Difference between a stop and a step in OpenEXR? - graphics

What is the difference between a stop and a step in OpenEXR? Steps seem like sub-units of stops but beyond that I;m not sure of their relation.

From the Technical Introduction:
The range of normalized 16-bit floating-point numbers can represent
thirty stops of information with 1024 steps per stop. We have eighteen
and a half stops above middle gray, and eleven and a half below.
Denormalized numbers provide an additional ten stops at the low end,
with gradually decreasing precision.

Related

To how many decimal places is bc accurate?

It is possible to print to several hundred decimal places a square root in bc, as it is in C. However in C it is only accurate to 15. I have checked the square root of 2 to 50 decimal places and it is accurate but what is the limit in bc? I can't find any reference to this.
To how many decimal places is bc accurate?
bc is an arbitrary precision calculator. Arbitrary precision just tells us how many digits it can represent (as many as will fit in memory), but doesn't tell us anything about accuracy.
However in C it is only accurate to 15
C uses your processor's built-in floating point hardware. This is fast, but has a fixed number of bits to represent each number, so is obviously fixed rather than arbitrary precision.
Any arbitrary precision system will have more ... precision than this, but could of course still be inaccurate. Knowing how many digits can be stored doesn't tell us whether they're correct.
However, the GNU implementation of bc is open source, so we can just see what it does.
The bc_sqrt function uses an iterative approximation (Newton's method, although the same technique was apparently known by the Babylonians in at least 1,000BC).
This approximation is just run, improving each time, until two consecutive guesses differ by less than the precision requested. That is, if you ask for 1,000 digits, it'll keep going until the difference is at most in the 1,001st digit.
The only exception is when you ask for an N-digit result and the original number has more than N digits. It'll use the larger of the two as its target precision.
Since the convergence rate of this algorithm is faster than one digit per iteration, there seems little risk of two consecutive iterations agreeing to some N digits without also being correct to N digits.

Segmenting License Plate Characters #ImageProcessing

For a university project I have to segment characters from a license plate using Python. This sounds reasonably simple. However, the thing is that we are not allowed to use any sophisticated library functions such as cv2.findContours(). The basics such as cv2.imread() cv2.resize() cv2.rectangle() are allowed.
I have written a function that localizes a license plate in an image and outputs a result as can be seen in the images Output 1 and Output 2 . These are binary images.
As one can see. Sometimes, the output of this function is relatively clean (Output 2). However, often it is also noisy (Output 1)
For a clean image (Output 2) I have tried finding the columns that contain less than x black pixels in order to segment the characters. However, this only works when the image is clean. This is often not the case. Changing the x parameter here does not make significant improvement.
Does anybody have suggestions on how I can approach this problem?
For an elementary solution, you can form a profile by counting the black pixels on all vertical lines. Then look for maximas and minimas of the average count in a sliding interval on this profile. The interval length should be a fraction of the expected width of a character. Only the extrema with sufficient contrast should be considered.
To avoid the effect of surrounding features in rotated plates, you can restrict the counting to just a slice of the image.
Once you have approximate vertical limits between the characters, you can repeat a similar processing to get the bottom and top limits of the characters (the sliding interval is no more necessary).
Finally, you can refine the boxing by finding the horizontal limits in the rectangles so formed.

