## 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 of both

**cosines**are

**different**! I wonder how this will affect the

**f(t)**plot and the

**rotating**trajectories

**F(jnω0t)**?

You will find that:

**– regardless**of the

**nω0**pulsation, the trajectory period

**F(jnω0t)**is

**T=6.28**

**sec**

**– for ω=0**the function does not rotate!

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

**R=1**will make

**1…8**turns

Will it be possible to extract the

**constant**component

**c0=1.3**and the harmonics

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

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

**Chapter 5.2 Trajectory F(0j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-0j1t) i.e. no rotation**

A problem similar to** Fig. 4-2** of the previous chapter. The radius **R** does not rotate, but changes according to **periodic** function **R(t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**. So the trajectory in the **complex** plane **Z** is**F(0j1t)=f(t)*exp(-j0)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**. By the way, you will see the** time chart** of the **f(t)** function.

**Fig. 5-1**

Function **F(0j1t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)** in **complex** and **real** version**Fig. 5-1a****F(0j1t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)** in **complex** version**Fig. 5-1b****F(0j1t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)** in **real** version**Conclusion**.

The average value of **f(t)** over the period **T=2π≈6.28sec** (and also **T=πsec≈3.14sec)** is **c0=1.3**.

It is also the** centroid** **sc0=(1.3,0)** of the “rocking” trajectory in** Fig.5-1a** when the plane rotation speed** ω=0**. Another name for parameter **c0=1.3** is the **constant** component of** f(t)**.**Note**.

In the **Fourier Series** of the periodic function** f(t)** the** first** element is the **constant** component – coefficient **c0=a0**. This is the **average f(t)** over period **T**.

We calculate it as an integral from the formula below.

**Fig.5-2**

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

This is the **average** of** f(t)** over period **T****Fig.5-2a**

General formula for** f(t)** with period** T****Fig.5-2b**

Formula for **f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)** and **T=2π****Fig.5-2c**

Formula for any **f(t)** and **T=2π**

**Chapter 5.3 Trajectory F(1j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-1j1t) i.e. with rotation 1ω0=-1/sec**

The radius **R=1** rotates with **1ω0=-1/sec**.

See the animation in **Fig. 5-1a**. The end of the vector moves “back and forth” around** c0=(+1.3, 0)** by formula **R(t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**. What if the vector started rotatimg at **1ω0=-1/sec**? So “clockwise” and with a period of **T≈6.28sec**. Then its motion in the complex** plane Z** is described by the complex function **F(1j1t)**.

**Fig.5-3****F(1j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-j1t)**

The animation takes **T=2π/ω0≈6.28sec**.**Fig.5-3a**

Radius** R=0.5** with** 1ω0=-1/sec** will make **1** revolution.

Notice that we chose** R=0.5** as the reference **constant** radius. Also in the next animations**Fig.5-3b**

During the rotation, the length of the **radius** changes according to the function **R(t)= f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**.

So the **complex** function is implemented as a **rotating** vector **F(1j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-j1t)****Fig.5-3c****Complex** function **F(1j1t)** as a trajectory.

It is drawn by the vector from **Fig.5-3b. **The **centroid** of the trajectory** F(1j1t)** is evidently** sc1=(0,0)=0**. So function** f(t)** has no harmonic with pulsation **1ω0=1/sec**.

**Chapter 5.4 Trajectory F(2j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-2j1t) i.e. with rotation 2ω0=-2/sec**

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

The animation lasts **T≈6.28sec**.**Fig. 5-4a**

Radius **R=0.5** will make **2** revolutions.**Fig.5-4b**

The **complex** function **F(2j1t)** as a** rotating** vector.

The variable-length vector** R(t)** will make **2** rotations.**Fig.5-4c**

The complex function **F(2j1t)** as a trajectory.

Second **rotation** on the same track. The **centroid** of the trajectory is now **sc2=(0.35,0)**. It is a vector and can also be written in the complex exponential form **sc2=R*exp(jϕ)=0.35*exp(j0°)**. In **Chapter 7** you will learn that from the **centroid** of the trajectory you can easily read the** 2nd** harmonic of **f(t)** as **h2(t)=0.7cos(2t)**. Look at** f(t)**. Correct!**Note.**

Don’t go into how we calculated the centroid **sc2=(0.35,0)**. Trust your intuition for now, although it seems like it should be a bit more to the right. Note, however, that the **velocity** of the** left** side of the **trajectory** is slightly **slower**. So the distances from **sc2** will be a little more **densely** spaced than on the **right** side. As if the trajectory was drawn with a** thicker line** here.

