I have written a function in MATLAB to count the number of zero crossings given a vector of signal data. If I find a zero crossing, I also check whether the absolute difference between the two vector indices involved is greater than a threshold value - this is to try to reduce the influence of signal noise.
zc = [];
thresh = 2;
for i = 1:length(v)-1
if ( (v(i)>0 && v(i+1)<0) || (v(i)<0 && v(i+1)>0) ) && abs(v(i)-v(i+1)) >= thresh
zc = [zc; i+1];
end
end
zcCount = length(zc);
I used the vector from the zero crossings function here to test it: http://hips.seas.harvard.edu/content/count-zero-crossings-matlab
A = [-0.49840598306643,
1.04975509964655,
-1.67055867973620,
-2.01437026154355,
0.98661592496732,
-0.06048256273708,
1.19294080740269,
2.68558025885591,
0.85373360483580,
1.00554850567375];
It seems to work fine but is there a more efficient way of achieving the same result? E.g. on the above webpage, they simply use the following line to calculate zero crossings:
z=find(diff(v>0)~=0)+1;
Is there a way to incorporate the threshold check into something similarly efficient?
How about
zeroCrossIndex=diff(v>0)~=0
threshholdIndex = diff(v) >= thresh;
zcCount = sum(zeroCrossIndex & threshholdIndex)
Related
I am using MATLAB R2020a on a MacOS. I am trying to remove outlier values in a while loop. This involves calculating an exponentially weighted moving mean and then comparing this a vector value. If the conditions are met, the vector input is then added to a separate vector of 'acceptable' values. The while loop then advances to the next input and calculates the new exponentially weighted moving average which includes the newly accepted vector input.
However, if the condition is not met, I written code so that, instead of adding the input sample, a zero is added to the vector of 'acceptable' values. Upon the next acceptable value being added, I currently have it so the zero immediately before is replaced by the mean of the 2 framing acceptable values. However, this only accounts for one past zero and not for multiple outliers. Replacing with a framing mean may also introduce aliaising errors.
Is there any way that the zeros can instead be replaced by linearly interpolating the "candidate outlier" point using the gradient based on the framing 2 accepted vector input values? That is, is there a way of counting backwards within the while loop to search for and replace zeros as soon as a new 'acceptable' value is found?
I would very much appreciate any suggestions, thanks in advance.
%Calculate exponentially weighted moving mean and tau without outliers
accepted_means = zeros(length(cycle_periods_filtered),1); % array for accepted exponentially weighted means
accepted_means(1) = cycle_periods_filtered(1);
k = zeros(length(cycle_periods_filtered),1); % array for accepted raw cycle periods
m = zeros(length(cycle_periods_filtered), 1); % array for raw periods for all cycles with outliers replaced by mean of framing values
k(1) = cycle_periods_filtered(1);
m(1) = cycle_periods_filtered(1);
tau = m/3; % pre-allocation for efficiency
i = 2; % index for counting through input signal
j = 2; % index for counting through accepted exponential mean values
n = 2; % index for counting through raw periods of all cycles
cycle_index3(1) = 1;
while i <= length(cycle_periods_filtered)
mavCurrent = (1 - 1/w(j))*accepted_means(j - 1) + (1/w(j))*cycle_periods_filtered(i);
if cycle_periods_filtered(i) < 1.5*(accepted_means(j - 1)) && cycle_periods_filtered(i) > 0.5*(accepted_means(j - 1)) % Identify high and low outliers
accepted_means(j) = mavCurrent;
k(j) = cycle_periods_filtered(i);
m(n) = cycle_periods_filtered(i);
cycle_index3(n) = i;
tau(n) = m(n)/3;
if m(n - 1) == 0
m(n - 1) = (k(j) + k(j - 1))/2;
tau(n - 1) = m(n)/3;
end
j = j + 1;
n = n + 1;
else
m(n) = 0;
n = n + 1;
end
i = i + 1;
end
% Scrap the tail
accepted_means(j - 1:end)=[];
k(j - 1:end) = [];
I am writing a Matlab Tool and certain processes have to be automated.
I am running a for loop, in which some decisions need to be made. Here is a piece of my code:
DecisionMatrix = [0.2 0.4; 0.5 0.7];
Beta =0:pi/20:pi;
Span_Loc = 0.5*(1-cos(Beta))';
for i=1:length(Span_Loc)
Position = Span(i)
% Check Clean of High Lift
if Position >= DecisionMatrix(1,1) && Position <= DecisionMatrix(1,2)
% HighLift run code here
elseif Position >= DecisionMatrix(2,1) && Position <= DecisionMatrix(2,2)
else
% Clean run code here
end
end
Herein, DecisionMatrix is a variable size matrix which is nx2 always. What I want to do is to determine when the value of Position is between the entries of any row of DecisionMatrix. This should be easy when DecisionMatrix is a constant matrix (as shown above). However, this matrix has a variable number of rows.
Hence, how would you do this?
Thanks in advance!!
To determine when the value of Position (scalar) is between the entries of any row of DecisionMatrix (2-column matrix):
result = any(Position>=DecisionMatrix(:,1) & Position<=DecisionMatrix(:,2));
The above gives a logical result (true or false). If you need to know the indices of the rows that fulfill the condition:
result = find(Position>=DecisionMatrix(:,1) & Position<=DecisionMatrix(:,2));
You can fix your code by introducing another loop and coming out of it when you find the required row.
DecisionMatrix = [0.2 0.4; 0.5 0.7];
Beta =0:pi/20:pi;
Span_Loc = 0.5*(1-cos(Beta))';
for p=1:length(Span_Loc)
Position = Span(p);
for q=1:n
if Position >= DecisionMatrix(q,1) && Position <= DecisionMatrix(q,2)
%do what you want when the condition is true
break
end
end
end
I've run an experiment where a machine exerts a force on a bridge until it breaks. I need to cycle through my entire data set and calculate the toughness until I've hit the breaking point (fdValue1).
The toughness is calculated by summing up all the rectangles under my load vs. distance curve. (Basically the integral)
However, I have not been able to find a reasonable way of doing so and my current loop is an infinite loop and I'm not sure why.
%Initializing variables
bridge1Data = xlsread('Bridge1Data.xlsx', 'A2:C2971');
bridge2Data = xlsread('Bridge2Data.xlsx', 'A2:C1440');
bridge1Load = bridge1Data(:, 2);
bridge2Load = bridge2Data(:, 2);
bridge1Dist = bridge1Data(:, 3);
bridge2Dist = bridge2Data(:, 3);
[row1, col1] = size(bridge1Dist);
[row2, col2] = size(bridge2Dist);
bridge1Disp = zeros(row1, col1);
bridge2Disp = zeros(row2, col2);
fdValue1 = 0.000407350000000029;
&Main code
%Displacement
for k = 2:length(bridge1Dist)
bridge1Disp(k-1, 1) = bridge1Dist(k, 1) - bridge1Dist(k-1, 1);
end
%Max Load Bridge 1
maxLoad1 = 0;
for n = 1:length(bridge1Load)
for k = 1
if bridge1Load(n, k) > maxLoad1
maxLoad1 = bridge1Load(n, k);
end
end
end
%Cycle through data till failure, change proj data
totalRect1 = 0;
for j = 2:length(bridge1Disp)
while bridge1Disp(j, 1) ~= fdValue1
rectangle = (bridge1Disp(j, 1) - bridge1Disp(j-1, 1))*...
((bridge1Load(j, 1) + bridge1Load(j-1, 1))/2);
totalRect1 = totalRect1 + rectangle;
end
end
Basically, I make an array for the load and distance the machine pushes down on the bridge, assign a 'Failure Distance' value (fdValue) which should be used to determine when we stop calculating toughness. I then calculate displacement, calculate the maximum load. Then through the variable 'rectangle', calculate each rectangle and sum them all up in 'totalRect1', and use that to calculate the toughness by finding the area under the curve. Is anyone able to see why the loop is an infinite loop?
