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+1 vote

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

0 votes

This will be the sequence after each clock cycle:

0001(in order Q0, Q1, Q2, Q3)

1000

1100

1110

0111

0011

0001

1000

After 7th clock cycle, we will get 1000.

0001(in order Q0, Q1, Q2, Q3)

1000

1100

1110

0111

0011

0001

1000

After 7th clock cycle, we will get 1000.

0

Do you find any mistake in the solution?? If not consider 7 as the answer. their solution might be wrong.

0

but brother may be your solution wrong i think i should wait for some better answer then i go to final decesion.

0

@Utkarsh Joshi you are correct but why you are taking given condition as clock 1...

Means your clock 1 is already exsisting pattern , no need to count it .

0

@utkarsh joshi

1000 (Q3, Q2, Q1, Q0) this is given in question

now you have taken initialy

1000 (which is right)

means after doing exor opration of Q3,Q2,Q0 =1 ,you put this value at the the position of Q3 and shifted the remaining bits and you got 1100

but according to figure it should be put at Q0 position

1000 (Q3, Q2, Q1, Q0) this is given in question

now you have taken initialy

1000 (which is right)

means after doing exor opration of Q3,Q2,Q0 =1 ,you put this value at the the position of Q3 and shifted the remaining bits and you got 1100

but according to figure it should be put at Q0 position

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