A X:Y decoder can be constructed using how many A:B decoder with enable?

Question

options are :

A)Y/B

B)X/A

3)X/B

4)Y/A

Answer 1

Here we follow the reverse process of MUX in finding the levels and deocder at each level

At last level Y/B=Output  

if ouput < B then we use 1 more level i.e. level 1 with one decoder

if output >B then keep on dividing with output and increasing the levels.

e.g. 4:16 realize by 3:8 

last level 16/8=2   3:8 decoder                          

level1 =1             3:8 decoder

total 2 level 3 3:8 decoder.