WebApr 13, 2024 · No. You cannot use fit to perform such a fit, where you place a constraint on the function values. And, yes, a polynomial is a bad thing to use for such a fit, but you don't seem to care. Regardless, you cannot put a constraint that the MAXIMUM value of the polynomial (or minimum) be any specific value. The problem is, the maximum is a rather ... WebFeb 8, 2024 · Now take that sine wave and add it to the polynomial you constructed earlier. By construction, the sine adds zero at every one of the given input points, and therefore at those points gives the same output as the polynomial. You now have two different models that predict exactly the same values at all input points.
FFT of a sinusoidal function - MATLAB Answers - MATLAB …
WebJul 26, 2024 · Select a Web Site. Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: . WebJan 2, 2024 · You can find a Matlab algorithm for a phase-locked loop on Wikipedia. I will suggest a slightly more sophisticated approach here that will identify the peaks and troughs in real-time: fitting a sine wave function to your data using moving least squares minimization with initial estimates from Fourier analysis. Here is my function to do that: derek traeger la county lifeguard
matlab - How to get the FFT of a sine wave - Signal …
WebExample #1. The below code is developed to generate sin wave having values for amplitude as ‘4’ and angular frequency as ‘5’. t = 0:0.01:2; w = 5; a = 4; st = a*sin (w*t); plot (t,st); Output: The resultant sine wave is displayed for the time duration of 0 to 2 attaining the peak amplitude +4 in the first half cycle and -4 in the second ... WebMar 4, 2024 · I'm trying to take a data sample which I know should fit a sin curve due to the nature of the data. The data is positive and negative on the y axis, and between 0 and 360 on the x axis. Hence, I need to fit one period of a sin curve to this plot. WebApr 24, 2012 · What goes wrong: by multiplying time vector t by 2*pi*60 your discrete step size becomes 0.1*2*pi*60=37.6991. But you need at least two samples per cycle (2*pi) to depict your sine wave. Otherwise you'll get an alias frequency, and in you special case the alias frequency is infinity as you produce a whole multiple of 2*pi as step size, thus ... derek toxicology