I'm working on aligning different measurements from sensors. Some of these are periodic and I just used the maximum of the cross correlations and it worked fine. Now I have a couple of non periodic signals similar to ramp/sigmoids/step/hill functions that I want to align, but for these the cross correlation fails miserably (giving me always the maximum at lag 0).
What is the approach for these kind of signals?
Ideal the approach would work for both signals without prior knowledge which one I'm encountering.
Here is an example (with noise)
One possible approach is to take your aperiodic signal and coerce it into a periodic signal.
One way to do this is to first normalize your signal, and then append an inverted version of your signal (1 - normalizedSignal) to your signal. This makes it a periodic signal which then should be able to be fed into cross-correlation analysis relatively easily.
Here is an example I whipped up using an inverted sigmoid shifted in time.
function aperiodicxcorr()
% Time step at which to sample the sigmoid
dt = 0.1;
t = -10:dt:5;
% Artificial lags to apply to the second and third signals
actualLag2 = 3;
actualLag3 = 5;
% Now create signals that are negative sigmoids with delays
S1 = -sigmoid(t);
S2 = -sigmoid(t + actualLag2);
S3 = -sigmoid(t + actualLag3);
% Normalize each sigmal
S1 = normalize(S1);
S2 = normalize(S2);
S3 = normalize(S3);
% Concatenate the inverted signal with signal to make it periodic
S1 = cat(2, 1-S1, S1);
S2 = cat(2, 1-S2, S2);
S3 = cat(2, 1-S3, S3);
% Retrieve lag (in samples)
[corr2, lag2] = computeLag(S1, S2);
[corr3, lag3] = computeLag(S1, S3);
% Convert lags to time by multiplying by time step
lag2 = lag2 * dt;
lag3 = lag3 * dt;
fprintf('Lag of S2: %0.2f (r = %0.2f)\n', lag2, corr2);
fprintf('Lag of S3: %0.2f (r = %0.2f)\n', lag3, corr3);
end
function [corr, lag] = computeLag(A, B)
[corr, lags] = xcorr(A, B, 'coeff');
[corr, ind] = max(corr);
lag = lags(ind);
end
function data = normalize(data)
data = data - min(data(:));
data = data ./ max(data(:));
end
function S = sigmoid(t)
S = 1 ./ (1 + exp(-t));
end
The modification to the signal that I discussed, looks like this for the above code.
And the result of the fprintf statements at the bottom are:
Lag of S2: 3.00 (r = 1.00)
Lag of S3: 5.00 (r = 1.00)
And these match up with the specified lags.
The drawback of this is that it won't work for signals which are already periodic. That being said, periodicity is relatively easy to check (particularly for a normalized signal) by comparing the first and last values of your signal and ensuring that they are within a specified tolerance of one another.
Related
I'm attempting to create a reverb function in MATLAB and would like to filter the repeats to make them sound more natural. I have the signal processing toolbox however I get an error telling me that I cannot use the lowpass() function on single samples and MATLAB wigs out. The idea is to filter the signal every time it enters the feedback loop to simulate the absorption of high frequencies every time the sound reflects off a surface. My code looks like this:
[x,Fs] = audioread('Clap.mp3');
%Calculate delay length
secOfDelay = 0.03; %units of seconds
samplesOfDelay = secOfDelay * Fs; %sec * samples/sec = samples
%Gain for wet mix
g = 0.9;
for n = 1:length(x)
if (n - samplesOfDelay) < 1 %For no past samples, don't add the delay yet.
out(n,1) = x(n,1);
else
a(n,1) = x(n,1);
b(n,1) = x(n,1) + (g * b(n-samplesOfDelay,1));
end
end
%Add wet mix and dry mix together
out = a+b;
sound(out,Fs);
Any help would be greatly appreciated!
JDB
I have the following Markov chain:
This chain shows the states of the Spaceship, which is in the asteroid belt: S1 - is serviceable, S2 - is broken. 0.12 - the probability of destroying the Spaceship by a collision with an asteroid. 0.88 - the probability of that a collision will not be critical. Need to find the probability of a serviceable condition of the ship after the third collision.
Analytical solution showed the response - 0.681. But it is necessary to solve this problem by simulation method using any modeling tool (MATLAB Simulink, AnyLogic, Scilab, etc.).
Do you know what components should be used to simulate this process in Simulink or any other simulation environment? Any examples or links.
First, we know the three step probability transition matrix contains the answer (0.6815).
% MATLAB R2019a
P = [0.88 0.12;
0 1];
P3 = P*P*P
P(1,1) % 0.6815
Approach 1: Requires Econometrics Toolbox
This approach uses the dtmc() and simulate() functions.
First, create the Discrete Time Markov Chain (DTMC) with the probability transition matrix, P, and using dtmc().
mc = dtmc(P); % Create the DTMC
numSteps = 3; % Number of collisions
You can get one sample path easily using simulate(). Pay attention to how you specify the initial conditions.
% One Sample Path
rng(8675309) % for reproducibility
X = simulate(mc,numSteps,'X0',[1 0])
% Multiple Sample Paths
numSamplePaths = 3;
X = simulate(mc,numSteps,'X0',[numSamplePaths 0]) % returns a 4 x 3 matrix
The first row is the X0 row for the starting state (initial condition) of the DTMC. The second row is the state after 1 transition (X1). Thus, the fourth row is the state after 3 transitions (collisions).
% 50000 Sample Paths
rng(8675309) % for reproducibility
k = 50000;
X = simulate(mc,numSteps,'X0',[k 0]); % returns a 4 x 50000 matrix
prob_survive_3collisions = sum(X(end,:)==1)/k % 0.6800
We can bootstrap a 95% Confidence Interval on the mean probability to survive 3 collisions to get 0.6814 ± 0.00069221, or rather, [0.6807 0.6821], which contains the result.
numTrials = 40;
ProbSurvive_3collisions = zeros(numTrials,1);
for trial = 1:numTrials
Xtrial = simulate(mc,numSteps,'X0',[k 0]);
ProbSurvive_3collisions(trial) = sum(Xtrial(end,:)==1)/k;
end
% Mean +/- Halfwidth
alpha = 0.05;
mean_prob_survive_3collisions = mean(ProbSurvive_3collisions)
hw = tinv(1-(0.5*alpha), numTrials-1)*(std(ProbSurvive_3collisions)/sqrt(numTrials))
ci95 = [mean_prob_survive_3collisions-hw mean_prob_survive_3collisions+hw]
maxNumCollisions = 10;
numSamplePaths = 50000;
ProbSurvive = zeros(maxNumCollisions,1);
for numCollisions = 1:maxNumCollisions
Xc = simulate(mc,numCollisions,'X0',[numSamplePaths 0]);
ProbSurvive(numCollisions) = sum(Xc(end,:)==1)/numSamplePaths;
end
For a more complex system you'll want to use Stateflow or SimEvents, but for this simple example all you need is a single Unit Delay block (output = 0 => S1, output = 1 => S2), with a Switch block, a Random block, and some comparison blocks to construct the logic determining the next value of the state.
Presumably you must execute the simulation a (very) large number of times and average the results to get a statistically significant output.
You'll need to change the "seed" of the random generator each time you run the simulation.
This can be done by setting the seed to be "now" (or something similar to that).
Alternatively you could quite easily vectorize the model so that you only need to execute it once.
If you want to simulate this, it is fairly easy in matlab:
servicable = 1;
t = 0;
while servicable =1
t = t+1;
servicable = rand()<=0.88
end
Now t represents the amount of steps before the ship is broken.
Wrap this in a for loop and you can do as many simulations as you like.
Note that this can actually give you the distribution, if you want to know it after 3 times, simply add && t<3 to the while condition.
%-------------------------------------------------------------------
% Function to Generate ECG of 1 heart beat signal
function [Heartbeat,t1] = ECG_Gen (HR,pulse_width,Amp)
Fs = 48000;
delay = ((60/HR)/2)-(0.5*pulse_width);
t1 = -delay:(1/Fs):delay;
Heartbeat = Amp*tripuls (t1,pulse_width);
%-------------------------------------------------------------------
%Test Circuit configuration
function [FECG_MECG,Mixed_ECG,fastTime] = Test_Circuit (FHR,MHR)
Fs = 48000;
%FHR = 150;
%MHR = 60;
Fpulse_width = 30e-3;
Mpulse_width = 60e-3;
FAmp = 0.2;
MAmp = 0.5;
% Fetal ECG Gen
%------------------------------------------------
[FECG,FHR_Delay]= ECG_Gen (FHR,Fpulse_width,FAmp);
% Maternal ECG Gen
%------------------------------------------------
[MECG,MHR_Delay]= ECG_Gen (MHR,Mpulse_width,MAmp);
% Composite signal implementation
%------------------------------------------------
% Set parameters for Composite signal Algorithms
if length (MECG) > length (FECG) % Check for time sequences for both ECG signal
slowECG = FECG; % Set interpolation to slower rate
fastECG = MECG;
timeSeg = length(MECG);
fastTime = MHR_Delay; % Set sampling times
slowTime = FHR_Delay;
else
slowECG = MECG;
fastECG = FECG;
timeSeg = length(FECG);
fastTime = FHR_Delay;
slowTime = MHR_Delay;
end
FECG_MECG = zeros (timeSeg,2); % To hold stereo output
FECG_MECG(:,2) = fastECG(1,:); % Assign higher rate signal to one channel
% Interpolation on the slower rater sampled ECG
slowECGInterp = interp1 (slowTime,slowECG,fastTime);
slowECG = num2cell(slowECGInterp); % Conversion to cell Array in order to remove NaN
slowECG(cellfun(#(slowECG) any(isnan(slowECG)),slowECG)) = [];
slowECG = cell2mat(slowECG);
j = 1;
for i = 1:timeSeg
FECG_MECG(i,1) = slowECG(1,j);
if j == length(slowECG)
j = 0;
end
j = j+1;
end
Mixed_ECG = FECG_MECG(:,1) + FECG_MECG(:,2); % to hold mono output
%while (1)
%sound(Mixed_ECG ,Fs);
%end
%-------------------------------------------------------------------
% Test Wave script
clear all
%clc
clc
Fs = 48000;
%for i = 1:3
%toc
MHR = 60;
FHR = 200;
% Obtain ECG interpolated signal and composite
[FECG_MECG,Mixed_ECG,fastTime] = Test_Circuit (FHR,MHR);
% for test purposes
[MECG,MHR_Delay]= ECG_Gen (60,60e-3,0.5);
%t = timer ('TimerFcn','stat=false','Period',2.0);
wavwrite (FECG_MECG(:,2),Fs,'ECGwav.wav');
i = 0;
a = 1;
tic
while (1)
while (toc < 20*a)
sound (MECG,Fs);
end
toc
a = a+1;
[MECG,MHR_Delay]= ECG_Gen (60*a,60e-3,0.5);
if a > 4
break
end
end
%start(t)
%tic
%t = cputime;
%y = wavread('ECGwav.wav');
%while (1)
% sound(y,Fs);
%end
%toc
Hey Thank you very much for getting back to me, I have made use of your interpolation but still have minor problems from the reading obtained from the Monitor. Fist of all, say I have a constant signal with fixed time period say, 0.8s and I want to add composite signal of say 0.3s, I have managed to use your interpolation method to sample the 0.3s signal at the rate of my 0.8s signal. I think I have solved this issue. Second issue deals with how to combine the two signals which I have managed to do somehow but when I use a while loop in order to repeat the composite signal say over 20s, the signals obtained from the sound output isn't quite what I expected since its sounding the array of stored composite signal which contain ( signal with 0.8s = slowInterp signal of 0.3s ). I have include all the codes and functions. Basically, I need the sound output in while loop to sync with the composite signal for example: if I have a signal that repeats every 1s, I would expect to hear a beep in a while loop that runs for 10s to produce 10 beeps, etc
It is hard to tell exactly what you need, but probably doing interpolation on the sequence that is sampled at the slower rate would work for your application.
If t1s are your times from your faster sequence, and t2s are times from your slower sequence, and slow is your slower sequence then do:
slowInterp = interp1(t2s, slow, t1s);
now you will have the sequency slow sampled at the faster rate.
This is only useful for displaying the time series. If you are doing any spectral analysis, this will introduce artifacts, but that is a more advanced topic.
Using the resample function in the signal processing toolbox could also be useful.
The final goal I am trying to achieve is the generation of a ten minutes time series: to achieve this I have to perform an FFT operation, and it's the point I have been stumbling upon.
Generally the aimed time series will be assigned as the sum of two terms: a steady component U(t) and a fluctuating component u'(t). That is
u(t) = U(t) + u'(t);
So generally, my code follows this procedure:
1) Given data
time = 600 [s];
Nfft = 4096;
L = 340.2 [m];
U = 10 [m/s];
df = 1/600 = 0.00167 Hz;
fn = Nfft/(2*time) = 3.4133 Hz;
This means that my frequency array should be laid out as follows:
f = (-fn+df):df:fn;
But, instead of using the whole f array, I am only making use of the positive half:
fpos = df:fn = 0.00167:3.4133 Hz;
2) Spectrum Definition
I define a certain spectrum shape, applying the following relationship
Su = (6*L*U)./((1 + 6.*fpos.*(L/U)).^(5/3));
3) Random phase generation
I, then, have to generate a set of complex samples with a determined distribution: in my case, the random phase will approach a standard Gaussian distribution (mu = 0, sigma = 1).
