Optimal detection of changepoints with a linear computational cost. diff(rising), and you can proceed with your jitter calculations.įor performing the segmentation, I used code from my thesis, but this simple case (piecewise constant) was already solved by Killick et. Legend('Low Time', 'High Time', 'Pulse Duration')ĭuty cycle is then hightime. Lowtime = rising(2:end) - falling(2:end) Hightime = falling(2:end) - rising(1:end-1) The differences between these two vectors give high and low duration of each pulse, and differences between rising edges gives per-pulse duration. Then, all the even times are rising edges and odd times are falling edges (or vice versa). Changepoint identification gives you the start and end time as well as magnitude of each segment. The basic approach is to decompose the signal piecewise (into a sequence of constant segments). ![]() ![]() If you needed to handle a little more noise, here's one way. If you want to understand why there is an abs, or what relevant info you are losing by not representing the phase of the fft, you may want to read a bit more about the DFT transform to understand exactly what you get.Your signal is quite clean, so you could just use simple thresholding. If by frequency you meant the frequency representation of your signal, then to a first approximation, you just want to plot the abs of the FFT to get an idea of where the energy is: plot(abs(fft)) UPDATE: I realize that I assumed you meant by "frequency" of your signal the pitch or base harmonic or frequency with the most energy, however you want to look at it. 95%, 99%, or some other number would depend on how much noise corrupts your signal. I need to generate a signal at any required crank angle say 200.3. So to account for that noise, you would take the absolute max of the autocorrelation (autocorrelation(length(autocorrelation)/2+1), and then find where the autocorrelation is larger than, say, 95% of that maximum value for the first time in the second half of the signal. In Simulink, I am getting a signal (a crank angle signal from an IC Engine) at a varying rate with in cycle of 720 degrees. Since the signal shifted by a multiple of its period will always look like itself, you need to make sure that the maximum you find indeed corresponds to the period of the signal and not one of its multiples.īecause of the noise in your signal, the absolute maximum could very well occur at a multiple of your period instead of the period itself. (The autocorrelation will be symmetric with its maximum in the middle.) By finding that maximum, you find the first place where the shifted signal looks more or less like itself. The signal frequency will then be: frequency = indexMax * Fs / L Īlternatively, faster and working fairly well too depending on the signal you have, take the autocorrelation of your signal: autocorrelation = xcorr(signal) Īnd find the first maximum occurring after the center point of the autocorrelation. Note: to get from indexMax to the actual frequency of interest, you will need to know the length L of the fft (same as the length of your signal), and the sampling frequency Fs. Where indexMax is the index where the max fft value can be found. Last, if your signal has an offset, as is the case with the one you show, you want to get rid of that offset before taking the fft so that you do not get a max at the origin representing the DC component.Įverything I described put in one line would be: = max(abs(fft(signal-mean(signal)))) ![]() The index will correspond to the normalized frequency with maximum energy. Since the fft gives you the frequency representation of the signal, you want to look for the maximum, and since the fft is a complex signal, you will want to take the absolute value first.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |