be 1 if majority of input is 1 else output is 0.

In other words, for an output to be 1, atleast 2 of the 3 inputs should be 1.

Tuth table will be:

A B C Z 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1