I'm struggling to find any win32 api call that can correctly identify the bit depth of a connected monitor in terms of bits per channel. The attached screenshot shows the setting visible.
With DXGI it is possible to enumerate all modes on an output and identify their bit depth easily, however there is no call to get the current mode.
All current win32 calls which return bitsPerPel always indicate 32bpp.
Does anybody know which call can get this data? I can detect this with NvAPI and ADL however have no solution for intel based machines.
I would additionally be interested in the active signal resolution vs desktop resolution field and how to retrieve these.
DXGI_OUTPUT_DESC1 structure gets you that:
BitsPerColor
Type: UINT
The number of bits per color channel for the active wire format of the display attached to this output.
…
The DXGI_OUTPUT_DESC1 structure is initialized by the IDXGIOutput6::GetDesc1 method.
Just an example of what sort of data you get there:
--
Adapter: Radeon RX 570 Series
Output: \.\DISPLAY4
Desktop Coordinates: (0, 0) - (3840, 2160); 3840 x 2160
Attached To Desktop: 1
Rotation: DXGI_MODE_ROTATION_IDENTITY
Monitor: 0x000100AF
Physical Monitors: LG ULTRA HD(DisplayPort) (0x00000000)
Bits Per Color: 10 <<----------------------------------
Color Space: DXGI_COLOR_SPACE_RGB_FULL_G22_NONE_P709
Primaries, White Point: R { 0.652, 0.335 }, G { 0.305, 0.637 }, B { 0.148, 0.062 }; { 0.313, 0.329 }
Luminance: Min 0.500, Max 270.000, Max Full Frame 270.000
Hardware Composition Support: DXGI_HARDWARE_COMPOSITION_SUPPORT_FLAG_FULLSCREEN | DXGI_HARDWARE_COMPOSITION_SUPPORT_FLAG_CURSOR_STRETCHED
In my opinion, the Bit depth in advanced display settings actually means bit depth for each color, it's just a naming confusion. Standard 32 bit color is 8 bits each for red, green, blue, the remaining 8 bits are either used for transparency or are just filled with zeros. That's why you always get 32 bits for each pixel.
Related
How can I reduce the number of bits from 24 bits to a number between 0 and 8 bits and distribute the bits for the three colors Red, Green and Blue
Any idea ?
This is called "Color Quantization". You have 16.777.216 colors and you want to map them to a smaller number (2 to 256).
Step 1: choose the colors you want to use. First their number, then the colors themselves. You need to chose if the colors are fixed for all images, or if they change based on the image (you will need to ship a palette of colors with every image).
Step 2: substitute the colors of your image with those in the selection.
If the colors are fixed and you want to stay very simple you can use 1 bit per channel (8 colors in total) or 2 bits per channel (64 colors in total).
Slightly more complex, use these values for each channel 0, 51, 102, 153, 204, 255, in any possible way, leading to 216 different color combinations. Make a table which associates every color combination with an index. That index requires 8 bits (with some spare). This is called a "web safe palette" (this brings me back to the late 1999). Now you are ready for substitution: take every pixel in your image and the quantized color can be found as x*6//256*51 (// is integer division).
If you want a better looking palette, look for the Median cut algorithm.
Keep only the most significant bits of the pixel's red channel. Do likewise for the green and blue channel. Then use C++'s bit manipulation operations to move those bits values into a single byte. There are multiple ways of doing so. For example, do an Internet search for "rgb332" for one example (where you keep 3 red bits, 3 green bits, and 2 blue bits).
I'm attempting to convert 12-bit RGGB color values into 8-bit RGGB color values, but with my current method it gives strange results.
Logically, I thought that simply dividing the 12-bit RGGB into 8-bit RGGB would work and be pretty simple:
// raw_color_array contains R,G1,G2,B in a bayer pattern with each element
// ranging from 0 to 4096
for(int i = 0; i < array_size; i++)
{
raw_color_array[i] /= 16; // 4096 becomes 256 and so on
}
However, in practice this actually does not work. Given, for example, a small image with water and a piece of ice in it you can see what actually happens in the conversion (right most image).
Why does this happen? and how can I get the same (or close to) image on the left, but as 8-bit values instead? Thanks!
EDIT: going off of #MSalters answer, I get a better quality image but the colors are still drasticaly skewed. What resources can I look into for converting 12-bit data to 8-bit data without a steep loss in quality?
