I tried this as well:
plot(x(bootsam(:,100)),y(bootsam(:,100)), 'r*') but it was exactly the same to my data! I want to resample my data in 95% confidence interval .
But it seems this command bootstrp doesn't work alone, it needs some function or other commands to combine. Would you help me to figure it out?
I would like to generate some data randomly but behave like my function around the original data, I attached a plot which original data which are red and resampled data are in blue and green colors.
Generally, I would like to use bootstrap to find error for my best-fit parameters. I read in this book:
http://books.google.de/books?id=ekyupqnDFzMC&lpg=PA131&vq=bootstrap&hl=de&pg=PA130#v=onepage&q&f=false
other methods for error analysis my fitted parameters are appreciated.
I suggest you start this way and then adapt it to your case.
% One step at a time.
% Step 1: Suppose you generate a simple linear deterministic trend with
% noise from the standardized Gaussian distribution:
N = 1000; % number of points
x = [(1:N)', ones(N, 1)]; % x values
b = [0.15, 157]'; % parameters
y = x * b + 10 * randn(N, 1); % linear trend with noise
% Step 2: Suppose you want to fit y with a linear equation:
[b_hat, bint1] = regress(y, x); % estimate parameters with linear regression
y_fit = x * b_hat; % calculate fitted values
resid = y - y_fit; % calculate residuals
plot(x(:, 1), y, '.') % plot
hold on
plot(x(:, 1), y_fit, 'r', 'LineWidth', 5) % fitted values
% Step 3: use bootstrap approach to estimate the confidence interval of
% regression parameters
N_boot = 10000; % size of bootstrap
b_boot = bootstrp(N_boot, #(bootr)regress(y_fit + bootr, x), resid); % bootstrap
bint2 = prctile(b_boot, [2.5, 97.5])'; % percentiles 2.5 and 97.5, a 95% confidence interval
% The confidence intervals obtained with regress and bootstrp are
% practically identical:
bint1
bint2
Related
I am trying to get the output of a Gaussian pulse going through a coax cable. I made a vector that represents a coax cable; I got attenuation and phase delay information online and used Euler's equation to create a complex array.
I FFTed my Gaussian vector and convoluted it with my cable. The issue is, I can't figure out how to properly iFFT the convolution. I read about iFFt in MathWorks and looked at other people's questions. Someone had a similar problem and in the answers, someone suggested to remove n = 2^nextpow2(L) and FFT over length(t) instead. I was able to get more reasonable plot from that and it made sense to why that is the case. I am confused about whether or not I should be using the symmetry option in iFFt. It is making a big difference in my plots. The main reason I added the symmetry it is because I was getting complex numbers in the iFFTed convolution (timeHF). I would truly appreciate some help, thanks!
clc, clear
Fs = 14E12; %1 sample per pico seconds
tlim = 4000E-12;
t = -tlim:1/Fs:tlim; %in pico seconds
ag = 0.5; %peak of guassian
bg = 0; %peak location
wg = 50E-12; %FWHM
x = ag.*exp(-4 .* log(2) .* (t-bg).^2 / (wg).^2); %Gauss. in terms of FWHM
Ly = x;
L = length(t);
%n = 2^nextpow2(L); %test output in time domain with and without as suggested online
fNum = fft(Ly,L);
frange = Fs/L*(0:(L/2)); %half of the spectrum
fNumMag = abs(fNum/L); %divide by n to normalize
% COAX modulation ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
%phase data
mu = 4*pi*1E-7;
sigma_a = 2.9*1E7;
sigma_b = 5.8*1E6;
a = 0.42E-3;
b = 1.75E-3;
er = 1.508;
vf = 0.66;
c = 3E8;
l = 1;
Lso = sqrt(mu) /(4*pi^3/2) * (1/(sqrt(sigma_a)*a) + 1/(b*sqrt(sigma_b)));
Lo = mu/(2*pi) * log(b/a);
%to = l/(vf*c);
to = 12E-9; %measured
phase = -pi*to*(frange + 1/2 * Lso/Lo * sqrt(frange));
%attenuation Data
k1 = 0.34190;
k2 = 0.00377;
len = 1;
mldb = (k1 .* sqrt(frange) + k2 .* frange) ./ 100 .* len ./1E6;
mldb1 = mldb ./ 0.3048; %original eqaution is in inch
tfMag = 10.^(mldb1./-10);
% combine to make in complex form
tfC = [];
for ii = 1: L/2 + 1
tfC(ii) = tfMag(ii) * (cosd(phase(ii)) + 1j*sind(phase(ii)));
