I am trying to plot a sine wave where the amplitude increases over time and the frequecy increases over time as well. I draw a normal sine wave as shown below but I couldn't change the amplitude and frequency. Any Ideas?
t = [ 0 : 1 : 40 ]; % Time Samples
f = 500; % Input Signal Frequency
fs = 8000; % Sampling Frequency
x = sin(2*pi*f/fs*t); % Generate Sine Wave
figure(1);
stem(t,x,'r'); % View the samples
figure(2);
stem(t*1/fs*1000,x,'r'); % View the samples
hold on;
plot(t*1/fs*1000,x); %
I believe what you are speaking of is amplitude modulation (AM) and frequency modulation (FM). Basically, AM refers to varying the amplitude of your sinusoidal signal and varying this uses a function that is time-dependent. FM is similar, except the frequency varies instead of the amplitude.
Given a time-varying signal A(t), AM is usually expressed as:
Minor note: The above is actually double sideband suppressed carrier (DSB-SC) modulation but if you want to achieve what you are looking for in your question, we actually need to do it this way instead. Also, the signal customarily uses cos instead of sin to ensure zero-phase shift when transmitting. However, because your original code uses sin, that's what I'll be using as well.
I'm putting this disclaimer here in case any communications theorists want to try and correct me :)
Similarly, FM is usually expressed as:
A(t) is what is known as the message or modulating signal as it is varying the amplitude or frequency of the sinusoid. The sinusoid itself is what is known as the carrier signal. The reason why AM and FM are used is due to communication theory. In analog communication systems, in order to transmit a signal from one point to another, the message needs to be frequency shifted or modulated to a higher range in the frequency spectrum in order to suit the frequency response of the channel or medium that the signal travels in.
As such, all you have to do is specify A(t) to be whichever signal you want, as long as your values of t are used in the same way as your sinusoid. As an example, let's say that you want the amplitude or frequency to increase linearly. In this case, A(t) = t. Bear in mind that you need to specify the frequency of the sinusoid f_c, the sampling period or sampling frequency of your data as well as the time frame that your signal is defined as. Let's call the sampling frequency of your data as f. Also bear in mind that this needs to be sufficiently high if you want the curve to be visualized properly. If you make this too low, what'll happen is that you will be skipping essential peaks and troughs of your signal and the graph will look poor.
Therefore, for AM your code may look something like this:
f = 24; %// Hz
f_c = 8; %// Hz
T = 1 / f; %// Sampling period from f
t = 0 : T : 5; %// Determine time values from 0 to 5 in steps of the sampling period
A = t; %// Define message
%// Define carrier signal
carrier = sin(2*pi*f_c*t);
%// Define AM signal
out = A.*carrier;
%// Plot carrier signal and modulated signal
figure;
plot(t, carrier, 'r', t, out, 'b');
grid;
The above code will plot the carrier as well as the modulated signal together.
This is what I get:
As you can see, the amplitude gets higher as the time increases. You can also see that the carrier signal is being bounded by the message signal A(t) = t. I've placed the original carrier signal in the plot as an aid. You can certainly see that the amplitude of the carrier is getting larger due to the message signal.
Similarly, if you want to do FM, most of the code is the same. The only thing that'll be different is that the message signal will be inside the carrier signal itself. Therefore:
f = 100; %// Hz
f_c = 1; %// Hz
T = 1 / f; %// Sampling period from f
t = 0 : T : 5; %// Determine time values from 0 to 5 in steps of the sampling period
A = t; %// Define message
%// Define FM signal
out = sin(2*pi*(f_c + A).*t);
%// Plot modulated signal
figure;
plot(t, out, 'b');
grid;
Bear in mind that I changed f_c and f in order for you to properly see the changes. I also did not plot the carrier signal so you don't get distracted and you can see the results more clearly.
This is what I get:
You can see that the frequency starts rather low, then starts to gradually increase itself due to the message signal A(t) = t. As time increases, so does the frequency.
You can play around with the different frequencies to get different results, but this should be enough to get you started.
Good luck!
Related
I'm trying to find the maximum frequency of a periodic signal in Matlab and as i know when you convert a periodic signal to the frequency spectrum you get only delta functions however i get a few curves between the produced delta functions. Here is the code :
t=[-0.02:10^-3:0.02];
s=5.*(1+cos(2*pi*10*t)).*cos(2*pi*100*t);
figure, subplot(211), plot(t,s);
y=fft(s);
subplot(212), plot(t,y);
Here is a code-snippet to help you understand how to get the frequency-spectrum using fft in matlab.
Things to remember are:
You need to decide on a sampling frequency, which should be high enough, as per the Nyquist Criterion (You need the number of samples, at least more than twice the highest frequency or else we will have aliasing). That means, fs in this example cannot be below 2 * 110. Better to have it even higher to see a have a better appearance of the signal.