In MATLAB I call
nn = complex(normrnd(0,1,Nfft/2),normrnd(0,1,Nfft/2));
4) Apply random phase
To apply the random phase, I just do this
Hu = Su*nn;
At this point start my pains!
So far, I only generated Nfft/2 = 2048 complex samples accounting for the fpos content. Therefore, the content accounting for the negative half of f is still missing. To overcome this issue, I was thinking to merge the real and imaginary part of Hu, in order to get a signal Huu with Nfft = 4096 samples and with all real values.
But, by using this merging process, the 0-th frequency order would not be represented, since the imaginary part of Hu is defined for fpos.
Thus, how to account for the 0-th order by keeping a procedure as the one I have been proposing so far?
I have two signals, let's call them 'a' and 'b'. They are both nearly identical signals (recorded from the same input and contain the same information) however, because I recorded them at two different 'b' is time shifted by an unknown amount. Obviously, there is random noise in each.
Currently, I am using cross correlation to compute the time shift, however, I am still getting improper results.
Here is the code I am using to calculate the time shift:
function [ diff ] = FindDiff( signal1, signal2 )
%FINDDIFF Finds the difference between two signals of equal frequency
%after an appropritate time shift is applied
% Calculates the time shift between two signals of equal frequency
% using cross correlation, shifts the second signal and subtracts the
% shifted signal from the first signal. This difference is returned.
length = size(signal1);
if (length ~= size(signal2))
error('Vectors must be equal size');
end
t = 1:length;
tx = (-length+1):length;
x = xcorr(signal1,signal2);
[mx,ix] = max(x);
lag = abs(tx(ix));
shifted_signal2 = timeshift(signal2,lag);
diff = signal1 - shifted_signal2;
end
function [ shifted ] = timeshift( input_signal, shift_amount )
input_size = size(input_signal);
shifted = (1:input_size)';
for i = 1:input_size
if i <= shift_amount
shifted(i) = 0;
else
shifted(i) = input_signal(i-shift_amount);
end
end
end
plot(FindDiff(a,b));
However the result from the function is a period wave, rather than random noise, so the lag must still be off. I would post an image of the plot, but imgur is currently not cooperating.
Is there a more accurate way to calculate lag other than cross correlation, or is there a way to improve the results from cross correlation?
Cross-correlation is usually the simplest way to determine the time lag between two signals. The position of peak value indicates the time offset at which the two signals are the most similar.
%// Normalize signals to zero mean and unit variance
s1 = (signal1 - mean(signal1)) / std(signal1);
s2 = (signal2 - mean(signal2)) / std(signal2);
%// Compute time lag between signals
c = xcorr(s1, s2); %// Cross correlation
lag = mod(find(c == max(c)), length(s2)) %// Find the position of the peak
Note that the two signals have to be normalized first to the same energy level, so that the results are not biased.
By the way, don't use diff as a name for a variable. There's already a built-in function in MATLAB with the same name.
Now there are two functions in Matlab:
one called finddelay
and another called alignsignals that can do what you want, I believe.
corr finds a dot product between vectors (v1, v2). If it works bad with your signal, I'd try to minimize a sum of squares of differences (i.e. abs(v1 - v2)).
signal = sin(1:100);
signal1 = [zeros(1, 10) signal];
signal2 = [signal zeros(1, 10)];
for i = 1:length(signal1)
signal1shifted = [signal1 zeros(1, i)];
signal2shifted = [zeros(1, i) signal2];
d2(i) = sum((signal1shifted - signal2shifted).^2);
end
[fval lag2] = min(d2);
lag2
It is computationally worse than cross-calculation which can be speeded up by using FFT. As far as I know you can't do this with euclidean distance.
UPD. Deleted wrong idea about cross-correlation with periodic signals
You can try matched filtering in frequency domain
function [corr_output] = pc_corr_processor (target_signal, ref_signal)
L = length(ref_signal);
N = length(target_signal);
matched_filter = flipud(ref_signal')';
matched_filter_Res = fft(matched_filter,N);
corr_fft = matched_filter_Res.*fft(target_signal);
corr_out = abs(ifft(corr_fft));
The peak of the matched filter maximum-index of corr_out above should give you the lag amount.