I have neural data collected across 16 different channels. This data was recorded over a 30 second period.
Over a 10s period (from 20 - 30s), I want to record the number of neural data points that are greater than or equal to a specified threshold. I would like to do this according to bins of 0.001s.
I am using MATLAB 2019b.
My code so far looks like this:
t1 = 20;
t2 = 30;
ind1 = find(tim_trl>=t1, 1);
ind2 = find(tim_trl>=t2, 1);
time1 = tim_trl(ind1:ind2); %10s window
sampRate = 24414; %sampling freq (Hz), samples per sec
muaWindow = 0.001; %1ms window
binWidth = round(muaWindow*sampRate); %samples per 1ms window
threshold = 0.018;
for jj = 1:16 %ch
data = AbData(ind1:ind2, jj); %10 sec of data
for kk = 1:10000
abDataBin = data(1:binWidth,jj); %data in 1 bin
dataThreshold = find(abDataBin >= threshold); %find data points >= threshold
mua(kk,jj) = sum(dataThreshold); %number of data pts over threshold per ch
end
end
So far, I'm just having a bit of trouble at this point:
abDataBin = data(1:binWidth,jj); %data in 1 bin
When I run the loop, the data in bin 1 gets overwritten, rather than shift to bin 2, 3...10000. I'd appreciate any feedback on fixing this.
Many thanks.
You forgot to use the running variable as index to access your data. Try
% create data with 16 channels
AbData = rand(10000,16);
binWidth = 24;
threshold = 0.001;
for channel=1:16
data = AbData(2001:3000,channel);
counter = 1; % needed for mua indexing
% looping over the bin starting indeces
for window=1:binWidth:length(data)-(binWidth)
% access data in one bin
bindata = data(window:window+binWidth);
% calculate ms above threshold
mua(counter, channel) = sum(bindata >= threshold);
counter = counter+1;
end
end
EDIT:
your data variable is of dimension nx1, therefore doenst need the column indexing with jj
I am struggling to plot the PDF and CDF graphs of where
Sn=X1+X2+X3+....+Xn
using central limit theorem where n = 1; 2; 3; 4; 5; 10; 20; 40
I am taking Xi to be a uniform continuous random variable for values between (0,3).
Here is what i have done so far -
close all
%different sizes of input X
%N=[1 5 10 50];
N = [1 2 3 4 5 10 20 40];
%interval (1,6) for random variables
a=0;
b=3;
%to store sum of differnet sizes of input
for i=1:length(N)
%generates uniform random numbers in the interval
X = a + (b-a).*rand(N(i),1);
S=zeros(1,length(X));
S=cumsum(X);
cd=cdf('Uniform',S,0,3);
plot(cd);
hold on;
end
legend('n=1','n=2','n=3','n=4','n=5','n=10','n=20','n=40');
title('CDF PLOT')
figure;
for i=1:length(N)
%generates uniform random numbers in the interval
X = a + (b-a).*rand(N(i),1);
S=zeros(1,length(X));
S=cumsum(X);
cd=pdf('Uniform',S,0,3);
plot(cd);
hold on;
end
legend('n=1','n=2','n=3','n=4','n=5','n=10','n=20','n=40');
title('PDF PLOT')
My output is nowhere near what I am expecting any help is much appreciated.
This can be done with vectorization using rand() and cumsum().
For example, the code below generates 40 replications of 10000 samples of a Uniform(0,3) distribution and stores in X. To meet the Central Limit Theorem (CLT) assumptions, they are independent and identically distributed (i.i.d.). Then cumsum() transforms this into 10000 copies of the Sn = X1 + X2 + ... where the first row is n = 10000copies of Sn = X1, the 5th row is n copies of S_5 = X1 + X2 + X3 + X4 + X5. The last row is n copies of S_40.
% MATLAB R2019a
% Setup
N = [1:5 10 20 40]; % values of n we are interested in
LB = 0; % lowerbound for X ~ Uniform(LB,UB)
UB = 3; % upperbound for X ~ Uniform(LB,UB)
n = 10000; % Number of copies (samples) for each random variable
% Generate random variates
X = LB + (UB - LB)*rand(max(N),n); % X ~ Uniform(LB,UB) (i.i.d.)