For a real signal, what you want is the power-spectrum obtained as the square of the absolute of the output of the fft() function. The imaginary part, which contains the phase should contain nothing but noise. (I didn't plot the phase here, but you can do this to check for yourself.)
Finally, we need to use fftshift to shift the signal such that we get the mirrored spectrum around the zero-frequency.
The peaks would be at the correct frequencies. Now considering only the positive frequencies, as you can see, we have the largest peak at 100Hz and two further lobs around 100Hz +- 10Hz i.e. 90Hz and 110Hz.
Apparently, 110Hz is the highest frequency, in your example.
The code:
fs = 500; % sampling frequency - Should be high enough! Remember Nyquist!
t=[-.2:1/fs:.2];
s= 5.*(1+cos(2*pi*10*t)).*cos(2*pi*100*t);
figure, subplot(311), plot(t,s);
n = length(s);
y=fft(s);
f = (0:n-1)*(fs/n); % frequency range
power = abs(y).^2/n;
subplot(312), plot(f, power);
Y = fftshift(y);
fshift = (-n/2:n/2-1)*(fs/n); % zero-centered frequency range
powershift = abs(Y).^2/n;
subplot(313), plot(fshift, powershift);
The output plots:
The first plot is the signal in the time domain
The signal in the frequency domain
The shifted fft signal
I'm working in the space of biosignal acquisition. I made a experiment as detailed below, and am now trying to obtain some results from the data.
I have a text file of a signal in Matlab. I loaded the signal onto a waveform generator, then I recorded the generator output on an oscilloscope.
I imported the recorded signal from the oscilloscope back into Matlab.
The Pearson's correlation coefficient between the original signal and the oscilloscope signal is 0.9958 (obtained using corrcoeff function).
I want to compute the SNR of the oscilloscope signal (what I'm calling my signal plus whatever noise is introduced through the digital-to-analog conversion and visa-versa). I have attached a snippet of the 2 signals for reference.
So my original signal is X and oscilloscope signal is X + N.
I used the snr function to compute SNR as follows.
snr(original, (oscilloscope - original))
The result I got was 20.44 dB.
This seems off to me as I would have thought with such a high correlation, that the SNR should be much higher?
Or is it not appropriate to try and compute SNR in this sort of situation?
All help is appreciated.
Thanks
Edit: Graph of a couple of results vs Sleutheye's simulated relationship
You might be surprised at just how even such moderate SNR can still result in fairly high correlations.
I ran an experiment to illustrate the approximate relation between correlation and signal-to-noise-ratio estimate. Since I did not have your specific EEG signal, I just used a reference constant signal and some white Gaussian noise. Keep in mind that the relationship could be affected by the nature of the signal and noise, but it should give you an idea of what to expect. This simulation can be executed with the following code:
SNR = [10:1:40];
M = 10000;
C = zeros(size(SNR));
for i=1:length(SNR)
x = ones(1,M);
K = sqrt(sum(x.*x)/M)*power(10, -SNR(i)/20);
z = x + K*randn(size(x));
C(i) = xcorr(x,z,0)./sqrt(sum(x.*x)*sum(z.*z));
end
figure(1);
hold off; plot(SNR, C);
corr0 = 0.9958;
hold on; plot([SNR(1) SNR(end)], [corr0 corr0], 'k:');
snr0 = 20.44;
hold on; plot([snr0 snr0], [min(C) max(C)], 'r:');
xlabel('SNR (dB)');
ylabel('Correlation');
The dotted black horizontal line highlights your 0.9958 correlation measurement, and the dotted red vertical line highlights your 20.44 dB SNR result.
I'd say that's a pretty good match!
In fact, for this specific case in my simulation (x = 1; z = x + N(0,σ)) if we denote C(x,z) to be the correlation between x and z, and σ as the noise standard deviation, we can actually show that:
Given a correlation value of 0.9958, this would yield an SNR of 20.79dB, which is consistent with your results.
Assume I have a smooth function (represented as a vector):
x=0:0.1:1000;
y=sin(2*x);
and I want to find its periodicity - pi (or even its frequency -2 ) .
I have tried the following:
nfft=1024;
Y=fft(y,nfft);
Y=abs(Y(1:nfft/2));
plot(Y);
but obviously it doesn't work (the plot does not give me a peak at "2" ).
Will you please help me find a way to find the value "2"?