end
%END ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
%convolute both h and signal
fNum = fNum(1:L/2+1);
convHF = tfC.*fNum;
convHFMag = abs(convHF/L);
timeHF = ifft(convHF, length(t), 'symmetric'); %this is the part im confused about
% Ignore,
% tfC(numel(fNum)) = 0;
% convHF = tfC.*fNum;
% convHFMag = abs(convHF/n);
% timeHF = ifft(convHF);
%% plotting
% subplot(2, 2, 1);
% plot(t, Ly)
% title('Gaussian input');
% xlabel('time in seconds')
% ylabel('V')
% grid
subplot(2, 2, 1)
plot(frange, abs(tfC(1: L/2 + 1)));
set(gca, 'Xscale', 'log')
title('coax cable model')
xlabel('Hz')
ylabel('|H(s)|V/V')
grid
ylim([0 1.1])
subplot(2, 2, 2);
plot(frange, convHFMag(1:L/2+1), '.-', frange, fNumMag(1:L/2+1)) %make both range and function the same lenght
title('The input signal Vs its convolution with coax');
xlabel('Hz')
ylabel('V')
legend('Convolution','Lorentzian in frequecuency domain');
xlim([0, 5E10])
grid
subplot(2, 2, [3, 4]);
plot(t, Ly, t, timeHF)
% plot(t, real(timeHF(1:length(t)))) %make both range and function the same lenght
legend('Input', 'Output')
title('Signal at the output')
xlabel('time in seconds')
ylabel('V')
grid
It's important to understand deeply the principles of the FFT to use it correctly.
When you apply Fourier transform to a real signal, the coefficients at negative frequencies are the conjugate of the ones at positive frequencies. When you apply FFT to a real numerical signal, you can show mathematically that the conjugates of the coefficients that should be at negative frequencies (-f) will now appear at (Fsampling-f) where Fsampling=1/dt is the sampling frequency and dt the sampling period. This behavior is called aliasing and is present when you apply fft to a discrete time signal and the sampling period should be chosen small enaough for those two spectra not to overlap Shannon criteria.
When you want to apply a frequency filter to a signal, we say that we keep the first half of the spectrum because the high frequencies (>Fsampling/2) are due to aliasing and are not characteristics of the original signal. To do so, we put zeros on the second half of the spectra before multiplying by the filter. However, by doing so you also lose half of the amplitude of the original signal that you will not recover with ifft. The option 'symmetric' enable to recover it by adding in high frequencis (>Fsampling/2) the conjugate of the coefficients at lower ones (<Fsampling/2).
I simplified the code to explain briefly what's happening and implemented for you at line 20 a hand-made symmetrisation. Note that I reduced the sampling period from one to 100 picoseconds for the spectrum to display correctly:
close all
clc, clear
Fs = 14E10; %1 sample per pico seconds % CHANGED to 100ps
tlim = 4000E-12;
t = -tlim:1/Fs:tlim; %in pico seconds
ag = 0.5; %peak of guassian
bg = 0; %peak location
wg = 50E-12; %FWHM
NT = length(t);
x_i = ag.*exp(-4 .* log(2) .* (t-bg).^2 / (wg).^2); %Gauss. in terms of FWHM
fftx_i = fft(x_i);
f = 1/(2*tlim)*(0:NT-1);
fftx_r = fftx_i;
fftx_r(floor(NT/2):end) = 0; % The removal of high frequencies due to aliasing leads to losing half the amplitude
% HER YOU APPLY FILTER
x_r1 = ifft(fftx_r); % without symmetrisation (half the amplitude lost)
x_r2 = ifft(fftx_r, 'symmetric'); % with symmetrisation
x_r3 = ifft(fftx_r+[0, conj(fftx_r(end:-1:2))]); % hand-made symmetrisation
figure();
subplot(211)
hold on
plot(t, x_i, 'r')
plot(t, x_r2, 'r-+')
plot(t, x_r3, 'r-o')
plot(t, x_r1, 'k--')
hold off
legend('Initial', 'Matlab sym', 'Hand made sym', 'No sym')
title('Time signals')
xlabel('time in seconds')
ylabel('V')
grid
subplot(212)
hold on
plot(f, abs(fft(x_i)), 'r')
plot(f, abs(fft(x_r2)), 'r-+')
plot(f, abs(fft(x_r3)), 'r-o')
plot(f, abs(fft(x_r1)), 'k--')
hold off
legend('Initial', 'Matlab sym', 'Hand made sym', 'No sym')
title('Power spectra')
xlabel('frequency in hertz')
ylabel('V')
grid
Plots the result:
Do not hesitate if you have further questions. Good luck!
---------- EDIT ----------
The amplitude of discrete Fourier transform is not the same as the continuous one. If you are interested in showing signal in frequency domain, you will need to apply a normalization based on the convention you have chosen. In general, you use the convention that the amplitude of the Fourier transform of a Dirac delta function has amplitude one everywhere.
A numerical Dirac delta function has an amplitude of one at an index and zeros elsewhere and leads to a power spectrum equal to one everywhere. However in your case, the time axis has sample period dt, the integral over time of a numerical Dirac in that case is not 1 but dt. You must normalize your frequency domain signal by multiplying it by a factor dt (=1picoseceond in your case) to respect the convention. You can also note that this makes the frequency domain signal homogeneous to [unit of the original multiplied by a time] which is the correct unit of a Fourier transform.
I'm trying to solve the following problem using MATLAB but I faced multiple issues. The plot I obtained doesn't seem right even though I tried to obtain the steady-state solution, I got a plot that doesn't look steady.
The problem I'm trying to solve
The incorrect plot I got.
and here is the code
% system parameters
m=1; k=1; c=.1; wn=sqrt(k/m); z=c/2/sqrt(m*k); wd=wn*sqrt(1-z^2);
% initial conditions
x0=0; v0=0;
%% time
dt=.001; tMax=8*pi; t=0:(tMax-0)/999:tMax;
% input
A=1
omega=(2*pi)/10
F=A/2-(4*A/pi^2)*cos(omega*t); Fw=fft(F);
F=k*A*cos(omega*t); Fw=fft(F);
% normalize
y = F/m;
% compute coefficients proportional to the Fourier series coefficients
Yw = fft(y);
% setup the equations to solve the particular solution of the differential equation
% by the method of undetermined coefficients
N=1000
T=10
k = [0:N/2];
w = 2*pi*k/T;
A = wn*wn-w.*w;
B = 2*z*wn*w;
% solve the equation [A B;-B A][real(Xw); imag(Xw)] = [real(Yw); imag(Yw)] equation
% Note that solution can be obtained by writing [A B;-B A] as a scaling + rotation
% of a 2D vector, which we solve using complex number algebra
C = sqrt(A.*A+B.*B);
theta = acos(A./C);
Ywp = exp(j*theta)./C.*Yw([1:N/2+1]);
% build a hermitian-symmetric spectrum
Xw = [Ywp conj(fliplr(Ywp(2:end-1)))];
% bring back to time-domain (function synthesis from Fourier Series coefficients)
x = ifft(Xw);
figure()
plot(t,x)
Your forcing function doesn't look like the triangle wave in the problem. I edited the %% time section of your code into the following and appeared to give a steady state response.
%% time
TP = 10; % forcing time period (10 s)
dt=.001;
tMax= 3*TP; % needs to be multiple of the time period
t=0:(tMax-0)/999:tMax;
% input
A=1; % Forcing amplitude
omega=(2*pi)/TP;
% forcing is a triangle wave
% generate a triangle wave with min/max values of 0/1.
F = 0*t;
for i = 1:length(t)
if mod(t(i), TP) <= TP/2
F(i) = mod(t(i), TP)/(TP/2);
else
F(i) = 2 - mod(t(i), TP)/(TP/2);
end
end
F = F*A; % scale triangle wave by amplitude
% you can also use MATLAB's sawtooth() function if you have the signal
% processing toolbox
I have been using the STK toolbox for a few days, for kriging of environmental parameter fields, i.e. in a geostatistical context.
I find the toolbox very well implemented and useful (big thanks to the authors!), and the kriging predictions I am getting through STK actually seem fine; however, I am finding myself unable to visualize a semivariogram model based on the STK output (i.e. estimated parameters for gaussian process / covariance functions).
I am attaching an example figure, showing the empirical semivariogram for a simple 1D test case and a Gaussian semivariogram model (as typically used in geostatistics, see also figure) fitted directly to that data. The figure further shows a semivariogram model based on STK output, i.e. using previously estimated model parameters (model.param from stk_param_estim) to get covariance K on a target grid of lag distances and then converting K to semivariance (according to the well-known relation semivar = K0-K where K0 is the covariance at zero lag). I am attaching a simple script to reproduce the figure and detailing the attempted conversion.
As you can see in the figure, this doesn’t do the trick. I have tried several other simple examples and STK datasets, but models obtained through STK vs direct fitting never agree, and in fact usually look much more different than in the example (i.e. the range often seems very different, in addition to the sill/sigma2; uncomment line 12 in the script to see another example). I have also attempted to input the converted STK parameters into the geostatistical model (also in the script), however, the output is identical to the result based on converting K above.
I’d be very thankful for your help!
Figure illustrating the lack of agreement between semivariograms based on direct fit vs conversion of STK output
% Code to reproduce the figure illustrating my problem of getting
% variograms from STK output. The only external functions needed are those
% included with STK.
% TEST DATA - This is simply a monotonic part of the normal pdf
nugget = 0;
X = [0:20]'; % coordinates
% X = [0:50]'; % uncomment this line to see how strongly the models can deviate for different test cases
V = normpdf(X./10+nugget,0,1); % observed values
covmodel = 'stk_gausscov_iso'; % covar model, part of STK toolbox
variomodel = 'stk_gausscov_iso_vario'; % variogram model, nested function
% GET STRUCTURE FOR THE SELECTED KRIGING (GAUSSIAN PROCESS) MODEL
nDim = size(X,2);
model = stk_model (covmodel, nDim);
model.lognoisevariance = NaN; % This makes STK fit nugget
% ESTIMATE THE PARAMETERS OF THE COVARIANCE FUNCTION
[param0, model.lognoisevariance] = stk_param_init (model, X, V); % Compute an initial guess for the parameters of the covariance function (param0)
model.param = stk_param_estim (model, X, V, param0); % Now model the covariance function
% EMPIRICAL SEMIVARIOGRAM (raw, binning removed for simplicity)
D = pdist(X)';
semivar_emp = 0.5.*(pdist(V)').^2;
% THEORETICAL SEMIVARIOGRAM FROM STK
% Target grid of lag distances
DT = [0:1:100]';
DT_zero = zeros(size(DT));
% Get covariance matrix on target grid using STK estimated pars
pairwise = true;
K = feval(model.covariance_type, model.param, DT, DT_zero, -1, pairwise);
% convert covariance to semivariance, i.e. G = C(0) - C(h)
sill = exp(model.param(1));
nugget = exp(model.lognoisevariance);
semivar_stk = sill - K + nugget; % --> this variable is then plotted
% TEST: FIT A GAUSSIAN VARIOGRAM MODEL DIRECTLY TO THE EMPIRICAL SEMIVARIOGRAM
f = #(par)mseval(par,D,semivar_emp,variomodel);
par0 = [10 10 0.1]; % initial guess for pars
[par,mse] = fminsearch(f, par0); % optimize
semivar_directfit = feval(variomodel, par, DT); % evaluate
% TEST 2: USE PARS FROM STK AS INPUT TO GAUSSIAN VARIOGRAM MODEL
par(1) = exp(model.param(1)); % sill, PARAM(1) = log (SIGMA ^ 2), where SIGMA is the standard deviation,
par(2) = sqrt(3)./exp(model.param(2)); % range, PARAM(2) = - log (RHO), where RHO is the range parameter. --- > RHO = exp(-PARAM(2))
par(3) = exp(model.lognoisevariance); % nugget
semivar_stkparswithvariomodel = feval(variomodel, par, DT);
% PLOT SEMIVARIOGRAM
figure(); hold on;
plot(D(:), semivar_emp(:),'.k'); % Observed variogram, raw
plot(DT, semivar_stk,'-b','LineWidth',2); % Theoretical variogram, on a grid
plot(DT, semivar_directfit,'--r','LineWidth',2); % Test direct fit variogram
plot(DT,semivar_stkparswithvariomodel,'--g','LineWidth',2); % Test direct fit variogram using pars from stk
legend('raw empirical semivariance (no binned data here for simplicity) ',...
