I am trying to generate a score or a number which represents how many cos and sin waves can generate my signal. For example, if the signal is a sine wave this means it is 100% pure as it can be generated by only one sine signal, if it consists of two sine wave .. this means it is not pure wave and if it consists 100 sine waves it is really unpure and so on .... I tried FFT and FS but it didn't work ... Can anyone help me ??
FFT will work.
You must process wave with Fourier Transform, then calculate the magnitude
sqrt(real*real + image*image). Counting the peaks of result will provide you number of sinewaves with different frequency.
Here you go:
x = s;
X = dct(x);
[XX,ind] = sort(abs(X),'descend');
i = 1;
while norm(X(ind(1:i)))/norm(X)<0.99
i = i + 1;
end
Needed = i;
Related
I have two vectors that represent two different signals, each being a sine wave with the same frequency. I've tried cross-correlation, Fourier transforms, Hilbert transforms, etc, but nothing returns the correct, theoretical value (in radians) at a specific frequency (should be negative). Is there any method in Matlab to calculate the phase difference of two sine waves with the same frequency?
Note: I have access to the frequency and amplitudes of both signals, and I can post some code if needed.
Assuming s1 and s2 are your isofrequential sine waves you can evaluate the phase difference (absolute value in radians) between them as easily as acos( dot(a,b) / (norm(a)*norm(b)) ).
x = 0:.001:100;
omega = 2*pi*100;
phi = pi/6;
s1 = sin(omega*x);
s2 = sin(omega*x - phi);
phase_diff = acos( dot(s1,s2) / (norm(s1)*norm(s2)) );
I'm writing a program on MATLAB that generates 13 waveforms of varying amplitude, duration, and frequency. Each waveform is repeated 5 times, which means I have 65 'trials' in total.
The total length of each trial = 1.5 ms. The sampling frequency = 4 kHz. I would like the wave to begin at 0.5 ms. Prior to the onset of the wave, and following its offset, I would like the amplitude to be zero (i.e. a 'flatline' prior to and following the wave).
I have created a 65x3 matrix where the columns denote the frequency ('hz'), amplitude ('a'), and duration (ms) of the 65 sine waves. Each row denotes a single wave.
I would like to use the information contained in this 65x3 matrix to generate 65 sine waves of amplitude 'a', frequency 'hz', and duration 'ms'. To be specific: each wave should be created using the parameters (hz,a,ms) specified in the nth row of the matrix. E.g. if row 1 = 100, 1, 50... this means I would like to generate a 100 Hz sine wave (amplitude = 1) lasting 50 ms.
I have attempted to construct a for loop to solve this problem. However, the loop returns a number of errors, and I'm not sure how to resolve them. I have adapted the code to the point where no errors are returned; however, my latest attempt seems to generate 65 waves of equal duration, when in fact the duration of each wave should be that which is stated in vector 'ms'.
Here is my latest, albeit newbie and still unsuccessful, attempt: (note that 'trials' represents the 65x3 matrix discussed above; mA = amplitude).
hz=trials(:,1); mA=trials(:,2); ms=trials(:,3);
trials_waves=zeros(65,500); % the max duration (= 500ms); unsure of this part?
for n = 1:size(order,1)
trials_waves = mA*sin(2*pi*hz*0:ms);
end
Apologies if the information provided is scarce. This is the first time I have asked a question on this website. I can provide more information if needed.
Thank you for your help.
Best,
H
Looks like you've got a good start, I'll try to help you get further towards your solution.
Make a Sine Wave
For starters, let's make a sine wave with variable rate, amplitude, and length.
Fs = 4e3; % sample rate of 4 kHz
Sr = 100; % example rate
Sa = 1; % amplitude
St = 10e-3; % signal duration is 10 ms
% To create a sine wave in MATLAB, I'm going to first create a vector of time,
% `t`, and then create the vector of sine wave samples.
N = St * Fs; % number of samples = duration times sample rate;
t = (1:N) * 1/Fs; % time increment is one over sample rate
% Now I can build my sine wave:
Wave = Sa * sin( 2 * pi * Sr * t );
figure; plot(t, Wave);
Note! This is barely enough time for a full wavelength, so be careful with slow rates and short time lengths.
