Counting procedure of a Johnson's Counter
- Starting from all 0's , each shift operation inserts 1's from the left until the register is filled with all 1's.
- When the register has all 1s,each shift operation inserts 0's from the left until the register is filled with all 0's.
- Go to step 1.
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
0000 $\rightarrow$ ${\color{Red} 1}$000 $\rightarrow$ ${\color{Red} 1}{\color{Red} 1}$00 $\rightarrow$ ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$0 $\rightarrow$ ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}$
OR
0 $\rightarrow$ 8 $\rightarrow$ 12 $\rightarrow$ 14 $\rightarrow$ 15 $\rightarrow$ 7 $\rightarrow$ 3 $\rightarrow$ 1 $\rightarrow$ 0
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
$\therefore$ Option D is correct answer.