Actual convention for the term Sinusoidal Phase "Shift" or "Offset"

Question

I have seen the convention for a sinusoid appear as:

• $x(\theta) = A \cdot \sin( B \cdot (\theta - \phi) \ ) + D$
• $y(\theta) = A \cdot \sin( B \cdot (\theta + \phi) \ ) + D$

Is "offset":

• $\phi$ from the $x$ equation (and thus $-\phi$ from the $y$ equation)?
• $\phi$ from the $y$ equation (and thus $-\phi$ from the $x$ equation)?

For example:

• $z(\theta) = 2 \cdot \sin( 3 \cdot (\theta - \frac{\pi}{2}) \ ) + 5$

Does this example have an offset of $+\frac{\pi}{2}$ or $-\frac{\pi}{2}$?

Further reading:

• $\phi$ from the $x$ equation (and thus $-\phi$ from the $y$ equation)?
• See References (1), (2), (3), (4), (5)
• $\phi$ from the $y$ equation (and thus $-\phi$ from the $x$ equation)?

Answer 1

$z(\theta) = 2 \cdot \sin( 3 \cdot (\theta - \frac{\pi}{2}) \ ) + 5$

Easiest way to see the answer is to plug in values of $\theta$ (0°, 30°, 60°, etc.), and sketch the resulting graph.

All sine waves look alike. The phase shift skews the starting point of the sinewave.

If $\theta$ = $\frac{\pi}{2}$, then $\frac{\pi}{2} - \frac{\pi}{2} = 0$, so $\sin (0) = 0$.

Start of sine wave or offset is $+ \frac{\pi}{2}$ or 90° after vertical axis.

+ve angle means sinewave starts before vertical axis. Offset would be -ve angle.

$\sin(\theta + \frac{\pi}{4})$ means a phase shift of -45°.

Typically phase shift is used for determining the phase relationship between two waveforms.

In your case, you have two identical sine waves (Amplitude A, DC offset D and frequency B). The only difference is $y(\theta)$ starts $\phi$ before vertical axis and $x(\theta)$ starts $\phi$ after.

Phase difference between the two is $\phi - (-\phi) = 2\phi$. Or $y(\theta)$ leads $x(\theta)$ by $2\phi$.