I would like to estimate parameters of a cubic phase signal. I have problems with the result from MATLAB.
I have signal if length l=1500 And n1=0,n2=0.11m.
I would to find a peak of CPF1(n1,omega) and CPF2(n2,omega).
A problem : m=? Omega=?:?:? And s(n-m+?) To avoid problem of s(value negative) .
For k=1 : ?
CP(k,:) = s(n+k)s(n-k)exp(-jomegak^2)``
I tried to add half of the signal size to s(n-m+l/2) and For k=1:l/2-1 Finally I have do not get the correct parameters.
Related
I've been running a variation of the doseResponse function downloaded from here to generate dose-response sigmoid curves. However, I've had trouble with one of my datasets generating a linear curve instead. By running the following code, I get the following error and produce the following graph. I also uploaded the data called dose2.csv and resp2.csv to google drive here. Does anyone know how I can fix this? Thanks.
Code to generate graph
% Plotting Dose-Response Curve
response = resp2;
dose = dose2;
% Deal with 0 dosage by using it to normalise the results.
normalised=0;
if (sum(dose(:)==0)>0)
%compute mean control response
controlResponse=mean(response(dose==0));
%remove controls from dose/response curve
response=response(dose~=0)/controlResponse;
dose=dose(dose~=0);
normalised=1;
end
%hill equation sigmoid
sigmoid=#(beta,x)beta(1)+(beta(2)-beta(1))./(1+(x/beta(3)).^beta(4));
%calculate some rough guesses for initial parameters
minResponse=min(response);
maxResponse=max(response);
midResponse=mean([minResponse maxResponse]);
minDose=min(dose);
maxDose=max(dose);
%fit the curve and compute the values
%[coeffs,r,J]=nlinfit(dose,response,sigmoid,[minResponse maxResponse midResponse 1]); % nlinfit doesn't work as well
beta_new = lsqcurvefit(sigmoid,[minResponse maxResponse midResponse 1],dose,response);
[coeffs,r,J]=nlinfit(dose,response,sigmoid, beta_new);
ec50=coeffs(3);
hillCoeff=coeffs(4);
%plot the fitted sigmoid
xpoints=logspace(log10(minDose),log10(maxDose),1000);
semilogx(xpoints,sigmoid(coeffs,xpoints),'Color',[1 0 0],'LineWidth',2)
hold on
%notate the EC50
text(ec50,mean([coeffs(1) coeffs(2)]),[' \leftarrow ' sprintf('EC_{50}=%0.2g',ec50)],'FontSize',20,'Color',[1 0 0]);
%plot mean response for each dose with standard error
doses=unique(dose);
meanResponse=zeros(1,length(doses));
stdErrResponse=zeros(1,length(doses));
for i=1:length(doses)
responses=response(dose==doses(i));
meanResponse(i)=mean(responses);
stdErrResponse(i)=std(responses)/sqrt(length(responses));
%stdErrResponse(i)=std(responses);
end
errorbar(doses,meanResponse,stdErrResponse,'o','Color',[1 0 0],'LineWidth',2,'MarkerSize',12)
Warning Message
Solver stopped prematurely.
lsqcurvefit stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 4.000000e+02.
Warning: Iteration limit exceeded. Returning results from final iteration.
Graph (looking to generate a sigmoid curve not linear)
You also need to optimize your initial value [minResponse maxResponse midResponse 1] for lsqcurvefit. Don't just simply start with minimum or maximum values of given values. Instead, you may first start with your equations to estimate your coefficients.
Given the sigmoid model of sigmoid=#(beta,x)beta(1)+(beta(2)-beta(1))./(1+(x/beta(3)).^beta(4)). As x gets arbitrarily close to inf, equation will return beta(2). And as x gets arbitrarily close to 0, equation will return beta(1). Therefore, initial estimation of minResponse, maxResponse, and midResponse seems reasonable enough. Actually your problem lies in your initial estimation of 1. beta(4) can be roughly estimated with the inclination of your log graph. To my rough sketch it was around 1/4 and therefore you may conclude that your initial estimation of 1 was too large for convergence.
beta_new = lsqcurvefit(sigmoid,[minResponse maxResponse midResponse 1/4],dose,response);
While trying to understand Fast Fourier Transform I encountered a problem with the phase. I have broken it down to the simple code below. Calculating one period of a 50Hz sinewave, and applying an fft algorithm:
fs = 1600;
dt = 1/fs;
L = 32;
t=(0:L-1)*dt;
signal = sin(t/0.02*2*pi);
Y = fft(signal);
myAmplitude = abs(Y)/L *2 ;
myAngle = angle(Y);
Amplitude_at_50Hz = myAmplitude(2);
Phase_at_50Hz = myAngle(2);
While the amplitude is ok, I don't understand the phase result. Why do I get -pi/2 ? As there is only one pure sinewave, I expected the phase to be 0. Either my math is wrong, or my use of Matlab, or both of them... (A homemade fft gives me the same result. So I guess I am stumbling over my math.)