Make many Sine Waves
To turn this into a loop, I need to index into vectors of input variables. Using my previous example:
Fs = 4e3; % sample rate of 4 kHz
Sr = [100 200 300]; % rates
Sa = [1 .8 .5]; % amplitudes
St = [10e-3 20e-3 25e-3]; % signal durations
nWaves = length(Sr);
N = max(St) * Fs; % number of samples = duration times sample rate;
t = (1:N) /Fs; % time increment is one over sample rate
% initialize the array
waves = zeros(nWaves, N);
for iWaves = 1:nWaves
% index into each variable
thisT = (1:St(iWaves) * Fs) * 1/Fs;
myWave = Sa(iWaves) * sin( 2 * pi * Sr(iWaves) * thisT );
waves(iWaves,1:length(myWave)) = myWave;
end
figure; plot(t, waves);
You still have one more piece, zero padding the front end of your signals, there's lots of ways to do it, one way would be to build the signal the way I've described and then concatenate an appropriate number of zeros to the front of your signal array. Feel free to ask a new question if you get stuck. Good luck!
I have a series of 2D measurements (time on x-axis) that plot to a non-smooth (but pretty good) sawtooth wave. In an ideal world the data points would form a perfect sawtooth wave (with partial amplitude data points at either end). Is there a way of calculating the (average) period of the wave, using OCTAVE/MATLAB? I tried using the formula for a sawtooth from Wikipedia (Sawtooth_wave):
P = mean(time.*pi./acot(tan(y./4))), -pi < y < +pi
also tried:
P = mean(abs(time.*pi./acot(tan(y./4))))
but it didn't work, or at least it gave me an answer I know is out.
An example of the plotted data:
I've also tried the following method - should work - but it's NOT giving me what I know is close to the right answer. Probably something simple and wrong with my code. What?
slopes = diff(y)./diff(x); % form vector of slopes for each two adjacent points
for n = 1:length(diff(y)) % delete slope of any two points that form the 'cliff'
if abs(diff(y(n,1))) > pi
slopes(n,:) = [];
end
end
P = median((2*pi)./slopes); % Amplitude is 2*pi
Old post, but thought I'd offer my two-cent's worth. I think there are two reasonable ways to do this:
Perform a Fourier transform and calculate the fundamental
Do a curve-fitting of the phase, period, amplitude, and offset to an ideal square-wave.
Given curve-fitting will likely be difficult because of discontinuities in saw-wave, so I'd recommend Fourier transform. Self-contained example below:
f_s = 10; # Sampling freq. in Hz
record_length = 1000; # length of recording in sec.
% Create noisy saw-tooth wave, with known period and phase
saw_period = 50;
saw_phase = 10;
t = (1/f_s):(1/f_s):record_length;
saw_function = #(t) mod((t-saw_phase)*(2*pi/saw_period), 2*pi) - pi;
noise_lvl = 2.0;
saw_wave = saw_function(t) + noise_lvl*randn(size(t));
num_tsteps = length(t);
% Plot time-series data
figure();
plot(t, saw_wave, '*r', t, saw_function(t));
xlabel('Time [s]');
ylabel('Measurement');
legend('measurements', 'ideal');
% Perform fast-Fourier transform (and plot it)
dft = fft(saw_wave);
freq = 0:(f_s/length(saw_wave)):(f_s/2);
dft = dft(1:(length(saw_wave)/2+1));
figure();
plot(freq, abs(dft));
xlabel('Freqency [Hz]');
ylabel('FFT of Measurement');
% Estimate fundamental frequency:
[~, idx] = max(abs(dft));
peak_f = abs(freq(idx));
peak_period = 1/peak_f;
disp(strcat('Estimated period [s]: ', num2str(peak_period)))
Which outputs a couple of graphs, and also the estimated period of the saw-tooth wave. You can play around with the amount of noise and see that it correctly gets a period of 50 seconds till very high levels of noise.
Estimated period [s]: 50
I have generated three identical waves with a phase shift in each. For example:
t = 1:10800; % generate time vector
fs = 1; % sampling frequency (seconds)
A = 2; % amplitude
P = 1000; % period (seconds), the time it takes for the signal to repeat itself
f1 = 1/P; % number of cycles per second (i.e. how often the signal repeats itself every second).
y1 = A*sin(2*pi*f1*t); % signal 1
phi = 10; % phase shift
y2 = A*sin(2*pi*f1*t + phi); % signal 2
phi = 15; % phase shift
y3 = A*sin(2*pi*f1*t + phi); % signal 3
YY = [y1',y2',y3'];
plot(t,YY)
I would now like to use a method for detecting this phase shift between the waves. The point of doing this is so that I can eventually apply the method to real data and identify phase shifts between signals.
So far I have been thinking of computing the cross spectra between each wave and the first wave (i.e. without the phase shift):
for i = 1:3;
[Pxy,Freq] = cpsd(YY(:,1),YY(:,i));
coP = real(Pxy);
quadP = imag(Pxy);
phase(:,i) = atan2(coP,quadP);
end
but I'm not sure if this makes any sense.
Has anyone else done something similar to this? The desired outcome should show a phase shift at 10 and 15 for waves 2 and 3 respectively.
Any advice would be appreciated.
There are several ways that you can measure the phase shift between signals. Between your response, the comments below your response, and the other answers, you've gotten most of the options. The specific choice of technique is usually based on issues such as:
Noisy or Clean: Is there noise in your signal?
Multi-Component or Single-Component: Are there more than one type of signal within your recording (multiple tones at multiple frequencies moving in different directions)? Or, is there just a single signal, like in your sine-wave example?
Instantaneous or Averaged: Are you looking for the average phase lag across your entire recording, or are you looking to track how the phase changes throughout the recording?
Depending on your answer to these questions, you could consider the following techniques:
Cross-Correlation: Use the a command like [c,lag]=xcorr(y1,y2); to get the cross-correlation between the two signals. This works on the original time-domain signals. You look for the index where c is maximum ([maxC,I]=max(c);) and then you get your lag value in units of samples lag = lag(I);. This approach gives you the average phase lag for the entire recording. It requires that your signal of interest in the recording be stronger than anything else in your recording...in other words, it is sensitive to noise and other interference.
Frequency Domain: Here you convert your signals into the frequency domain (using fft or cpsd or whatever). Then, you'd find the bin that corresponds to the frequency that you care about and get the angle between the two signals. So, for example, if bin #18 corresponds to your signal's frequency, you'd get the phase lag in radians via phase_rad = angle(fft_y1(18)/fft_y2(18));. If your signals have a constant frequency, this is an excellent approach because it naturally rejects all noise and interference at other frequencies. You can have really strong interference at one frequency, but you can still cleanly get your signal at another frequency. This technique is not the best for signals that change frequency during the fft analysis window.
Hilbert Transform: A third technique, often overlooked, is to convert your time-domain signal into an analytic signal via the Hilbert transform: y1_h = hilbert(y1);. Once you do this, your signal is a vector of complex numbers. A vector holding a simple sine wave in the time domain will now be a vector of complex numbers whose magnitude is constant and whose phase is changing in sync with your original sine wave. This technique allows you to get the instantaneous phase lag between two signals...it's powerful: phase_rad = angle(y1_h ./ y2_h); or phase_rad = wrap(angle(y1_h) - angle(y2_h));. The major limitation to this approach is that your signal needs to be mono-component, meaning that your signal of interest must dominate your recording. Therefore, you may have to filter out any substantial interference that might exist.
For two sinusoidal signal the phase of the complex correlation coefficient gives you what you want. I can only give you an python example (using scipy) as I don't have a matlab to test it.
x1 = sin( 0.1*arange(1024) )
x2 = sin( 0.1*arange(1024) + 0.456)
x1h = hilbert(x1)
x2h = hilbert(x2)
c = inner( x1h, conj(x2h) ) / sqrt( inner(x1h,conj(x1h)) * inner(x2h,conj(x2h)) )
phase_diff = angle(c)
There is a function corrcoeff in matlab, that should work, too (The python one discard the imaginary part). I.e. c = corrcoeff(x1h,x2h) should work in matlab.
