I have two audio recordings of a same signal by 2 different microphones (for example, in a WAV format), but one of them is recorded with delay, for example, several seconds.
It's easy to identify such a delay visually when viewing these signals in some kind of waveform viewer - i.e. just spotting first visible peak in every signal and ensuring that they're the same shape:
(source: greycat.ru)
But how do I do it programmatically - find out what this delay (t) is? Two digitized signals are slightly different (because microphones are different, were at different positions, due to ADC setups, etc).
I've digged around a bit and found out that this problem is usually called "time-delay estimation" and it has myriads of approaches to it - for example, one of them.
But are there any simple and ready-made solutions, such as command-line utility, library or straight-forward algorithm available?
Conclusion: I've found no simple implementation and done a simple command-line utility myself - available at https://bitbucket.org/GreyCat/calc-sound-delay (GPLv3-licensed). It implements a very simple search-for-maximum algorithm described at Wikipedia.
The technique you're looking for is called cross correlation. It's a very simple, if somewhat compute intensive technique which can be used for solving various problems, including measuring the time difference (aka lag) between two similar signals (the signals do not need to be identical).
If you have a reasonable idea of your lag value (or at least the range of lag values that are expected) then you can reduce the total amount of computation considerably. Ditto if you can put a definite limit on how much accuracy you need.
Having had the same problem and without success to find a tool to sync the start of video/audio recordings automatically,
I decided to make syncstart (github).
It is a command line tool. The basic code behind it is this:
import numpy as np
from scipy import fft
from scipy.io import wavfile
r1,s1 = wavfile.read(in1)
r2,s2 = wavfile.read(in2)
assert r1==r2, "syncstart normalizes using ffmpeg"
fs = r1
ls1 = len(s1)
ls2 = len(s2)
padsize = ls1+ls2+1
padsize = 2**(int(np.log(padsize)/np.log(2))+1)
s1pad = np.zeros(padsize)
s1pad[:ls1] = s1
s2pad = np.zeros(padsize)
s2pad[:ls2] = s2
corr = fft.ifft(fft.fft(s1pad)*np.conj(fft.fft(s2pad)))
ca = np.absolute(corr)
xmax = np.argmax(ca)
if xmax > padsize // 2:
file,offset = in2,(padsize-xmax)/fs
else:
file,offset = in1,xmax/fs
A very straight forward thing todo is just to check if the peaks exceed some threshold, the time between high-peak on line A and high-peak on line B is probably your delay. Just try tinkering a bit with the thresholds and if the graphs are usually as clear as the picture you posted, then you should be fine.
Related
So, I have a model of a tube with pressure loss, where the unknown is the mass flow rate. Normally, and on most models of this problem, the conservation equations are used to calculate the mass flow rate, but such models have lots of convergence issues (because of the blocked flow at the end of the tube which results in an infinite pressure derivative at the end). See figure below for a representation of the problem on the left and the right a graph showing the infinite pressure derivative.
Because of that I'm using a model which is more robust, though it outputs not the mass flow rate but the tube length, which is known. Therefore an iterative loop is needed to determine the mass flow rate. Ok then, I coded a function length that given the tube geometry, mass flow rate and boundary conditions it outputs the calculated tube length and made the equations like so:
parameter Real L;
Real m_flow;
...
equation
L = length(geometry, boundary, m_flow)
It simulates fine, but it takes ages... And it shouldn't because the mass flow rate is rather insensitive to the tube length, e.g. if L=3 I could say that m_flow has converged if the output of length is within L ± 0.1. On the other hand the default convergence tolerance of DASSL in Dymola is 0.0001, which is fine for all other variables, but a major setback to my model here...
