# Rotating Fourier Series

**Chapter 4. How to filter out the harmonic with f(t)=0.5*cos(4t)?**

**Chapter 4.1 Introduction**Each periodic function

**f(t)**is a sum of

**sinusoids/cosines**, i.e. the so-called

**harmonics**with pulsations

**1ω0, 2ω0,3ω0…**Constructing

**f(t)**when we know the harmonics is simple. Just add them. Conversely, i.e. finding

**harmonics**with pulsations

**1ω0, 2ω0,…nω0**, when we know

**f(t)**, is a more interesting problem. For example, how to extract the

**3rd**harmonic, i.e. for

**3ω0=3*1/sec**, from a

**square wave**with a pulsation of

**1ω0=1/sec**, i.e. a period

**T≈6.28sec**? The

**two-dimensional**and

**complex**version of the

**f(t)**function will be helpful here, i.e. the trajectory*

**F(njω0t)=f(t)*exp(-njω0t)**. It is actually

**n**trajectories for

**n**different harmonics with

**ω=n*ω0**pulsations. From each trajectory

**F(njω0t)**we will

**extract**successive

**harmonics**. This is more

**intuitive**than extracting from the actual

**f(t)**function.

In

**Chapter 4.4**we will only analyze the function

**f(t)=c0+A*cos(nω0t**) for

**c0=0**,

**A=0.5, n=4**and

**ω0=1/sec**. So the function consists of only one harmonic, i.e.

**0.5*cos(4t)**. As in the riddle “The duck floats on the pool, what is the name of the animal?” Duck of course. What is it for? Well, to get acquainted with the

**rotating plane**method. You will see how for almost every rotation pulsation

**ω**nothing is spins. Or in other words, the harmonic

**A*cos(nω0t)**with amplitude

**A=0**will spin off. That is nothing. But for one

**particular**pulsation, i.e.

**ω=4*1/sec**, the harmonic

**h4(t)=0.5*cos(4t)**spins off. Note that here the spin and harmonic

**pulsations**are the

**same**!

***trajectory**-path of a point moving along the

**Z**plane.

**Chapter 4.2 F(1j1t)=1*exp(-1j1t) i.e. rotating radius R=1 at a speed of ω=1/sec**

The complex function itself, **F(njω0t)=f(t)*exp(-njω0t)**, looks quite exotic and you don’t know how to do it. Therefore, we will start with the simplest case, i.e. when **f(t)=1, n=1** and **ω0=1/sec**. The function** f(t)** is a **constant**, and it couldn’t be easier! So the general **F(njω0t)=f(t)*exp(-njω0t)** becomes the specific **F(1j1t)=1*exp(-1j1t)**. We know from **Chapter 2** that this is a radius **R** of length **1** rotating at **-ω0=-1/sec** (i.e. in the “clockwise” direction). Its **initial** state is vector **(1,0)**.

**F****ig.4-1**

**Fig.4-1a**

The function

**F(1j1t)=1*exp(-1j1t)**as a rotating vector

**Fig.4-1b**

The function

**F(1j1t)=1*exp(-1j1t)**as a rotating trajectory.

In the animation you see only one rotation lasting

**T=2π/ω0=2πsec≈6.28sec**. Then the situation repeats itself along the same track. So even if the animation lasted longer than

**T**, the trajectory

**f(1j1t)**, unlike the rotating vector, would be

**stationary**!

Here was the speed of rotation of the plane

**-1ω0=-1/sec**

For

**F(2j1t)=1*exp(-2j1t)**the rotation speed will be

**twice**as high.

For

**F(3j1t)=1*exp(-3j1t)**will be

**3 times…**

For

**F(1j1t)=7*exp(-1j1t)**the circle will be

**7**times bigger… here

**f(t)=7**

Nothing more, nothing less.

For each spin speed

**nω0**, the center of gravity of this trajectory

**scn=0**. Here

**n=1**, i.e.

**sc1=0**. This means that the constant function

**f(t)=1**has no harmonics. You won’t get the Nobel Prize for this, but you’ve gotten used to the easiest case of the trajectory

**F(njω0t)=f(t)*exp(-njω0t)**, when

**f(t)=1**.

**Fig.4-1c**.

The function

**F(0j1t)=(1,0)**as a vector in the initial state

**t=0**.

Previously, we studied the complex function

**F(njω0t)**for

**n=1,2,3**… And what will happen for

**n=0**?

Pure mathematics shows that

**F(0jω0t)=f(0)=1*exp(0)=1**. Otherwise, the rotating radius will stop and it will be vector

**(1,0)**. You will admit that the expression

**F(0)=(1,0)**looks quite strange, although it is true. So let’s agree that we will write this situation as

**F(0j1t)=(1,0)**. The same as

**F(0)=(1,0)**but you can see that the rotating vector has stopped for

**n=0**!

**Fig.4-1d**.

The function

**F(0j1t)=(1,0)**as a trajectory in the initial state

**t=0**. Here, the trajectory degenerated to a point

**(1.0)**on the

**Re z**axis.

**Chapter 4.3 Trajectory F(njω0t)=f(t)*exp(-njω0t) for f(t)=0.75+0.5cos(4t), n=0 and n=1 when ω0=1/sec.****Chapter 4.3.1 Introduction**

A slightly more complicated example.