It appears that your raw 12 bits data isn't on a linear scale. That is quite common for images. For a non-linear scale, you can't use a linear transformation like dividing by 16.
A non-linear transform like sqrt(x*16) would also give you an 8 bits value. So would std::pow(x, 12.0/8.0)
A known problem with low-gradient images is that you get banding. If your images has an area where the original value varies from say 100 to 200, the 12-to-8 bit reduction will shrink that to less than 100 different values. You get rounding , and with naive (local) rounding you get bands. Linear or non-linear, there will then be some inputs x that all map to y, and some that map to y+1. This can be mitigated by doing the transformation in floating point, and then adding a random value between -1.0 and +1.0 before rounding. This effectively breaks up the band structure.
After you clarified that this 12bit data is only for one color, here is my simple answer:
Since you want to convert its value to its 8 bit equivalent, it obviously means you lost some of the data (4bits). This is the reason why you are not getting the same output.
After clarification:
If you want to retain the actual colour values!
Apply de-mosaicking in the 12 Bit image and then scale the resultant data to 8 - Bit. So that the colour loss due to de-mosaicking will be less compared to the previous approach.
You say that your 12-bits represent 2^12 bits of one colour. That is incorrect. There are reds, greens and blues in your image. Look at the histogram. I made this with ImageMagick at the command line:
convert cells.jpg histogram:png:h.png
If you want 8-bits per pixel, rather than trying to blindly/statically apportion 3 bits to Green, 2 bits to Red and 3 bits to Blue, you would probably be better off going with an 8-bit palette so you can have 250+ colours of all variations rather than restricting yourself to just 8 blue shades, 4 reds an 8 green. So, like this:
convert cells.jpg -colors 254 PNG8:result.png
Here is the result of that beside the original:
The process above is called "quantisation" and if you want to implement it in C/C++, there is a writeup here.
I've been under the assumption that my gamma correction pipeline should be as follows:
Use sRGB format for all textures loaded in (GL_SRGB8_ALPHA8) as all art programs pre-gamma correct their files. When sampling from a GL_SRGB8_ALPHA8 texture in a shader OpenGL will automatically convert to linear space.
Do all lighting calculations, post processing, etc. in linear space.
Convert back to sRGB space when writing final color that will be displayed on the screen.
Note that in my case the final color write involves me writing from a FBO (which is a linear RGB texture) to the back buffer.
My assumption has been challenged as if I gamma correct in the final stage my colors are brighter than they should be. I set up for a solid color to be drawn by my lights of value { 255, 106, 0 }, but when I render I get { 255, 171, 0 } (as determined by print-screening and color picking). Instead of orange I get yellow. If I don't gamma correct at the final step I get exactly the right value of { 255, 106, 0 }.
According to some resources modern LCD screens mimic CRT gamma. Do they always? If not, how can I tell if I should gamma correct? Am I going wrong somewhere else?
Edit 1
I've now noticed that even though the color I write with the light is correct, places where I use colors from textures are not correct (but rather far darker as I would expect without gamma correction). I don't know where this disparity is coming from.
Edit 2
After trying GL_RGBA8 for my textures instead of GL_SRGB8_ALPHA8, everything looks perfect, even when using the texture values in lighting computations (if I half the intensity of the light, the output color values are halfed).
My code is no longer taking gamma correction into account anywhere, and my output looks correct.
This confuses me even more, is gamma correction no longer needed/used?
Edit 3 - In response to datenwolf's answer
After some more experimenting I'm confused on a couple points here.
1 - Most image formats are stored non-linearly (in sRGB space)
I've loaded a few images (in my case both .png and .bmp images) and examined the raw binary data. It appears to me as though the images are actually in the RGB color space, as if I compare the values of pixels with an image editing program with the byte array I get in my program they match up perfectly. Since my image editor is giving me RGB values, this would indicate the image stored in RGB.
I'm using stb_image.h/.c to load my images and followed it all the way through loading a .png and did not see anywhere that it gamma corrected the image while loading. I also examined the .bmps in a hex editor and the values on disk matched up for them.
If these images are actually stored on disk in linear RGB space, how am I supposed to (programatically) know when to specify an image is in sRGB space? Is there some way to query for this that a more featured image loader might provide? Or is it up to the image creators to save their image as gamma corrected (or not) - meaning establishing a convention and following it for a given project. I've asked a couple artists and neither of them knew what gamma correction is.
