identifying phase shift between signals - matlab

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).

Related

Matlab cos signal from defined frequency course

how come the spectrogram of this code is at its max at approximately 4400Hz instead of 2400Hz in the last timestep of the stft? (frequencyCourse(end) = 100, meshingOrder = 24 -> f = 2400)
startTime = 0; %s
endTime = 30; %s
startIAS = 15; %Hz
endIAS = 100; %Hz
meshingOrder = 24;
fs = 100000; %Hz
t = startTime:1/fs:endTime-1/fs;
frequencyCourse = linspace(startIAS, endIAS, length(t));
signal = cos(2*pi*meshingOrder*frequencyCourse.*t);
spectrogram(signal, hanning(2^13), 0, 2^14, fs, 'yaxis')
Here's a picture:
It works fine as long as I use chirp instead of my self constructed signal, it's not an option though, since more specific courses are to come.
Summary
The problem is that the instantaneous phase is the integral of the instantaneaous frequency with respect to time, not the instantaneous frequency multiplied by time.
You should compute the signal as
signal = cos(2*pi*meshingOrder*cumtrapz(t, frequencyCourse));
What your code does
In your example, you seem to want to generate a linear chirp with initial frequency meshingOrder*startIAS and final frequency meshingOrder*endIAS. But that's not the code is doing.
In your computed signal, the instantaneous phase is the argument to the cos function:
2*pi*meshingOrder*frequencyCourse.*t
Since the variable frequencyCourse increases from meshingOrder*startIAS at start time (which is 0) to meshingOrder*endIAS at end time, this can be expressed as
2*pi*(A+B*t).*t
where A = meshingOrder*startIAS and B = meshingOrder*(endIAS-startIAS)/endTime. Differentiating the instantanous phase with respect to t gives an instantaneous frequency
A + 2*B*t
that is
meshingOrder*startIAS + 2*meshingOrder*(endIAS-startIAS)/endTime * t
As you can see, the problem is the factor 2 here. At end time the instantaneous frequency is
meshingOrder*startIAS + 2*meshingOrder*(endIAS-startIAS)
that is
2*meshingOrder*endIAS - meshingOrder*startIAS
In your example this is 4440 Hz, which is in accordance with your observed value.
What the code should do
For a linear chirp (or a chirp with any other simple frequency variation, such as a quadratic or exponential) you could compute the correct instantaneous phase that gives rise to the desired instantaneous frequency. See for example here. This is also what the chirp function internally does.
But you seem to be want to deal with arbitrary frequency courses. To do that, given arbitrary t, just compute the argument of the cos as the cumulative integral of frequencyCourse with respect to t. This is easily done with cumtrapz:
signal = cos(2*pi*meshingOrder*cumtrapz(t, frequencyCourse));
Changing this line in your example gives the following figure, which has the expected frequency variation from 360 Hz to 2400 Hz:

period of sawtooth from measurements

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

changing frequency using fft and ifft not using whole numbers

I know I can change frequency by whole numbers by changing the variable shift but how can I change the frequency using numbers with decimal places like .754 or 1.2345 or 67.456. If I change the variable 'shift' to a non-whole like number like 5.1 I get an error subscript indices must be either positive integers less than 2^31 or logicals from line mag2s = [mag2(shift+1:end), zeros(1,shift)];
Example Code below from question increase / decrease the frequency of a signal using fft and ifft in matlab / octave works with changing the variable shift (but it only works with whole numbers, I need it to work with decimals numbers also).
PS: I'm using octave 3.8.1 which is like matlab and I know I could change the frequency by adjusting the formula in the variable ya but ya will be a signal taken from an audio source (human speech) so it won't be an equation. The equation is just used to keep the example simple. And yes Fs is large due to the fact that signal files used are around 45 seconds long which is why I can't use resample because I get a out of memory error when used.
Here's a animated youtube video example of what I'm trying to get when I use the test equation ya= .5*sin(2*pi*1*t)+.2*cos(2*pi*3*t) and what I'm trying to get happen if I varied the variable shift from (0:0.1:5) youtu.be/pf25Gw6iS1U please keep in mind that ya will be an imported audio signal so I won't have an equation to easily adjust
clear all,clf
Fs = 2000000;% Sampling frequency
t=linspace(0,1,Fs);
%1a create signal
ya = .5*sin(2*pi*2*t);
%2a create frequency domain
ya_fft = fft(ya);
mag = abs(ya_fft);
phase = unwrap(angle(ya_fft));
ya_newifft=ifft(mag.*exp(i*phase));
% ----- changes start here ----- %
shift = 5; % shift amount
N = length(ya_fft); % number of points in the fft
mag1 = mag(2:N/2+1); % get positive freq. magnitude
phase1 = phase(2:N/2+1); % get positive freq. phases
mag2 = mag(N/2+2:end); % get negative freq. magnitude
phase2 = phase(N/2+2:end); % get negative freq. phases
% pad the positive frequency signals with 'shift' zeros on the left
% remove 'shift' components on the right
mag1s = [zeros(1,shift) , mag1(1:end-shift)];
phase1s = [zeros(1,shift) , phase1(1:end-shift)];
% pad the negative frequency signals with 'shift' zeros on the right
% remove 'shift' components on the left
mag2s = [mag2(shift+1:end), zeros(1,shift)];
phase2s = [phase2(shift+1:end), zeros(1,shift) ];
% recreate the frequency spectrum after the shift
% DC +ve freq. -ve freq.
magS = [mag(1) , mag1s , mag2s];
phaseS = [phase(1) , phase1s , phase2s];
x = magS.*cos(phaseS); % change from polar to rectangular
y = magS.*sin(phaseS);
yafft2 = x + i*y; % store signal as complex numbers
yaifft2 = real(ifft(yafft2)); % take inverse fft
plot(t,ya,'-r',t,yaifft2,'-b'); % time signal with increased frequency
legend('Original signal (ya) ','New frequency signal (yaifft2) ')
You can do this using a fractional delay filter.
First, lets make the code ore workable by letting Matlab handle the conjugate symmetry of the FFT. Just make mag1 and phase1 go to the end . . .
mag1 = mag(2:end);
phase1 = phase(2:end);
Get rid of mag2s and phase2s completely. This simplifies lines 37 and 38 to . .
magS = [mag(1) , mag1s ];
phaseS = [phase(1) , phase1s ];
Use the symmetric option of ifft to get Matlb to handle the symmetry for you. You can then drop the forced real, too.
yaifft2 = ifft(yafft2, 'symmetric'); % take inverse fft
With that cleaned up, we can now think of the delay as a filter, e.g.
% ----- changes start here ----- %
shift = 5;
shift_b = [zeros(1, shift) 1]; % shift amount
shift_a = 1;
which can be applied as so . . .
mag1s = filter(shift_b, shift_a, mag1);
phase1s = filter(shift_b, shift_a, phase1);
In this mindset, we can use an allpass filter to make a very simple fractional delay filter
The code above gives the 'M Samples Delay' part of the circuit. You can then add on the fraction using a second cascaded allpass filter . .
shift = 5.5;
Nw = floor(shift);
shift_b = [zeros(1, Nw) 1];
shift_a = 1;
Nf = mod(shift,1);
alpha = -(Nf-1)/(Nf+1);
fract_b = [alpha 1];
fract_a = [1 alpha];
%// now filter as a cascade . . .
mag1s = filter(shift_b, shift_a, mag1);
mag1s = filter(fract_b, fract_a, mag1s);
Ok so the question as I understand it is "how do I shift my signal by a specific frequency?"
First let's define Fs which is our sample rate (ie samples per second). We collect a signal which is N samples long. Then the frequency change between samples in the Fourier domain is Fs/N. So taking your example code Fs is 2,000,000 and N is 2,000,000 so the space between each sample is 1Hz and shifting your signal 5 samples shifts it 5Hz.
