How can I properly plot a peak in Matlab with a given height? My current method:
x = linspace(0,500,500);
peaks = zeros(size(x));
peaks(50) = 5;
peaks(300) = 20;
peaks(302) = 17;
peaks(375) = 15;
plot(x,peaks)
which gives
But this is ugly, confusing when multiple lines are close and leads to problems when converting it to a log scale since log(0)=-Inf. Is there a proper way of plotting peaks?
Context: I'm trying to analyse a spectogram (EDXS) to identify which material I'm working with.
Scaling the scale to log, you are asking Matlab to draw a line between -Inf and a positive value. Such lines are skipped.
To get a usable log plot, use points instead of lines:
plot(x,peaks,'x')
Another nice possibility to plot peaks is a stem-plot
I think the reason why you only get single pixels in your log-plot is because most of your values are 0. As log(0) = -Inf, this can't be plotted.
I know nothing of spectogram analyses, but if you absolutely want to plot this with a logarithmic scale, is it possible to use a "base line" equal 1 (0 in a log scale). If so:
x = linspace(0,100);
peaks = ones(size(x));
peaks(10) = 5;
peaks(60) = 20;
peaks(75) = 15;
semilogy(x,peaks)
I guess that this will look more reasonable with your real data. Note: If you have values that are less than 1, I would not go with this approach!
Related
I'm trying to make a plot with density of the zeros of random quadratic monic polynomials in the complex plane. In fact, I am plotting! But there is just a little detail which is bugging me: the values of the axes are not matching with the points in the plot. Here is my code.
n=2;
p = [1 random('Uniform', -1, 1, [1,n])]
roots(p)
z = zeros(0);
n = 2;
for j=1:10000
p = [1 random('Uniform', -1, 1, [1,n])];
R = roots(p);
z = [ z, R.' ];
end
Re = real(z);
Im = imag(z);
[values, centers] = hist3([Im(:) Re(:)],[1000 1000]);
imagesc(centers{:}, values,[0,10]);
colorbar
axis equal
axis xy
cmap = summer(max(values(:)));
cmap(1:1,:) = 0;
colormap(cmap);
Now here is the plot generated by this code.
You can try this code and check max(Re) and max(Im), which correspond to the maximum values of the real part (x axis) and imaginary part (y axis). I have max(Re) = 1.6076 (it's always something close to 1.5) and max(Im) = 0.9993 (it's always something close to 1). These values doesn't match the plot, which seems to be the opposite.
If I try the scatter function (losing density and all the nice visual), I have the right values. The following command generates the figure below.
scatter(Re(1,:), Im(1,:),'.')
This clearly shows that the first plot is actually(not rotated as I was thinking first) correct, except for the axes values. I need a help to fix this. Thanks.
PS: The commands to make this plot I got in the answer here. Note the comments there. I explicitly asked for a solution for this problem and got one. The given solution actually worked in some cases but failed in this one, I don't know why.
I believe what you are looking for should be:
imagesc(centers{[2,1]}, values,[0,10]);
Incidently, you did not spot the problem in the other post is because the sample images all happen to be more or less square.
I created the following code in order to find the natural frequencies of a test sample which is excited by use of an impact hammer and has an accelerometer attached to it. However, I got stuck at interp_accelerance_dB_first. This interpolation creates a set of NaN values and I don't know why. I find it strange since interp_accelerance worked fine. I hope someone can help me!
N = 125000;
fs = 1/(x(2)-x(1));
ts = 1/fs;
f = -fs/2:fs/(N-1):fs/2;
% Set x-axis of graph
x_max = (N-1)*ts;
x_axis=0:ts:x_max;
% find the first natural frequency between these boundaries
First_lower_boundary = 15;
First_upper_boundary = 30;
Input = abs(fft(y)); %FFT input force
Output = abs(fft(o)); %FFT output acceleration
Accelerance = Output./Input;
bin_vals = [0 : N-1];
fax_Hz = bin_vals*fs/N;
N_2 = ceil(N/2);
% Interpolate accelerance function in order to be able to average all accelerance functions
Interp_accelerance = interp1(fax_Hz(1:N_2),Accelerance(1:N_2),x_axis);
% --- Find damping ratio of first natural frequency
% Determine the x-axis (from the boundries at the beginning of this script)
x_axis_first_peak = First_lower_boundary:ts:First_upper_boundary;
% Accelerance function with a logarithmic scale [dB]
Accelerance_dB_first = 20*log10(Accelerance(First_lower_boundary:First_upper_boundary));
% Interpolate the accelerance function [dB]
Interp_accelerance_dB_first = interp1(fax_Hz(First_lower_boundary:First_upper_boundary),Accelerance_dB_first,x_axis_first_peak);
Hard to say for sure without knowing what x,y,o are, but generally interp1 returns NaN when you try to interpolate outside of the bounds of the data's x-axis. Append the following at the end of your code:
[min(fax_Hz(First_lower_boundary:First_upper_boundary)),max(fax_Hz(First_lower_boundary:First_upper_boundary))]
[min(x_axis_first_peak),max(x_axis_first_peak)]
If the second segment doesn't fall inside of the first segment, then you've found your problem.
Incidentally, I think that interp_accelerance may be susceptible to the same error, again depending on the input parameters' exact nature.
I want to plot the area above and below a particular value in x axis.
The problem i am facing is with discrete values. The code below for instance has an explicit X=10 so i have written it in such a way that i can find the index and calculate the values above and below that particular value but if i want to find the area under the curve above and below 4 this program will now work.
Though in the plot matlab does a spline fitting(or some sort of fitting for connecting discrete values) there is a value for y corresponding to x=4 that matlab computes i cant seem to store or access it.
%Example for Area under the curve and partial area under the curve using Trapezoidal rule of integration
clc;
close all;
clear all;
x=[0,5,10,15,20];% domain
y=[0,25,50,25,0];% Values
LP=log2(y);
plot(x,y);
full = trapz(x,y);% plot of the total area
I=find(x==10);% in our case will be the distance value up to which we want
half = trapz(x(1:I),y(1:I));%Plot of the partial area
How can we find the area under the curve for a value of ie x = 2 or 3 or 4 or 6 or 7 or ...
This is an elaboration of patrik's comment, "first interpolate and then integrate".
For the purpose of this answer I'll assume that the area in question is the area that can be seen in the plot, and since plot connects points by straight lines I assume that linear interpolation is adequate. Moreover, since the trapezoidal rule itself is based on linear interpolation, we only need interpolated values at the beginning and end of the interval.
Starting from the given points
x = [0, 5, 10, 15, 20];
y = [0, 25, 50, 25, 0];
and the integration interval limits, say
xa = 4;
xb = 20;
we first select the data points within the limits
ind = (x > xa) & (x < xb);
xw = x(ind);
yw = y(ind);
and then complete them by interpolation values at the edges:
ya = interp1(x, y, xa);
yb = interp1(x, y, xb);
xw = [xa, xw, xb];
yw = [ya, yw, yb];
Now we can simply apply trapezoidal integration:
area = trapz(xw, yw);
I think that you either need more samples, or to interpolate the data. Another alternative is to use a function handle. Then you need to know the function though. Example using linear interpolation follows.
x0 = [0;5;10;15;20];
y0 = [0,25,50,25,0];
x1 = 0:20;
y1 = interp1(x0,y0,x1,'linear');
xMax = 4;
partInt = trapz(x1(x1<=xMax),y1(x1<=xMax));
Some other kind of interpolation may be suitable, but that is hard to say without more information. Also, this interpolates from the beginning to x. However, I guess figuring out how to change the limits should be easy from here. This solution is different than the former, since it is less depending on the pyramid shape of the data. So to say, it is more general.
Hi i'm having a problem where I have a dataset which ranges between -10^3 to 10^3
I need to be able to plot this as with a log scale but semilogy cannot plot negative values
Say for example my data is:
x = [-3,-2,-1,0,1,2,3];
y = [-1000,-100,-10,1,10,100,1000];
(or in general y=sign(x).*10.^abs(x);)
How can I plot this in MATLAB with a log scale? If possible It would be great if the log scale ticks could be on the Y-axis too
Use your actual data as labels, but scale the plotted data with log10.
% data
x = -3:0.1:3;
y = sign(x).*10.^abs(x);
% scaling function
scale = #(x) sign(x).*log10(abs(x));
N = 7; % number of ticks desired
% picking of adequate values for the labels
TickMask = linspace(1,numel(y),N);
YTickLabels = y(TickMask);
% scale labels and plotdata, remove NaN ->inconsistency, do you really want that?
YTick = scale( YTickLabels );
Y = scale(y);
YTick(isnan(YTick)) = 0;
Y(isnan(Y)) = 0;
% plot
plot(x,Y)
set(gca,'YTick',YTick,'YTickLabels',YTickLabels)
grid on
For N = 7:
For N = 11
How to find a valid value for N?
The following function (thanks to gnovice) will return all possible values you could choose for N:
n = numel(x);
N = find(rem(n./(1:n), 1) == 0) + 1;
about the semilogy-style labels: by adding the following line before the plot:
YTickLabels = cellfun(#(x) ['10^' num2str(x)], num2cell(YTick),'UniformOutput',false)
you could at least achieve something like this:
not beautiful and not generic, but a good point to start for you.
The reason you can't make a logarithmic axis that crosses zero, is that it doesn't make sense!