Look at the animation **Fig.4-1a** where the radius **R=1** rotates with a speed of **1ω0**. These are the laws of complex functions that every “creation” in the complex plane **Z** multiplied by **exp(-1jω0t)** will rotate “clockwise” by the angle** α=-1ω0t** around **z=(0,0)** . So the angle **α** rotates with a speed of **-1ω0**, therefore the “creature” will also rotate “clockwise” with a speed of **-ω0**. **Note**

In each experiments there will appear the **n-th** trajectory **F(njω0t)** rotating around the point **scn**. Let’s call it the **center of gravity** of the trajectory, although we calculate it differently than the **center of gravity–>centroid** of a plane figure in mechanics! Most often it will be **scn=(0,0)**, which is the center of the complex plane Z. But sometimes at some rotation speeds **ω=n*ω0** the point** scn** will be **non-zero**. I wonder at what velocities **ω** and what will be the coordinates **scn=(a,b)** of this point?

The **scn-****centroid** of the trajectory will often be **intuitive**. E.g. **Fig. 4-4b**. But that’s not always the case. Then you just have to believe me. In **Chapter 7** you will learn how to accurately calculate the **scn-centroid** for the** n-th** trajectory **F(njω0t)**, and how it relates directly to the spin harmonic **ω=njω0**.**Chapter 4.3.2 Trajectory F(njω0t)=f(t)*exp(-njω0t) for n=0 and ω0=4/sec **

**that is F(0j1t)=0.75+0.5cos(4t) that is non-spinning radius R(t)=0.75+0.5cos(4t)**

The radius

**R**changes according to periodic function

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

**f(t)=R(t)=0.75+0.5cos(4t)**and does not rotate

**(0j1t=0)**. Instead, it pulsates around

**c0=+0.75**on the

**Re z**axis according to the function

**f(t)=R(t)**. Animation shows it best.

**Fig. 4-2**

Trajectory **F(0j1t)=0.75+0.5cos(4t)** in** complex** and **real** version.

Like the number** x=1**, it can be in the **complex** version **x=1+0j** or only real **x=1**.**Fig.4-2a****F(0j1t)=0.75+0.5cos(4t)** in the **complex** version because in the** Z** plane

Indeed, every time **t** corresponds to a **point** in the **complex** plane **Z**, here on the **real** axis **Re z**. Remember that every **real** number is** complex**, but not every **complex** number is **real**! The function **f(t)** is a vector that is constantly “swinging” on the **real** axis **Re z** left and right around the vector** c0=(+0.75,0)**.

The **average** value of **F(0j1t)** in the period **T=2πsec≈6.28sec** is **c0=(+0.75.0)**.** T=2πsec** is the period of **F(0jω0t)=f(t)** but not its base period **T=πsec/2≈1.57sec**! –>**Note** for **Fig.4-3b**.**Fig.4-2b****F(0j1t)= 0.75+cos(4t)** in the **rea**l version, i.e. the classic function **f(t)= 0.75+0.5cos(4t)**.

The average value of f(t) over the period **T≈6.28sec** is also **c0=+0.75**.**Note**

In the **Fourier Series** of any **periodic** function, the **first** element is the **constant** component** f(t)**, the coefficient **c0**.

We calculatet it as an **integral** from the formula below.

