How can i calculate network load of this camera? - compression

It's part of an assignment that i am trying to solve
Determine the network load from a security camera given the
following camera specs:
a. Resolution: 1200x800
b. Color: 8-bit color gamut
c. Speed: 30 frames pr. second.
d. Compression: 3/1000
How would you go about it? Is it a matter of calculating total bits pr frame?

Based on the information it is not so complicated to calculate:
To get the total number of pixels: 1200*800=960000
Then its unclear about the bits depth if its total number of colours
or per channel (RGB). I assume second (so 3 bytes): 960000*3=2880000
For second we need to multiply by 30 (frames per second)
86400000*30=86400000
To get in consideration compression we multiply by 3 and delete to
1000: 86400000*3/1000=259200 bytes per second
And 259200/1024=253.12 kilobytes per second

Related

Representing a Roughness texture with a 16 to one compression ratio using functions or Compression losslessly

I have a 4 by 4 texture of bits black and white texture and I want to use a lossless compression or Function to represent the Bits with less values accurately at least 16 to 1
Bit orientation
F[100][120]=1 function
That means compressed or Equation it must average 4bits
Here is an example function that takes low Additional bits due to constants which each take 1 bit and do the whole texture. It need to be for 60 bitwise Ops piecewise and max 16 functions for the whole texture(4bit). It must do the same.
for (q=0;q<122880;q++){
screen[122879-q]=((127-(((q>>10)%128)+Math.abs(((q%256)%128)-64)>>1))>>4)
}
for (q=0;q<122880;q++){
screen[q]=(q>>8)%128+Math.abs(((q%256)%128)-64)>>1
}
Good Bumpy
for (q=0;q<122880;q++){
screen[q]=(((q>>8)%25+Math.abs(((q%256)%128)-64)>>1)^2)
}
This is an attempt for a bumpy and terrain texture at 256*480

ITU-R 2k filter implementation

I have an array coming from a digitizer. I do an fft on it and then I calculate the frequency bins and apply a 20kHz low pass filter. The next step would be to apply an ITU-R 2k filter on this array and the filter behaves like the curve in the picture. I know I am supposed to do a multiplication one by one of the samples but I am not sure how to start with it. I know the 0 dB point is at 2 kHz and the maximum of 6 dB is located at 7 kHz. The implementation has to done in C++.
itu-r 468 filter behavior
An LTI filter like this is a straightforward multiplication in the frequency domain. Put the filter coefficients in an array of the same length, multiply the two: std::transform(std::begin(fftbins), std::end(fftbins), std::begin(filtercoeff), std::multiplies<std::complex<double>>()); and perform the IFFT.

Understanding SDL frame animation

I am working through the book SDL game development. In the first project, there is a bit of code meant to move the coords of the rendered frame of a sprite sheet:
void Game::update()
{
m_sourceRectangle.x = 128 * int((SDL_GetTicks()/100)%6);
}
I am having trouble understanding this... I know that it moves m_sourceRectangle 128 pixels along the x axis every 100 ms... but how does it actually work? Can somebody breakdown each element of this code to help me understand?
I don't understand why SDL_GetTicks() needs to be called to do this...
I also know that %6 is there because there are 6 frames in the animation... but how does it actually do that?
The book says:
Here we have used SDL_GetTicks() to find out the amount of milliseconds since SDL was initialized. We then divide this by the amount of time (in ms) we want between frames and then use the modulo operator to keep
it in range of the amount of frames we have in our animation. This code will (every 100 milliseconds) shift the x value of our source rectangle by
128 pixels (the width of a frame), multiplied by the current frame we want, giving us the correct position. Build the project and you should see the animation displayed.
But I am not sure I understand why getting the amount of milliseconds since SDL was initialized works.
The modulo operator takes the rest of a division. So for example if GetTicks() is 2600, first dividing by 100 makes it 26 and modulo 6 of 26 is 2. Therefore it's frame 2.
if GetTicks() is 3300; you divide by 100 and get 33; modulo 6 of 33 is 3; frame 3.
Each frame will be displayed for 100ms, so at T=0ms it's Frame 0, t=100ms it's Frame 100/100, at T=200ms it's Frame 200/100 and so on. So at T=SDL_GetTicks() ms, it's Frame SDL_GetTicks()/100. But than you only have 6 frames all together and cycling, therefore at T=SDL_GetTicks() ms it's in face Frame (SDL_GetTicks()/100) % 6.
There is an assumption here is that when the program start, Frame 0 is displayed, which may not be true because there are lots of things to do at starting which take time. But for simple demo to illustrate cycling of frames, it is good enough.
Hope this helps.