Now say we want to shift our signal by 5.25Hz instead. Well if our signal was 8,000,000 samples then the spacing would be Fs/N = 0.25Hz and we would shift our signal 11 samples. So how do we get an 8,000,000 sample signal from a 2,000,000 sample signal? Just zero pad it! literally append zeros until it is 8,000,000 samples long. Why does this work? Because you are in essence multiplying your signal by a rectangular window which is equivalent to sinc function convolution in the frequency domain. This is an important point. By appending zeros you are interpolating in the frequency domain (you don't have any more frequency information about the signal you are just interpolating between the previous DTFT points).
We can do this down to any resolution you want, but eventually you'll have to deal with the fact that numbers in digital systems aren't continuous so I recommend just choosing an acceptable tolerance. Lets say we want to be within 0.01 of our desired frequency.
So lets get to actual code. Most of it doesn't change luckily.
clear all,clf
Fs = 44100; % lets pick actual audio sampling rate
tolerance = 0.01; % our frequency bin tolerance
minSignalLen = Fs / tolerance; %minimum number of samples for our tolerance
%your code does not like odd length signals so lets make sure we have an
%even signal length
if(mod(minSignalLen,2) ~=0 )
minSignalLen = minSignalLen + 1;
end
t=linspace(0,1,Fs); %our input signal is 1s long
%1a create 2Hz signal
ya = .5*sin(2*pi*2*t);
if (length(ya) < minSignalLen)
ya = [ya, zeros(1, minSignalLen - length(ya))];
end
df = Fs / length(ya); %actual frequency domain spacing;
targetFreqShift = 2.32; %lets shift it 2.32Hz
nSamplesShift = round(targetFreqShift / df);
%2a create frequency domain
ya_fft = fft(ya);
mag = abs(ya_fft);
phase = unwrap(angle(ya_fft));
ya_newifft=ifft(mag.*exp(i*phase));
% ----- changes start here ----- %
shift = nSamplesShift; % shift amount
N = length(ya_fft); % number of points in the fft
mag1 = mag(2:N/2+1); % get positive freq. magnitude
phase1 = phase(2:N/2+1); % get positive freq. phases
mag2 = mag(N/2+2:end); % get negative freq. magnitude
phase2 = phase(N/2+2:end); % get negative freq. phases
% pad the positive frequency signals with 'shift' zeros on the left
% remove 'shift' components on the right
mag1s = [zeros(1,shift) , mag1(1:end-shift)];
phase1s = [zeros(1,shift) , phase1(1:end-shift)];
% pad the negative frequency signals with 'shift' zeros on the right
% remove 'shift' components on the left
mag2s = [mag2(shift+1:end), zeros(1,shift)];
phase2s = [phase2(shift+1:end), zeros(1,shift) ];
% recreate the frequency spectrum after the shift
% DC +ve freq. -ve freq.
magS = [mag(1) , mag1s , mag2s];
phaseS = [phase(1) , phase1s , phase2s];
x = magS.*cos(phaseS); % change from polar to rectangular
y = magS.*sin(phaseS);
yafft2 = x + i*y; % store signal as complex numbers
yaifft2 = real(ifft(yafft2)); % take inverse fft
%pull out the original 1s of signal
plot(t,ya(1:length(t)),'-r',t,yaifft2(1:length(t)),'-b');
legend('Original signal (ya) ','New frequency signal (yaifft2) ')
The final signal is a little over 4Hz which is what we expect. There is some distortion visible from the interpolation, but that should be minimized with a longer signal with a smother frequency domain representation.
Now that I've gone through all of that you may be wondering if there is an easier way. Fortunately for us, there is. We can take advantage of the hilbert transform and fourier transform properties to achieve a frequency shift without ever worrying about Fs or tolerance levels or bin spacing. Namely we know that a time shift leads to a phase shift in the Fourier domain. Well time and frequency are duals so a frequency shift leads to a complex exponential multiplication in the time domain. We don't want to just do a bulk shift of all frequencies because that that will ruin our symmetry in Fourier space leading to a complex time series. So we use the hilbert transform to get the analytic signal which is composed of only the positive frequencies, shift that, and then reconstruct our time series assuming a symmetric Fourier representation.
Fs = 44100;
t=linspace(0,1,Fs);
FShift = 2.3 %shift our frequency up by 2.3Hz
%1a create signal
ya = .5*sin(2*pi*2*t);
yaHil = hilbert(ya); %get the hilbert transform
yaShiftedHil = yaHil.*exp(1i*2*pi*FShift*t);
yaShifted = real(yaShiftedHil);
figure
plot(t,ya,'-r',t,yaShifted,'-b')
legend('Original signal (ya) ','New frequency signal (yaifft2) ')
Band-limited interpolation using a windowed-Sinc interpolation kernel can be used to change sample rate by arbitrary ratios. Changing the sample rate changes the frequency content of the signal, relative to the sample rate, by the inverse ratio.

