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Difference Between Phase and Time-Delay

  1. Sep 14, 2011 #1
    Hi,
    I understand what phase and time-delay are, but can they be converted from one to the other?

    For example is the working out below correct?:

    Sin(2[itex]π[/itex]Bt + [itex]\frac{π}{2}[/itex]) = Sin(2[itex]π[/itex]B(t + [itex]\frac{1}{4B}[/itex]))

    So basically a time delay of (t + [itex]\frac{1}{4B}[/itex]) = a phase of [itex]\frac{π}{2}[/itex] ?
     
  2. jcsd
  3. Sep 14, 2011 #2

    Hootenanny

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    You have the right idea, but you have just made a small conceptual error). A time delay is equivalent to a negative temporal shift; or a translation in the positive direction along the time axis. If your original signal [itex]\sin(2\pi B t)[/itex], then after a time delay of 1/(4B), your new signal will be

    [tex]\sin\left(2\pi B\left[t-\frac{1}{4B}\right]\right) = \sin\left(2\pi Bt - \frac{\pi}{2}\right)[/tex]

    Does that make sense?
     
  4. Sep 14, 2011 #3
    Thanks for your help, yeh it makes sense.
    So basically it's true that a phase change has a corresponding time delay and vice versa?
     
  5. Sep 14, 2011 #4

    Hootenanny

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    Yes, assuming the shift is linear, i.e. [itex]t\mapsto t+\text{const.}[/itex] and [itex]\omega\mapsto \omega+\text{const.}[/itex].
     
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