I have previous experience with MATLAB, but the problem that I face is some problems in applying a problem in (DSP: Digital signal processing) which is not my study field, but I must finish those problems in days to complete my project.
All I want is help with a method and steps of solving this problem in MATLAB and then I can write the code with myself.
The problem is about the signal
x(t) = exp(-a*t);
1) What's the discrete Fourier transform of the sampled signal with sample rate fs?
2) If a=1 and fs=1, plot the amplitude spectrum of the sampled signal
3) Fix the sampling frequency at fs = 1 (Hz) [what does it mean?] and plot the magnitude of the Fourier Transform of the sampled signal at various values of a
The DFT is used to bring a discrete (i.e. sampled) signal from the time domain to the frequency domain. It's an extension of the Fourier transform. It is used when you are interested in the frequency content of your data. The DFT { x(t) } yields an expression X(F); sample rate (fs) is a term in its expression... This will be your answer to part 1.
You can find tables for the example expression exp(-a*t), but you really should study the electrical engineering (and perhaps math) background for this concept. If you have ability in MATLAB AND you get the basics of the Fourier transform/discrete Fourier transform, then these problems will be straightforward.
MATLAB actually has a tutorial for this subject matter: http://www.mathworks.com/help/techdoc/math/brenr5t-1.html. Good luck...
Related
I'm trying to plot a simple signal in fourier domain using Matlab. It's not plotting the correct signal. Here is my code:
clc;
clear all;
close all;
x=1:0.001:10;
f1=sin(2*pi*10*x);
f2=sin(2*pi*15*x);
f3=sin(2*pi*30*x);
f=f1+f2+f3;
plot(2*pi*x,fft(f1));
figure
plot(x,fft(f1));
I've expected a peak at 10 since the frequency is 10. But it is giving a peak at some other point
Here are the two plot images:
This is the image for plot(x,fft(f1))
This is the image for plot(2*pi*x,fft(f1))
It is not showing the peak at 10.I even tried using abs(fft(f1)). No luck :/
Isn't it the correct way to plot signal in fourier domain?
The fft function assumes unit time step. In order to correct for non unit time step you need to define the frequency component based on the nyquist rate. The following code plots the magnitude of the fft with the correct frequency axis.
clc;
clear all;
close all;
x=1:0.001:10;
% ^ this is your sampling time step
f1=sin(2*pi*10*x);
f2=sin(2*pi*15*x);
f3=sin(2*pi*30*x);
% bounds of fourier transform based on sampling rate
Fs = 1/0.001;
ff = linspace(-Fs/2,Fs/2,numel(x));
F1 = fftshift(fft(f1)/numel(x));
F2 = fftshift(fft(f2)/numel(x));
F3 = fftshift(fft(f3)/numel(x));
figure();
plot(ff,abs(F1),'-r'); hold on;
plot(ff,abs(F2),'-b');
plot(ff,abs(F3),'-k');
Edit: To answer OPs question in the comment.
Speaking in normalized frequency units (assuming sampling rate of 1). The fft function returns the frequency response from 0 to 2*pi radians, but due to some signal processing properties and the way that discrete signals are interpreted when performing an FFT, the signal is actually periodic so the pi to 2*pi section is identical to the -pi to 0 section. To display the plot with the DC component (0 frequency) in the center we use fftshift which does a circular shift equal to 1/2 the length of the signal on the data returned by fft. Before you take the ifft make sure you use ifftshift to put it back in the right place.
Edit2: The normalization term (/numel(x)) is necessary to estimate the continuous time fourier transform using the discrete fourier transform. I don't remember the precise mathematical reason off the top of my head but the examples in the MATLAB documentation also imply the necessity of this normalization.
Edit 3: The original link that I had is down. I may come back to add a more detailed answer but in the mean time I definitely recommend that anyone interested in understanding the relationship between the fundamentals of the FS, FT, DTFT, and DFT watch Professor Oppenheim's hilariously old, but amazingly informative and straightforward lectures on MIT OpenCourseWare.
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 have a time series which is a linear combination of damped waves. The data is real.
Y(t) =SUM_w exp(- gamma t) sin(omega t)
There is no analytic form but this is a closest guess. I want to fourier analyze (FFT) such data and get the real frequencies and damping rates.