If I specify my images are sRGB, they are too dark unless I gamma correct in the end (which would be understandable if the monitor output using sRGB, but see point #2).
2 - "On most computers the effective scanout LUT is linear! What does this mean though?"
I'm not sure I can find where this thought is finished in your response.
From what I can tell, having experimented, all monitors I've tested on output linear values. If I draw a full screen quad and color it with a hard-coded value in a shader with no gamma correction the monitor displays the correct value that I specified.
What the sentence I quoted above from your answer and my results would lead me to believe is that modern monitors output linear values (i.e. do not emulate CRT gamma).
The target platform for our application is the PC. For this platform (excluding people with CRTs or really old monitors), would it be reasonable to do whatever your response to #1 is, then for #2 to not gamma correct (i.e. not perform the final RGB->sRGB transformation - either manually or using GL_FRAMEBUFFER_SRGB)?
If this is so, what are the platforms on which GL_FRAMEBUFFER_SRGB is meant for (or where it would be valid to use it today), or are monitors that use linear RGB really that new (given that GL_FRAMEBUFFER_SRGB was introduced 2008)?
--
I've talked to a few other graphics devs at my school and from the sounds of it, none of them have taken gamma correction into account and they have not noticed anything incorrect (some were not even aware of it). One dev in particular said that he got incorrect results when taking gamma into account so he then decided to not worry about gamma. I'm unsure what to do in my project for my target platform given the conflicting information I'm getting online/seeing with my project.
Edit 4 - In response to datenwolf's updated answer
Yes, indeed. If somewhere in the signal chain a nonlinear transform is applied, but all the pixel values go unmodified from the image to the display, then that nonlinearity has already been pre-applied on the image's pixel values. Which means, that the image is already in a nonlinear color space.
Your response would make sense to me if I was examining the image on my display. To be sure I was clear, when I said I was examining the byte array for the image I mean I was examining the numerical value in memory for the texture, not the image output on the screen (which I did do for point #2). To me the only way I could see what you're saying to be true then is if the image editor was giving me values in sRGB space.
Also note that I did try examining the output on monitor, as well as modifying the texture color (for example, dividing by half or doubling it) and the output appeared correct (measured using the method I describe below).
How did you measure the signal response?
Unfortunately my methods of measurement are far cruder than yours. When I said I experimented on my monitors what I meant was that I output solid color full screen quad whose color was hard coded in a shader to a plain OpenGL framebuffer (which does not do any color space conversion when written to). When I output white, 75% gray, 50% gray, 25% gray and black the correct colors are displayed. Now here my interpretation of correct colors could most certainly be wrong. I take a screenshot and then use an image editing program to see what the values of the pixels are (as well as a visual appraisal to make sure the values make sense). If I understand correctly, if my monitors were non-linear I would need to perform a RGB->sRGB transformation before presenting them to the display device for them to be correct.
I'm not going to lie, I feel I'm getting a bit out of my depth here. I'm thinking the solution I might persue for my second point of confusion (the final RGB->sRGB transformation) will be a tweakable brightness setting and default it to what looks correct on my devices (no gamma correction).
First of all you must understand that the nonlinear mapping applied to the color channels is often more than just a simple power function. sRGB nonlinearity can be approximated by about x^2.4, but that's not really the real deal. Anyway your primary assumptions are more or less correct.
If your textures are stored in the more common image file formats, they will contain the values as they are presented to the graphics scanout. Now there are two common hardware scenarios:
The scanout interface outputs a linear signal and the display device will then internally apply a nonlinear mapping. Old CRT monitors were nonlinear due to their physics: The amplifiers could put only so much current into the electron beam, the phosphor saturating and so on – that's why the whole gamma thing was introduced in the first place, to model the nonlinearities of CRT displays.
Modern LCD and OLED displays either use resistor ladders in their driver amplifiers, or they have gamma ramp lookup tables in their image processors.
Some devices however are linear, and ask the image producing device to supply a proper matching LUT for the desired output color profile on the scanout.
