Rotating Fourier Series

Chapter 6. How to filter harmonic from
f(t)=0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)

Chapter 6.1 Introduction
The farther into the forest, the more trees. In Chapter 4 we extracted the harmonic  0.5cos(4t) from the function f(t)=0.5cos(4t). It’s a bit like taking a rabbit out of a bag with only one rabbit in it-harmonic 0.5cos (4t).  A cliche, but it was only an excuse to learn about the
rotating plane Z method.
In Chapter 5 there is already a more interesting function f (t)=1.3+0.7cos (2t)+0.5cos(4t). Exrtracting made more sense. There are 2 or even 3 rabbits in the bag-two harmonics and a one constant component c0=+1.3. Now we have a bag with 4 rabbits. The function f(t) has 3 harmonics and a constant component c0=+ 0.5. In addition, harmonics are cosines with phase shifts ϕ!

Chapter 6.2 Three equivalent versions of the f(t)=0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)
There are:
1. f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)
2. f(t)=0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t)
3. f(t)=Re {0.5+(0.9-j0.6)*exp(+j1t)+(0.6+j0.4)*exp(+j3t)+(0.4-j0.2)*exp(+j5t)}
Ad 1. f (t) as in the chapter title.
Ad 2. f (t) where each harmonic is broken down into sine and cosine components.
Ad 3. This is the real part (“Re”) of the complex function (what is between the braces {}
Associate the appropriate factors, e.g. Ad,2 (0.9-j0.6)*exp (+j1t) with the Ad.3 factors 0.9*cos (1t) + 0.6*sin (1t) .
The diagram f (t) in Fig. 6-1b was created on the basis of Ad.1. Based on Ad.2 and Ad.3 would of course be the same.

Chapter 6.3 Centrifugation of successive harmonics. In other words, the centrifuge starts
Centrifuge starts from ω=0*ω0 to ω=8*ω0

Chapter 6.3.1 Trajectory F (0j1t) =[0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(0j1t) that is, without rotatingg
Radius R does not rotate-->exp(0j1t)=1, but changes in Fig. 6-1a along the real axis Re Z from acc. periodic function.
R(t)=F(0j1t)=f(t)=0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°).

Fig. 6-1
F(0j1t)=f (t)=0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°) in the complex and real version
a
F(0j1t) in the complex version. It’s a pulsating line along c0=(0.5,0)=0.5
b
F(0j1t)in the real version ie time chart

Fig. 6-2
Formula for the constant component of the periodic function f(t), i.e. for c0 of the Fourier Series
This is the mean of the function f(t) over the period T
a
General formula for f(t) with period T. Another look at c0=0.5. It is also the center of gravity sc = (0.5) of the “swinging” trajectory.
b
Formula for any f(t) when T=2π.

Chapter 6.3.2 Trajectory F(1j1t) =[0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(1j1t) that is, with rotation ω0=-1/sec
Note
I chose R=0.5 as the reference radius of the rotating plane Z. Coincidentally, the constant component is also c0=0.5

Fig. 6-3
F(1j1t)=F(0j1t)=f (t)=[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-1j1t)
a
Radius R=0.5 when will make 1 rotation
b
During rotation, the length of the radius changes according to the function
R(t)=f(t)=0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°).
So the complex function F(1j1t) = [0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)] * exp (-1j1t) is realized.
c
Complex function F(1j1t) as a trajectory.
The trajectory F(1j1t) goes around sc1=(0.45, -0.3) at a speed of 1ω0=-1/sec. It is a vector and can also be written as exponential sc1=0.54exp (j-33.7 °). In Chapter 7 you will learn that from the center of gravity sc1=(0.45,-0.3) or sc1=0.54*exp(-j33.7°) you can easily read the first harmonic of the function as 1.08cos (1t-33.7 °) or 0.9cos (1t)+0.6sin (1t). 