That being said, I'd like to know if there's a (hacky) way of setting a specific tolerance L (from annotations or something). I was unable to find any solution online or in Dymola's user manual... So far I managed a workaround by making a second function which uses a Newton-Raphson method to determine the mass flow rate, something like:
function massflowrate
input geometry, boundary, m_flow_start, tolerance;
output m_flow;
protected
Real error, L, dL, dLdm_flow, Delta_m_flow;
algorithm
error = geometry.L;
m_flow = m_flow_start;
while error>tolerance loop
L = length(geometry, boundary, m_flow);
error = abs(boundary.L - L);
dL = length(geometry, boundary, m_flow*1.001);
dLdm_flow = dL/(0.001*m_flow);
Delta_m_flow = (geometry.L - L)/dLdm_flow;
m_flow = m_flow + Delta_m_flow;
end while;
end massflowrate;
And then I use it in the equations section:
parameter Real L;
Real m_flow;
...
equation
m_flow = massflowrate(geometry, boundary, delay(m_flow,10), tolerance)
Nevertheless, this solutions is not without it's problems, the real equations are very non-linear and depending on the boundary conditions the solver reaches a never-ending loop... =/
PS: I'm sorry for the long post and the lack of a MWE, the real equations are very long and with loads of thermodynamics which I believe not to be of any help, be that as it may, if necessary, I'm able to provide the real model.
Is the length-function smooth? To me that it being non-smooth seems like a likely cause for problems, and the suggestions by #Phil might also be good ideas.
However, it should also be possible to do what you want as follows:
Real m_flow(nominal=1e9);
Explanation: The equations are normally solved to a certain tolerance in unknowns - in this case m_flow.
The tolerance for each variable is a relative/absolute tolerance taking into the nominal value, and Dymola does not allow you to set different tolerances for different variables.
Thus the simple way to compute m_flow less accurately is by setting a high nominal value for it, since the error tolerance will be tol*(abs(m_flow)+abs(nominal(m_flow))) or something like that.
The downside is that it may be too inaccurate, e.g. causing additional events, or that the error is so random that the solver is still slowed down.
I've been experimenting with the GPU support of Matlab (v2014a). The notebook I'm using to test my code has a NVIDIA 840M build in.
Since I'm new to GPU computing with Matlab, I started out with a few simple examples and observed a strange scalability behavior. Instead of increasing the size of my problem, I simply put a forloop around my computation. I expected the time for the computation, to scale with the number of iterations, since the problem size itself does not increase. This was also true for smaller numbers of iterations, however at a certain point the time does not scale as expected, instead I observe a huge increase in computation time. Afterwards, the problem continues to scale again as expected.
The code example started from a random walk simulation, but I tried to produce an example that is easy and still shows the problem.
Here's what my code does. I initialize two matrices as sin(alpha)and cos(alpha). Then I loop over the number of iterations from 2**1to 2**15. I then repead the computation sin(alpha)^2 and cos(alpha)^2and add them up (this was just to check the result). I perform this calculation as often as the number of iterations suggests.
function gpu_scale
close all
NP = 10000;
NT = 1000;
ALPHA = rand(NP,NT,'single')*2*pi;
SINALPHA = sin(ALPHA);
COSALPHA = cos(ALPHA);
GSINALPHA = gpuArray(SINALPHA); % move array to gpu
GCOSALPHA = gpuArray(COSALPHA);
PMAX=15;
for P = 1:PMAX;
for i=1:2^P
GX = GSINALPHA.^2;
GY = GCOSALPHA.^2;
GZ = GX+GY;
end
end
The following plot, shows the computation time in a log-log plot for the case that I always double the number of iterations. The jump occurs when doubling from 1024 to 2048 iterations.
The initial bump for two iterations might be due to initialization and is not really relevant anyhow.
I see no reason for the jump between 2**10 and 2**11 computations, since the computation time should only depend on the number of iterations.
My question: Can somebody explain this behavior to me? What is happening on the software/hardware side, that explains this jump?
Thanks in advance!
EDIT: As suggested by Divakar, I changed the way I time my code. I wasn't sure I was using gputimeit correctly. however MathWorks suggests another possible way, namely
gd= gpuDevice();
tic
% the computation
wait(gd);
Time = toc;
Using this way to measure my performance, the time is significantly slower, however I don't observe the jump in the previous plot. I added the CPU performance for comparison and keept both timings for the GPU (wait / no wait), which can be seen in the following plot
It seems, that the observed jump "corrects" the timining in the direction of the case where I used wait. If I understand the problem correctly, then the good performance in the no wait case is due to the fact, that we do not wait for the GPU to finish completely. However, then I still don't see an explanation for the jump.
Any ideas?
