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

Question No. 21

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.