There is a similar post here: MATLAB FFT Phase plot. However, the suggested 'unwrap' command doesn't solve my problem.
Thanks and best regards,
DanK
The default waveform for an FFT phase angle of zero is a cosine wave which starts and ends in the FFT window at 1.0 (not a sinewave which starts and ends in the FFT window at 0.0, or at its zero crossings.) This is because the common nomenclature is to call the cosine function components of the FFT basis vectors (the complex exponentials) the "real" components. The sine function basis components are called "imaginary", and thus infer a non-zero complex phase.
That is what it should be. If you used cosine, you would have found a phase of zero.
Ignoring numerical Fourier transforms for a moment and taking a good old Fourier transform of sin(x), which I am too lazy to walk through, we get a pair of purely imaginary deltas.
As for an intuitive reason, recall that a discrete Fourier transform is averaging a bunch of points along a curve in the complex plane while turning at the angular frequency of the bin you're computing and using the amplitude corresponding to the sample. If you sample a sine curve while turning at its own frequency, the shape you get is a circle centered on the imaginary axis (see below). The average of that is of course going to be right on the imaginary axis.
Plot made with wolfram alpha.
Fourier transform of a sine function such as A*sin((2*pi*f)*t) where f is the frequency will yield 2 impulses of magnitude A/2 in the frequency domain at +f and -f where the associated phases are -pi/2 and pi/2 respectively.
You can take a look at its proof here:
http://mathworld.wolfram.com/FourierTransformSine.html
So the code is working fine.
I am trying to use the ifft function in MATLAB on some experimental data, but I don't get the expected results.
I have frequency data of a logarithmic sine sweep excitation, therefore I know the amplitude [g's], the frequency [Hz] and the phase (which is 0 since the point is a piloting point).
I tried to feed it directly to the ifft function, but I get a complex number as a result (and I expected a real result since it is a time signal). I thought the problem could be that the signal is not symmetric, therefore I computed the symmetric part in this way (in a 'for' loop)
x(i) = conj(x(mod(N-i+1,N)+1))
and I added it at the end of the amplitude vector.
new_amp = [amplitude x];
In this way the new amplitude vector is symmetric, but now I also doubled the dimension of that vector and this means I have to double the dimension of the frequency vector also.
Anyway, I fed the new amplitude vector to the ifft but still I don't get the logarithmic sine sweep, although this time the output is real as expected.
To compute the time [s] for the plot I used the following formula:
t = 60*3.33*log10(f/f(1))/(sweep rate)
What am I doing wrong?
Thank you in advance
If you want to create identical time domain signal from specified frequency values you should take into account lots of details. It seems to me very complicated problem and I think it need very strength background on the mathematics behind it.
But I think you may work on some details to get more acceptable result:
1- Time vector should be equally spaced based on sampling from frequency steps and maximum.
t = 0:1/fs:N/fs;
where: *N* is the length of signal in frequency domain, and *fs* is twice the
highest frequency in frequency domain.
2- You should have some sort of logarithmic phases on the frequency bins I think.
3- Your signal in frequency domain must be even to have real signal in time domain.
I hope this could help, even for someone to improve it.
I'm trying to find a factor using matlab that requires me to compute the Fourier transform of an input signal. The problem was stated to me this way:
fbin = 50HZ
0 <= n <= 1999
alpha = F {Blackman[2000] . cos[-2pi . fbin . n/2000]} (f)
where F is the Continous Time Fourier Transform operator.