On most computers the effective scanout LUT is linear! What does this mean though? A little detour:
For illustration I quickly hooked up my laptop's analogue display output (VGA connector) to my analogue oscilloscope: Blue channel onto scope channel 1, green channel to scope channel 2, external triggering on line synchronization signal (HSync). A quick and dirty OpenGL program, deliberately written with immediate mode was used to generate a linear color ramp:
#include <GL/glut.h>
void display()
{
GLuint win_width = glutGet(GLUT_WINDOW_WIDTH);
GLuint win_height = glutGet(GLUT_WINDOW_HEIGHT);
glViewport(0,0, win_width, win_height);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glOrtho(0, 1, 0, 1, -1, 1);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
glBegin(GL_QUAD_STRIP);
glColor3f(0., 0., 0.);
glVertex2f(0., 0.);
glVertex2f(0., 1.);
glColor3f(1., 1., 1.);
glVertex2f(1., 0.);
glVertex2f(1., 1.);
glEnd();
glutSwapBuffers();
}
int main(int argc, char *argv[])
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_RGBA | GLUT_DOUBLE);
glutCreateWindow("linear");
glutFullScreen();
glutDisplayFunc(display);
glutMainLoop();
return 0;
}
The graphics output was configured with the Modeline
"1440x900_60.00" 106.50 1440 1528 1672 1904 900 903 909 934 -HSync +VSync
(because that's the same mode the flat panel runs in, and I was using cloning mode)
gamma=2 LUT on the green channel.
linear (gamma=1) LUT on the blue channel
This is how the signals of a single scanout line look like (upper curve: Ch2 = green, lower curve: Ch1 = blue):
You can clearly see the x⟼x² and x⟼x mappings (parabola and linear shapes of the curves).
Now after this little detour we know, that the pixel values that go to the main framebuffer, go there as they are: The OpenGL linear ramp underwent no further changes and only when a nonlinear scanout LUT was applied it altered the signal sent to the display.
Either way the values you present to the scanout (which means the on-screen framebuffers) will undergo a nonlinear mapping at some point in the signal chain. And for all standard consumer devices this mapping will be according to the sRGB standard, because it's the smallest common factor (i.e. images represented in the sRGB color space can be reproduced on most output devices).
Since most programs, like webbrowsers assume the output to undergo a sRGB to display color space mapping, they simply copy the pixel values of the standard image file formats to the on-screen frame as they are, without performing a color space conversion, thereby implying that the color values within those images are in sRGB color space (or they will often merely convert to sRGB, if the image color profile is not sRGB); the correct thing to do (if, and only if the color values written to the framebuffer are scanned out to the display unaltered; assuming that scanout LUT is part of the display), would be conversion to the specified color profile the display expects.
But this implies, that the on-screen framebuffer itself is in sRGB color space (I don't want to split hairs about how idiotic that is, lets just accept this fact).
How to bring this together with OpenGL? First of all, OpenGL does all it's color operations linearly. However since the scanout is expected to be in some nonlinear color space, this means, that the end result of the rendering operations of OpenGL somehow must be brougt into the on-screen framebuffer color space.
This is where the ARB_framebuffer_sRGB extension (which went core with OpenGL-3) enters the picture, which introduced new flags used for the configuration of window pixelformats:
New Tokens
Accepted by the <attribList> parameter of glXChooseVisual, and by
the <attrib> parameter of glXGetConfig:
GLX_FRAMEBUFFER_SRGB_CAPABLE_ARB 0x20B2
Accepted by the <piAttributes> parameter of
wglGetPixelFormatAttribivEXT, wglGetPixelFormatAttribfvEXT, and
the <piAttribIList> and <pfAttribIList> of wglChoosePixelFormatEXT:
WGL_FRAMEBUFFER_SRGB_CAPABLE_ARB 0x20A9
Accepted by the <cap> parameter of Enable, Disable, and IsEnabled,
and by the <pname> parameter of GetBooleanv, GetIntegerv, GetFloatv,
and GetDoublev:
FRAMEBUFFER_SRGB 0x8DB9
So if you have a window configured with such a sRGB pixelformat and enable sRGB rasterization mode in OpenGL with glEnable(GL_FRAMEBUFFER_SRGB); the result of the linear colorspace rendering operations will be transformed in sRGB color space.
Another way would be to render everything into an off-screen FBO and to the color conversion in a postprocessing shader.
But that's only the output side of rendering signal chain. You also got input signals, in the form of textures. And those are usually images, with their pixel values stored nonlinearly. So before those can be used in linear image operations, such images must be brought into a linear color space first. Lets just ignore for the time being, that mapping nonlinear color spaces into linear color spaces opens several of cans of worms upon itself – which is why the sRGB color space is so ridiculously small, namely to avoid those problems.
