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If a clock with time period $“T”$ is used with $n$ stage shift register, then output of final stage will be delayed by

  1. $nT$ sec
  2. $(n-1)T$ sec
  3. $n/T$ sec
  4. $(2n-1)T$ sec
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3 Answers

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1 vote
ANS : B

(n−1)T(n−1)T sec
by
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take 4 register and initial data is 1011

ans bit are moving from MSB to LSB

1'st  2'nd  3'rd  4'rt

1     0      1       1  -initial data

1     1      0       1  -first shift

1     1      1       0  -second shift

0     1      1       1  -third shift

1     0      1       1  -initial data

option B

n-1 shift required
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Ans b is correct

Because  shift register

Serial output n-1 times required
Answer: