I am currently trying to decode OpenLR binary Strings (version 3) as specified in the OpenLR Whitepaper (version 1.5, rev 2) which can be found at the OpenLR Association website.
Unfortunately, while comparing the results with the TomTom Demo Tool.
I had to find some key differences, mistakes, and missing information.
For example, decoding this binary String with the Demo Tool (enable show decoding steps)
Cwl3syVRnAELHgor/6YBBw0DgP61AQwnDGz94AEIGQe1/j4BCj0TSv9NAXYZJw==
shows that:
Negative byte values for relative coordinates have to be flipped and 1 added to obtain the actual value. The OpenLR Whitepaper is missing this step.
is this step also necessary for absolute coordinates or only for relative coordinates?
The calculation of relative coordinates is described to be the same calculation for absolute coordinates
(int - (sgn(int) *0.5)) * 360 / 2^resolution
(resolution = 24 for absolute, 16 for relative), but it gets obvious that this equation does not lead to the correct value.
The calculation would lead to a value of -0.49 not -0.0009 using the values and formula as shown. Instead, for relative coords, the (eventually flipped and 1 added) concatenated byte value has to be divided by 10^5 to obtain -0.0009.
For Location Reference Points n>0,n<2 (the second LRP, thus first relative coordinate) the final addition of the values is somehow correct, but for n >=2 the differences get bigger, and thus the result is wrong, it can be easily seen in the demo tools calculation:
the addition of the correct decoded byte values is simply wrong:
This leads to big differences in the final location. The resulting value of the demo tool is correct, as the locations described follow the streets while using the correct sum would shift them off the street. But the equation is missing some key aspects.
Also, the OpenLR Whitepaper describes adding the relative coordinate value to the previous LRP.
(comparing the values used in the demo tool shows that the first LRP is beeing used instead of the previous LRP)
Which formula is the correct one? The Demo Tool generated correct values but uses wrong calculations.
Edit, for the third LRP I found using the previous LRP leads to the value calculated by the online tool (which shows the first LRP value to be used).
For reference and comparison some examples:
Binary String of above example:
Cwl3syVRnAELHgor/6YBBw0DgP61AQwnDGz94AEIGQe1/j4BCj0TSv9NAXYZJw==
Differences:
Using the correct sum of relative coord value and LRP 0 (first 2 LRPs are correct, then it gets worse, can also be checked by verifying the sum of the demo tool for LRPs 3-6):
The demo tool uses a wrong calculation but the final values shown are correct as they follow the street. It seems to be mirrored along a horizontal line going through the second LRP (first relative coordinate):
I'd be very thankful for any hints on how to solve this correctly.
Steps done:
I wrote a decoder according to the whitepaper and contacted TomTom Support who asked me to discuss this issue here. I double-checked the calculations and found mistakes in the demo tool as well as in the OpenLR white paper specification.
I solved it.
The calculation for relative Coords:
If the first bit of the concatinated bytes (of lat or lon) is 1 (negative value):
byteValueRel = bytevalue - 2^16
In any case divide it by 10^5:
byteValueRel = bytevalue/(10^5)
The resulting relative coordinate is the sum of the previous LRP value and the calculated relative value:
previousLrpValue + byteValueRel
Related
I want to fit a curve to data obtained from an FFT. While working on this, I remembered that an FFT gives binned data, and therefore I wondered if I should treat this differently with curve-fitting.
If the bins are narrow compared to the structure, I think it should not be necessary to treat the data differently, but for me that is not the case.
I expect the right way to fit binned data is by minimizing not the difference between values of the bin and fit, but between bin area and the area beneath the fitted curve, for each bin, such that the energy in each bin matches the energy in the range of the bin as signified by the curve.
So my question is: am I thinking correctly about this? If not, how should I go about it?
Also, when looking around for information about this subject, I encountered the "Maximum log likelihood" for example, but did not find enough information about it to understand if and how it applied to my situation.
PS: I have no clue if this is the right site for this question, please let me know if there is a better place.
For an unwindowed FFT, the correct interpolation between bins is by using a Sinc (sin(x)/x) or periodic Sinc (Dirichlet) interpolation kernel. For an FFT of samples of a band-limited signal, thus will reconstruct the continuous spectrum.
A very simple and effective way of interpolating the spectrum (from an FFT) is to use zero-padding. It works both with and without windowing prior to the FFT.
