What is the physical interpretation of Power Spectral Density (PSD) in optical tweezer?

Question

For the calibration of an optical tweezer, the PSD of position values are plotted. If it follows a certain behavior having two regions - free diffusion and potential limited regions, then the trap is considered working fine.

I read that $\text{PSD} = \left\lvert \text{DFT}(X) \right \rvert^2/T_m$ where $X$ is the signal, $T_m$ is the time over which measurement is made, and $\text{DFT}$ means the discrete Fourier transform.,

In the above equation, why do we have time of measurement in the denominator? Does the value of $\left\lvert \text{DFT}(X) \right \rvert^2$ increase with $T_m$? If so, to make PSD independent of time variable are we doing the same?

What is the density in PSD? What information do we get from PSD given in $\text{V}^2/\text{Hz}$?

Answer 1

In the above equation, why do we have time of measurement in the denominator?

PSD = Power Spectral Density.

It is a measure of the effective power of a given input. I say effective because the units are $(data \ units)^{2} \ (unit \ frequency)^{-1}$. I think you incorrectly typed your equation for PSD too. The units of $|DFT(X)|^{2}/T$ would be $(data \ units)^{2} \ (unit \ frequency)^{+1}$.

The correct form should be something like (this is for the IDL programming language... see my notes below about language-specific normalizations): $$ PSD = \frac{2 N}{f_{o}} \lvert DFT\left(X\right) \rvert^{2} $$ where $N$ is the number of samples provided to create the discrete Fourier transform (DFT) and $f_{o}$ is the sample rate of the data [units of # of samples per unit time].

What is the density in PSD?

In any case, the PSD gives you a measure of the power at any given frequency up to the Nyquist frequency. It is used to determine whether power is isolated within a specific frequency range, called a frequency peak, or if the data are a form of noise. It is called a density because it is a measure of the power per unit frequency, so kind of like a linear mass density with units of mass per unit length (very loose analogy here).

What information do we get from PSD...

The PSD is very useful for wave mode identification and is often used in conjunction with other measures like polarization and/or wavelet transforms.

Does the value of $|DFT(X)|^{2}$ increase with $T_{m}$?

No, $T_{m}$ should be a scalar constant that depends upon the input time series $X$. As I stated above, the correct normalization should be dependent upon the sample rate and the number of elements in the time series array $X$.

Side Note
The correct normalization factor is actually dependent upon the computer language used to calculate the DFT (which is often synonymous with an FFT). Some languages will only divide by $N$ on the forward transform (i.e., going from time-space to frequency-space) while other languages will divide by $\sqrt{N}$ frequency-space. So read the documentation to ensure you have the correct normalization factor.

Answer 2

It is power because the energy is measured as a rate, i.e. power. They are dividing by $T$ because the measurement was energy. Taking a longer measurement increases the signal, but that doesn't mean the light was brighter.

Any spectrum is given as a density with respect to either frequency or wavelength. The units tell you which. The reason for the difference is because of the chain rule of derivatives. Reconstructing the total power is an integral, and the change of variables between frequency and wavelength is more complicated than just a unit conversion.