Numerical Integration

Generally speaking when you are numerically evaluating and integral, say in MATLAB do I just pick a large number for the bounds or is there a way to tell MATLAB to "take the limit?"
I am assuming that you just use the large number because different machines would be able to handle numbers of different magnitudes.
I am just wondering if their is a way to improve my code. I am doing lots of expected value calculations via Monte Carlo and often use the trapezoid method to check my self of my degrees of freedom are small enough.
Strictly speaking, it's impossible to evaluate a numerical integral out to infinity. In most cases, if the integral in question is finite, you can simply integrate over a reasonably large range. To converge at a stable value, the integral of the normal error has to be less than 10 sigma -- this value is, for better or worse, as equal as you are going to get to evaluating the same integral all the way out to infinity.
It depends very much on what type of function you want to integrate. If it is "smooth" (no jumps - preferably not in any derivatives either, but that becomes progressively less important) and finite, that you have two main choices (limiting myself to the simplest approach):
1. if it is periodic, here meaning: could you put the left and right ends together and the also there have no jumps in value (and derivatives...): distribute your points evenly over the interval and just sample the functionvalues to get the estimated average, and than multiply by the length of the interval to get your integral.
2. if not periodic: use Legendre-integration.
Monte-carlo is almost invariably a poor method: it progresses very slow towards (machine-)precision: for any additional significant digit you need to apply 100 times more points!
The two methods above, for periodic and non-periodic "nice" (smooth etcetera) functions gives fair results already with a very small number of sample-points and then progresses very rapidly towards more precision: 1 of 2 points more usually adds several digits to your precision! This far outweighs the burden that you have to throw away all parts of the previous result when you want to apply a next effort with more sample points: you REPLACE the previous set of points with a fresh new one, while in Monte-Carlo you can just simply add points to the existing set and so refine the outcome.