I'm using SVMLib to train a simple SVM over the MNIST dataset. It contains 60.000 training data. However, I have several performance issues: the training seems to be endless (after a few hours, I had to shut it down by hand, because it doesn't respond). My code is very simple, I just call ovrtrain on the dataset without any kernel and any special constants:
function features = readFeatures(fileName)
[fid, msg] = fopen(fileName, 'r', 'ieee-be');
header = fread(fid, 4, "int32" , 0, "ieee-be");
if header(1) ~= 2051
fprintf("Wrong magic number!");
end
M = header(2);
rows = header(3);
columns = header(4);
features = fread(fid, [M, rows*columns], "uint8", 0, "ieee-be");
fclose(fid);
return;
endfunction
function labels = readLabels(fileName)
[fid, msg] = fopen(fileName, 'r', 'ieee-be');
header = fread(fid, 2, "int32" , 0, "ieee-be");
if header(1) ~= 2049
fprintf("Wrong magic number!");
end
M = header(2);
labels = fread(fid, [M, 1], "uint8", 0, "ieee-be");
fclose(fid);
return;
endfunction
labels = readLabels("train-labels.idx1-ubyte");
features = readFeatures("train-images.idx3-ubyte");
model = ovrtrain(labels, features, "-t 0"); % doesn't respond...
My question: is it normal? I'm running it on Ubuntu, a virtual machine. Should I wait longer?
I don't know whether you took your answer or not, but let me tell you what I predict about your situation. 60.000 examples is not a lot for a power trainer like LibSVM. Currently, I am working on a training set of 6000 examples and it takes 3-to-5 seconds to train. However, the parameter selection is important and that is the one probably taking long time. If the number of unique features in your data set is too high, then for any example, there will be lots of zero feature values for non-existing features. If the tool is implementing data scaling on your training set, then most probably those lots of zero feature values will be scaled to a certain non-zero value, leaving you astronomic number of unique and non-zero valued features for each and every example. This is very very complicated for a SVM tool to get in and extract efficient parameter values.
Long story short, if you had enough research on SVM tools and understand what I mean, you either assign parameter values in the training command before executing it or find a way to decrease the number of unique features. If you haven't, go on and download the latest version of LibSVM, read the ReadME files as well as the FAQ from the website of the tool.
If non of these is the case, then sorry for taking your time:) Good luck.
It might be an issue of convergence given the characteristics of your data.
Check the kernel you have as default selection and change it. Also, check the stopping criterion of the package. Additionally, if you are looking for faster implementation, check MSVMpack which is a parallel implementation of SVM.
Finally, feature selection in your case is desired. You can end up with a good feature subset of almost half of what you have. In addition, you need only a portion of data for training e.g. 60~70 % are sufficient.
First of all 60k is huge data for training.Training that much data with linear kernel will take hell of time unless you have a supercomputing. Also you have selected a linear kernel function of degree 1. Its better to use Gaussian or higher degree polynomial kernel (deg 4 used with the same dataset showed a good tranning accuracy). Try to add the LIBSVM options for -c cost -m memory cachesize -e epsilon tolerance of termination criterion (default 0.001). First run 1000 samples with Gaussian/ polynomial of deg 4 and compare the accuracy.
I'm not sure StackOverflow is the right place to ask this question, because this question is half-programming and half-mathematics. And also really sorry if my question is stupid ^_^
I'm studying about Monte Carlo simulations via the "Monte Carlo Methods" book. One of the first thing I must learn is about Random Number Generator. The basic algorithm of RNG is:
1. Initialize: Draw the seed S0 from the distribution µ on S. Set t = 1.
2. Transition: Set St = f(St−1).
3. Output: Set Ut = g(St).
4. Repeat: Set t = t+ 1 and return to Step 2.
(µ is a probability distribution on the finite set of states S, the input is S0 and the random number we desire it the output Ut)
It is not hard to understand, but the problem here is I don't see the random factor which lie in the number of repeat. How can we decide when to stop the loop of the RNG? All examples I read which implement a RNG are loop for 100 times, and they returns the same value for a specific seed. It is not random at all >_<
Can someone explain what I'm missing here? Any help will be appreciated. Thanks everyone
You can't get a true sequence of random numbers on a computer, without specialized hardware. (Such specialized hardware performs the equivalent of an initial roll of the dice using physics to provide the randomness. Electronic ones often use the electronic noise of specialized diodes at constant temperatures; others use radioactive decay events.)
