first i want to calibrate my Beacons, so for this i go 1 meter away and get 60 rssi values and take the average of them. Then I have the receiving signal power at 1m distance from my beacon.
Now I want to calculate the distance based on the following formula:
A represents the receiving signal power at 1 meter distance
K represents the exponent of the path loss
d represents the distance
K depends on the room, in which i want to calculate the distance. What is the best course of action to calculate the variable K for this solution?
Essentially, you need to solve for K and A. To do this, you need to repeat the calibration procedure for other distances to get more data points so you have multiple d values and multiple RSSI values. Then you need to run a regression to find the best fit values for K and A.
That said, I doubt you will have much success with this formula. I have not been able to use it to accurately predict distance. I have found this formula to be a better predictor.
Related
Suppose I have a continuous probability distribution, e.g., Normal, on a support A. Suppose that there is a Matlab code that allows me to draw random numbers from such a distribution, e.g., this.
I want to build a Matlab code to "approximate" this continuous probability distribution with a probability mass function spanning over r points.
This means that I want to write a Matlab code to:
(1) Select r points from A. Let us call these points a1,a2,...,ar. These points will constitute the new discretised support.
(2) Construct a probability mass function over a1,a2,...,ar. This probability mass function should "well" approximate the original continuous probability distribution.
Could you help by providing also an example? This is a similar question asked for Julia.
Here some of my thoughts. Suppose that the continuous probability distribution of interest is one-dimensional. One way to go could be:
(1) Draw 10^6 random numbers from the continuous probability distribution of interest and store them in a column vector D.
(2) Suppose that r=10. Compute the 10-th, 20-th,..., 90-th quantiles of D. Find the median point falling in each of the 10 bins obtained. Call these median points a1,...,ar.
How can I construct the probability mass function from here?
Also, how can I generalise this procedure to more than one dimension?
Update using histcounts: I thought about using histcounts. Do you think it is a valid option? For many dimensions I can use this.
clear
rng default
%(1) Draw P random numbers for standard normal distribution
P=10^6;
X = randn(P,1);
%(2) Apply histcounts
[N,edges] = histcounts(X);
%(3) Construct the new discrete random variable
%(3.1) The support of the discrete random variable is the collection of the mean values of each bin
supp=zeros(size(N,2),1);
for j=2:size(N,2)+1
supp(j-1)=(edges(j)-edges(j-1))/2+edges(j-1);
end
%(3.2) The probability mass function of the discrete random variable is the
%number of X within each bin divided by P
pmass=N/P;
%(4) Check if the approximation is OK
%(4.1) Find the CDF of the discrete random variable
CDF_discrete=zeros(size(N,2),1);
for h=2:size(N,2)+1
CDF_discrete(h-1)=sum(X<=edges(h))/P;
end
%(4.2) Plot empirical CDF of the original random variable and CDF_discrete
ecdf(X)
hold on
scatter(supp, CDF_discrete)
I don't know if this is what you're after but maybe it can help you. You know, P(X = x) = 0 for any point in a continuous probability distribution, that is the pointwise probability of X mapping to x is infinitesimal small, and thus regarded as 0.
What you could do instead, in order to approximate it to a discrete probability space, is to define some points (x_1, x_2, ..., x_n), and let their discrete probabilities be the integral of some range of the PDF (from your continuous probability distribution), that is
P(x_1) = P(X \in (-infty, x_1_end)), P(x_2) = P(X \in (x_1_end, x_2_end)), ..., P(x_n) = P(X \in (x_(n-1)_end, +infty))
:-)
Hi I am using tensorflow at my university to try to classify steering angles of a simulation program using only the images the simulation produces.
The Steering angles are values from -1 to 1 and I separated them into 50 "buckets". So the first value of my prediction vector would mean that the predicted steering angle is between -1 and -0.96.
The following shows the classification and optimization functions I am using.
cost = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(prediction, y))
optimizer = tf.train.AdamOptimizer(0.001).minimize(cost)
y is a vector that with 49 zeros and a single 1 for the correct bucket. My question now is.
How do I take into account if e.g. the correct bucket is at index 25, that the a prediction of 26 is much better than a prediction of 48.
I didn't post the actual network since it is just a couple of conv2d and maxpool layers with a fully connected layer at the end.
Since you are applying Cross entropy or negative log likelihood. you are penalizing the system given the predicted output and the ground truth.
So saying that your system predicted different numbers on your 50 classes output and the highest one was the class number 25 but your ground truth is class 26. So your system will take the value predicted on 26 and adapt the parameters to produce the highest number on this output the next time it sees this input.
You could do two basic things:
Change your y and prediction to be scalars in the range -1..1; make the loss function be (y-prediction)**2 or something. A very different model, but perhaps more reasonable that the one-hot.
Keep the one-hot target and loss, but have y = target*w, where w is a constant matrix, mostly zeros, 1s on the diagonal, and smaller values on the next diagonal, elements (e.g. y(i) = target(i) * 1. + target(i-1) * .5 + target(i+1) * .5 + ...); kind of gross, but it should converge to something reasonable.
