As per question we have arrangement as shown in fig .
We have arranged the registers are Q0,Q1,Q2,Q3 input to Q0 is ((Q2 ex-or Q3) Ex-Or Q0)
as per current values we have Q0,Q1,Q2,Q3 as 1000 now ,
CLOCK NUMBER |
Q0 |
Q1 |
Q2 |
Q3 |
INPUT At D (Q2 exor Q3 exor Q0) |
its already loaded input |
1 |
0 |
0 |
0 |
1
|
CLOCK 1 |
1 |
1 |
0 |
0 |
1 |
CLOCK 2 |
1 |
1 |
1 |
0 |
0 |
CLOCK 3 |
0 |
1 |
1 |
1 |
0 |
CLOCK 4 |
0 |
0 |
1 |
1 |
0 |
CLOCK 5 |
0 |
0 |
0 |
1 |
1 |
CLOCK 6 |
1 |
0 |
0 |
0 |
DESIRED OUTPUT GOT. |
Hence in 6th clock we got output as 1000