Let's assume two sine waves
$$\begin{align}
u(t) &= \hat u \sin(\omega t + \phi_u)\\
y(t) &= \hat y \sin(\omega t + \phi_y)
\end{align}$$
whose phase shift can be calculated as
$$ \Delta\phi = \phi_y – \phi_u.$$
How can I now derive the time delay between them?
I found somewhere that
$$\Delta t = \frac{\Delta\phi_\text{rad}}{\omega}.$$
Is that correct? And how can I derive it?
Application-wise, I want to convert the phase diagram of a traditional Bode plot which is usually given in degree or radians vs frequency $f = \frac{\omega}{2\pi}$ into a time delay vs frequency diagram.
Any help would be greatly appreciated.
Best Answer
Yes, this is correct. To derive it, note that $u$ is $0$ and increasing when $\omega t+\phi_u=0$ and similarly for $y$. The time between these crossing is just as you have said.