So to address this an extension EXT_texture_sRGB was introduced, which turned out to be so vital, that it never went through being ARB, but went straight into the OpenGL specification itself: Behold the GL_SRGB… internal texture formats.
A texture loaded with this format undergoes a sRGB to linear RGB colorspace transformation, before being used to source samples. This gives linear pixel values, suitable for linear rendering operations, and the result can then be validly transformed to sRGB when going to the main on-screen framebuffer.
A personal note on the whole issue: Presenting images on the on-screen framebuffer in the target device color space IMHO is a huge design flaw. There's no way to do everything right in such a setup without going insane.
What one really wants is to have the on-screen framebuffer in a linear, contact color space; the natural choice would be CIEXYZ. Rendering operations would naturally take place in the same contact color space. Doing all graphics operations in contact color spaces, avoids the opening of the aforementioned cans-of-worms involved with trying to push a square peg named linear RGB through a nonlinear, round hole named sRGB.
And although I don't like the design of Weston/Wayland very much, at least it offers the opportunity to actually implement such a display system, by having the clients render and the compositor operate in contact color space and apply the output device's color profiles in a last postprocessing step.
The only drawback of contact color spaces is, that there it's imperative to use deep color (i.e. > 12 bits per color channel). In fact 8 bits are completely insufficient, even with nonlinear RGB (the nonlinearity helps a bit to cover up the lack of perceptible resolution).
Update
I've loaded a few images (in my case both .png and .bmp images) and examined the raw binary data. It appears to me as though the images are actually in the RGB color space, as if I compare the values of pixels with an image editing program with the byte array I get in my program they match up perfectly. Since my image editor is giving me RGB values, this would indicate the image stored in RGB.
Yes, indeed. If somewhere in the signal chain a nonlinear transform is applied, but all the pixel values go unmodified from the image to the display, then that nonlinearity has already been pre-applied on the image's pixel values. Which means, that the image is already in a nonlinear color space.
2 - "On most computers the effective scanout LUT is linear! What does this mean though?
I'm not sure I can find where this thought is finished in your response.
This thought is elaborated in the section that immediately follows, where I show how the values you put into a plain (OpenGL) framebuffer go directly to the monitor, unmodified. The idea of sRGB is "put the values into the images exactly as they are sent to the monitor and build consumer displays to follow that sRGB color space".
From what I can tell, having experimented, all monitors I've tested on output linear values.
How did you measure the signal response? Did you use a calibrated power meter or similar device to measure the light intensity emitted from the monitor in response to the signal? You can't trust your eyes with that, because like all our senses our eyes have a logarithmic signal response.
Update 2
To me the only way I could see what you're saying to be true then is if the image editor was giving me values in sRGB space.
That's indeed the case. Because color management was added to all the widespread graphics systems as an afterthought, most image editors edit pixel values in their destination color space. Note that one particular design parameter of sRGB was, that it should merely retroactively specify the unmanaged, direct value transfer color operations as they were (and mostly still are done) done on consumer devices. Since there happens no color management at all, the values contained in the images and manipulated in editors must be in sRGB already. This works for so long, as long images are not synthetically created in a linear rendering process; in case of the later the render system has to take into account the destination color space.
I take a screenshot and then use an image editing program to see what the values of the pixels are
Which gives you of course only the raw values in the scanout buffer without the gamma LUT and the display nonlinearity applied.
I wanted to give a simple explanation of what went wrong in the initial attempt, because although the accepted answer goes in-depth on colorspace theory, it doesn't really answer that.
The setup of the pipeline was exactly right: use GL_SRGB8_ALPHA8 for textures, GL_FRAMEBUFFER_SRGB (or custom shader code) to convert back to sRGB at the end, and all your intermediate calculations will be using linear light.
The last bit is where you ran into trouble. You wanted a light with a color of (255, 106, 0) - but that's an sRGB color, and you're working with linear light. To get the color you want, you need to convert that color to the linear space, the same way that GL_SRGB8_ALPHA8 is doing for your textures. For your case, this would be a vec3 light with intensity (1, .1441, 0) - this is the value after applying gamma-compression.
Is there any way to clamp out of range texture addresses to a certain value? In my case, I want them to be set to a simple zero, but the address mode I need doesn't seem to exist.
Thanks.
Edit: Any idea what the cudaAddressModeBorder setting does?
