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+2 votes
71 views

Problem:- here Johnson counter initial state is 101 but 3-bit jhonson counter isequence is like this...(000 100 110 111 011 001)

asked in Digital Logic by Loyal (7.8k points) | 71 views
+1
i am getting 7 .. is it correct
+2
getting $7$ as answer
0
Solution please,

@saxena @Anu sir.
0
yes 7 is correct...explain please..

1 Answer

+4 votes
$T$ Flipflop sequence $1\rightarrow0\rightarrow1\rightarrow0\rightarrow1\rightarrow0 $ so $Q'=1$ enabled

$Johnson$ Counter : $101\rightarrow010\rightarrow101\rightarrow010\rightarrow101\rightarrow010(Q_2Q_1Q_0)$

$Ring$ Counter : $1010\rightarrow0101\rightarrow1010\rightarrow0101\rightarrow1010\rightarrow0101(Q_3Q_2Q_1Q_0)$

$\{f,b,d\}$ are enabled. making $7$ as decimal value.
answered by Boss (11.4k points)
+1
thank u


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