The Matlab code to find relative phase using cross-correlation:
fr = 20; % input signal freq
timeStep = 1e-4;
t = 0:timeStep:50; % time vector
y1 = sin(2*pi*t); % reference signal
ph = 0.5; % phase difference to be detected in radians
y2 = 0.9 * sin(2*pi*t + ph); % signal, the phase of which, is to be measured relative to the reference signal
[c,lag]=xcorr(y1,y2); % calc. cross-corel-n
[maxC,I]=max(c); % find max
PH = (lag(I) * timeStep) * 2 * pi; % calculated phase in radians
>> PH
PH =
0.4995
With the correct signals:
t = 1:10800; % generate time vector
fs = 1; % sampling frequency (seconds)
A = 2; % amplitude
P = 1000; % period (seconds), the time it takes for the signal to repeat itself
f1 = 1/P; % number of cycles per second (i.e. how often the signal repeats itself every second).
y1 = A*sin(2*pi*f1*t); % signal 1
phi = 10*pi/180; % phase shift in radians
y2 = A*sin(2*pi*f1*t + phi); % signal 2
phi = 15*pi/180; % phase shift in radians
y3 = A*sin(2*pi*f1*t + phi); % signal 3
The following should work:
>> acos(dot(y1,y2)/(norm(y1)*norm(y2)))
>> ans*180/pi
ans = 9.9332
>> acos(dot(y1,y3)/(norm(y1)*norm(y3)))
ans = 0.25980
>> ans*180/pi
ans = 14.885
Whether or not that's good enough for your "real" signals, only you can tell.
Here is the little modification of your code: phi = 10 is actually in degree, then in sine function, phase information is mostly expressed in radian,so you need to change deg2rad(phi) as following:
t = 1:10800; % generate time vector
fs = 1; % sampling frequency (seconds)
A = 2; % amplitude
P = 1000; % period (seconds), the time it takes for the signal to repeat itself
f1 = 1/P; % number of cycles per second (i.e. how often the signal repeats itself every second).
y1 = A*sin(2*pi*f1*t); % signal 1
phi = deg2rad(10); % phase shift
y2 = A*sin(2*pi*f1*t + phi); % signal 2
phi = deg2rad(15); % phase shift
y3 = A*sin(2*pi*f1*t + phi); % signal 3
YY = [y1',y2',y3'];
plot(t,YY)
then using frequency domain method as mentioned chipaudette
fft_y1 = fft(y1);
fft_y2 = fft(y2);
phase_rad = angle(fft_y1(1:end/2)/fft_y2(1:end/2));
phase_deg = rad2deg(angle(fft_y1(1:end/2)/fft_y2(1:end/2)));
now this will give you a phase shift estimate with error = +-0.2145
If you know the frequency and just want to find the phase, rather than use a full FFT, you might want to consider the Goertzel algorithm, which is a more efficient way to calculate the DFT for a single frequency (an FFT will calculate it for all frequencies).
For a good implementation, see: https://www.mathworks.com/matlabcentral/fileexchange/35103-generalized-goertzel-algorithm and https://asp-eurasipjournals.springeropen.com/track/pdf/10.1186/1687-6180-2012-56.pdf
If you use an AWGN signal with delay and apply your method it works, but if you are using a single tone frequency estimation will not help you. because there is no energy in any other frequency but the tone. You better use cross-correlation in the time domain for this - it will work better for a fixed delay. If you have a wideband signal you can use subbands domain and estimate the phase from that (it is better than FFT due to low cross-frequency dependencies).
How do I generate a simple sine wave in matlab?
I would like to generate a wave which represents a temperature signal with an amplitude of 15 degrees during a 24 hour period, how can I do this?
t = 1:24
x = 15.*sin(pi*t)
plot(t,x)
where 15 is the amplitude. This does not generate a sine wave as I expected. I was expecting to see one wave which extends over a 24 hour period with an amplitude of 15, say with the lowest value of 5 and a maximum of 20 (how do I include these in the equation?).
Add a constant and adjust frequency:
x = 5 + 15*sin(2*pi*t/24);
In your code the frequency is incorrect, and the sampling period is too large for that frequency: you have aliasing. That's why you don't see a sine wave.
This doesn't have to with Matlab really.
If you'd like to generate a wave with fixed period of, say, T = 24hours you'll have to calculate the sine-function accordingly.
E.g.
t = 1:24;
y = 15 * sin(2*pi*t / T);