in Digital Logic edited by
8,340 views
12 votes
12 votes

How many $2$-input multiplexers are required to construct a $2^{10}$-input multiplexer?

  1. $1023$
  2. $31$
  3. $10$
  4. $127$
in Digital Logic edited by
8.3k views

4 Answers

16 votes
16 votes
Best answer
1023
1st level 512 MUX
2nd level 256 MUX
.
.
So on
Total= 512+256+128+64+32+16+8+4+2+1=1023 MUX
selected by

1 comment

0
0
14 votes
14 votes

The No of 2 input Mux required =2n -1

                                                 =210 -1

                                                 =1023

2 Comments

is 2^n -1 can be applied to all cases to find no of MUX or decoders needed?
0
0
2$\times$1 MUX only
1
1
8 votes
8 votes

Number of N input MUXs required to make M input MUX is = ceil (M-1 / N-1)

In given question,

no of MUX needed = 2 10 -1 /2-1 = 1023/1 =1023

by
0 votes
0 votes

Levels of Multiplexers:

  • Level 1: Use $2^9$ 2-input multiplexers to select between 2^9 pairs of inputs.
  • Level 2: Use $2^8$ 2-input multiplexers to select between the outputs of pairs of Level 1 multiplexers, resulting in $2^8$ selected inputs.
  • Level 3: Repeat the process with $2^7$ multiplexers, then $2^6$, and so on, until reaching a single multiplexer at the top level.

Total Multiplexers:

  • Sum up the number of multiplexers at each level:
    • $2^9$ + $2^8$ + $2^7$ + ... + $2^1$ + $2^0$ $=$ $2^{10}$ - $1$ $=$ $1024 - 1$ $=$ $1023$