Multiplexer
Question 1 |
Consider the two cascaded 2-to-1 multiplexers as shown in the figure.
The minimal sum of products form of the output X is
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Question 1 Explanation:
Output of 1st MUX is
Now
Now
Question 2 |
Consider a 4-to-1 multiplexer with two select lines S1 and S0, given below
The minimal sum-of-products form of the Boolean expression for the output F of the multiplexer is
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Question 2 Explanation:
F(P,Q,R) = P’Q’(0) + P’Q (1) + PQ’(R) + PQ(R’)
= P’Q + PQ’R + PQR’
= Q(P’ + P R’) + PQ’R
= Q(P’ + R’) + PQ’R
= P’Q + QR’ + PQ’R
= P’Q + PQ’R + PQR’
= Q(P’ + P R’) + PQ’R
= Q(P’ + R’) + PQ’R
= P’Q + QR’ + PQ’R
Question 3 |
The Boolean expression for the output 'f' of the multiplexer shown below is
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P⊕Q⊕R | |
P+Q+R | |
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Question 3 Explanation:
f = P’Q’ R + P’Q R’ + PQ’ R’ + PQR
= (P’Q’ + PQ)R + (P’Q+PQ’)R’
= (P⊕Q)’R + (P⊕Q)R’
= (P⊕Q⊕R)
= (P’Q’ + PQ)R + (P’Q+PQ’)R’
= (P⊕Q)’R + (P⊕Q)R’
= (P⊕Q⊕R)
Question 4 |
Suppose only one multiplexer and one inverter are allowed to be used to implement any Boolean function of n variables. What is the minimum size of the multiplexer needed?
2n line to 1 line | |
2n+1 line to 1 line | |
2n-1 line to 1 line
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2n-2 line to 1 line |
Question 4 Explanation:
Both true and complement forms of all variables are necessary to implement any function of n variables.
A 2n X 1 multiplexer can implement any function of n variables. As n variables are given to select lines, so that true and complement forms of all variables get generated inside the MUX.
As one inverter is available, we can generate complement of one variable outside of the Multiplexer. And remaining (n-1) variables are given to select lines. With this we have true and complement form of all n variables.
So, the answer is 2n-1 X 1 MUX.
A 2n X 1 multiplexer can implement any function of n variables. As n variables are given to select lines, so that true and complement forms of all variables get generated inside the MUX.
As one inverter is available, we can generate complement of one variable outside of the Multiplexer. And remaining (n-1) variables are given to select lines. With this we have true and complement form of all n variables.
So, the answer is 2n-1 X 1 MUX.
There are 4 questions to complete.