Take your input vector of length N and extend it to length M*N, where M is an integer
Set all values beyond the original N values to zeros
Perform an FFT of length (N*M)
Calculate the magnitude of the ouput bins
What you get is the interpolated spectrum.
Best regards,
Jens
This can be done by using maximum log likelihood estimation. This is a method that finds the set of parameters that is most likely to have yielded the measured data - the technique originates in statistics.
I have finally found an understandable source for how to apply this to binned data. Sadly I cannot enter formulas here, so I refer to that source for a full explanation: slide 4 of this slide show.
EDIT:
For noisier signals this method did not seem to work very well. A method that was a bit more robust is a least squares fit, where the difference between the area is minimized, as suggested in the question.
I have not found any literature to defend this method, but it is similar to what happens in the maximum log likelihood estimation, and yields very similar results for noiseless test cases.
The documentation for both of these methods are both very generic wherever I look. I would like to know what exactly I'm looking at with the returned arrays I'm getting from each method.
For getByteTimeDomainData, what time period is covered with each pass? I believe most oscopes cover a 32 millisecond span for each pass. Is that what is covered here as well? For the actual element values themselves, the range seems to be 0 - 255. Is this equivalent to -1 - +1 volts?
For getByteFrequencyData the frequencies covered is based on the sampling rate, so each index is an actual frequency, but what about the actual element values themselves? Is there a dB range that is equivalent to the values returned in the returned array?
getByteTimeDomainData (and the newer getFloatTimeDomainData) return an array of the size you requested - its frequencyBinCount, which is calculated as half of the requested fftSize. That array is, of course, at the current sampleRate exposed on the AudioContext, so if it's the default 2048 fftSize, frequencyBinCount will be 1024, and if your device is running at 44.1kHz, that will equate to around 23ms of data.
The byte values do range between 0-255, and yes, that maps to -1 to +1, so 128 is zero. (It's not volts, but full-range unitless values.)
If you use getFloatFrequencyData, the values returned are in dB; if you use the Byte version, the values are mapped based on minDecibels/maxDecibels (see the minDecibels/maxDecibels description).
Mozilla 's documentation describes the difference between getFloatTimeDomainData and getFloatFrequencyData, which I summarize below. Mozilla docs reference the Web Audio
experiment ; the voice-change-o-matic. The voice-change-o-matic illustrates the conceptual difference to me (it only works in my Firefox browser; it does not work in my Chrome browser).
TimeDomain/getFloatTimeDomainData
TimeDomain functions are over some span of time.
We often visualize TimeDomain data using oscilloscopes.
In other words:
we visualize TimeDomain data with a line chart,
where the x-axis (aka the "original domain") is time
and the y axis is a measure of a signal (aka the "amplitude").
Change the voice-change-o-matic "visualizer setting" to Sinewave to
see getFloatTimeDomainData(...)
Frequency/getFloatFrequencyData
Frequency functions (GetByteFrequencyData) are at a point in time; i.e. right now; "the current frequency data"
We sometimes see these in mp3 players/ "winamp bargraph style" music players (aka "equalizer" visualizations).
In other words:
we visualize Frequency data with a bar graph
where the x-axis (aka "domain") are frequencies or frequency bands
and the y-axis is the strength of each frequency band
Change the voice-change-o-matic "visualizer setting" to Frequency bars to see getFloatFrequencyData(...)
Fourier Transform (aka Fast Fourier Transform/FFT)
Another way to think about "time domain vs frequency" is shown the diagram below, from Fast Fourier Transform wikipedia
getFloatTimeDomainData gives you the chart on on the top (x-axis is Time)
getFloatFrequencyData gives you the chart on the bottom (x-axis is Frequency)
a Fast Fourier Transform (FFT) converts the Time Domain data into Frequency data, in other words, FFT converts the first chart to the second chart.
cwilso has it backwards.
the time data array is the longer one (fftSize), and the frequency data array is the shorter one (half that, frequencyBinCount).
fftSize of 2048 at the usual sample rate of 44.1kHz means each sample has 1/44100 duration, you have 2048 samples at hand, and thus are covering a duration of 2048/44100 seconds, which 46 milliseconds, not 23 milliseconds. The frequencyBinCount is indeed 1024, but that refers to the frequency domain (as the name suggests), not the time domain, and it the computation 1024/44100, in this context, is about as meaningful as adding your birth date to the fftSize.
