# Why is the sample interferogram given by the following integral of the intensity pattern?

I am trying to understand equation (2) in the paper by J. E. Greivenkamp with the title Generalized data reduction for heterodyne interferometry from year 1984. If you dont have acces to the article, don't worry; the equation appears in the post below.

The general expression for the intensity pattern of a sinusoidal fringe is given by $$I(x,y,\delta) = I'(x,y) + I''(x,y) \cos(\phi(x,y) - \delta)$$ $$\qquad$$ where $$I'$$ is the average background Intensity, $$I''$$ is the modulation coefficient, $$\phi$$ is the unknown phase and $$\delta$$ is the phase shift.

The data required for heterodyne interferometry (or equivalently for digital fringe profilometry) are a series of interferograms (or fringe images) with different phase shifts. To be more specific we need atleast 3 phase-shifts since there are 3 unknowns $$I', I''$$ and $$\phi$$ in the above equation.

In Fig.1 below is a camera which captures images of the fringes displayed on the LCD monitor. For total number of 4 phase-shifts as shown in fig.2, one obtains 4 images $$I_1,\dots, I_4$$, where the phase-shifts are $$\delta_1 = 0, \delta_2 = \dfrac{\pi}{2}, \delta_3 = \pi, \delta_4 = \dfrac{3\pi}{2}$$ respectively.

Now comes the part which I dont fully grasp. The Author mentions:

"In practice, the phase shift is often allowed to vary linearly over a range of $$\Delta$$ during the measurement, so that a sample interferogram is the integral of the intensity pattern from $$\delta_i - \frac{\Delta}{2}$$ to $$\delta_i + \frac{\Delta}{2}:$$ $$I_i (x,y) = \int_{\delta_i - \frac{\Delta}{2}}^{\delta_i + \frac{\Delta}{2}} I(x,y,\delta) \, \mathrm{d}\delta\,.$$ This is the equation (2) in the paper I mentioned in the begining. I don't fully understand

• what exactly is $$\Delta$$- is it $$2\pi$$?
• Why is the lower and upper bound of the integral chosen this way or in other words what is the intuition behind the upper and lower bounds?

Evaluating the integral and exploiting trigonometric identities gives $$I_i(x,y) = I'(x,y) + I''(x,y) \, \mathrm{sinc}\bigg(\dfrac{\Delta}{2}\bigg)\cos(\phi(x,y) - \delta_i).$$

I am familiar with the $$\mathrm{sinc}$$ function but I can't grasp why its appearance makes sense in this context. Maybe I am missing a key point here. Any comments will be kindly appreciated.

• From the 4-phase description, I would guess that $\Delta$ would be anything up to $2\pi$ , but if you have sufficient camera resolution, it could be considerably smaller since you're effectively taking a sample every time your record an image. Nov 4, 2021 at 15:14
• Pretty sure the sinc function falls out of the antiderivative of the $I(x,y,\delta)$ function Nov 4, 2021 at 15:15
• Yes indeed the sinc function falls out of the antiderivate. In the general expression the modulation term is $I''$ which can take all values between 0 and 1. After the evaluation of the antiderivative, the new modulation term is $I'' \mathrm{sinc}(\Delta / 2)$ which means it can take negative values. I dont grasp the intuition behind it. Nov 4, 2021 at 15:51

Without having read the paper, it looks like what they mean is that when you take a measurement, you are not getting $$I(x,y,\delta)$$ with infinite precision in $$\delta$$, rather you choose some fixed phase parameter $$\Delta$$ (either fixed by you or your equipment) and sample across the range $$\delta-\Delta/2$$ to $$\delta+\Delta/2$$. In this sense, $$\Delta$$ is a free parameter.
The signal you get out is then given by the integral in your Eq. (2), rather than just $$I(x,y,\delta)$$.
The intuition behind this is that sampling over the range given by $$\Delta$$ can have an effect on the signal you get out, so it is better to have a theoretical expression which takes it into account. The appearance of the sinc function is just a consequence of evaluating the integral.
A better way of understanding $$\Delta$$ could also be to reverse the experiment: If you assume that $$\Delta$$ is a parameter representing the accuracy of your equipment, you can measure the signal and fit it to your Eq. (2) to find $$\Delta$$.