Sn = cumsum(X);
You can see from the image that the n = 2 case, the sum is indeed a Triangular(0,3,6) distribution. For the n = 40 case, the sum is approximately Normally distributed (Gaussian) with mean 60 (40*mean(X) = 40*1.5 = 60). This shows the convergence in distribution for both the probability density function (PDF) and the cumulative distribution function (CDF).
Note: The CLT is often stated with convergence in distribution to a Normal distribution with zero mean as it has been shifted. Shifting the results by subtracting mean(Sn) = n*mean(X) = n*0.5*(LB+UB) from Sn gets this done.
Code below isn't the gold standard but it produced the image.
figure
s(11) = subplot(6,2,1) % n = 1
histogram(Sn(1,:),'Normalization','pdf')
title(s(11),'n = 1')
s(12) = subplot(6,2,2)
cdfplot(Sn(1,:))
title(s(12),'n = 1')
s(21) = subplot(6,2,3) % n = 2
histogram(Sn(2,:),'Normalization','pdf')
title(s(21),'n = 2')
s(22) = subplot(6,2,4)
cdfplot(Sn(2,:))
title(s(22),'n = 2')
s(31) = subplot(6,2,5) % n = 5
histogram(Sn(5,:),'Normalization','pdf')
title(s(31),'n = 5')
s(32) = subplot(6,2,6)
cdfplot(Sn(5,:))
title(s(32),'n = 5')
s(41) = subplot(6,2,7) % n = 10
histogram(Sn(10,:),'Normalization','pdf')
title(s(41),'n = 10')
s(42) = subplot(6,2,8)
cdfplot(Sn(10,:))
title(s(42),'n = 10')
s(51) = subplot(6,2,9) % n = 20
histogram(Sn(20,:),'Normalization','pdf')
title(s(51),'n = 20')
s(52) = subplot(6,2,10)
cdfplot(Sn(20,:))
title(s(52),'n = 20')
s(61) = subplot(6,2,11) % n = 40
histogram(Sn(40,:),'Normalization','pdf')
title(s(61),'n = 40')
s(62) = subplot(6,2,12)
cdfplot(Sn(40,:))
title(s(62),'n = 40')
sgtitle({'PDF (left) and CDF (right) for Sn with n \in \{1, 2, 5, 10, 20, 40\}';'note different axis scales'})
for tgt = [11:10:61 12:10:62]
xlabel(s(tgt),'Sn')
if rem(tgt,2) == 1
ylabel(s(tgt),'pdf')
else % rem(tgt,2) == 0
ylabel(s(tgt),'cdf')
end
end
Key functions used for plot: histogram() from base MATLAB and cdfplot() from the Statistics toolbox. Note this could be done manually without requiring the Statistics toolbox with a few lines to obtain the cdf and then just calling plot().
There was some concern in comments over the variance of Sn.
Note the variance of Sn is given by (n/12)*(UB-LB)^2 (derivation below). Monte Carlo simulation shows our samples of Sn do have the correct variance; indeed, it converges to this as n gets larger. Simply call var(Sn(40,:)).
% with n = 10000
var(Sn(40,:)) % var(S_40) = 30 (will vary slightly depending on random seed)
(40/12)*((UB-LB)^2) % 29.9505
You can see the convergence is very good by S_40:
step = 0.01;
Domain = 40:step:80;
mu = 40*(LB+UB)/2;
sigma = sqrt((40/12)*((UB-LB)^2));
figure, hold on
histogram(Sn(40,:),'Normalization','pdf')
plot(Domain,normpdf(Domain,mu,sigma),'r-','LineWidth',1.4)
ylabel('pdf')
xlabel('S_n')
Derivation of mean and variance for Sn:
For the expectation (mean), the second equality holds by linearity of expectation. The third equality holds since X_i are identically distributed.
The discrete version of this is posted here.
In the below code, each y is of 8 min duration. Once I generate the Powerr matrix I have what I Need but How do i do 3 hour averages? Should i do within the for loop or once I generate the Powerr matrix can i still be able to find out the 3 hourly averages of my data? thanks in advance.
