I am stuck at something in MATLAB and would appreciate if someone can help. Here is what I would like to do. I have a velocity field data w.r.t. time (see figure attached).
I would like to count number of time intervals (T1,T2,T3,etc) for which value of velocity is greater than threshold. Please note that I would like to count time intervals and not number of velocity points above threshold.
After doing step 1, I would like to somehow calculate average of time duration (e.g. average of T1, T2, T3, etc).
Finally, I want to calculate average value of velocity data for intervals T1, T2, T3, etc.
I think step 2 and 3 should be straightforward if I can get step 1 done. I would really appreciate any help. I am doing this in MATLAB but I am open to use any other tools if required.
I'd suggest the use of bwconncomp:
A = rand(1,100); %your 1d list of speeds
T = 0.5;
B = A > T;
tmp = bwconncomp(B);
pxlist = tmp.PixelIdxList; %connected components are above thresh
N = zeros(1,length(pxlist));
V = zeros(1,length(pxlist));
for li = 1:length(pxlist)
N(li) = length(pxlist{li});
V(li) = mean(A(pxlist{li})); %avg velocity in interval li
end
avg = mean(N); %this is the avg. "time", i.e. length of connected components
Related
I am trying to analyze timeseries of wheel turns that were sampled at 1 minute intervals for 10 days. t is a 1 x 14000 array that goes from .1666 hours to 240 hours. analysis.timeseries.(grp).(chs) is a 1 x 14000 array for each of my groups of interest and their specific channels that specifize activity at each minute sampled. I'm interested in collecting the maximum power and the frequency it occurs at. My problem is I'm not sure what units f is coming out in. I would like to have it return in cycles per hour and span to a maximum period of 30 hours. I tried to use the Galileo example in the documentation as a guide, but it didn't seem to work.
Below is my code:
groups = {'GFF' 'GMF' 'SFF' 'SMF'};
chgroups = {chnamesGF chnamesGM chnamesSF chnamesSM};
t1 = (t * 3600); %matlab treats this as seconds so convert it to an hour form
onehour = seconds(hours(1));
for i = 1:4
grp = groups{1,i};
chn = chgroups{1,i};
for channel = 1:length(chn)
chs = chn{channel,1};
[pxx,f]= plomb(analysis.timeseries.(grp).(chs),t, 30/onehour,'normalized');
analysis.pxx.(grp).(chs) = pxx;
analysis.f.(grp).(chs) = f;
analysis.lsp.power.(grp).(chs) = max(pxx);
[row,col,v] = find(analysis.pxx.(grp).(chs) == analysis.lsp.power.(grp).(chs));
analysis.lsp.tau.(grp).(chs) = analysis.f.(grp).(chs)(row);
end
end
Not really an answer but it is hard to put a image in a comment.
Judging by this (plomb manual matlab),
I think that pxx is without dimension as for f is is the frequency so 1/(dimension of t) dimension. If your t is in hours I would say h^-1.
So I'd rather say try
[pxx,f]= plomb(analysis.timeseries.(grp).(chs),t*30.0/onehour,'normalized');
I want to define a matrix (ag,t) that can take the values of an arrival process that follows Binomial Distribution. Its values should be different for each timeslot. I am not sure how to follow the problem and would really appreciate help in this regards. The attached picture explains the query.
This is what I have tried but I don't know how to define it such that it different for each time slot.
cluster_nbrs = 3;
G = zeros(1,cluster_nbrs);
p = 0.97; % Probability of packet being transferred
n = 1000; % # of time slots
arrival_gt = zeros(1,cluster_nbrs); % Arrival Process during slot t
for q = 1:cluster_nbrs
% Probability that arrival = n
arrival_gt(q) = binopdf(G(q),n,p);
end
I can calculate daily averages of a data set like so:
Jday = datenum('2010-11-01 00:00','yyyy-mm-dd HH:MM'):60/(60*24):...
datenum('2011-02-31 23:00','yyyy-mm-dd HH:MM');
Dat = rand(length(Jday),1);
DateV = datevec(Jday);
[~,~,b] = unique(DateV(:,1:3),'rows');
AvDat = abs(accumarray(b,Dat,[],#nanmean));
AvJday = abs(accumarray(b,Jday,[],#nanmean));
However, I would like to take an average of a data set given a number for the output resolution. For example, if I wrote
outRes = 86400; % in seconds
I would like to average the values so that the output resolution is equal to 86400 seconds, and if the outRes defined is shorter than the resolution of the data then no averaging will take place.
How can this be done?
You should be able to detect the resolution of the data by:
CurrentRes=mean(diff(Jday))*86400;
if (~all(diff(Jday)==CurrentRes))
error('Inequally distributed sampling times');
end
if (CurrentRes>outRes)
return;
Then the amount you need to downsample by is:
AveragingFactor=outRes/CurrentRes;
Then you can average by discarding some samples at the end (or start) and averaging:
Dat = reshape(Dat(1:end-mod(length(Dat),AveragingFactor),:),AveragingFactor,[]);
AvDat = mean(Dat,1)'; % Transpose to keep it as a column vector if necessary.
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.