Without that specialized hardware, what you can generate are pseudorandom numbers which, as you've observed, always generate the same sequence of numbers for the same initial seed. For simple applications, you can often get away with generating an initial seed from the time of invocation, which is effectively random.
And when I say "simple applications," I am excluding cryptography. (Not just that, but especially that.)
Sometimes when you are trying to debug a simulation, you actually want to have a reproducible stream of "random" numbers so you might specifically sent a stream to start with a specific seed.
For instance in the answer Creating a facet_wrap plot with ggplot2 with different annotations in each plot rcs starts the answer by creating a reproducible set of data using the R code
set.seed(1)
df <- data.frame(x=rnorm(300), y=rnorm(300), cl=gl(3,100)) # create test data
before going on to demonstrate how to answer the actual question.
I apologize for being a bit verbose in advance: if you want to skip all the background mumbo jumbo you can see my question down below.
This is pretty much a follow up to a question I previously posted on how to compare two 1D (time dependent) signals. One of the answers I got was to use the cross-correlation function (xcorr in MATLAB), which I did.
Background information
Perhaps a little background information will be useful: I'm trying to implement an Independent Component Analysis algorithm. One of my informal tests is to (1) create the test case by (a) generate 2 random vectors (1x1000), (b) combine the vectors into a 2x1000 matrix (called "S"), and multiply this by a 2x2 mixing matrix (called "A"), to give me a new matrix (let's call it "T").
In summary: T = A * S
(2) I then run the ICA algorithm to generate the inverse of the mixing matrix (called "W"), (3) multiply "T" by "W" to (hopefully) give me a reconstruction of the original signal matrix (called "X")
In summary: X = W * T
(4) I now want to compare "S" and "X". Although "S" and "X" are 2x1000, I simply compare S(1,:) to X(1,:) and S(2,:) to X(2,:), each which is 1x1000, making them 1D signals. (I have another step which makes sure that these vectors are the proper vectors to compare to each other and I also normalize the signals).
So my current quandary is how to 'grade' how close S(1,:) matches to X(1,:), and likewise with S(2,:) to X(2,:).
So far I have used something like: r1 = max(abs(xcorr(S(1,:), X(1,:)))
My question
Assuming that using the cross correlation function is a valid way to go about comparing the similarity of two signals, what would be considered a good R value to grade the similarity of the signals? Wikipedia states that this is a very subjective area, and so I defer to the better judgment of those who might have experience in this field.
As you might realize, I'm not coming from a EE/DSP/statistical background at all (I'm a medical student) so I'm going through a sort of "baptism through fire" right now, and I appreciate all the help I can get. Thanks!
(edit: as far as directly answering your question about R values, see below)
One way to approach this would be to use cross-correlation. Bear in mind that you have to normalize amplitudes and correct for delays: if you have signal S1, and signal S2 is identical in shape, but half the amplitude and delayed by 3 samples, they're still perfectly correlated.
For example:
>> t = 0:0.001:1;
>> y = #(t) sin(10*t).*exp(-10*t).*(t > 0);
>> S1 = y(t);
>> S2 = 0.4*y(t-0.1);
>> plot(t,S1,t,S2);
These should have a perfect correlation coefficient. A way to compute this is to use maximum cross-correlation:
>> f = #(S1,S2) max(xcorr(S1,S2));
f =
#(S1,S2) max(xcorr(S1,S2))
>> disp(f(S1,S1)); disp(f(S2,S2)); disp(f(S1,S2));
12.5000
2.0000
5.0000
The maximum value of xcorr() takes care of the time-delay between signals. As far as correcting for amplitude goes, you can normalize the signals so that their self-cross-correlation is 1.0, or you can fold that equivalent step into the following:
ρ2 = f(S1,S2)2 / (f(S1,S1)*f(S2,S2);
In this case ρ2 = 5 * 5 / (12.5 * 2) = 1.0
You can solve for ρ itself, i.e. ρ = f(S1,S2)/sqrt(f(S1,S1)*f(S2,S2)), just bear in mind that both 1.0 and -1.0 are perfectly correlated (-1.0 has opposite sign)
Try it on your signals!
