## 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°)

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

F(0j1t)

**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**

**Genera**l 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 c

**enter 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)=0.45**

**cos(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**.