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I am given the distance from the source of radiation of the harmonic oscillation to the point of observation $R$ and signal frequency $f$.

I need to find the phase shift due to this distance.

When calculated using this formula, the signal phase shift exceeds the maximum $2\pi$. $$\Delta\varphi=\frac{R\cdot 2\pi}{\lambda}$$

How to write the formula correctly so that the phase incursion is reset to zero when the maximum value is reached?

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  • $\begingroup$ use the modulo function $\Delta\phi = \frac{R 2\pi}{\lambda} mod 2 \pi$ $\endgroup$
    – Ben
    Feb 7, 2022 at 13:12
  • $\begingroup$ Or $\Delta \phi = \frac{R 2 \pi}{\lambda} \mod 2\pi$. Because LaTeX recognizes \mod and treats it properly when typesetting ($\Delta \phi = \frac{R 2 \pi}{\lambda} \mod 2\pi$). $\endgroup$
    – TimWescott
    Feb 7, 2022 at 16:00
  • $\begingroup$ I have replaced the inline math and image with MathJax. (edit may be pending). Please check that I did so correctly and consider using MathJax in the future. $\endgroup$ Feb 8, 2022 at 0:28

1 Answer 1

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You can use the the modulo function. Some ways to write this would be

$$\Delta \varphi = \mod \left(\frac{ 2 \pi R}{\lambda},2\pi\right)$$

or

$$ \newcommand{\floor}[1]{\lfloor #1 \rfloor}\Delta \varphi = \frac{ 2 \pi R}{\lambda} -2\pi\floor{ \frac{R}{\lambda}} $$

where $\lfloor \rfloor$ is the truncation symbol.

In many cases it's actually preferable NOT to reset the phase as it creates a discontinuity.

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