Question #205051

Realize the following Boolean expression using only 8x1 multiplexers.

y = A’B’C + A’BC” + AB’C + ABC

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2021-06-11T06:10:36-0400 It's simple to find an equivalent Boolean statement by looking at this circuit:

"I_3 \\bar{S_0} \\bar{S_1}+I_2S_0\\bar{S_1}+I_1 \\bar{S_0} S_1+I_0 S_0S_1"

When comparing your expression to yours, it appears that A and C correspond to S0 and S1, respectively, because they appear as both themselves and negations. And B will take over the roles of certain of I0.3.

Now, if we swap A for S0 and C for S1, we can see that the I2 term ( I2AC ) is missing from your expression. As a result, since 0x=0 and 0+x=x, we may set I2=0.

Because B appears in other words in your equation, we may argue that I0=I1=I3=B.

Other options are available. Essentially, A and C go into the selector, and all inputs receive B input, except the one picked with A=1 and C=0, which receives 0 input.

X is the inversion of x, a+b is the disjunction of a and b, and ab is the conjunction of a and b.

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