MadeEasy Test Series: Digital Logic - Multiplexer

Question

A 32 : 1 MUX has to be designed using a 16 : 1 MUX. It was found that for this task we require ‘X ’ number of 16 : 1 MUX and “Y ” number of two input OR gates, then the value of X + Y = __________.

Comments

Answer is 3 ?

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$2+1$..rgt?

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I got 3 MUX

first 2 for 16, 16 lines and next 1 is for next level, 2 lines.{though it is not used at its full capacity}

Don't know where they used OR ?

Hence 3 + 0 = 3

:(

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In the solution they used 2 MUX at first level and a OR gate at second level.
but I dont get it, because in this case both MUX at 1st level can output their value simultaneously then how we detect them uniquely.
Also what about the 5th select line? both 16x1 MUX used 4 select lines.
If anyone find a solution please post it.

Answer 1

two mux can be used and it will give two o/p .so we can use 1 or gate so total we need 3.

one of the selection line can be used as enable for both the mux.so only one of them will give the o/p other will give 0.

and by using or gate(0+o/p)=o/p..

Comments

GREAT
Thank you sir :D

just one more doubt ...

if the 5th select line is used as an enable input then in one MUX enable input it should be given in TRUE form and in the other one as COMPLEMENTED form. Then how do we perform the complement. We are not provided with any NOT gate. Only 2 MUX and 1 OR gate

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In the question it doesn't state that these are the only two to be used.

unless it is written in the question i think we can use not gate for complement.(the question is not properly stated ,i think so.)As there are no other way to solve it,using the 3rd mux will be only a fix.

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Ok :)

thank you sir

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OR gate doesn't necessarily work as selector for 2 lines.

Then why use it on 2nd level?

We can use 2:1 MUX or some NOR gates should be available.