## Rotating Fourier Series

### Chapter 5. How to extract harmonics from f(t)=1.3+0.7cos(2t)+0.5cos(4t)?

**Chapter 5.1 Introduction**

In the previous chapter, we extracted the harmonic **0.5cos(4t)** from the functions** f(t)=0.5cos(4t)** and **f(t)=0.5cos(4t-30 °)** using the

**rotating Z plane** method. Now we will do the same, but with the function **f (t)=1.3+0.7cos (2t)+0.5cos (4t)**. The pulsations of both cosines are different and the constant **co=1.3** component was added! I wonder how this will affect the graph of** f(t)** and the rotating trajectories of **F(jnω0t)**?

**Chapter 5.2 How to extract harmonics from the function f (t)=1.3+0.7cos (2t)+0.5 cos (4t)
using a plane rotating at speed ω=nω0? **

**Chapter 5.2.1 Introduction**

You will find that regardless of the pulsation of

**nω0**, the period of the trajectory

**F(jnω0t)**is (also)*,

**T=6.28sec**

and for:

**– ω = 0ω0=0/sec**the

**trajectory**doesn’t rotate

**– ω=1*ω0=1/sec …8 ω0 =8/se**c the radius

**R=1**will make

**1…8**turns

Will it be possible to extract the constant

**c0=1.3**and harmonics

**h2(t)=0.7cos (2t)**and

**h4(t)=0.5cos (4t)**?

*Period

**T**may be less. Hence “also”.

**Chapter 5.2.2 Trajectory F (0j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-0j1t), i.e. without rotation
**The problem is similar to

**Fig. 4-2**of the previous

**chapter**. Radius

**R**doen’t rotate, but varies according to periodic function

**R(t)=f(t)=1.3+0.7cos(2t)+0.5cos(4t)**. So the trajectory on the complex plane Z is

**F (0j1t) = f (t) * exp (-0j1t)**

**=1.3+0.7cos(2t)+0.5cos(4t)**.

By the way, you’ll see a time graph of the

**f(t)**.

**Fig. 5-1
**

**F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t)**in the

**complex**and real

**version**

**a**

**F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t)**in the

**complex**

**version**

**b**

F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t)in the

F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t)

**real version**

**Conclusion.**

The average value of

**f(t)**in the period

**T=2π≈6.28sec**(and also

**T= πsek≈3.14sec**) is

**c0=1.3**. It is also the center of gravity

**sc0=(1.3,0)**of the “swinging” trajectory in

**Fig.5-1a**, when the plane rotattion speed

**ω=0**. Another name for the parameter

**c0=1.3**is the constant component of the function

**f (t)**.

**Note.**

In the

**Fourier Series**of the periodic function

**f(t)**, the first element is the constant component-the coefficient c0. This is the mean of

**f(t)**over the period of

**T**. We calculate it as the integral of the formula below.

**Fig.5-2**

Formula for the constant component of the periodic function

**f (t)**, i.e. for

**c0**

This is the

**mean**of the function

**f(t)**over the period

**T**

**a**

**Genera**l formula for

**f(t)**with period

**T**

**b**

Formula for

**f (t)=1.3+0.7cos (2t)+0.5cos (4t)**and

**T=2π**

**c**

Formula for any

**f(t)**and

**T=2π**

**Chapter 5.2.3 Trajectory F (1j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-1j1t) i.e. with rotation 1ω0=-1/sec**

Radius** R=1** spins at **1ω0=-1/sec**.

See the animation in **Fig. 5-1a**. The end of the vector moves “there and back again” around **c0=(+1.**** 3, 0) **acc. to

formula **R(t)=f(t)=1.3+0.7 cos (2t) + 0.5 cos (4t)**. What if the vector started to spin at **1ω0 =-1/sec**? So “clockwise” and

with the period **T≈6.28sec**. Then its motion on the complex plane **Z** is described by the complex function **F(1j1t)**.

**Fig.5-3
**

**F(1j1t)=[1.3+0.7cos (2t)+0.5cos (4t)]*exp(-j1t)**

The animation lasts

**T=2π/ω0≈6.28sec**.

**a**

The radius

**R = 0.5**with

**1ω0=-1/sec**will make

**1**revolution.

Notice that we chose

**R=0.5**as the reference for the

**constant**radius. Also in the next animations

**b**

During rotation, the length of the radius changes according to the function

**R(t)=**

**f(t)=1.3+0.7cos (2t)+0.5cos (4t)**.

So the complex function is realized as a rotating vector

**F(1j1t)=[1.3+0.7cos (2t)+0.5cos (4t)]*exp(-j1t)**

**c**

Complex function

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

It is drawn by the vector from

**b**

The

**center of gravity**of the trajectory

**F(1j1t)**is

**sc1 = (0,0)= 0**. So the

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

**1ω0=1/sec**.

**Chapter 5.2.4 Trajectory F (2j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-2j1t) i.e. with rotation 2ω0=-2/sec
**