Thanks
The problem with the condition while bridge1Disp(j, 1) ~= fdValue1 is that you need to check for <= and not for (in)equality, since double numbers will almost never evaluate to be equal even if they seem so. To read more about this you can look here and also google for matlab double comparison. Generally it has something to do with precision issues.
Usually when checking for double equality you should use something like if abs(val-TARGET)<1E-4, and specify some tolerance which you are willing to permit.
Regardless,
You don't need to use loops for what you're trying to do. I'm guessing this comes from some C-programming habits, which are not required in MATLAB.
The 1st loop (Displacement), which computes the difference between every two adjacent elements can be replaced by the function diff like so:
bridge1Disp = diff(bridge1Dist);
The 2nd loop construct (Max Load Bridge 1), which retrieves the maximum element of bridge1Load can be replaced by the command max as follows:
maxLoad1 = max(bridge1Load);
For the last loop construct (Cycle ...) consider the functions I've mentioned above, and also find.
In the code section
%Cycle through data till failure, change proj data
totalRect1 = 0;
for j = 2:length(bridge1Disp)
while bridge1Disp(j, 1) ~= fdValue1
rectangle = (bridge1Disp(j, 1) - bridge1Disp(j-1, 1))*...
((bridge1Load(j, 1) + bridge1Load(j-1, 1))/2);
totalRect1 = totalRect1 + rectangle;
end
end
the test condition of the while loop is
bridge1Disp(j, 1) ~= fdValue1
nevertheless, in the while loop the value of bridge1Disp(j, 1) does not change so if at the first iteratiion of the while loop bridge1Disp(j, 1) is ~= fdValue1 the loop will never end.
Hope this helps.
if (pbcg(k+M) > pbcg(k-1+M) && pbcg(k+M) > pbcg(k+1+M) && pbcg(k+M) > threshold)
peaks_y(Counter) = pbcg(k+M);
peaks_x(Counter) = k + M;
py = peaks_y(Counter);
px = peaks_x(Counter);
plot(px,py,'ro');
Counter = (Counter + 1)-1;
fid = fopen('y1.txt','a');
fprintf(fid, '%d\t%f\n', px, py);
fclose(fid);
end
end
this code previously doesn't have any issue on finding the peak..
the main factor for it to find the only peak is this
if (pbcg(k+M) > pbcg(k-1+M) && pbcg(k+M) > pbcg(k+1+M) && pbcg(k+M) > threshold)
but right now it keep show me all the peak that is above the threshold instead of the particular highest peak..
UPDATE: what if the highest peaks have 4nodes that got the same value?
EDIT:
If multiple peaks with the same value surface, I will take the value at the middle and plot.
What I mean by that is for example [1,1,1,4,4,4,2,2,2]
I will take the '4' at the 5th position, so the plot will be at the center of the graph u see
It will be much faster and much more readable to use the built-in max function, and then test if the max value is larger than the threshold.
[C,I] = max(pbcg);
if C > threshold
...
%// I is the index of the maximal value, and C is the maximal value.
end
As alternative solution, you may evaluate the idea of using the built-in function findpeaks, which encompasses several methods to ascertain the existance of peaks within a given signal. Within thos methods you may call
findPeaks = findpeaks(data,'threshold',threshold_resolution);
The only limit I see is that findpeaks is only available with the Signal Processing Toolbox.
EDIT
In case of multiple peaks over the defined threshold, I would just call max to figure the highest peak, as follows
max(peaks);
Assuming you have a vector with peaks pbcg
Here is how you can get the middle one:
highestPeakValue = max(pbcg)
f = find(pbcg == highestPeakValue);
middleHighestPeakLocation = f(ceil(length(f)/2))
Note that you can still make it more robust for cases where you have no peaks, and can adjust it to give different behavior when there are two middle peaks (now it will take the second one)
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.