inverse fast fourier transform for frequency range

My problem is to obtain original signal from amplitude spectrum (fft) based on inverse fft but only for some frequency range ex. 8-12 Hz. Could anyone help me? I try to used:
xdft=fft(x);
ixdft=ifft(xdft(a:b)), %where xdft(a:b) is |Y(f)| for freq 8-12 Hz.
But it doesn't want to work.
You can set all the values of xdft to zero except those you want, i.e.,
xdft = fft(x);
xdft = xdft(1:ceil(length(xdft) / 2));
xdft(1:a) = 0;
xdft(b+1:end) = 0;
ixdft = ifft(xdft, 'symmetric');
The reason I have taken only half of the original FFT'd data is that your result will be symmetric about Fs / 2 (where Fs is the sample rate), and if you don't do the same thing to the frequencies either side of the centre, you will get a complex signal out. Instead of doing the same thing to both sides manually, I've just taken one side, modified it, and told ifft that it has to reconstruct the data for the full frequency range by appending a mirror image of what you pass it; this by done by calling it with the 'symmetric' option.
If you need to figure out what a and b should be for some frequency, you can first create a vector of the frequencies at which you've performed the FFT, then find those frequencies that are within your range, like so:
xdft = fft(x);
xdft = xdft(1:ceil(length(xdft) / 2));
f = linspace(0, Fs / 2, length(xdft));
keepInd = f >= 8 & f <= 12; % Keep frequencies between 8 and 12 Hz
xdft(~keepInd) = 0;
Note that I've actually omitted the use of the two variables a and b altogether in this example and opted for logical indexing, and that Fs is the sample rate.

Difficulty understanding the phase calculated by the FFT. Short matlab demo to illustrate

I'm testing the phase output of an fft of a sin signal and a cos signal.
The script below creates the signals and performs an FFT on them. Bins who's amplitude is below a threshold are zeroed for the phase spectrum because I am only interested in the phase of the signals.
% 10khz 10 second long time interval
t = 0:1 / 10000:10;
%1khz cos
c = cos(2 * pi * 1000 .* t);
%1khz sin
s = sin(2 * pi * 1000 .* t);
%ffts
C = fft(c)/length(c);
S = fft(s)/length(s);
%magnitude and phases of ffts
CA = abs(C); %cos magnitude
SA = abs(S); %sin magnitude
Cthresh = max(CA) * 0.5;
Sthresh = max(SA) * 0.5;
%find all indeces below the threshold
Crange = find(CA < Cthresh);
Srange = find(SA < Sthresh);
%set the indeces below the threshold to 0 - phase will be meaningless for
%noise values
CP = angle(C);
CP(Crange) = 0;
SP = angle(S);
SP(Srange) = 0;
If you plot CP - the phase of the cos - you will get a phase of 0.3142 in the bins corresponding to the frequency of the cos signal and zeros elsewhere. This is pi/10. I'm expecting to get pi. Why is this?
If you plot SP you get values of 1.2566. I'm expecting to get pi/2 or 1.5708. 80% of the expected value. What is causing these errors?
If your input signal is not perfectly periodic in the FFT aperture length (an exact integer number of full periods), the sinusoids will be discontinuous across the ends of the FFT aperture. Thus you will get a phase that is the average of the two different phases at both ends of the FFT input vector.
If you want a more sensible phase, reference the phase of your sinusoids to the center of the FFT input vector, and do an fft shift before the FFT. This will result in a continuous sinusoid at the zero phase reference position, with a single phase instead of a weird average value.
Also note that matlab may reference the phase to the second point in a sampled sinusoid, e.g. vectorElement[i=1], not the first, vectorElement[i=0]. This will have a phase of pi/10 for a sinusoid of period = 20 samples.
The issue you have is exactly what hotpaw2 has stated. You have 100001 samples in t, so you do not have a perfectly periodic signal, therefore you have leakage. That means that you've got a sin()/sin() function from your implicit rectangular window convolved with your solution. That's what changes your phase.
If instead you try the following:
t = 0:1 / 10000:9.9999;
c = cos(2 * pi * 1000 .* t);
%1khz sin
s = sin(2 * pi * 1000 .* t);
%ffts
C = fft(c)/length(c);
S = fft(s)/length(s);
you would find that the phase of the cosine is zero (which is what you would expect) and that the phase of sine is pi/2.
Performing a linear shift in the time domain (using fftshift) will merely introduce a linear phase term in the frequency domain, and will not resolve the original problem.
In practice, rather than trying to set the length of the sequence precisely to match the period of the signal, windowing should be applied if the signal is to be examined in the frequency domain. In that case you really should make sure that your signals are appropriately aligned so that the window attenuates the end points, thus smoothing out the discontinuity. This has the effect of broadening the main lobe of the FFT, but it also controls the leakage.