## 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°)?

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)**. A cliché, but it was just an excuse to learn the rotating plane method. In

**Chapter 5**there is a more interesting function

**f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**. The extraction made more sense. There are

**2**or even

**3**rabbits in the bag –

**two**harmonics and also a constant component

**c0=+1.3**. Now we have a bag with

**4**rabbits. The

**f(t)**function has

**three**harmonics and a constant component

**c0=+0.5**. In addition, harmonics are

**cosines**with phase

**shifts ϕ**!

**Chapter 6.2 How to extract harmonics from a function using a plane rotating at ω=n*ω0f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)?Chapter 6.2.1 Introduction**We will check the rotating function

**F(jnω0t)**at different speeds

**n*ω0**for

**n=0…8**and

**ω0=1/sec**. Our goal is to extract the constant component

**c0=0.5**and

**three**harmonics. We know

**f(t)**as a formula, so we know

**c0**and

**3**harmonics. Let’s assume that “Someone” knows only the time chart in

**Fig. 6-1b**. So what? All that’s left is a rotating plane.

The

**f(t)**function has

**3**equivalent forms:

**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**was 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 coefficients, e.g.

**(0.9-j0.6)*exp(+j1t)**with

**cos(1t)+0.6*sin(1t)**from Ad.

**2**.

Time chart

**f(t)**in

**Fig. 6-1b**was created on the basis of Ad.

**1.**Based on Ad.

**2**and Ad.

**3**would of time chart be the same.

**Chapter 6.2.2 Trajectory F(0j1t)=f(t)*exp(-0j1t)=f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45* cos(5t-26.6°) i.e. no rotating**

The radius

**R**does not rotate, but changes in

**Fig. 6-1a**along the real axis

**Re z**acc.

**periodic**function

**R(t)=F(0j1t)=f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°).**

So the trajectory in the complex Z plane in

**Fig. 6-1**a will be a pulsating line.

**Fig. 6-1**

Function **f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)** in complex and real version**Fig. 6-1a****F(0j1t)=f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)** in the **complex** version.**Fig.6-1b****F(0j1t)=f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)** in the **real** version, i.e. the time chart** f(t)**.

Fig.6-2

The formula for the constant component of the periodic function **f(t)**, i.e. **c0 **of the **Fourier Series**

This is the average of the function** f(t)** in period **T****Fig.6-2a**

General formula for** f(t)** with period** T**

Another look at **c0=0.5**. It is also the centroid** sc0=(0.5,0)** of the “rocking” trajectory in **Fig. 6-1a**.**Fig.6-2b**

The formula for our **f(t)** when **T=2π**.**Fig.6-2c**

The formula for any **f(t)** when **T=2π**.**Chapter 6.2.3 Trajectory F(1j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-1j1t)****i.e. with rotating 1ω0=-1/sec**

In this and the following subsections, the pulsating radius **R(t)** from **Fig. 6-1a** will begin to spin at speeds** -1/sec … -8/sec**. Here the animation will last **T=2π/ω0≈6.28sec** and radius **R=0.5** from **Fig.6-3a** will make **1** rotation. At the next rotation speeds, i.e. **-ω0=-2/sec**, **-3ω0=3/sec…-8ω0=-8/sec**, the time of each animation will also be **T≈6.28sec**. This means that the radius** R=0.5*** will make **2.3…8** turns. Or in other words, the** complex** plane **Z** will make** 2.3…8** revolutions.***Note**

I chose **R=0.5** as the reference **radius** of the rotating plane. By chance, the constant component is also **c0=0.5**

**Fig. 6-3****F(1j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-1j1t)****Fig. 6-3a**

Radius **R=0.5** **1ω0=-1/sec** will make **1** rotation**Fig.6-3b**

During rotation, the **length** of the **radius** changes according to the function**R(t)=f(t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)**.

So the complex function **F(1j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-1j1t)** is realized**Fig.6-3c**

Complex function **F(1j1t)** as a trajectory.

Trajectory **F(1j1t)** orbits around the centroid **sc1=(0.45,-0.3)** with a speed of **1ω0=-1/sec**. It is a vector and can also be written in exponential form **sc1=0.54*exp(j-33.7°)**. In **Chapter 7**, you will learn that with **sc1=(0.45,-0.3)**, otherwise **sc1=0.54*exp(-j33.7°)**, you can easily read the **first harmonic** of the function as** 1.08*cos(1t-33.7°)** or** 0.9* cos(1t)+0.6*sin(1t)**.**Chapter 6.2.4 Trajectory f(2j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-2j1t)****i.e. with spinning 2ω0=-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)**.