A little math illustrating what's happening: Fourier transform is a 'vector space isomorphism', that is, a mapping going bijectively (i.e., reversible) between 2 vector spaces of the same dimension; the 'time domain' and the 'frequency domain.' The vector space dimension we have here (in both cases) is fftSize.
So where does the 'half' come from? The frequency domain coefficients 'count double'. Either because they 'actually are' complex numbers, or because you have the 'sin' and the 'cos' flavor. Or, because you have a 'magnitude' and a 'phase', which you'll understand if you know how complex numbers work. (Those are 3 ways to say the same in a different jargon, so to speak.)
I don't know why the API only gives us half of the relevant numbers when it comes to frequency - I can only guess. And my guess is that those are the 'magnitude' numbers, and the 'phase' numbers are thrown out. The reason that this is my guess is that in applications, magnitude is far more important than phase. Still, I'm quite surprised that the API throws out information, and I'd be glad if some expert who actually knows (and isn't guessing) can confirm that it's indeed the magnitude. Or - even better (I love to learn) - correct me.
I can get maxima to solve an equation but can't figure out why it won't show it's numerical value without typing the extra command/step of float(%). Is there away to automatically convert a solved variable to a numerical format.
Example of equation below:
kill(all); alpha:float(.0014931); endfreq:50; dursec:1200; solve(alpha=log(startfreq/endfreq)/dursec,float(startfreq));
what comes back is
startfreq=50%e(44793/25000)
I would like it to say 299.988 instead
Well, Maxima prefers exact results (i.e., integers, rational numbers, and symbolic constants) instead of inexact (i.e., float and bigfloat numbers). If you want to work only with numerical solutions, take a look at find_root. E.g.:
(%i1) [alpha, endfreq, dursec] : [0.0014931, 50, 1200] $
(%i2) find_root (alpha = log(startfreq / endfreq)/dursec, startfreq, 1, 500);
(%o2) 299.9881594652534
Note that to use find_root you must know an interval (here 1 to 500) which contains a root of the equation.
I have a series of data and need to detect peak values in the series within a certain number of readings (window size) and excluding a certain level of background "noise." I also need to capture the starting and stopping points of the appreciable curves (ie, when it starts ticking up and then when it stops ticking down).
The data are high precision floats.
Here's a quick sketch that captures the most common scenarios that I'm up against visually:
One method I attempted was to pass a window of size X along the curve going backwards to detect the peaks. It started off working well, but I missed a lot of conditions initially not anticipated. Another method I started to work out was a growing window that would discover the longer duration curves. Yet another approach used a more calculus based approach that watches for some velocity / gradient aspects. None seemed to hit the sweet spot, probably due to my lack of experience in statistical analysis.
Perhaps I need to use some kind of a statistical analysis package to cover my bases vs writing my own algorithm? Or would there be an efficient method for tackling this directly with SQL with some kind of local max techniques? I'm simply not sure how to approach this efficiently. Each method I try it seems that I keep missing various thresholds, detecting too many peak values or not capturing entire events (reporting a peak datapoint too early in the reading process).
Ultimately this is implemented in Ruby and so if you could advise as to the most efficient and correct way to approach this problem with Ruby that would be appreciated, however I'm open to a language agnostic algorithmic approach as well. Or is there a certain library that would address the various issues I'm up against in this scenario of detecting the maximum peaks?
my idea is simple, after get your windows of interest you will need find all the peaks in this window, you can just compare the last value with the next , after this you will have where the peaks occur and you can decide where are the best peak.
I wrote one simple source in matlab to show my idea!
My example are in wave from audio file :-)
waveFile='Chick_eco.wav';
[y, fs, nbits]=wavread(waveFile);
subplot(2,2,1); plot(y); legend('Original signal');
startIndex=15000;
WindowSize=100;
endIndex=startIndex+WindowSize-1;
frame = y(startIndex:endIndex);
nframe=length(frame)
%find the peaks
peaks = zeros(nframe,1);
k=3;
while(k <= nframe - 1)
y1 = frame(k - 1);
y2 = frame(k);
y3 = frame(k + 1);
if (y2 > 0)
if (y2 > y1 && y2 >= y3)
peaks(k)=frame(k);
end
end
k=k+1;
end
peaks2=peaks;
peaks2(peaks2<=0)=nan;
subplot(2,2,2); plot(frame); legend('Get Window Length = 100');
subplot(2,2,3); plot(peaks); legend('Where are the PEAKS');
subplot(2,2,4); plot(frame); legend('Peaks in the Window');
hold on; plot(peaks2, '*');
for j = 1 : nframe
if (peaks(j) > 0)
fprintf('Local=%i\n', j);
fprintf('Value=%i\n', peaks(j));
end
end
%Where the Local Maxima occur
[maxivalue, maxi]=max(peaks)
you can see all the peaks and where it occurs
Local=37
Value=3.266296e-001
Local=51
Value=4.333496e-002
Local=65
Value=5.049438e-001
Local=80
Value=4.286804e-001
Local=84
Value=3.110046e-001
I'll propose a couple of different ideas. One is to use discrete wavelets, the other is to use the geographer's concept of prominence.