Consider a Normal distribution with mean 0 and standard deviation 1. I would like to divide this distribution into 9 regions of equal probability and take a random sample from each region.
It sounds like you want to find the values that divide the area under the probability distribution function into segments of equal probability. This can be done in matlab by applying the norminv function.
In your particular case:
segmentBounds = norminv(linspace(0,1,10),0,1)
Any two adjacent values of segmentBounds now describe the boundaries of segments of the Normal probability distribution function such that each segment contains one ninth of the total probability.
I'm not sure exactly what you mean by taking random numbers from each sample. One approach is to sample from each region by performing rejection sampling. In short, for each region bounded by x0 and x1, draw a sample from y = normrnd(0,1). If x0 < y < x1, keep it. Else discard it and repeat.
It's also possible that you intend to sample from these regions uniformly. To do this you can try rand(1)*(x1-x0) + x0. This will produce problems for the extreme quantiles, however, since the regions extend to +/- infinity.
I have two sensors seperated by some distance which receive a signal from a source. The signal in its pure form is a sine wave at a frequency of 17kHz. I want to estimate the TDOA between the two sensors. I am using crosscorrelation and below is my code
x1; % signal as recieved by sensor1
x2; % signal as recieved by sensor2
len = length(x1);
nfft = 2^nextpow2(2*len-1);
X1 = fft(x1);
X2 = fft(x2);
X = X1.*conj(X2);
m = ifft(X);
r = [m(end-len+1) m(1:len)];
[a,i] = max(r);
td = i - length(r)/2;
I am filtering my signals x1 and x2 by removing all frequencies below 17kHz.
I am having two problems with the above code:
1. With the sensors and source at the same place, I am getting different values of 'td' at each time. I am not sure what is wrong. Is it because of the noise? If so can anyone please provide a solution? I have read many papers and went through other questions on stackoverflow so please answer with code along with theory instead of just stating the theory.
2. The value of 'td' is sometimes not matching with the delay as calculated using xcorr. What am i doing wrong? Below is my code for td using xcorr
[xc,lags] = xcorr(x1,x2);
[m,i] = max(xc);
td = lags(i);
One problem you might have is the fact that you only use a single frequency. At f = 17 kHz, and an estimated speed-of-sound v = 340 m/s (I assume you use ultra-sound), the wavelength is lambda = v / f = 2 cm. This means that your length measurement has an unambiguity range of 2 cm (sorry, cannot find a good link, google yourself). This means that you already need to know your distance to better than 2 cm, before you can use the result of your measurement to refine the distance.
Think of it in another way: when taking the cross-correlation between two perfect sines, the result should be a 'comb' of peaks with spacing equal to the wavelength. If they overlap perfectly, and you displace one signal by one wavelength, they still overlap perfectly. This means that you first have to know which of these peaks is the right one, otherwise a different peak can be the highest every time purely by random noise. Did you make a plot of the calculated cross-correlation before trying to blindly find the maximum?
This problem is the same as in interferometry, where it is easy to measure small distance variations with a resolution smaller than a wavelength by measuring phase differences, but you have no idea about the absolute distance, since you do not know the absolute phase.
The solution to this is actually easy: let your source generate more frequencies. Even using (band-limited) white-noise should work without problems when calculating cross-correlations, and it removes the ambiguity problem. You should see the white noise as a collection of sines. The cross-correlation of each of them will generate a comb, but with different spacing. When adding all those combs together, they will add up significantly only in a single point, at the delay you are looking for!
White Noise, Maximum Length Sequency or other non-periodic signals should be used as the test signal for time delay measurement using cross correleation. This is because non-periodic signals have only one cross correlation peak and there will be no ambiguity to determine the time delay. It is possible to use the burst type of periodic signals to do the job, but with degraded SNR. If you have to use a continuous periodic signal as the test signal, then you can only measure a time delay within one period of the periodic test signal. This should explain why, in your case, using lower frequency sine wave as the test signal works while using higher frequency sine wave does not. This is demonstrated in these videos: https://youtu.be/L6YJqhbsuFY, https://youtu.be/7u1nSD0RlwY .
If I have wind speed measurements for 4 different locations within a geographical radius of approximately 400km for one year, is there a method for determining which wind speed measurement best fits all of the location i.e. does one of the locations have a similar wind speed to all other locations? Can this be achieved?
I suppose you could find the one that provides minimum e.g. quadratic loss from all the others:
speeds is an N-by-4 matrix, with N windspeed measurements for each location
%Loss will find the squared loss for location i. It subtracts column i from each column in speeds, and squares this difference (for column i, this will always be 0). Then average over all rows and the 3 non-zero columns.
loss = #(i)sum(mean((speeds - repmat(speeds(:, i), 1, 4)).^2)) ./ 3;
%Apply loss to each of the 4 locations, find the minimum.
[v i] = min(arrayfun(loss, 1:4));
The loss function takes the average squared difference between each windspeed and the speeds at all other locations. Then we use arrayfun to calculate this loss for each location.