**Fig5-4
**

**F(2j1t)=[1.3+0.7cos (2t)+0.5cos (4t)]*exp(-2j1t)**

The animation lasts

**T=2π/ω0≈6.28sec**.

**a**

The radius

**R = 0.5**with

**1ω0=-1/sec**will make

**2**rotations.

**b**

Complex function

**F(2j1t)**as a

**rotating vector**.

The vector

**R(t)**of variable length will make

**2**rotations

**c**

Complex function

**F(2j1t)**as a

**trajectory**.

**Second**rotation on the same track. The

**center of gravity**of the trajectory is now

**sc2=(0.35.0)**. It is a

**vector**and can also be written as a complex exponential

**sc2=R*exp(jϕ) = 0.35*exp (j0 °)**. In

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

**center of gravity**of the trajectory, you can easily read the

**2nd**harmonic of

**f(t)**as

**h2 (t)=0.7cos (2t)**. Look at

**f(t)**. That’s right!

**Note**.

Don’t go into detail as we found the center of gravity

**sc2=(0.35.0)**. Trust your gut for now, although it seems like it should be a bit more to the right. Note, however, that the speed on the

**left**side of the trajectory is slightly

**slowe**r. So the distances from the

**sc2**will be a little more

**densely**spaced than on the

**righ**t side. As if the trajectory was drawn with a

**thicker**line here.

**Chapter 5.2.5 Trajectory F 3j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-3j1t) i.e. with rotation 3ω0=-3/sec
**

**Fig.5-5
**

**F(3j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-3j1t)**

**The animation lasts**

**T≈6.28sek**.

**a**

The radius

**R=0.5**with

**1ω0=-1/se**c will make

**3**rotations.

**b**

Complex function

**F(3j1t)**as a

**rotating vector.**

The vector

**R(t)**of variable length will make

**3**rotations

**c**

Complex function

**F(3j1t)**as a

**trajectory**.

The

**center of gravity**of the trajectory

**F(3j1t)**is

**sc3=(0,0)**.So the

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

**3ω0=3/sec.**

**Chapter 5.2.6 Trajectory F 4j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-4j1t) i.e. with rotation 4ω0=-4/sec
**

**Rys.5-6
**

**F(4j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-4j1t)**

**The animation lasts**

**T≈6.28sek**.

**a**

The radius

**R=0.5**with

**4ω0=-4/se**c will make

**4**rotations.

**b**

Complex function

**F(4j1t)**as a

**rotating vector.**

The vector

**R(t)**of variable length will make

**4**rotations

**c**

Complex function

**F(4j1t)**as a

**trajectory**.

**3rd**and

**4th**rotation along the same track.

**Non-zero**center of gravity

**sc4 = (0.25.0)**! From the

**center of gravity**, the

**4th**harmonic of the function

**f(t)**can be read as

**0.5cos (4t)**.

**Chapter 5.2.7 Trajectory F 5j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-5j1t) i.e. with rotation 5ω0=-5/sec
**

**Fig.5-7
**

**F(5j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-5j1t)**

**The animation lasts**

**T≈6.28sek**.

**a**

The radius

**R=0.5**with

**5ω0=-5/se**c will make

**5**rotations.

**b**

Complex function

**F(5j1t)**as a

**rotating vector.**

The vector

**R(t)**of variable length will make

**5**rotations

**c**

Complex function

**F(5j1t)**as a

**trajectory**.

The

**center of gravity**of the trajectory

**F(5j1t)**is

**sc5=(0,0)**.So the

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

**5ω0=5/sec.**

**Chapter 5.2.8 Trajectory F 6j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-6j1t) i.e. with rotation 6ω0=-6/sec
**

**Fig.5-8
**

**F(6j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-6j1t)**

**The animation lasts**

**T≈6.28sek**.

**a**

The radius

**R=0.5**with

**6ω0=-6/se**c will make

**6**rotations.

**b**

Complex function

**F(6j1t)**as a

**rotating vector.**

The vector

**R(t)**of variable length will make

**6**rotations

**c**

Complex function

**F(5j1t)**as a

**trajectory**.

The

**center of gravity**of the trajectory

**F(6j1t)**is

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

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

**6ω0=6/sec.**

**Chapter 5.2.9 Trajectory F 7j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-7j1t) i.e. with rotation 7ω0=-7/sec
**

**Fig.5-9
**

**F(7j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-7j1t)**

**The animation lasts**

**T≈6.28sek**.

**a**

The radius

**R=0.5**with

**6ω0=-7/se**c will make

**7**rotations.

**b**

Complex function

**F(7j1t)**as a

**rotating vector.**

The vector

**R(t)**of variable length will make

**7**rotations

**c**

Complex function

**F(7j1t)**as a

**trajectory**.

The

**center of gravity**of the trajectory

**F(7j1t)**is

**sc7=(0,0)**.So the

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

**7ω0=7/sec.**

**Chapter 5.2.10 Trajectory F 8j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-8j1t) i.e. with rotation 8ω0=-8/sec
**

**Fig.5-10
**

**F(8j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-8j1t)**

**The animation lasts**

**T≈6.28sek**.

**a**

The radius

**R=0.5**with

**8ω0=-8/se**c will make

**8**rotations.

**b**

Complex function

**F(8j1t)**as a

**rotating vector.**

The vector

**R(t)**of variable length will make

**8**rotations

**c**

Complex function

**F(8j1t)**as a

**trajectory**.

The

**center of gravity**of the trajectory

**F(8j1t)**is

**sc8=(0,0)**.So the

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

**8ω0=8/sec.**