Thanks in advance
You have several issues here:
You are computing the fft of x when your actual signal is y
x should be in radians
You need to define a sampling rate and use that to determine the frequency values along the x axis
So once we correct all of these things, we get:
samplingRate = 1000; % Samples per period
nPeriods = 10;
nSamples = samplingRate * nPeriods;
x = linspace(0, 2*pi*nPeriods, nSamples);
y = sin(2*x);
F = fft(y);
amplitude = abs(F / nSamples);
f = samplingRate / nSamples*[0:(nSamples/2-1),-nSamples/2:-1];
plot(f, amplitude)
In general, you can't use an FFT alone to find the period of a periodic signal. That's because an FFT does sinusoidal basis decomposition (or basis transform), and lots of non-sinusoidal waveforms (signals that look absolutely nothing like a sinewave or single sinusoidal basis vector) can be repeated to form a periodic function, waveform, or signal. Thus, it's quite possible for the frequency of a periodic function or waveform to not show up at all in an FFT result (it's called the missing fundamental problem).
Only in the case of a close or near sinusoidal signal will an FFT reliably report the reciprocal of the period of that periodic function.
There are lots of pitch detection/estimation algorithms. You can use an FFT as a sub-component of some composite methods, including cepstrums or cepstral analysis, and Harmonic Product Spectrum pitch detection methods.
I'm having problems plotting the FFT of a wav file. I managed to plot the magnitude and phase spectrums of the signal, however I need to repeat this in range -fs/2:fs/2.
%read sound files
%'y' is the vector holding the original samples & 'fs' refers to the sampling frequency
[y,fs] = wavread('handel.wav');
ydft = fft(y); %fft to transform the original signal into frequency domain
n = length (y); %length of the original signal
% y has even length
ydft = ydft(1:length(y)/2+1);
% create a frequency vector
freq = 0:fs/length(y):fs/2;
shiftfreq = fftshift(freq);
%plot original signal in time domain;
figure;
plot ((1:n)/fs, y);
title('handel.wav in time domain');
xlabel ('second');
grid on;
% plot magnitude in frequency domain
figure;
plot(freq,abs(ydft));
title('handel.wav in frequency domain');
xlabel ('Hz');
ylabel('Magnitude');
grid on;
% plot phase in frequency domain
figure;
plot(freq,unwrap(angle(ydft)));
title ('handel.wav in frequency domain');
xlabel ('Hz');
ylabel ('Phase');
grid on;
What you are currently doing now is plotting the half spectrum, so from 0 <= f < fs/2 where fs is the sampling frequency of your signal, and so fs/2 is the Nyquist frequency. Take note that considering the half spectrum is only valid if the signal is real. This means that the negative spectra is symmetric to the positive spectra and so you don't really need to consider the negative spectra here.
However, you would like to plot the full spectrum of the magnitude and phase. Take note that when calculating the fft using MATLAB, it uses the Cooley-Tukey algorithm so when computing the N point FFT, half of result is for the frequencies from 0 Hz inclusive up to fs/2 Hz exclusive and the other half is for the frequencies from -fs/2 Hz inclusive up to 0 Hz exclusive.
As such, to plot the full spectrum, simply perform a fftshift on the full signal so that the right half and left half of the spectrum is swapped so that the 0 Hz frequency is located in the centre of the signal. Also, you must generate frequencies between -fs/2 to fs/2 to cover the full spectrum. Specifically, you need to generate N points linearly spaced between -fs/2 to fs/2. However, take note that the Nyquist frequency at fs/2 Hz is being excluded at the end, so you need to generate N+1 points between -fs/2 to fs/2 and remove the last point in order for the right step size between each frequency bin to be correct. The easiest way to generate this linear array of points is by using the linspace command where the start frequency is -fs/2, the ending frequency is fs/2 and you want N+1 points between this range and remove the last point:
freq = linspace(-fs/2, fs/2, n+1);
freq(end) = [];
As such, borrowing some parts of your code, this is what the modified code looks like to plot the full spectrum of the magnitude and phase:
%// Read in sound file
[y,fs] = wavread('handel.wav');
%// Take N-point FFT where N is the length of the signal
ydft = fft(y);
n = numel(y); %// Get N - length of signal
%// Create frequency vector - make sure you remove last point
freq = linspace(-fs/2, fs/2, n+1);
freq(end) = [];
%// Shift the spectrum
shiftSpectrum = fftshift(ydft);
%//plot original signal in time domain;
figure;
plot ((0:n-1)/fs, y); %// Note you should start from time = 0, not time = 1/fs
title('handel.wav in time domain');
xlabel ('second');
grid on;
%// plot magnitude in frequency domain
figure;
plot(freq,abs(shiftSpectrum));
title('handel.wav in frequency domain');
xlabel ('Hz');
ylabel('Magnitude');
grid on;
%// plot phase in frequency domain
figure;
plot(freq,unwrap(angle(shiftSpectrum)));
title('handel.wav in frequency domain');
xlabel('Hz');
ylabel('Phase');
grid on;
I don't have access to your handel.wav file, but I'll be using the one provided with MATLAB. You can load this in with load handel;. The sampling frequency is stored in a variable called Fs, so I had to do fs = Fs; before the code I wrote above could work. The sampling frequency for this particular file is 8192 Hz, and this is approximately a 9 second long file (numel(y) / fs = 8.9249 seconds). With that file, this is the magnitude and phase that I get:
For the discrete Fourier transform (DFT) as well as its fast implementations (FFTs), the frequencies are normalized with the sampling frequency fs, i.e., the original range -fs/2:fs/2 is changed to -pi:pi.