**Fig.4-3**

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.4-3a**

General formula for constant **co** for **f(t)****Fig.4-3b **

Special formula for **f(t)= 0.75+0.5cos(4t)**

**Fig.4-3c**

Special formula for

**f(t)**when

**T=2π**

**Note.**

Here the basic period is

**T=π/2**. Refer to

**Fig.4-2a**. But

**T=2π**is also a period! In both cases we get the same result

**c0=+0.75**. Also for e.g.

**f(t)= +0.75+2.8cos(27t)**the formula is true for

**T=2π**. The mean

**c0=+0.75**is obvious without calculus. The surfaces above and below the green line

**c0=+0.75**are equal.

Another look at

**c0=+0.75**. It is also the center of gravity

**sc1=(0.75,0)**of the “swinging” trajectory

**F(0j1t)**

in

**Fig.4-2a**, when the plane rotation speed

**ω=0**.

**Chapter 4.3.3 Trajectory F(njω0t)=f(t)*exp(-njω0t) for n=1 and ω0=4/sec.**

**that is F(1j1t)=[0.75+0.5cos(4t)]*exp(-1j1t) that is rotating ray R(t)=0.75+0.5cos(4t)**

See the animation in

**Fig. 4-2a**. The end of the vector moves “back and forth” around the point

**c0=(+0.75, 0)**by formula

**f(t)= 0.75+0.5cos(4t)**.

What if, in addition, the vector rotated at a speed of

**–ω0=-1/sec**? So “clockwise” with period

**T≈6.28sec**. Then its motion in the complex plane can be described as in the title of the chapter.

**Fig. 4-4****F(1j1t)=[0.75+0.5cos(4t)]*exp(-1j1t)**

The animation shows **one** period **T=2π/ω≈6.28sec** of the complex function **F(1j1t)**.**Fig.4-4a****F(1j1t)** as modulated by** f(t)=0.75+0.5cos(4t)** radius **R(t)** whose rotation takes **T=2π/ω0≈6.28sec**.

As radius **R** rotates, its **length** changes according to the function **R(t)=f(t)=0.75+0.5cos(4t)**.**Fig.4-4b****F(1j1t)** as a **trajectory**.

The** centroid** (centre of mass) of the trajectory** F(njω0t)** for **n=1** is clearly **sc1=(0,0)**. Conclusion–> **f(t)=0.75+0.5cos(4t)** there is no harmonic with **1ω0=1/sec**! Directly from the formula, this is not the discovery of America. But we got it with **F(1j1t)=[0.75+0.5cos(4t)]*exp(-j1t)**! So we rotated** f(t)** “clockwise” at a speed of **-1ω0=-1/sec**.**Note**

Instead of the description of with a modulated radius** R**, one can more generally.

It is a combination of **2** moves

– **harmonic** along the **Re z** axis in **Fig. 4-2a** described by the equation **f(t)=0.75+0.5cos(4t)**, i.e. with the speed** ω=4/sec****– rotation** of the complex plane around **z=(0,0)** with a speed of **–ω0=-1/sec**.

The entire plane with its contents rotates (also with radius modulation **R**, but the **Rez/Imz** axes remain stationary!

**Chapter 4.4 How, using a plane rotating with velocity ω=n*ω0, extract the harmonic 0.5cos(4t) from the function f(t)=0.5cos(4t)?****Chapter Introduction 4.4.1**

We have the function **f(t)=0.5cos(4t)**. We know it, but “Someone” only knows that it’s a **cosine** function. He does not know the amplitude **A=0.5**, the pulsation **ω=4/sec**. It only sees the harmonic motion in **Fig.4-5a**. It’s roughly some kind of **sine/cosine** wave, but it’s not enough to determine the exact formula for **f(t)**. Instead, it has the ability to rotate the** complex plane** at any speed **ω**. However, let’s make it easier for “Someone” and let this speed be **n*ω0** for **n=0…8** and **ω0=1/sec**. He knows that the pulsation **f(t)** is some multiple of** ω0=1/sec**. We know that this multiple is** n=4**, “Someone” is not. What is the function **f(t)**? More specifically, what are the parameters of **f(t)=c0+Acos(nω0t)**? So we are looking for **c0**, **A** and **n**. We know the parameter **ω0–> ω0=1/sec**.**Chapter 4.4.2 Trajectory F(0j1t)=0.5cos(4t)*exp(-0j1t) i.e. no rotation**

“Someone” scratched his head and “laid”** f(t)** on the complex **Z** plane. First, without **z**-rotating, i.e. **n=0**, i.e. **0jω0=0**. An animation similar to **Fig. 4-2** will be created, only without the **c0** constant component.

**Fig.4-5**

Trajectory **F(0j1t)=f(t)=0.5cos(4t)** in **complex** and **real** version**Fig.4-5a****F(0j1t)=f(t)=0.5+cos(4t)** in the **complex** version**Fig.4-5b****F(0j1t)=f(t)=0.5+cos(4t)** in the **real** version, i.e. **f(t)=0.5cos(4t)**.

It’s just a function** f(t)**! Its average** f(t)** over the period **T=2π≈6.28sec** (and also T**=πsec/2≈1.57sec**) is **c0=0**.

