Test by Bikram | Digital Logic | Test 2 | Question: 21

Question

On the fifth clock pulse, a $4$-bit Johnson sequence is $Q0 = 0, \ Q1 = 1, \ Q2 = 1$, and $Q3 = 1$. On the sixth clock pulse, the sequence is ________.

• A. $Q0 = 1, \ Q1 = 0, \ Q2 = 0, \ Q3 = 0$
• B. $Q0 = 1, \ Q1 = 1, \ Q2 = 1, \ Q3 = 0$
• C. $Q0 = 0, \ Q1 = 0, \ Q2 = 1, \ Q3 = 1$
• D. $Q0 = 0, \ Q1 = 0, \ Q2 = 0, \ Q3 = 1$

Comments

Why option C?

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Becoz last bit complement will take and it will make

0 1 1 1 -> 0 0 1 1 (MSB to LSB)

Answer 1

https://en.wikipedia.org/wiki/Ring_counter#Johnson_counter

Ans is C

Comments

No answer b D

let check and tell me .

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Yes answer should be d how it  can be c

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only last flip flop inverted output given in feedback so c is correct answer

Answer 2

Set all the bits to 0	0000
Flip the first bit	1000
Flip the second bit	1100
Flip the third bit	1110
Flip the fourth bit	1111
Again flip the first bit	0111
Again flip the second bit	0011
Again flip the third bit	0001

Loop this.

This is how a 4-bit Johnson Counter counts. You can extend this to n-bits.

Clearly, after 0111, 0011 follows.

Option C

Answer 3

Johnson counter inverts the last bit and feedbacks it to the 1st bit causing 1 right shift. Seems difficult to read but thats how concept originated.