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Digital Electronics - Shift Registers - Discussion

Discussion Forum : Shift Registers - General Questions (Q.No. 4)
Shift Registers
4.
On the third clock pulse, a 4-bit Johnson sequence is Q0 = 1, Q1 = 1, Q2 = 1, and Q3 = 0. On the fourth clock pulse, the sequence is ________.
Q0 = 1, Q1 = 1, Q2 = 1, Q3 = 1
Q0 = 1, Q1 = 1, Q2 = 0, Q3 = 0
Q0 = 1, Q1 = 0, Q2 = 0, Q3 = 0
Q0 = 0, Q1 = 0, Q2 = 0, Q3 = 0
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.
Swapnil said:   7 years ago
0000- 0stage clock pulse.
1000-1stage.
1100
1110
1111-4th clock pulse.
0111
0011
0001
0000-8stage c'pulse.

ANS = 1111.
(1)
Khushi said:   1 decade ago
Anusha, just check the representation of 14 in digital format. Its 1110.
Shilpa said:   1 decade ago
In Johnson counter, complement o/p of last f/f is connected to i/p of 1st f/f as q3 is 0 its complement 1 is applied as i/p to the 1st f/f. So o/p is 1111.
Anushree Saha said:   1 decade ago
After 3rd ck pulse Q3=0.so, i/p of Qo is 1.O/p Qo=1,Q1=1 as prev state o/p for Q0=1. same reason for Q2 and Q3 o/p.

So, answer is 1111
Mukesh said:   1 decade ago
Given 3rd clock pulse for jhoson counter you need to find next clock pulse that will b 1 so there for 1111 is required solution.
Anusha said:   1 decade ago
0011 1110
3 13=10
4 + 10=14=1111
Kiran said:   1 decade ago
How?
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