My matlab code looks like this:
blackman_v = blackman(2000);
signal_x = cos(-2 * pi() .* fbin * (0:(1999)) ./ 2000) .* blackman_v';
fft_real = abs(fft(signal_x, 2000));
alpha = fft_real(51); %51 is the bin for 50hz => or {(f * N/Fs)+1}==51
My problem is that I'm supposed to get a value of around 412 for 49hz but I get about 250 (I'm actually verifying some previous results). Did I wrongly translate the problem? I've been battling for quite a while and I really don't see anything wrong here. Thought the value a 50Hz (430) is ok.
Would really appreciate any hint!
EDIT
blackman_v = blackman(2000);
signal_x = cos(-2 * pi() .* fbin * (0:(1999)) ./ 2000) .* blackman_v';
alpha = abs(freqz(signal_x , 1, 2*pi*50/10000))
Do you know what the freqz is? I read matlab doc and it is still not to clear in my head.
Maybe I misinterpreted your question but Matlab is not for continuous time analysis. It's for numerical analysis only, with discrete values. You can however calculate the discrete time fourier transform (DFT) of your signal, the resolution of which will depend on the length of your signal. Are you using a Blackmann window because your signal is non-periodic?
How to calculate the FFT (DFT) in Matlab: http://www.mathworks.se/help/matlab/ref/fft.html
Any discrete Fourier transform will assume that your signal is periodic. If it isn't you will obtain spectral leakage where certain frequency peaks "leak" their energy to the sides resulting in less defined peak with smeared out frequency values. Thus, the time-domain signal is preferably made periodic before calculting the DFT - periodic to the extent that a general pattern is repeated, values does not have to exact between periods since noise can/will be inherent in the signal. Applying a window function to the time-domain signal before calculating the DFT will make the signal periodic but you will have change amplitude values and introduced a low frequency component.
I am using the FFT function in Matlab in an attempt to analyze the output of a Travelling Wave Laser Model.
The of the model is in the time domain in the form (real, imaginary), with the idea being to apply the FFT to the complex output, to obtain phase and amplitude information in the frequency domain:
%load time_domain field data
data = load('fft_data.asc');
% Calc total energy in the time domain
N = size(data,1);
dt = data(2,1) - data (1,1);
field_td = complex (data(:,4), data(:,5));
wavelength = 1550e-9;
df = 1/N/dt;
frequency = (1:N)*df;
dl = wavelength^2/3e8/N/dt;
lambda = -(1:N)*dl +wavelength + N*dl/2;
%Calc FFT
FT = fft(field_td);
FT = fftshift(FT);
counter=1;
phase=angle(FT);
amptry=abs(FT);
unwraptry=unwrap(phase);
Following the unwrapping, a best fit was applied to the phase in the region of interest, and then subtracted from the phase itself in an attempt to remove wavelength dependence of phase in the region of interest.
for i=1:N % correct phase and produce new IFFT input
bestfit(i)=1.679*(10^10)*lambda(i)-26160;
correctedphase(i)=unwraptry(i)-bestfit(i);
ReverseFFTinput(i)= complex(amptry(i)*cos(correctedphase(i)),amptry(i)*sin(correctedphase(i)));
end
Having performed the best fit manually, I now have the Inverse FFT input as shown above.
pleasework=ifft(ReverseFFTinput);
from which I can now extract the phase and amplitude information in the time domain:
newphasetime=angle(pleasework);
newamplitude=abs(pleasework);
However, although the output for the phase is greatly different compared to the input in the time domain
the amplitude of the corrected data seems to have varied little (if at all!),
despite the scaling of the phase. Physically speaking this does not seem correct, as my understanding is that removing wavelength dependence of phase should 'compress' the pulsed input i.e shorten pulse width but heighten peak.
My main question is whether I have failed to use the inverse FFT correctly, or the forward FFT or both, or is this something like a windowing or normalization issue?
Sorry for the long winded question! And thanks in advance.
You're actually seeing two effects.
First the expected one goes. You're talking about "removing wavelength dependence of phase". If you did exactly that - zeroed out the phase completely - you would actually get a slightly compressed peak.
What you actually do is that you add a linear function to the phase. This does not compress anything; it is a well-known transformation that is equivalent to shifting the peaks in time domain. Just a textbook property of the Fourier transform.
Then goes the unintended one. You convert the spectrum obtained with fft with fftshift for better display. Thus before using ifft to convert it back you need to apply ifftshift first. As you don't, the spectrum is effectively shifted in frequency domain. This results in your time domain phase being added a linear function of time, so the difference between the adjacent points which used to be near zero is now about pi.