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Consider the two cascade $2$ to $1$ multiplexers as shown in the figure .

The minimal sum of products form of the output $X$ is

1. $\overline{P} \ \overline {Q}+PQR$
2. $\overline{P} \ {Q}+QR$
3. $PQ +\overline{P} \ \overline{Q}R$
4. $\overline{Q} \ \overline{R} + PQR$

edited | 3.1k views

For $2 :1$ MUX, output $Y=S'I_o+SI_1$

So, output of MUX$1$ ,$f_1=P'0+PR =PR$

Output of MUX$2$ , $f_2=Q'R'+Qf_1=Q'R'+PQR$

which is option D

by Loyal (9.5k points)
selected
Ans d
by (173 points)
Let the output of 1st mux be S

S=P'.0+PR=>PR

now the 2nd Mux would give the output =>

Q'R'+QPR

so option D is the right answer..
by Active (1.9k points)

Lets say output of 1st  multiplexer = m

table for 1st multiplexer

 P m 0 0 1 R

m = (P' * 0) + (P * R ) = PR ...(1)

table for 2nd multiplexer

 Q X 0 R' 1 m

X = (Q' * R') + (Q * m) ...(2)

put (1) in (2)

X=(Q' * R') + (Q * PR)

X= Q' R'+PQR

by Junior (987 points)