Difference Between Phase and Time-Delay

[frenzal_dude]

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$π$Bt + $\frac{π}{2}$) = Sin(2$π$B(t + $\frac{1}{4B}$))

So basically a time delay of (t + $\frac{1}{4B}$) = a phase of $\frac{π}{2}$ ?

 

[Hootenanny]

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$π$Bt + $\frac{π}{2}$) = Sin(2$π$B(t + $\frac{1}{4B}$))

So basically a time delay of (t + $\frac{1}{4B}$) = a phase of $\frac{π}{2}$ ?

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 $\sin(2\pi B t)$, then after a time delay of 1/(4B), your new signal will be

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

Does that make sense?

 

[frenzal_dude]

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 $\sin(2\pi B t)$, then after a time delay of 1/(4B), your new signal will be

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

Does that make sense?

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?

 

[Hootenanny]

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?

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