Applied Test Series

Question

What will be the output ?

 

Answer 1

Output of the first multiplexer will be:

=> (1)’ . 0 + 1.1 

=> 0.0 + 1.1

=> 1

Output of the second multiplexer will be:

=> (0)’. P + 0 . 1

=> 1.P + 0

=> P

 

Output of the third multiplexer will be:

=> (1)’ . Q + 1.P

=> 0 + P

=> P

 

Hence, the output of the given circuit is P

Comments

Got it

Answer 2

For $2*1$ multiplexer, output equation is : $Y=S’I_0+SI_1$

So output of first multiplexer $(f_1):\bar1*0+1*1=1$

Now output of $f_1$ will send to as input for multiplexer $f_2$

So output of second multiplexer $(f_2):\bar0*P+0*1 \implies1*P=P$

Now $f_2$ will send as input to $F$,so output is: $F=\bar1*Q+1*P\implies 0*Q+P\implies P$

So final output is $F=P.$