Chapter 6.3.3 Trajectory F(2j1t) =[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(2j1t) that is, with rotation ω0=-2/sec

Fig.6-4
F(2j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-2j1t) f(j2ω0t).
a
Radius R=0.5 will make 2 turns.
b
Complex function F(2j1t) as a rotating vector.
c
Complex function F(2j1t) as a trajectory.
The center of gravity of the trajectory F(2j1t) is sc2=(0,0). So the function f(t) has no harmonic with pulsation 2ω0=2/sec. This zero sc2 is not very convincing. Seems like it should be slightly shifted to the left. Note, however, that the speed of the trajectory on the left is on average slower than on the right. 

Chapter 6.3.4 Trajectory F(3j1t) =[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(3j1t) that is, with rotation ω0=-3/sec

Fig.6-5
F(3j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-3j1t).
a
Radius R=0.5 will make 3 turns.
b |
Complex function F(3j1t) as a rotating vector.
c
Complex function F(3j1t) as a trajectory.
The trajectory F(3j1t) orbits sc3=(0.3,0.2) at a speed of 3ω0=-3/sec. It is a vector and can also be written as exponential sc3=0.36exp(+j33.7 °). In Chapter 7 you will learn that with the center of gravity sc3=(0.3,0.2) or sc3 =0.36exp (+j33.7 °) you can easily read the third harmonic of the function as 0.72cos (3t+33.7°) or 0.6cos ( 3t)-0.4sin (3t). You will also learn how to calculate the center of gravity sc3. 

Chapter 6.3.5 Trajectory F(4j1t) =[[0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(4j1t) that is, with rotation 4ω0=-4/sec

Fig.6-6
F(4j1t)=[[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-4j1t)
a
Radius R=0.5 will make 4 turns.
b |
Complex function F(4j1t) as a rotating vector.
c
Complex function F(4j1t) as a trajectory.
The center of gravity of the trajectory F(4j1t) is sc4=(0,0), so the function f(t) has no harmonic with a pulsation of 4ω0=4/sec. As in Figure 6-4c, the center of gravity sc4=(0,0) seems too shifted to the left.

Chapter 6.3.6 Trajectory F(5j1t) =[[0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(5j1t) that is, with rotation 5ω0=-5/sec

Fig.6-7
 F(5j1t) =[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(5j1t).
a
Radius R=0.5 will make 5 turns.
b |
Complex function F(5j1t) as a rotating vector.
c
Complex function F(5j1t) as a trajectory.
The trajectory F(5j1t) orbits sc5=(0.2,-0.1)= at a speed of 5ω0=-5/sec. This means that the f(t) has a harmonic 0.4cos(t)+0.2sin(t)=0.45cos(5t-26.6°)

Chapter 6.3.7 Trajectory F(6j1t) =[0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(6j1t) that is, with rotation 6ω0=-6/sec

Fig.6-8
F(6j1t) =[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(6j1t).
a
Radius R=0.5 will make 6 turns.
b |
Complex function F(6j1t) as a rotating vector.
c
Complex function F(6j1t) as a trajectory.
The trajectory F(6j1t) orbits sc6=(0,0) at a speed of 6ω0=-6/sec. This means that the f(t) has a no harmonic with  6ω0=6/sec.

Chapter 6.3.8 Trajectory F(7j1t) =[[0.5+1.08cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(7j1t) that is, with rotation 7ω0=-7/sec

Fig.6-9
F(7j1t)=[0.5+1.08*cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(-7j1t)
a
Radius R=0.5 will make 7 turns.
b |
Complex function F(7j1t) as a rotating vector.
c
Complex function F(7j1t) as a trajectory.
The trajectory F(7j1t) orbits sc7=(0,0) at a speed of 7ω0=-7/sec. This means that the f(t) has a no harmonic with  7ω0=7/sec.

Chapter 6.3.9 Trajectory F(8j1t) =[0.5+1.08cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(8j1t) that is, with rotation 8ω0=-8/sec

Fig.6-10
F(8j1t)=[0.5+1.08*cos(1t-33.7°)+0.72cos(3t+33.7°)+0.45cos(5t-26.6°)]*exp(-8j1t)
a
Radius R=0.5 will make 8 turns.
b |
Complex function F(8j1t) as a rotating vector.
c
Complex function F(8j1t) as a trajectory.
The trajectory F(8j1t) orbits sc8=(0,0) at a speed of 8ω0=-7/sec. This means that the f(t) has a no harmonic with 8ω0=8/sec.