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How many illegitimate states does synchronous mod-6 counter have?
1) 3

2) 2

3) 1

4) 0
in Digital Logic by Active (4.9k points)
edited by | 150 views
0
how we defined the illegitimate states ??

I heard this term ist time

can anyone explain me properly :3
0
What I felt after reading that word was the states which doesn't participate in counting loop are considered as illegitimate states.

Its my assumption
0

Hemanth_13

yes ...but answer given is 3 ..my doubt is how 3 states are illegitimate states here 

if we need mod 6 counter

we need 3 flip flips [if we use Johnson counter]

therefore total 8 states we have (0 to 7)

In which states are required to make 6 mod counter

remaining 2 are left

therefore answer should be 2 right

but I search in the Google ....and find correct answer is  3

if you find any solution for this  or know how to solve this question

let me know ok ???

+1

Magma 

Yes what you said was right in case of Johnson counter if we have n states we get mod 2n counter, so we have two illegitimate states.

But we can use ring counter as well where we get mod n counter. Its huge taking to check with 6 bits 2^6=64 combinations

I tried with 4 bits and i got 3 illegitimate states.

0

Hemanth_13  thanks !!

you give the example of mod 4

for mod 6 also it gives 3 illegitimate states

therefore  it means In mod n counter  using ring counter where n >= 3 and  n is any arbitrary number  gives 3  illegitimate states ??

0
But how to know which counter to consider?Its just said synchronous counter..
0
Hmm point !

but if it's given that it's a synchronous counter (ring counter) ..and mod n (n>=3) then it always gives 3 illegitimate states
0

Magma 

For n=3 we only get 2. Plz check once

0
hmmm 2 :/
0
Multiple Sources are saying are different answers.

When I googled this question, answer was mentioned as 3 in 1 website, 1 in another website, 0 in another website. Now here in this website, you guys are saying 2 :\

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