I am using matlab but any tool would be fine
Thanks!
Your question would be a better fit for http://math.stackexchange.com, where LaTeX rendering is available for formula. Instead, you have to use e.g. this bookmarklet for proper display:
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First of all, I assume your function is more precisely of the kind
$\sum\limits_k e^{-\gamma_k t}\sin(\omega_k t)$ for $t>0$ and $0$ for $t<0$ (otherwise the function would tend to infinity for $t\to-\infty$). Since a damped function is not periodic, you cannot use Fourier analysis but have to use the Fourier transform, which yields an amplitude for continuous frequencies instead of discrete ones. Using the complex representation $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$, each term in the sum can be Fourier transformed individually, yielding
$\frac1{2\pi}\int_0^\infty e^{-\gamma_k t}\sin(\omega_k t)e^{-i\omega t}\,dt = \frac{\omega_k}{2\pi[(\gamma+i\omega)^2+\omega_k^2]} = \frac{\omega_k}{2\pi}\frac{(\gamma^2-\omega^2+\omega_k^2) + 2i\gamma\omega\omega_k}{[\gamma^2-\omega^2+\omega_k^2]^2+4\gamma^2\omega^2}$
Since you state the damped sine is only a guess, you have to discretize this unlimited integral somehow, though due to the damping a cutoff at some sufficiently large time $t$ should be in order. If the damping $\gamma_k$ is actually the same for all summands, your life becomes easier: Multiply the data by $e^{+\gamma t}$ to obtain a periodic signal which you can now really FFT on.
First digitize your function:
t=0:dt:T; % define sampling interval dt and duration T according to your needs
Y=sum(exp(-gamma*t).*sin(omega*t));
Then do fft and plot:
Y_f=fft(Y);
plot(abs(Y_f));
So I have the samples (hex values) of a sinus signal and I know the sampling frequency. Using this I can plot an fft or periodogram but then I would like to find out the SNR ratio. What would be the most accurate way to calculate the noise and signal power? I would prefer doing it in frequency domain. Is there a way to do this also in time domain?
Thanks a lot in advance!!!
So if there is noise on your signal and you know that your underlying signal is a sine wave, you can easily get your signal parameters i.e. amplitude,frequency and phase by least squares. If y(t) is your signal just minimize the L2 norm of (y(t)-A.sin(wt+b)) over A,w and b. Then you can easily get signal power from the underlying signal and the noise power from the error signal (y(t)-A.sin(wt+b)).
I'm working on a control system that measures the movement of a vibrating robot arm. Because there is some deadtime, I need to look into the future of the somewhat noisy signal.
My idea was to use the frequencies in the sampled signal and produce a fourier function that could be used for extrapolation.
My question: I already have the FFT of the signal vector (containing 60-100 values e.g.) and can see the main frequencies in the amplitude spectrum. Now I want to have a function f(t) which fits to the signal, removes some noise, and can be used to predict the near future of the signal. How do I calculate the coefficients for the sine/cosine functions out of the complex FFT data?
Thank you so much!
AFAIR FFT essentially produces output as a sum of sine functions with different frequencies. The importance of each frequency is the height of each peak. So what you really want to do here is filter out some frequencies (ie. high frequencies for the arm to move gently) and then come back to the time domain.
In matlab this should be like going through the vector of what you got from fft, setting some values to 0 (or doing something more complex to it) and then use ifft to come back to time domain and make the prediction based on what you get.
There's also one thing you should consider while doing this - Nyquist frequency - this means that the highest frequency that you get on your fft is half of the sampling frequency.
If you use an FFT for data that isn't periodic within the FFT aperture length, then you may need to use a window to reduce spurious frequencies due to "spectral leakage". Frequency estimation techniques to better estimate "between bin" frequency content may also be appropriate. The phase of each cosine sinusoid, relative to the edge of the window, is usually atan2(imag[i], real[i]). The frequency depends on the sample rate and bin number versus the length of the FFT.
You might also want to look into using a Kalman filter instead of an FFT.
Added: If your signal isn't exactly integer periodic in the FFT length, then you may want to do an fftshift before the FFT to move the resulting phase measurement reference point to the center of your data vector, instead of a possibly discontinuous circular edge.