Wavelets: Apply some sort of wavelet decomposition to your data. There are multiple choices, with Daubechies wavelets being the most widely used. You want the low frequency peaks. Zero out the high frequency wavelet elements, reconstruct your data, and look for local extrema.
Prominence: Those noisy peaks and valleys are of key interest to geographers. They want to know exactly which of a mountain's multiple little peaks is tallest, the exact location of the lowest point in the valley. Find the local minima and maxima in your data set. You should have a sequence of min/max/min/max/.../min. (You might want to add an arbitrary end points that are lower than your global minimum.) Consider a min/max/min sequence. Classify each of these triples per the difference between the max and the larger of the two minima. Make a reduced sequence that replaces the smallest of these triples with the smaller of the two minima. Iterate until you get down to a single min/max/min triple. In your example, you want the next layer down, the min/max/min/max/min sequence.
Note: I'm going to describe the algorithmic steps as if each pass were distinct. Obviously, in a specific implementation, you can combine steps where it makes sense for your application. For the purposes of my explanation, it makes the text a little more clear.
I'm going to make some assumptions about your problem:
The windows of interest (the signals that you are looking for) cover a fraction of the entire data space (i.e., it's not one long signal).
The windows have significant scope (i.e., they aren't one pixel wide on your picture).
The windows have a minimum peak of interest (i.e., even if the signal exceeds the background noise, the peak must have an additional signal excess of the background).
The windows will never overlap (i.e., each can be examined as a distinct sub-problem out of context of the rest of the signal).
Given those, you can first look through your data stream for a set of windows of interest. You can do this by making a first pass through the data: moving from left to right, look for noise threshold crossing points. If the signal was below the noise floor and exceeds it on the next sample, that's a candidate starting point for a window (vice versa for the candidate end point).
Now make a pass through your candidate windows: compare the scope and contents of each window with the values defined above. To use your picture as an example, the small peaks on the left of the image barely exceed the noise floor and do so for too short a time. However, the window in the center of the screen clearly has a wide time extent and a significant max value. Keep the windows that meet your minimum criteria, discard those that are trivial.
Now to examine your remaining windows in detail (remember, they can be treated individually). The peak is easy to find: pass through the window and keep the local max. With respect to the leading and trailing edges of the signal, you can see n the picture that you have a window that's slightly larger than the actual point at which the signal exceeds the noise floor. In this case, you can use a finite difference approximation to calculate the first derivative of the signal. You know that the leading edge will be somewhat to the left of the window on the chart: look for a point at which the first derivative exceeds a positive noise floor of its own (the slope turns upwards sharply). Do the same for the trailing edge (which will always be to the right of the window).
Result: a set of time windows, the leading and trailing edges of the signals and the peak that occured in that window.
It looks like the definition of a window is the range of x over which y is above the threshold. So use that to determine the size of the window. Within that, locate the largest value, thus finding the peak.
If that fails, then what additional criteria do you have for defining a region of interest? You may need to nail down your implicit assumptions to more than 'that looks like a peak to me'.
in xxxx.mxml (from flex) i have called the remote remote method (of java) the method return type is float
in the xxxx.mxml's remote objects result handler i need get the float values as Numeric.....or String..i tried with string...i did Alert.show to see the value some times i get exact value for eg, 0.5 is the value returning from java methid but here it will show 0.50000454...so on..how get the exact value?
It is because of the way floating point numbers are stored; basically they can't be stored precisely. A quick search in SO would reveal a lot of threads about this. Also read "What Every Computer Scientist Should Know About Floating-Point Arithmetic"
Thus the problem of getting the exact value boils down to what you define exact to be. Try rounding it to a given number of floating points at java end, convert the rounded number to a string (I'm not sure if this conversion would preserve the precision) and send that string.