quantization of dct image for steganography

I hav a greyscale image. I did 8x8 blocks and computed each of their DCTs. I want to quantize the DCT coefficients and then replace their LSBs with my secret message bits. How exactly do I quantize the coefficients? Should I use the quantization matrix used by JPEG? How to determine the values of such a quantization matrix?
You will probably want to set the quality level to the highest (smallest values in the quantization matrix) so that the modified LSB of each coefficient perturbs the image data the least.
For encoding:
You will need access to the DCT values after quantization and before entropy coding. There you can modify the LSB's. You should probably only modify the non-zero coefficient values or you will make the compressed image file much larger and more distorted. This way, you will probably be able to encode 20-30 bits per DCT block.
For decoding:
You will need to do the reverse and get access to the DCT values immediately after the entropy decode and before the dequantization step.
To calculate the total number of bits available for your message, use the following example:
For a VGA sized image (640x480) which is encoded as 4:2:0 (subsampled color in both dimensions), you will have 40 x 30 = 1200 MCUs. Each MCU has 6 DCT blocks (4Y, 1Cr, 1Cb). This is a total of 7200 DCT blocks. If each block encodes an average of 25 coefficients (a reasonable quality level), then your message can be a total of 7200x25 = 180000 bits.

compact representation and delivery of point data

I have an array of point data, the values of points are represented as x co-ordinate and y co-ordinate.
These points could be in the range of 500 upto 2000 points or more.
The data represents a motion path which could range from the simple to very complex and can also have cusps in it.
Can I represent this data as one spline or a collection of splines or some other format with very tight compression.
I have tried representing them as a collection of beziers but at best I am getting a saving of 40 %.
For instance if I have an array of 500 points , that gives me 500 x and 500 y values so I have 1000 data pieces.
I around 100 quadratic beziers from this. each bezier is represented as controlx, controly, anchorx, anchory.
which gives me 100 x 4 = 400 pcs of data.
So input = 1000pcs , output = 400pcs.
I would like to further tighen this, any suggestions?
By its nature, spline is an approximation. You can reduce the number of splines you use to reach a higher compression ratio.
You can also achieve lossless compression by using some kind of encoding scheme. I am just making this up as I am typing, using the range example in previous answer (1000 for x and 400 for y),
Each point only needs 19 bits (10 for x, 9 for y). You can use 3 bytes to represent a coordinate.
Use 2 byte to represent displacement up to +/- 63.
Use 1 byte to represent short displacement up to +/- 7 for x, +/- 3 for y.
To decode the sequence properly, you would need some prefix to identify the type of encoding. Let's say we use 110 for full point, 10 for displacement and 0 for short displacement.
The bit layout will look like this,
Coordinates: 110xxxxxxxxxxxyyyyyyyyyy
Dislacement: 10xxxxxxxyyyyyyy
Short Displacement: 0xxxxyyy
Unless your sequence is totally random, you can easily achieve high compression ratio with this scheme.
Let's see how it works using a short example.
3 points: A(500, 400), B(550, 380), C(545, 381)
Let's say you were using 2 byte for each coordinate. It will take 16 bytes to encode this without compression.
To encode the sequence using the compression scheme,
A is first point so full coordinate will be used. 3 bytes.
B's displacement from A is (50, -20) and can be encoded as displacement. 2 bytes.
C's displacement from B is (-5, 1) and it fits the range of short displacement 1 byte.
So you save 10 bytes out of 16 bytes. Real compression ratio is totally depending on the data pattern. It works best on points forming a moving path. If the points are random, only 25% saving can be achieved.
If for example you use 32-bit integers for point coords and there is range limit, like x: 0..1000, y:0..400, you can pack (x, y) into a single 32-bit variable.
That way you achieve another 50% compression.
You could do a frequency analysis of the numbers you are trying to encode and use varying bit lengths to represent them, of course here I am vaguely describing Huffman coding
Firstly, only keep enough decimal points in your data that you actually need. Removing these would reduce your accuracy, but its a calculated loss. To do that, try converting your number to a string, locating the dot's position, and cutting of those many characters from the end. That could process faster than math, IMO. Lastly you can convert it back to a number.
150.234636746 -> "150.234636746" -> "150.23" -> 150.23
Secondly, try storing your data relative to the last number ("relative values"). Basically subtract the last number from this one. Then later to "decompress" it you can keep an accumulator variable and add them up.
A A A A R R
150, 200, 250 -> 150, 50, 50