8 bit audio samples to 16 bit

This is my "weekend" hobby problem.
I have some well-loved single-cycle waveforms from the ROMs of a classic synthesizer.
These are 8-bit samples (256 possible values).
Because they are only 8 bits, the noise floor is pretty high. This is due to quantization error. Quantization error is pretty weird. It messes up all frequencies a bit.
I'd like to take these cycles and make "clean" 16-bit versions of them. (Yes, I know people love the dirty versions, so I'll let the user interpolate between dirty and clean to whatever degree they like.)
It sounds impossible, right, because I've lost the low 8 bits forever, right? But this has been in the back of my head for a while, and I'm pretty sure I can do it.
Remember that these are single-cycle waveforms that just get repeated over and over for playback, so this is a special case. (Of course, the synth does all kinds of things to make the sound interesting, including envelopes, modulations, filters cross-fading, etc.)
For each individual byte sample, what I really know is that it's one of 256 values in the 16-bit version. (Imagine the reverse process, where the 16-bit value is truncated or rounded to 8 bits.)
My evaluation function is trying to get the minimum noise floor. I should be able to judge that with one or more FFTs.
Exhaustive testing would probably take forever, so I could take a lower-resolution first pass. Or do I just randomly push randomly chosen values around (within the known values that would keep the same 8-bit version) and do the evaluation and keep the cleaner version? Or is there something faster I can do? Am I in danger of falling into local minimums when there might be some better minimums elsewhere in the search space? I've had that happen in other similar situations.
Are there any initial guesses I can make, maybe by looking at neighboring values?
Edit: Several people have pointed out that the problem is easier if I remove the requirement that the new waveform would sample to the original. That's true. In fact, if I'm just looking for cleaner sounds, the solution is trivial.
You could put your existing 8-bit sample into the high-order byte of your new 16-bit sample, and then use the low order byte to linear interpolate some new 16 bit datapoints between each original 8-bit sample.
This would essentially connect a 16 bit straight line between each of your original 8-bit samples, using several new samples. It would sound much quieter than what you have now, which is a sudden, 8-bit jump between the two original samples.
You could also try apply some low-pass filtering.
Going with the approach in your question, I would suggest looking into hill-climbing algorithms and the like.
http://en.wikipedia.org/wiki/Hill_climbing
has more information on it and the sidebox has links to other algorithms which may be more suitable.
AI is like alchemy - we never reached the final goal, but lots of good stuff came out along the way.
Well, I would expect some FIR filtering (IIR if you really need processing cycles, but FIR can give better results without instability) to clean up the noise. You would have to play with it to get the effect you want but the basic problem is smoothing out the sharp edges in the audio created by sampling it at 8 bit resolutions. I would give a wide birth to the center frequency of the audio and do a low pass filter, and then listen to make sure I didn't make it sound "flat" with the filter I picked.
It's tough though, there is only so much you can do, the lower 8 bits is lost, the best you can do is approximate it.
It's almost impossible to get rid of noise that looks like your signal. If you start tweeking stuff in your frequency band it will take out the signal of interest.
For upsampling, since you're already using an FFT, you can add zeros to the end of the frequency domain signal and do an inverse FFT. This completely preserves the frequecy and phase information of the original signal, although it spreads the same energy over more samples. If you shift it 8bits to be a 16bit samples first, this won't be a too much of a problem. But I usually kick it up by an integer gain factor before doing the transform.
Pete
Edit:
The comments are getting a little long so I'll move some to the answer.
The peaks in the FFT output are harmonic spikes caused by the quantitization. I tend to think of them differently than the noise floor. You can dither as someone mentioned and eliminate the amplitude of the harmonic spikes and flatten out the noise floor, but you loose over all signal to noise on the flat part of your noise floor. As far as the FFT is concerned. When you interpolate using that method, it retains the same energy and spreads over more samples, this reduces the amplitude. So before doing the inverse, give your signal more energy by multipling by a gain factor.
Are the signals simple/complex sinusoids, or do they have hard edges? i.e. Triangle, square waves, etc. I'm assuming they have continuity from cycle to cycle, is that valid? If so you can also increase your FFT resolution to more precisely pinpoint frequencies by increasing the number of waveform cycles fed to your FFT. If you can precisely identify the frequencies use, assuming they are somewhat discrete, you may be able to completely recreate the intended signal.
The 16-bit to 8-bit via truncation requirement will produce results that do not match the original source. (Thus making finding an optimal answer more difficult.) Typically you would produce a fixed point waveform by attempting to "get the closest match" that means rounding to the nearest number (trunking is a floor operation). That is most likely how they were originally generated. Adding 0.5 (in this case 0.5 is 128) and then trunking the output would allow you to generate more accurate results. If that's not a worry then ok, but it definitely will have a negative effect on accuracy.
UPDATED:
Why? Because the goal of sampling a signal is to be able to as close a possible reproduce the signal. If conversion threshold is set poorly on the sampling all you're error is to one side of signal and not well distributed and centered about zero. On such systems you typically try to maximize the use the availiable dynamic range, particularly if you have low resolution such as an 8-bit ADC.
Band limited versions? If they are filtered at different frequencies, I'd suspect it was to allow you to play the same sound with out distortions when you went too far out from the other variation. Kinda like mipmapping in graphics.
I suspect the two are the same signal with different aliasing filters applied, this may be useful in reproducing the original. They should be the same base signal with different convolutions applied.
There might be a simple approach taking advantange of the periodicity of the waveforms. How about if you:
Make a 16-bit waveform where the high bytes are the waveform and the low bytes are zero - call it x[n].
Calculate the discrete Fourier transform of x[n] = X[w].
Make a signal Y[w] = (dBMag(X[w]) > Threshold) ? X[w] : 0, where dBMag(k) = 10*log10(real(k)^2 + imag(k)^2), and Threshold is maybe 40 dB, based on 8 bits being roughly 48 dB dynamic range, and allowing ~1.5 bits of noise.
Inverse transform Y[w] to get y[n], your new 16 bit waveform.
If y[n] doesn't sound nice, dither it with some very low level noise.
Notes:
A. This technique only works in the original waveforms are exactly periodic!
B. Step 5 might be replaced with setting the "0" values to random noise in Y[w] in step 3, you'd have to experiment a bit to see what works better.
This seems easier (to me at least) than an optimization approach. But truncated y[n] will probably not be equal to your original waveforms. I'm not sure how important that constraint is. I feel like this approach will generate waveforms that sound good.

Identifying common periodic waveforms (square, sine, sawtooth, ...)