I don't think there's a way to specify the clamp but you can do the obvious and add a 1 pixel black (zero) border around the edge and offset your addressing by 1. It shouldn't be much more data and it'll get you the clamping for free.
If you have a maximum size 2D texture (for CUDA 2.x it is 64k x 64k) with 16 bytes per pixel (worst case) then you're looking at only 4 MB of extra data for the 1 pixel border which for a PCIe x16 card will take about 500 microseconds to copy to the card--hardly anything even in the worst case.
You can set the boundary mode to return zero when accessing to textures using Surface functions. I can not test it right now as you need a device of compute capability 2.0+ but you can check the reference in the NVIDIA CUDA C Programming Guide (version 3.2), Section B.9 p.114.
We can also clamp the boundary and trap it (make kernel fail) what is the default when using the surface memory.
Regards!
Update: This only seems to be a problem at some computers. The normal, intuitive code seems to work fine one my home computer, but the computer at work has trouble.
Home computer: (no problems)
Windows XP Professional SP3
AMD Athlon 64 X2 3800+ Dual Core 2.0 GHz
NVIDIA GeForce 7800 GT
2 GB RAM
Work computer: (this question applies to this computer)
Windows XP Professional SP3
Intel Pentium 4 2.8 Ghz (dual core, I think)
Intel 82945G Express Chipset Family
1 GB RAM
Original post:
I'm trying to apply a very simple texture to a part of the screen using Psychtoolbox in Matlab with the following code:
win = Screen('OpenWindow', 0, 127); % open window and obtain window pointer
tex = Screen('MakeTexture', win, [255 0;0 255]); % get texture pointer
% draw texture. Args: command, window pointer, texture pointer, source
% (i.e. the entire 2x2 matrix), destination (a 100x100 square), rotation
% (none) and filtering (nearest neighbour)
Screen('DrawTexture', win, tex, [0 0 2 2], [100 100 200 200], 0, 0);
Screen('Flip', win); % flip the buffer so the texture is drawn
KbWait; % wait for keystroke
Screen('Close', win); % close screen
Now I would expect to see this (four equally sized squares):
But instead I get this (right and bottom sides are cut off and top left square is too large):
Obviously the destination rectangle is a lot bigger than the source rectangle, so the texture needs to be magnified. I would expect this to happen symmetrically like in the first picture and this is also what I need. Why is this not happening and what can I do about it?
I have also tried using [128 0 1152 1024] as a destination rectangle (as it's the square in the center of my screen). In this case, all sides are 1024, which makes each involved rectangle a power of 2. This does not help.
Increasing the size of the checkerboard results in a similar situation where the right- and bottommost sides are not showed correctly.
Like I said, I use Psychtoolbox, but I know that it uses OpenGL under the hood. I don't know much about OpenGL either, but maybe someone who does can help without knowing Matlab. I don't know.
Thanks for your time!
While I don't know much (read: any) Matlab, I do know that textures are very picky in openGL. Last I checked, openGL requires texture files to be square and of a power of two (i.e. 128 x 128, 256 x 256, 512 x 512).
If they aren't, openGL is supposed to pad the file with appropriate white pixels where they're needed to meet this condition, although it could be a crapshoot depending on which system you are running it on.
I suggest making sure that your checkerboard texture fits these requirements.
Also, I can't quite make sure from your code posted, but openGL expects you to map the corners of your texture to the corners of the object you are intending to texture.
Another bit of advice, maybe try a linear filter instead of nearest neighbor. It's heavier computationally, but results in a better image. This probably won't matter in the end.
While this help is not Matlab specific, hope it is useful.
Without knowing a lot about the Psychtoolbox, but having dealt with graphics and user interfaces a lot in MATLAB, the first thing I would try would be to fiddle with the fourth input to Screen (the "source" input). Try shifting each corner by half-pixel and whole-pixel values. For example, the first thing I would try would be:
Screen('DrawTexture', win, tex, [0 0 2.5 2.5], [100 100 200 200], 0, 0);
And if that didn't seem to do anything, I would next try:
Screen('DrawTexture', win, tex, [0 0 3 3], [100 100 200 200], 0, 0);
My reasoning for this advice: I've noticed sometimes that images or GUI controls in my figures can appear to be off by a pixel, which I can only speculate is some kind of round-off error when scaling or positioning them.
That's the best advice I can give. Hope it helps!