GATE Papers >> CSE >> 2015 >> Question No 21

Question No. 21 CSE | GATE 2015

Consider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is


Answer : (D) 0, 8, 12, 14, 15, 7, 3, 1, 0


Solution of Question No 21 of GATE 2015 CSE Paper

$ \boxed{\begin{array}{c}{\mathrm D}_3\;{\mathrm Q}_3\\{\mathrm{FF}}_3\\{\overline{\mathrm Q}}_3\end{array}}\rightarrow\boxed{\begin{array}{c}{\mathrm D}_2\;{\mathrm Q}_2\\{\mathrm{FF}}_2\\{\overline{\mathrm Q}}_2\end{array}}\rightarrow\boxed{\begin{array}{c}{\mathrm D}_1\;{\mathrm Q}_1\\{\mathrm{FF}}_1\\{\overline{\mathrm Q}}_1\end{array}}\rightarrow\boxed{\begin{array}{c}{\mathrm D}_0\;{\mathrm Q}_0\\{\mathrm{FF}}_0\\{\overline{\mathrm Q}}_0\end{array}}\rightarrow $

Clock D3 = $ {\overline{\mathrm Q}}_0 $ Q3  Q2  Q1  Q4
    0    0    0    0
1 1 1    0    0    0
2 1 1    1    0    0
3 1 1    1    1    0
4 1 1    1    1    1
5 0 0    1    1    1
6 0 0    0    1    1
7 0 0    0    0    1
8 0 0    0    0    0

 

Hence the switching sequence is : 0, 8, 12, 14, 15, 7, 3, 1, 0.

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