'Gaussian cov model from STK, i.e. exp(Sigma2) - K + exp(lognoisevar)',...
'Gaussian semivariogram model (fitted directly to semivariance)',...
'Gaussian semivariogram model (using transformed params from STK)');
xlabel('Lag distance','Fontweight','b');
ylabel('Semivariance','Fontweight','b');
% NESTED FUNCTIONS
% Objective function for direct fit
function [mse] = mseval(par,D,Graw,variomodel)
Gmod = feval(variomodel, par, D);
mse = mean((Gmod-Graw).^2);
end
% Gaussian semivariogram model.
function [semivar] = stk_gausscov_iso_vario(par, D) %#ok<DEFNU>
% D : lag distance, c : sill, a : range, n : nugget
c = par(1); % sill
a = par(2); % range
if length(par) > 2, n = par(3); % nugget optional
else, n = 0; end
semivar = n + c .* (1 - exp( -3.*D.^2./a.^2 )); % Model
end
There is nothing wrong with the way you compute the semivariogram.
To understand the figure that you obtain, consider that:
The parameters of the model are estimated in STK using the (restricted) maximum likelihood method, not by least-squares fitting on the semi-variogram.
For very smooth stationary random fields observed over short intervals, you should not expect that the theoretical semivariogram will agree with the empirical semivariogram, with or without binning. The reason for this is that the observations, and thus the squared differences, are very correlated in this case.
To convince yourself of the second point, you can run the following script repeatedly:
% a smooth GP
model = stk_model (#stk_gausscov_iso, 1);
model.param = log ([1.0, 0.2]); % unit variance
x_max = 20; x_obs = x_max * rand (50, 1);
% Simulate data
z_obs = stk_generate_samplepaths (model, x_obs);
% Empirical semivariogram (raw, no binning)
h = (pdist (double (x_obs)))';
semivar_emp = 0.5 * (pdist (z_obs)') .^ 2;
% Model-based semivariogram
x1 = (0:0.01:x_max)';
x0 = zeros (size (x1));
K = feval (model.covariance_type, model.param, x0, x1, -1, true);
semivar_th = 1 - K;
% Figure
figure; subplot (1, 2, 1); plot (x_obs, z_obs, '.');
subplot (1, 2, 2); plot (h(:), semivar_emp(:),'.k'); hold on;
plot (x1, semivar_th,'-b','LineWidth',2);
legend ('empirical', 'model'); xlabel ('lag'); ylabel ('semivar');
Further questions on parameter estimation for Gaussian process models should probably be asked on Cross-Validated rather than Stack Overflow.
I have my data in arrays which are exponential like e^(ax+c)+d. I'm trying to draw a fit to them.
a = data1 (:,1);
b = data1 (:,2);
log(b);
p = polyfit (a,log(b),1);
But I don't know what to do now. I found an equation by polyfit and I was hoping to take the exponential of the equation I got from polyfit with
exp (0.5632x+2.435)
But I figured out that it doesn't work like that. Does anyone have any suggestions?