with respect to what threshold to use for acceptance/rejection, that really depends on what kind of signals you have. 0.9 and above is fairly good but can be misleading. I would consider looking at the residual signal you get after you subtract out the correlated version. You could do this by looking at the time index of the maximum value of xcorr():
>> t = 0:0.001:1;
>> y = #(a,t) sin(a*t).*exp(-a*t).*(t > 0);
>> S1=y(10,t);
>> S2=0.4*y(9,t-0.1);
>> f(S1,S2)/sqrt(f(S1,S1)*f(S2,S2))
ans =
0.9959
This looks pretty darn good for a correlation. But let's try fitting S2 with a scaled/shifted multiple of S1:
>> [A,i]=max(xcorr(S1,S2)); tshift = i-length(S1);
>> S2fit = zeros(size(S2)); S2fit(1-tshift:end) = A/f(S1,S1)*S1(1:end+tshift);
>> plot(t,[S2; S2fit]); % fit S2 using S1 as a basis
>> plot(t,[S2-S2fit]); % residual
Residual has some energy in it; to get a feel for how much, you can use this:
>> S2res=S2-S2fit;
>> dot(S2res,S2res)/dot(S2,S2)
ans =
0.0081
>> sqrt(dot(S2res,S2res)/dot(S2,S2))
ans =
0.0900
This says that the residual has about 0.81% of the energy (9% of the root-mean-square amplitude) of the original signal S2. (the dot product of a 1D signal with itself will always be equal to the maximum value of cross-correlation of that signal with itself.)
I don't think there's a silver bullet for answering how similar two signals are with each other, but hopefully I've given you some ideas that might be applicable to your circumstances.
A good starting point is to get a sense of what a perfect match will look like by calculating the auto-correlations for each signal (i.e. do the "cross-correlation" of each signal with itself).
THIS IS A COMPLETE GUESS - but I'm guessing max(abs(xcorr(S(1,:),X(1,:)))) > 0.8 implies success. Just out of curiosity, what kind of values do you get for max(abs(xcorr(S(1,:),X(2,:))))?
Another approach to validate your algorithm might be to compare A and W. If W is calculated correctly, it should be A^-1, so can you calculate a measure like |A*W - I|? Maybe you have to normalize by the trace of A*W.
Getting back to your original question, I come from a DSP background, so I get to deal with fairly noise-free signals. I understand that's not a luxury you get in biology :) so my 0.8 guess might be very optimistic. Perhaps looking at some literature in your field, even if they aren't using cross-correlation exactly, might be useful.
Usually in such cases people talk about "false acceptance rate" and "false rejection rate".
The first one describes how many times algorithm says "similar" for non-similar signals, the second one is the opposite.
Selecting a threshold thus becomes a trade-off between these criteria. To make FAR=0, threshold should be 1, to make FRR=0 threshold should be -1.
So probably, you will need to decide which trade-off between FAR and FRR is acceptable in your situation and this will give the right value for threshold.
Mathematically this can be expressed in different ways. Just a couple of examples:
1. fix some of rates at acceptable value and minimize other one
2. minimize max(FRR,FAR)
3. minimize aFRR+bFAR
Since they should be equal, the correlation coefficient should be high, between .99 and 1. I would take the max and abs functions out of your calculation, too.
EDIT:
I spoke too soon. I confused cross-correlation with correlation coefficient, which is completely different. My answer might not be worth much.
I would agree that the result would be subjective. Something that would involve the sum of the squares of the differences, element by element, would have some value. Two identical arrays would give a value of 0 in that form. You would have to decide what value then becomes "bad". Make up 2 different vectors that "aren't too bad" and find their cross-correlation coefficient to be used as a guide.
(parenthetically: if you were doing a correlation coefficient where 1 or -1 would be great and 0 would be awful, I've been told by bio-statisticians that a real-life value of 0.7 is extremely good. I understand that this is not exactly what you are doing but the comment on correlation coefficient came up earlier.)