tic
profile on
warning('off','MATLAB:audiovideo:wavread:functionToBeRemoved');
neglect_the_first_few= 5;
lastfile = 10;
Powerr = zeros(2049,10);
count =[];
parfor count = neglect_the_first_few + 1 :lastfile
% time_spectrum(timeindex) = timecount;
warning('off','MATLAB:audiovideo:wavread:functionToBeRemoved');
% infiles = find(timestamp >= timecount & timestamp <= (timecount + timestep));
% if ~isempty(infiles)
% if isnan(time_spectrum)
% time_spectrum(timeindex) = NaN
% end
y= [];
% for inputfile = infiles'
fn = [directory '/001/' file_names(count,:)];
[y,fs,nbits] = wavread(fn,'native');
disp(['sampling freq is ',num2str(fs),' hz. length of y is ',num2str(size(y)),'filenumber is ',num2str(count)])
% end
y = y- mean(y);
Y = double(y).*(1.49365435131571e-07);%CONVERSION - volts per bit for 24bit channel
% clear y;
nooverlap =[];%number of overlapped segments set to zero on Nov20th 2017
[Poww,fr] = pwelch(Y,hanning_window,nooverlap,nfft,fs);%no overlap taken into account%
Powerr(:,count) = Poww;
end
fr is independent from the parfor iteration count. So,this variable did not get transferred to the host from workers. Write fr(count)=...
You've done it correctly for Powerr.
I have to simulate a simple queue in Matlab using Lindley's equation:
W_{n+1}^Q = max(0, W_n^Q + S_n - X_{n+1}
I think I have done so, with the following code, but I am trying to run it several times and cannot save the information correct. The variables I want to save at the end of running the simulation are only saving the information from the last attempt (here for m=3).. while I clearly would like to see this for all runs (m=1,2,3).
for m=1:3
l = 1.1; % try this value for lambda
N = 10000; % let 1000 people arrive
X = exprnd(l,[1,N]); % make 1000 exponential interarrivals
S = 2*rand(1,N); % uniform on [0,2]
w = zeros(1,N);
sum1 = zeros(1,m);
avg1 = zeros(1,m);
max1 = zeros(1,m);
for i=1:N
if i==1 % first customer doesn't have to wait
w(i) = 0;
else % following customers follow lindley's equation
w(i) = max(w(i-1) + S(i-1) - X(i), 0); % n-th customer's waiting time
count(i) = w(i) > 15; % count number of times greater than 15
end
end
max1(m) = max(w);
sum1(m) = sum(count); % sum number of times greater than 15
avg1(m) = sum1(m)/N; % divide by 1000 to get probability delay is greater than 15
end
You are initializing sum1, avg1 and max1 inside the for loop so in every iteration, these variables are set to zero (i.e. by initialization). This is the reason you loose your previous iteration value. To avoid this, initialize sum1, avg1 and max1 before you for loop. Refer below code for reference. HTH
sum1 = zeros(1,m);
avg1 = zeros(1,m);
max1 = zeros(1,m);
for m=1:3
l = 1.1; % try this value for lambda
N = 10000; % let 1000 people arrive
X = exprnd(l,[1,N]); % make 1000 exponential interarrivals
S = 2*rand(1,N); % uniform on [0,2]
w = zeros(1,N);
for i=1:N
if i==1 % first customer doesn't have to wait
w(i) = 0;
else % following customers follow lindley's equation
w(i) = max(w(i-1) + S(i-1) - X(i), 0); % n-th customer's waiting time
count(i) = w(i) > 15; % count number of times greater than 15
end
end
max1(m) = max(w);
sum1(m) = sum(count); % sum number of times greater than 15
avg1(m) = sum1(m)/N; % divide by 1000 to get probability delay is greater than 15
end
I'm trying to get volume-time graph of .wav file. First, I recorded sound (patient exhalations) via android as .wav file, but when I read this .wav file in MATLAB it has negative values. What is the meaning of negative values? Second, MATLAB experts could you please check if the code below does the same as written in my comments? Also another question. Y = fft(WindowArray);
p = abs(Y).^2;
I took the power of values returned from fft...is that correct and what is the goal of this step??
