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Some manufacturers of Digital Light Processing (DLP) projectors offers the possibility to purchase development kits with a SDK that allow to directly control the DMD chipset with a computer. The only one widely available is the Texas Instrument DLP Discovery 4100 Development Kit compatible with two different chipsets; - the DLP7000 with a resolution of 1024x768, a 0.7 inch chipset and a maximum refresh rate of 32.5 kHz - the DLP9500 with a resolution of 1920x1080, a 0.95 inch chipset and a maximum refresh rate of 23.1 kHz.

This technology based on micro-mirrors allows a binary amplitude modulation, meaning that the amplitude of one pixel can only be set to 0 or 1, with no control on the phase. To get phase modulation using an amplitude modulator, the idea is to filter an amplitude signal in the Fourier plane to obtain a complex field modulation. Let $\dpi{300}&space;\phi(x,y)$ be the phase hologram to display. The Lee method [W.H. Lee, Progress in Optics (1978)] consists in generating the amplitude pattern:

$\dpi{300}&space;f(x,y)&space;=&space;\frac{1}{2}\left[&space;1+cos(2\pi&space;(x-y)\nu_0&space;-\phi(x,y))\right]$

This function can also be written:

$\dpi{300}&space;f(x,y)&space;=&space;\frac{1}{2}&space;+&space;\frac{1}{4}e^{i&space;2\pi&space;(x-y)&space;\nu_0}&space;e^{-i&space;\phi(x,y)}&space;+&space;\frac{1}{4}e^{i&space;2\pi&space;(y-x)&space;\nu_0}&space;e^{i&space;\phi(x,y)}$

The three terms of this expression have respective center spatial frequencies $\dpi{300}&space;(\nu_x=0,\nu_y=0)$$\dpi{300}&space;(\nu_x=\nu_0,\nu_y=-\nu_0)$ and $\dpi{300}&space;(\nu_x=-\nu_0,\nu_y=\nu_0)$. If $\inline&space;\dpi{300}&space;\nu_0$ is higher than the highest spatial frequency of $\dpi{300}&space;\phi(x,y)$, this three terms correspond to three distinct orders of diffraction, that we call 0, +1 and -1. We see that the last component of the previous expression corresponds to the desired phase modulation tilted by an angle corresponding to the diffraction. To select only this term, one can use a 4-f system with an iris or a pinhole in the Fourier plane [see Fig. 1.].

Figure 1. Schematic of the setup.

The resulting field after the 4-f system is proportional to $\dpi{300}&space;e^{i\phi(x,y)}$.

Since DLP devices can only perform a binary amplitude modulation, we generate a binary amplitude hologram by thresholding the previous amplitude hologram:

$\dpi{300}&space;g(x,y)=\left\{&space;\begin{array}{l}&space;1&space;\,\,\text{if}&space;\,\,f(x,y)>1/2&space;\\&space;0&space;\,\,\text{otherwise}&space;\end{array}&space;\right.$

We show in Fig. 2 an example of binary hologram used to generate a phase pattern.

Figure 2. Example of a desired phase pattern (a.) and the corresponding binary Lee hologram wich encodes the same phase distribution. Image from [D.B. Conkey et al., Opt. Express (2012)].

The phase of the desired field is encoded in the phase shift of the cosinus function in $\dpi{300}&space;f(x,y)$. For a given spatial sampling, the approximation $\dpi{300}&space;g(x,y)$ for the binary amplitude modulation is less efficient to render the desired phase pattern since the phase shift is partially blurred by the square shape of the signal. To obtain a correct approximation of the phase profile, it is necessary to have a sufficient number of oscillations inside one macro-pixel. This requires a great number of pixels per macro-pixel, which drastically decreases the resolution of the system.

The three parameters to carefully tune to obtain a good trade-off between resolution and fidelity of the generated phase pattern are the carrier spatial frequency $\inline&space;\dpi{300}&space;\nu_0$, the spatial sampling (how many pixels per macro-pixel) and the diameter of the iris or pinhole (that defines the cutoff of the spatial frequency filtering).

#3 SANJIV 2018-07-02 16:42
Quoting Sébastien Popoff:
Quoting SANJIV:
Hi Sebastsin
in the paper D.B. Conkey et al., Opt. Express (2012)]. They have taken the -1st order instead of 1st order as in your Tutorial. Shouldn't it be the -1st order as the Signal is in the -1st order ?

Sanjiv

Hi Sanjiv,

The only difference between +1 and -1 order is that the phase is conjugated. So, yes, if you defined the mask as in the first equation, the right phase is in -1 order, I have the opposite of this phase in the +1 order.

Best,

Sebastien

thanks Sebastin!
I have one more doubt, the second lens in the 4f setup should be placed off axis or in the same axis of the first lens. ?

#2 Sébastien Popoff 2018-06-29 21:10
Quoting SANJIV:
Hi Sebastsin
in the paper D.B. Conkey et al., Opt. Express (2012)]. They have taken the -1st order instead of 1st order as in your Tutorial. Shouldn't it be the -1st order as the Signal is in the -1st order ?

Sanjiv

Hi Sanjiv,

The only difference between +1 and -1 order is that the phase is conjugated. So, yes, if you defined the mask as in the first equation, the right phase is in -1 order, I have the opposite of this phase in the +1 order.

Best,

Sebastien

#1 SANJIV 2018-06-28 11:49
Hi Sebastsin
in the paper D.B. Conkey et al., Opt. Express (2012)]. They have taken the -1st order instead of 1st order as in your Tutorial. Shouldn't it be the -1st order as the Signal is in the -1st order ?

Sanjiv