Besides, the DFT/FFT always starts with 0, and you can use fftshift() to shift the 0 frequency to the center. Therefore, after fftshift(), the range is -pi:pi, then, you can scale to -fs/2:fs/2.
look at the following Matlab function, it can calculate phase spectrum as well as amplitude spectrum with a perfect accuracy:
https://www.mathworks.com/matlabcentral/fileexchange/63965-amplitude-and-phase-spectra-of-a-signal--fourier-transform-
This program calculates amplitude and phase spectra of an input signal with acceptable accuracy especially in the calculation of phase spectrum.The code does three main jobs for calculation amplitude and phase spectra. First of all, it extends the input signal to infinity; because for calculation Fourier transform(FT) (fft function in Matlab), we consider our signal is periodic with an infinite wavelength, the code creates a super_signal by putting original signal next to itself until the length of super_signal is around 1000000 samples, why did I choose 1000000 samples? Actually, it is just based on try and error!! For most signals that I have tried, a supper signal with 1000000 samples has the best output.
Second, for calculating fft in Matlab you can choose different resolutions, the Mathwork document and help use NFFT=2^nextpow2(length(signal)), it definitely isn't enough for one that wants high accuracy output. Here, I choose the resolution of NFFT=100000 that works for most signals.
Third, the code filters result of FT by thresholding, it is very important step! For calculating phase spectrum, its result is very noisy because of floating rounding off error, it causes during calculation "arctan" even small rounding off error produces significant noise in the result of phase spectrum, for suppressing this kind of noise you can define a threshold value. It means if amplitude of specific frequency is less than predefined threshold value (you must define it) it put zero instead of it.
These three steps help to improve the result of amplitude and phase spectra significantly.
IF YOU USE THIS PROGRAM IN YOUR RESEARCH, PLEASE CITE THE FOLLOWING PAPER:
Afshin Aghayan, Priyank Jaiswal, and Hamid Reza Siahkoohi (2016). "Seismic denoising using the redundant lifting scheme." GEOPHYSICS, 81(3), V249-V260. https://doi.org/10.1190/geo2015-0601.1
I have a signal of wave song and I have a problem with amplitudes.
%graph 1 time domain
song2 = song(1:size(song));
fs = 44100;
dt = 1/fs;
t = 0:dt:(length(song2)*dt)-dt;
figure();
plot(t,abs(song2)); xlabel('Seconds'); ylabel('Amplitude');
Then I put signal in FFT, because I want to get amplitude of detecting peaks(for example; 164Hz).
%graph 2 fft
L = length(song2);
NFFT = 2^nextpow2(L);
f = fs/2*linspace(0,1,NFFT/2+1);
X = fft(song2,NFFT)/L;
X = 2*abs(X(1:NFFT/2+1));
figure();
plot(f,X);
The problem appear when I get the amplitude of the signal (for example; 0.0103) but if I compare with the amplitude of the (time domain) is not the same.
My question is How in the time domain(graph 1) I detect amplitude of the of a frequency(for example; 164 with amplitude 0.0103)?
EDIT:
Hm, I will rather ask in this way. I detect frequency in frequency domain spectrum as the graph link
For example Let us take the the first signal (82hz)(amplitude:0.0075) And my question if is posible to detect position of this first signal in time-domain as the graph in link
Any help would be helpful.
Have you tried the code with a sinusoidal input?
Make a sinusoid of length 512 (any reasonable length) with amplitude 1 and frequency equal to 164 Hz.
When you run the program, I'm sure you'll see the gain at the frequency bin corresponding to 164 Hz close to one, as long as your frequency resolution isn't too bad (meaning you haven't used too few FFT points).
If the above works, then your code should work for a song too. How you're judging/verifying time-domain amplitude in the case of a multi-tonal time domain signal like music is something I don't know.
If I understand your question correctly, you can find what you want by taking the inner product of the time domain signal with a complex sinusoid at the frequency of interest. For example, using the information from your post we would something like the following:
% graph 1 time domain
song2 = song(1:size(song));
fs = 44100;
dt = 1/fs;
t = 0:dt:(length(song2)*dt)-dt;
x_164 = exp(-1j * 2*pi * 164 * t);
f_164 = x_164(:).' * song2(:); % Spectral content of the signal at 164(Hz).
If you think about what a DFT is, you'll see it's just the above process but with the frequencies chosen such that they from an orthogonal basis. However, you can use the same process to estimate the spectral content at any frequency you want.