# Control Unit - Q2

 $C_1$ $C_2$ $C_3$ $C_4$ $C_5$ $C_6$ 1 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0

Minimum and maximum no of bits used to generate these control signals is

$(A)5, 6$

$(B)5, 20$

$(C)6, 20$

$(D)20, 20$

reshown Jun 9

+1 vote
Maximum bit in horizontal case so for 6 control signal max 6 bit used

For minimum we used vertical but in this problem at a same time at most 4 signal active so if we used decoder for each active signal than for 1 decoder 3 bit for 4 decoder 12 bit it is not the case because no option is there

Total deffrent 19 combination co if we used topically connection in decoder so log 19=4. Something so 5 bit is required

So minimum 5,

Max 6

5,6 is correct one .
answered Jun 9 by (5,940 points)
Please explain me regarding minimum bits because not getting it properly.
Total 19 combination here for select one in 19 we used minimum 5 bit
Or for maximum 6 horizontal microprograming