**Chapter 5.5 Trajectory f(3j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-3j1t) i.e. with spin 3ω0=-3/sec**

**Fig.5-5****F(3j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-3j1t)**

The animation lasts **T≈6.28sec**.**Fig. 5-5a**

The radius **R=0.5** will make **3** revolutions.**Fig.5-5b**

The complex function **F(3j1t)** as a rotating vector**Fig.5-5c****Complex** function **F(3j1t)** as a trajectory.

The **centroid** of the trajectory **F(3j1t) is sc3=(0,0)=0**. So the function **f(t)** has no harmonic with pulsation **3ω0=3/sec**.

**Chapter 5.6 Trajectory F(4j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-4j1t) i.e. with spin 4ω0=-4/sec**

**Fig.5-6****F(4j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-4j1t)**

The animation lasts** T≈6.28sec**.**Fig. 5-6a**

The radius **R=0.5** will make **4** turns.**Fig.5-6b**

The **complex** function **F(4j1t)** as a rotating vector.

The** variable-length** vector **R(t)** will make **4** rotations.**Fig.5-6c**

Complex function **F(4j1t)** as a trajectory.**3rd** and **4th** rotation along the same track. **Non-zero** centroid **sc4=(0.25,0)**!

From the centroid** sc4**, the **4th** harmonic of the** f(t)** function can be read as **0.5cos(4t)**.

**Chapter 5.7 Rotating radius 5ω0=5/sec, so the trajectory F(j5ω0t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-j5ω0t) for 5ω0=-5/sec.**

**Fig.5-7****F(j5ω0t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-j5t)**

The animation lasts **T≈6.28sec**.**Fig. 5-7a**

The radius **R=0.5** will make **5** revolutions.**Fig.5-7b**

The complex function **F(5j1t)** as a rotating vector**Fig.5-7c**

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

The centroid of the trajectory **F(5j1t) is sc5=(0,0)=0**. So the function **f(t)** has no **harmonic** with a pulsation of **5ω0=5/sec**.

**Chapter 5.8 Trajectory F(6j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-6j1t) so with rotation 6ω0=-6/sec**

**Fig.5-8****F(6j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-j2t)**

The animation lasts** T≈6.28sec**.**Fig.5-8a**

The radius **R=0.5** will make **6** turns.**Fig.5-8b**

The complex function **F(6j1t)** as a rotating vector**Fig.5-8c**

Complex function **F(6j1t)** as a trajectory. **3** revolutions along the same path, which is why it seemingly stopped.

The centroid of the** F(6j1t)** trajectory is **sc6=(0,0)=0***. So the function **f(t)** has no harmonic with a pulsation of **6ω0=6/sec**.***sc6=(0,0)** is not as obvious as, for example, for **Fig.5-7c**. It seems too far to the left. Note, however, that the trajectory velocities on the left are lower than those on the right.

**Chapter 5.9 Trajectory F(7j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-7j1t) i.e. with spin 7ω0=-7/sec**

**Fig.5-9****F(7j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-7j1t)**

Complex function **F(j7ω0t)** for **7ω0=7/sec**.

The animation lasts **T≈6.28sec**.**Fig. 5-9a**

The radius **R=0.5** will make **7** revolutions.**Fig.5-9b**

The complex function **F(7j1t)** as a rotating vector**Fig.5-9c**

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

The **centroid** of the trajectory **F(7j1t) is sc7=(0,0)=0**. So the function **f(t)** has no harmonic with a pulsation of **7ω0=7/sec**.

**Chapter 5.10 Trajectory F(8j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-8j1t) i.e. with spin 8ω0=-8/sec**

**Fig.5-10****F(8j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-8j1t)**

The animation lasts **T≈6.28sec**. **Fig. 5-10a**

The radius **R=0.5** will make **8** turns.**Fig.5-10b**

The complex function **F(8j1t)** as a rotating vector.**Fig.5-10c**

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

The centroid of the trajectory **F(8j1t) is sc8=(0,0)=0**. So the function** f(t)** has no harmonic with a pulsation of **8ω0=8/sec**.