Rotating Fourier Series Chapter 12. Fourier Series Classically Chaptere 12.1 IntroductionActually, all the formulas related to the Fourier Series are presented in Fig. 7.2 in Chapter 7. They were based on the fact that the nth harmonic was a doubled vector indicating the nth centroid scn of the trajectory F(njω0t)=f(t)*exp(njω0t). More precisely, it was the …
Rotating Fourier Series Chapter 11. Checking the Fourier Series formulas with the Wolfram Alfa program Chapter 11.1 IntroductionMy main goal is to convince the reader of the following formulas, thoroughly discussed in chapter. 7.2. They are not easy, but I hope that the large number of examples with animations helped a bit. It is known …
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Rotating Fourier Series Chapter 10. Fourier Series of a square wave with a shift -30º Chapter 10.1 Introduction Fig.10-1 Square wave f(t) A=1, ω=1/sec and ϕ=-30° or -π/6. The parameter ϕ=-30° of the wave means that it is shifted by -30° relative to the even wave from Chapter 8. Therefore, it is not an even …
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Rotating Fourier Series Chapter 9. Fourier series of an odd square wave. Chapter 9.1 Introduction This is the wave from the previous chapter shifted to the right (i.e. delayed) by a quarter of a period (i.e. by 90° or π/2). And just as the first harmonic of the previous one was a cosine, this one …
Rotating Fourier Series Chapter 8. Fourier series of an even square wave Chapter 8.1 Introduction So far, we have studied the function f(t), in which all harmonics were visible in the formula, e.g. f(t)=1.3+0.7cos(2t)+05.cos(4t). What if harmonics are not visible in the function f(t)? As, for example, in a square wave f(t) with amplitude A=1, …
Rotating Fourier Series Chapter 7. How to calculate centroids of scn trajectories and harmonics detector. Chapter 7.1 Introduction In chapter 4 (4.5, 4.6), we extracted from the rotating trajectory F(njω0t) its “centre of gravity” or “centroid” scn as a complex number. And this is almost the nth harmonic with the pulsation ω=n*ω0 of the function …
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Rotating Fourier Series Chapter 6. How to filter out the harmonics with f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)? Chapter 6.1 Introduction The further into the forest, the more… In Chapter 4, we extracted from the function f(t)=0.5*cos(4t) the harmonic 0.5*cos(4t). It’s a bit like pulling a rabbit out of a bag with only one rabbit in it – harmonic 0.5*cos(4t). …
Rotating Fourier Series Chapter 5. How to filter out the harmonics with 1.3+0.7*cos(2t)+0.5*cos(4t)? Chapter 5.1 Introduction In the previous chapter, we extracted the harmonic 0.5*cos(4t) from the function f(t)=0.5*cos(4t) and from the function f(t)=0.5*cos(4t-30°) using a rotating speed of ω=n*ω0 Z planes. Now we will do the same, but with the function f(t)=1.3+0.7*cos(2t)+0.5*cos(4t). The pulsations …
5. How to filter out the harmonics with f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)? Read More »
Rotating Fourier Series Chapter 4. How to filter out the harmonic with f(t)=0.5*cos(4t)? Chapter 4.1 IntroductionEach periodic function f(t) is a sum of sinusoids/cosines, i.e. the so-called harmonics with pulsations 1ω0, 2ω0,3ω0… Constructing f(t) when we know the harmonics is simple. Just add them. Conversely, i.e. finding harmonics with pulsations 1ω0, 2ω0,…nω0, when we know …
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Rotating Fourier Series Chapter 3 Summation of rotating vectors exp(jωt) Chapter 3.1 What’s that for? To make it easier to understand the summation of the cos(ωt) function, more broadly – sinusoidal type functions. And how he found it fits Fourier Series, which is the sum of sines and cosines. Chapter 3.2 General summation of rotating …