try with nonlinear fitting:
%% PARAMETERS (you need this part)
clear all;
clc, clf;
N = 128; % number of datapoints
Nint = N*10; % number of datapoints for curve interpolation
fun = #(prms,x) prms(4).^(prms(1)*x+prms(2))+prms(3); % write your function
iniPrm = rand(4,1); % find some initial values for the parameters (choose meaningful values for better results)
%% SIMULATE DATA (this is only for testing purposes)
SNR = .01; % signal to noise ratio for simulated data
noise = (rand(1,N)-.5)*SNR; % create some random noise
x = linspace(0,10,N); % create the x axis
y = fun(iniPrm,x) + noise; % simulate a dataset that follows the given function
x = x(:); % reshape as a vector
y = y(:); % reshape as a vector
X = linspace(x(1),x(end),Nint); % interpolate the output to plot it smoothly
plot(x,y,'.r','markersize',10); hold on; % plot the dataset
%% FIT AND INTERPOLATE YOUR MODEL
[out.BETA,out.RESID,out.J,out.COVB,out.MSE] = nlinfit(x,y,fun,iniPrm,[]); % model your data
[out.YPRED,out.DELTA] = nlpredci(fun,X,out.BETA,out.RESID,'Covar',out.COVB); % interpolate your model
out.YPREDLOWCI = out.YPRED - out.DELTA; % find lower confidence intervals of your fitting
out.YPREDUPCI = out.YPRED + out.DELTA; % find upper confidence intervals of your fitting
out.X = X; % store the interpolated X
%% PLOT FITTING
plotCI = #(IO,spec) patch([IO.X(:);flipud(IO.X(:))],[IO.YPREDLOWCI(:);flipud(IO.YPREDUPCI(:))],spec{:}); % create patches: IE: patch(0:10,10:-1:0,ones(10,1)-1,1,{'r','facealpha',0.2})
plot(X,out.YPRED,'-b','linewidth',3);
plotCI(out,{'r','facealpha',.3,'edgealpha',0})
I have some measurements done and It should be a damped sine wave but I can't find any information on how to make (if possible) a good damped sine wave with Matlab's curve fitting tool.
Here's what I get using a "Smoothing spline":
Image http://s21.postimg.org/yznumla1h/damped.png.
Edit 1:
Here's what I got using the "custom equation" option:
Edit 2:
I've uploaded the data to pastebin in csv format where the first column is the amplitude and the second is the time.
The damped sin function can be created using the following code:
f=f*2*pi;
t=0:.001:1;
y=A*sin(f*t + phi).*exp(-a*t);
plot(t,y);
axis([0 1 -2.2 2.2]);
Now you can use "cftool" from matlab and load your data then set the equation type to custom and enter the formula of the damped sin function. Here you can see what I found so far...
I think the distribution of the data makes it hard for the fitting tool to do a good fit. It may need more data or more distributed data. In addition, there are other tools and methods as well, for instance check this for now: docstoc.com/docs/74524947/Mathcad-Damped-sine-fit-mcd
For all these methods that actually trying to search and find a local optimum (for each fitting parameters), the most important thing is the initial condition. I believe if you choose a good initial condition (initial guess), the fitting tool works well.
Good Luck
I wouldn't use the curve fitting toolbox for this, I'd use a curve-fitting function, e.g. lsqcurvefit. Here is an example taken from something I did a while back:
% Define curve model functions
expsin = #(a, f, phi, tau, t)a * sin(omega * t + phi) .* exp(-tau * t);
lsqexpsin = #(p, t)expsin(p(1), p(2), p(3), p(4), t);
% Setup data params
a = 1; % gain
f = 10; % frequency
phi = pi/2; % phase angle
tau = 0.9252523;% time constant
fs = 100; % sample rate
N = fs; % length
SNR = 10; % signal to noise ratio
% Generate time vector
dt = 1/fs;
t = (0:N-1)*dt;
omega = 2 * pi * f; % angular freq
noiseGain = 10^(-SNR/20); % gain for given SNR
% Generate dummy data: decaying sinusoid plus noise
x = expsin(a, omega, phi, tau, t);
noise = noiseGain * rand(size(x));
noise = noise - mean(noise);
x = x + noise;
close all; figure; hold on;
plot(t, x, 'k-', 'LineWidth', 2);
% Count zero crossings to find frequency
zCross = find(x(1:end-1) .* x(2:end) < 0);
T = mean(diff(zCross) * dt) * 2;
fEstimate = 1 / T;
omegaEstimate = 2 * pi * fEstimate;
% Fit model to data
init = [0.5, omegaEstimate, 0, 0.5];
[newparams, err] = lsqcurvefit(lsqexpsin, init, t, x);
plot(t, lsqexpsin(newparams, t))
Here some data with known parameters is generated, and some random noise added; the data is plotted. The parameters [a, phi, tau] are estimated from the data and a curve with the estimated parameters plotted on top.