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How to create a Boolean expression with MUX circuits?
If the inputs are 0,1,1,0 respectively, I would make a simple 2 variable k-map and construct a boolean expression with it, but I’m kinda confused how to build one with the variable inputs. (Only with this expression you get C as output) when A=B=1, therefore f2=C. Similarly if A=1,B=0 then C0 is selected at output so
Can a Boolean expression be implemented with 2 multiplexers?
If the boolean expression has n+1 variables, we take only n variables as the selection line of the multiplexer. The remaining single variable is used as the inputs of the mux. In this way, any boolean expression having n variables can be implemented with 2 n-1 to 1 multiplexer.
What is the difference between a multiplexer and A mux?
The multiplexer or MUX is a digital switch, also called as data selector. It is a Combinational Logic Circuit with more than one input line, one output line and more than one select line.
How many selection lines are there in the MUX?
So, the mux has 8 input lines, 3 selection lines, and one output. The inputs, corresponding to the minterms (2, 3, 6, 7) are connected to the logic 1 and remaining terms to the logic 0 (grounded). The given input variables are connected as three selection lines.
How to simply this 2 to 1 Boolean expression?
I’m confused how they made the simplification though. I see that the expression can be simplified to I 0 ∗ I 1 ∗ ( A ¯ + A) by simplifying the double negatives, but I don’t see how they got any farther than that Thanks for contributing an answer to Electrical Engineering Stack Exchange!
How many variables are in a Boolean function?
In the given boolean expression, there are 3 variables. We should use 2 3 : 1 = 8 : 1 multiplexer. But as per the question, it is to be implemented with 4 : 1 mux. For 4 : 1 multiplexer, there should be 2 selection lines. So from the given 3 variables, the 2 least significant variables (B, C) are used as selection line inputs.