Without any user interaction, how would a program identify what type of waveform is present in a recording from an ADC?
For the sake of this question: triangle, square, sine, half-sine, or sawtooth waves of constant frequency. Level and frequency are arbitrary, and they will have noise, small amounts of distortion, and other imperfections.
I'll propose a few (naive) ideas, too, and you can vote them up or down.
You definitely want to start by taking an autocorrelation to find the fundamental.
With that, take one period (approximately) of the waveform.
Now take a DFT of that signal, and immediately compensate for the phase shift of the first bin (the first bin being the fundamental, your task will be simpler if all phases are relative).
Now normalise all the bins so that the fundamental has unity gain.
Now compare and contrast the rest of the bins (representing the harmonics) against a set of pre-stored waveshapes that you're interested in testing for. Accept the closest, and reject overall if it fails to meet some threshold for accuracy determined by measurements of the noisefloor.
Do an FFT, find the odd and even harmonic peaks, and compare the rate at which they decrease to a library of common waveform.. peak... ratios.
Perform an autocorrelation to find the fundamental frequency, measure the RMS level, find the first zero-crossing, and then try subtracting common waveforms at that frequency, phase, and level. Whichever cancels out the best (and more than some threshold) wins.
This answer presumes no noise and that this is a simple academic exercise.
In the time domain, take the sample by sample difference of the waveform. Histogram the results. If the distribution has a sharply defined peak (mode) at zero, it is a square wave. If the distribution has a sharply defined peak at a positive value, it is a sawtooth. If the distribution has two sharply defined peaks, one negative and one positive,it is a triangle. If the distribution is broad and is peaked at either side, it is a sine wave.
arm yourself with more information...
I am assuming that you already know that a theoretically perfect sine wave has no harmonic partials (ie only a fundamental)... but since you are going through an ADC you can throw the idea of a theoretically perfect sine wave out the window... you have to fight against aliasing and determining what are "real" partials and what are artifacts... good luck.
the following information comes from this link about csound.
(*) A sawtooth wave contains (theoretically) an infinite number of harmonic partials, each in the ratio of the reciprocal of the partial number. Thus, the fundamental (1) has an amplitude of 1, the second partial 1/2, the third 1/3, and the nth 1/n.
(**) A square wave contains (theoretically) an infinite number of harmonic partials, but only odd-numbered harmonics (1,3,5,7,...) The amplitudes are in the ratio of the reciprocal of the partial number, just as sawtooth waves. Thus, the fundamental (1) has an amplitude of 1, the third partial 1/3, the fifth 1/5, and the nth 1/n.
I think that all of these answers so far are quite bad (including my own previous...)
after having thought the problem through a bit more I would suggest the following:
1) take a 1 second sample of the input signal (doesn't need to be so big, but it simplifies a few things)
2) over the entire second, count the zero-crossings. at this point you have the cps (cycles per second) and know the frequency of the oscillator. (in case that's something you wanted to know)
3) now take a smaller segment of the sample to work with: take precisely 7 zero-crossings worth. (so your work buffer should now, if visualized, look like one of the graphical representations you posted with the original question.) use this small work buffer to perform the following tests. (normalizing the work buffer at this point could make life easier)
4) test for square-wave: zero crossings for a square wave are always very large differences, look for a large signal delta followed by little to no movement until the next zero crossing.
5) test for saw-wave: similar to square-wave, but a large signal delta will be followed by a linear constant signal delta.
6) test for triangle-wave: linear constant (small) signal deltas. find the peaks, divide by the distance between them and calculate what the triangle wave should look like (ideally) now test the actual signal for deviance. set a deviance tolerance threshold and you can determine whether you are looking at a triangle or a sine (or something parabolic).
First find the base frequency and the phase. You can do that with FFT. Normalize the sample. Then subtract each sample with the sample of the waveform you want to test (same frequency and same phase). Square the result add it all up and divide it by the number of samples. The smallest number is the waveform you seek.

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