**Fig. 6-4a**

The radius

**R=0.5**will make

**2**revolutions.

**Fig.6-4b**

The complex function

**F(2j1t)**as a rotating vector

**Fig.6-4c**

Complex function

**F(2j1t)**as a trajectory.

The

**centroid**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. It seems like it should be slightly shifted to the

**right**. Note, however, that the speed of the trajectory on the

**left**is on average

**lower**than on the

**right**.

**Chapter 6.2.5 TrajectoryF(3j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-3j1t)**

**i.e. with spinning 3ω0=-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)**.

**Fig.6-5a**

The radius

**R=0.5**will make

**3**turns.

**Fig.6-5b**

The complex function

**F(3j1t)**as a rotating vector.

**Fig.6-5c**

Complex function

**F(3j1t)**as a trajectory.

The trajectory

**F(3j1t)**revolves around

**sc3=(0.3,0.2)**with a speed of

**3ω0=-3/sec**. It is a vector and can also be written in exponential form

**sc3=0.36*exp(+j33.7°)**. In

**Chapter 7,**you will learn that from the centroid

**sc3=(0.3,0.2)**or

**sc3=0.36*exp(+j33.7°)**you can easily read the

**third**harmonic of the function as

**0.72*cos(3t+33.7°)**or

**0.6*cos( 3t)-0.4*sin(3t)**.

**Chapter 6.2.6 Trajectory F(4j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-4j1t)**

**i.e. with rotation 4ω0=-4/sec**

**Fig.6-6**

**F(4j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-j26.6°)]*exp(-4j1t)**

Fig. 6-6a

Fig. 6-6a

The radius

**R=0.5**will make

**4**turns.

**Fig.6-6b**

The complex function

**F(4j1t)**as a rotating vector

**Fig.6-6c**

Complex function

**F(4j1t)**as a trajectory.

The

**centroid**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

**Fig. 6-4c**, the centroid

**sc4=(0,0)**seems too shifted to the

**left**.

**Chapter 6.2.7 Trajectory F(5j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-j26.6°)]*exp(-5j1t )**

**i.e. with rotating 5ω0=-5/sec**

**Fig. 6-7**

**F(5j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-j26.6°)]*exp(-5j1t)**.

**Fig.6-7a**

The radius

**R=0.5**will make

**5**revolutions.

**Fig.6-7b**

The complex function

**F(5j1t)**as a rotating vector.

**Fig.6-7c**

Complex function

**F(5j1t)**as a trajectory.

The trajectory

**F(5j1t)**revolves around

**sc5=(0.2,-0.1)**with a speed of

**5ω0=-5/sec**. It is a vector and can also be written in exponential form

**sc5=0.36*exp(+j33.7°)**. In

**Chapter 7**, you will learn that from the centroid

**sc5=(0.2,-0.1)**or

**sc5=0.36*exp(-j26.6°)**you can easily read the fifth harmonic of the function as

**0.45*cos(5t-33.7°)**or

**0.4*cos (5t)+0.2*sin(5t)**.

**Chapter 6.2.8 Trajectory F(6j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-6j1t)**

i.e. with rotation 6ω0=-6/sec

i.e. with rotation 6ω0=-6/sec

**Fig.6-8**

**F(4j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-6j1t)**

Fig. 6-8a

Fig. 6-8a

The radius

**R=0.5**will make

**6**turns.

**Fig.6-8b**

The complex function

**F(6j1t)**as a rotating vector

**Fig.6-8c**

Complex function

**F(6j1t)**as a trajectory.

The centroid of the trajectory

**F(6j1t)**is

**sc6=(0,0)**. So the function

**f(t)**has no harmonic with a pulsation of

**6ω0=6/sec**.

**Chapter 6.2.9 Trajectory F(7j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-7j1t)**

**i.e. with spinning 7ω0=-7/sec**

**Fig.6-9**

**F(7j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-7j1t)**

Fig. 6-9a

Fig. 6-9a

The radius

**R=0.5**will make

**7**revolutions.

**Fig.6-9b**

The complex function

**F(7j1t)**as a rotating vector

**Fig.6-9c**

Complex function

**F(7j1t)**as a trajectory.

The centroid of the trajectory

**F(7j1t)**is

**sc7=(0,0)**. That is, the function

**f(t)**has no harmonic with a pulsation of

**7ω0=7/sec**.

**Chapter 6.2.10 Trajectory F(8j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-8j1t)**

i.e. with rotation 8ω0=-8/sec

i.e. with rotation 8ω0=-8/sec

**Fig.6-10**

**F(8j1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-8j1t)**

Fig. 6-10a

Fig. 6-10a

The radius

**R=0.5**will make

**8**turns.

**Fig.6-10b**

The complex function

**F(8j1t)**as a rotating vector

**Fig.6-10c**

Complex function

**F(8j1t)**as a trajectory.

The centroid of the trajectory

**F(8j1t)**is

**sc8=(0,0)**. That is, the function

**f(t)**has no harmonic with a pulsation of

**8ω0=8/sec**.