Control Unit - Q2

0 votes
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

asked Jun 8 in Computer Organization by gbmentor (6,810 points)
reshown Jun 9 by gbmentor

2 Answers

+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 tsnikhilsharmagate2018 (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
0 votes
Same logic apply in problem no 4

Total number of bit 10 in that one problem

So correct one is a minimum 5 max 6
answered Jun 9 by tsnikhilsharmagate2018 (5,940 points)
Answer:
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