Since a logarithmic scale is generally displayed as eg. 100 - 10 - 1 - 1/10 - 1/100 - ..., you would need an infinite amount of space to make the axis cross zero.
How about this:
x=logspace(-3,3);
y=sign(x).*10.^abs(x);
loglog(x,y)
#thewaywewalk has already given a beautiful solution to it. The one I'm suggesting is an epsilon improvement on it. If you make two changes
(a) Define a new MATLAB function signia that basically extracts the sign before a number.
function value = signia(x)
if(x>=0)
value = '';
else
value = '-';
end
and (b) make this little change that instead of
YTickLabels = cellfun(#(x) ['10^' num2str(x)], num2cell(YTick),'UniformOutput',false)
you use
YTickLabels = cellfun(#(x) [signia(x) '10^{' num2str(x) '}'], num2cell(YTick),'UniformOutput',false);
(notice the presence of curly braces), you'll get an improvement in the Y ticks display. I got the following.
enter image description here
I have a vector of data, which contains integers in the range -20 20.
Bellow is a plot with the values:
This is a sample of 96 elements from the vector data. The majority of the elements are situated in the interval -2, 2, as can be seen from the above plot.
I want to eliminate the noise from the data. I want to eliminate the low amplitude peaks, and keep the high amplitude peak, namely, peaks like the one at index 74.
Basically, I just want to increase the contrast between the high amplitude peaks and low amplitude peaks, and if it would be possible to eliminate the low amplitude peaks.
Could you please suggest me a way of doing this?
I have tried mapstd function, but the problem is that it also normalizes that high amplitude peak.
I was thinking at using the wavelet transform toolbox, but I don't know exact how to reconstruct the data from the wavelet decomposition coefficients.
Can you recommend me a way of doing this?
One approach to detect outliers is to use the three standard deviation rule. An example:
%# some random data resembling yours
x = randn(100,1);
x(75) = -14;
subplot(211), plot(x)
%# tone down the noisy points
mu = mean(x); sd = std(x); Z = 3;
idx = ( abs(x-mu) > Z*sd ); %# outliers
x(idx) = Z*sd .* sign(x(idx)); %# cap values at 3*STD(X)
subplot(212), plot(x)
EDIT:
It seems I misunderstood the goal here. If you want to do the opposite, maybe something like this instead:
%# some random data resembling yours
x = randn(100,1);
x(75) = -14; x(25) = 20;
subplot(211), plot(x)
%# zero out everything but the high peaks
mu = mean(x); sd = std(x); Z = 3;
x( abs(x-mu) < Z*sd ) = 0;
subplot(212), plot(x)
If it's for demonstrative purposes only, and you're not actually going to be using these scaled values for anything, I sometimes like to increase contrast in the following way:
% your data is in variable 'a'
plot(a.*abs(a)/max(abs(a)))
edit: since we're posting images, here's mine (before/after):
You might try a split window filter. If x is your current sample, the filter would look something like:
k = [L L L L L L 0 0 0 x 0 0 0 R R R R R R]
For each sample x, you average a band of surrounding samples on the left (L) and a band of surrounding samples on the right. If your samples are positive and negative (as yours are) you should take the abs. value first. You then divide the sample x by the average value of these surrounding samples.
y[n] = x[n] / mean(abs(x([L R])))
Each time you do this the peaks are accentuated and the noise is flattened. You can do more than one pass to increase the effect. It is somewhat sensitive to the selection of the widths of these bands, but can work. For example:
Two passes:
What you actually need is some kind of compression to scale your data, that is: values between -2 and 2 are scale by a certain factor and everything else is scaled by another factor. A crude way to accomplish such a thing, is by putting all small values to zero, i.e.
x = randn(1,100)/2; x(50) = 20; x(25) = -15; % just generating some data
threshold = 2;
smallValues = (abs(x) <= threshold);
y = x;
y(smallValues) = 0;
figure;
plot(x,'DisplayName','x'); hold on;
plot(y,'r','DisplayName','y');
legend show;
Please do not that this is a very nonlinear operation (e.g. when you have wanted peaks valued at 2.1 and 1.9, they will produce very different behavior: one will be removed, the other will be kept). So for displaying, this might be all you need, for further processing it might depend on what you are trying to do.
To eliminate the low amplitude peaks, you're going to equate all the low amplitude signal to noise and ignore.
If you have any apriori knowledge, just use it.
if your signal is a, then
a(abs(a)<X) = 0
where X is the max expected size of your noise.
If you want to get fancy, and find this "on the fly" then, use kmeans of 3. It's in the statistics toolbox, here:
http://www.mathworks.com/help/toolbox/stats/kmeans.html
Alternatively, you can use Otsu's method on the absolute values of the data, and use the sign back.
Note, these and every other technique I've seen on this thread is assuming you are doing post processing. If you are doing this processing in real time, things will have to change.