[data, fs] = wavread('newF2');
% read exhalation audio wav file (1 channel, mono)
% frequency is 44100 HZ
% windows of 0.1 s and overlap of 0.05 seconds
WINDOW_SIZE = fs*0.1; %4410 = fs*0.1
array_size = length(data); % array size of data
numOfPeaks = (array_size/(WINDOW_SIZE/2)) - 1;
step = floor(WINDOW_SIZE/2); %step size used in loop
transformed = data;
start =1;
k = 1;
t = 1;
g = 1;
o = 1;
% performing fft on each window and finding the peak of windows
while(((start+WINDOW_SIZE)-1)<=array_size)
j=1;
i =start;
while(j<=WINDOW_SIZE)
WindowArray(j) = transformed(i);
j = j+1;
i = i +1;
end
Y = fft(WindowArray);
p = abs(Y).^2; %power
[a, b] = max(abs(Y)); % find max a and its indices b
[m, i] = max(p); %the maximum of the power m and its indices i
maximum(g) = m;
index(t) = i;
power(o) = a;
indexP(g) = b;
start = start + step;
k = k+1;
t = t+1;
g = g+1;
o=o+1;
end
% low pass filter
% filtering noise: ignor frequencies that are less than 5% of maximum frequency
for u=1:length(maximum)
M = max(maximum); %highest value in the array
Accept = 0.05* M;
if(maximum(u) > Accept)
maximum = maximum(u:length(maximum));
break;
end
end
% preparing the time of the graph,
% Location of the Peak flow rates are estimated
TotalTime = (numOfPeaks * 0.1);
time1 = [0:0.1:TotalTime];
if(length(maximum) > ceil(numOfPeaks));
maximum = maximum(1:ceil(numOfPeaks));
end
time = time1(1:length(maximum));
% plotting frequency-time graph
figure(1);
plot(time, maximum);
ylabel('Frequency');
xlabel('Time (in seconds)');
% plotting volume-time graph
figure(2);
plot(time, cumsum(maximum)); % integration over time to get volume
ylabel('Volume');
xlabel('Time (in seconds)');
(I only answer the part of the question which I understood)
Per default Matlab normalizes the audio wave to - 1...1 range. Use the native option if you want the integer data.
First, in your code it should be p = abs(Y)**2, this is the proper way to square the values returned from the FFT. The reason why you take the absolute value of the FFT return values is because those number are complex numbers with a Real and Imaginary part, therefore the absolute value (or modulus) of an imaginary number is the magnitude of that number. The goal of taking the power could be for potentially obtaining an RMS value (root mean squared) of your overall amplitude values, but you could also have something else in mind. When you say volume-time I assume you want decibels, so try something like this:
def plot_signal(file_name):
sampFreq, snd = wavfile.read(file_name)
snd = snd / (2.**15) #Convert sound array to floating point values
#Floating point values range from -1 to 1
s1 = snd[:,0] #left channel
s2 = snd[:,1] #right channel
timeArray = arange(0, len(snd), 1)
timeArray = timeArray / sampFreq
timeArray = timeArray * 1000 #scale to milliseconds
timeArray2 = arange(0, len(snd), 1)
timeArray2 = timeArray2 / sampFreq
timeArray2 = timeArray2 * 1000 #scale to milliseconds
n = len(s1)
p = fft(s1) # take the fourier transform
m = len(s2)
p2 = fft(s2)
nUniquePts = ceil((n+1)/2.0)
p = p[0:nUniquePts]
p = abs(p)
mUniquePts = ceil((m+1)/2.0)
p2 = p2[0:mUniquePts]
p2 = abs(p2)
'''
Left Channel
'''
p = p / float(n) # scale by the number of points so that
# the magnitude does not depend on the length
# of the signal or on its sampling frequency
p = p**2 # square it to get the power
# multiply by two (see technical document for details)
# odd nfft excludes Nyquist point
if n % 2 > 0: # we've got odd number of points fft
p[1:len(p)] = p[1:len(p)] * 2
else:
p[1:len(p) -1] = p[1:len(p) - 1] * 2 # we've got even number of points fft
plt.plot(timeArray, 10*log10(p), color='k')
plt.xlabel('Time (ms)')
plt.ylabel('LeftChannel_Power (dB)')
plt.show()
'''
Right Channel
'''
p2 = p2 / float(m) # scale by the number of points so that
# the magnitude does not depend on the length
# of the signal or on its sampling frequency
p2 = p2**2 # square it to get the power
# multiply by two (see technical document for details)
# odd nfft excludes Nyquist point
if m % 2 > 0: # we've got odd number of points fft
p2[1:len(p2)] = p2[1:len(p2)] * 2
else:
p2[1:len(p2) -1] = p2[1:len(p2) - 1] * 2 # we've got even number of points fft
plt.plot(timeArray2, 10*log10(p2), color='k')
plt.xlabel('Time (ms)')
plt.ylabel('RightChannel_Power (dB)')
plt.show()
I hope this helps.