I think everyone can see it, even without integrating. Another look at **c0=0**. It is also the centroid **sc(0,0)** of the “swinging” trajectory in** Fig. 4-5a** when the plane spin speed **ω=0**.**Conclusion**

We already know the first parameter **f(t)=c0+Acos(nω0t)**. It is the constant component **c0=a0=0**.**Chapter 4.4.3 Trajectory F(1j1t)=0.5cos(4t)*exp(-1j1t) i.e. with spin -1ω0=-1/sec**

In this and the following subsections the **pulsating** radius **R(t)=0.5cos(4t)** will start rotating. We start with the slowest speed **–1ω0=-1/sec**. Here the animation will last **T=2π/ω0≈6.28sec** and radius **R=0.5** from **Fig.4-6a** will make **1** rotation. At the following speeds, i.e. **–2ω0=-2/sec, -3ω0=-3/sec…-8ω0=-8/sec,** the duration of each animation will be the same **T≈6.28sec**. Then radius **R=0.5** will make **2,3…8** turns. Or in other words, the complex plane** Z** will make **2.3…8** revolutions

**Fig.4-6**

**F(1j1t)=0.5cos(4t)*exp(-1j1t)**

The animation lasts T=2π/ω0≈6.28sec.

You will admit that the animation, especially

**Fig.4-6c**gives more information than a bare drawing.

**Fig.4-6a**

The radius

**R=0.5**out of

**1ω0=1/sec**will make

**1**revolution.

**Fig.4-6b**

During the rotation, the length of the

**radius**changes according to the function

**R(t)= f(t)=0.5cos(4t)**. The radius becomes “negative” at times. This is not mathematical because the

**radius**is always

**positive**! This means a situation when the rotating radius

**R**passes through

**(0,0)**another

**quadrant**of the plane. Thus, the

**complex**function is implemented as a

**rotating**vector, i.e. the trajectory

**F(1j1t)=0.5cos(4t)*exp(-1j1t)**. Note that the lack of precision (“negative radius”) makes it easier to understand the problem. I’d get slapped by a real mathematician.

**Fig.4-6c**

The complex function

**F(1j1t)**as a trajectory drawn by a rotating vector.

The center of mass (centroid) of the trajectory

**F(1j1t)**is clearly

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

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

**1ω0=1/sec**.

***Note**

The function

**F(1j1t)=0.5cos(4t)*exp(-j1t)**is simpler than the previous one with a constant component c0. i.e. from

**F(1j1t)=[0.75+0.5cos(4t)]*exp(-1j1t)**. But harder to imagine! Why? Because sometimes the radius

**R**becomes “negative” just like every

**cosine**.

Therefore, earlier in

**Chapter 4.3.3,**we studied the “easier” radius

**R**, which is still “positive”.

**Chapter 4.4.4 Trajectory F(2j1t)=0.5cos(4t)*exp(-2j1t) i.e. with spin -2ω0=-2/sec**

** **

**Fig.4-7****F(2j1t)=0.5cos(4t)]*exp(-2j1t)**

The animation lasts **T≈6.28sec**.**Fig.4-7a**

Radius **R=0.5** will make **2** revolutions.**Fig.4-7b**

The complex function **F(2j1t)** as a** rotating vector****Fig.4-7c**

The complex function** F(2j1t)** as a **trajectory**. The second turn on the same track, which is why it seemingly stopped. The **centroid** of the trajectory **F(2j1t)** is **sc2=(0,0)**. So function **f(t)** has no harmonic with pulsation **2ω0=2/sec**.**Chapter 4.4.5 Trajectory F(3j1t)=0.5cos(4t)*exp(-3j1t) i.e. with rotating -3ω0=-3/sec**

**F****ig.4-8**

**F(3j1t)=0.5cos(4t)]*exp(-3j1t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-8a**

The radius

**R=0.5**will make

**3**turns.

**Fig.4-8b**

The complex function

**F(3j1t)**as a

**rotating vector**.

**Fig.4-8c**

**Complex**function

**F(3j1t)**as a

**trajectory**.

The centroid of the trajectory

**F(3j1t) is sc3=(0,0)**. So the function

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

**3ω0=3/sec**.

**Note**

The faintly visible

**red**horizontal line will be useful later for comparison purposes with

**Fig.4-16c**.

**Chapter 4.4.6 Trajectory F(4j1t)=0.5cos(4t)*exp(-4j1t) i.e. with spinning -4ω0=-4/sec**

**Fig****.4-9**

**F(4j1t)=0.5cos(4t)*exp(-4j1t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-9a**

The radius

**R=0.5**will make

**4**turns.

**Fig.4-9b**

The complex function

**F(4j1t)**as a

**rotating vector**. The variable-length

**vector**

**R(t)**will make

**8**revolutions in a

**circle**. That is

**2**times more than

**R**in

**Fig.4-9a**!

**Fig.4-9c**

Complex function

**F(4j1t)**as a

**trajectory**.

You only see the

**first**of

**eight**turns of the

**trajectory**.

Something interesting is happening with the

**sc4**

**centroid**of the

**trajectory**. It is not as before (and later too!)

**Zero**, but

**sc4=(0.25,0)**. It is a vector and can also be written in exponential form

**sc4=R*exp(jϕ)=0.25*exp(j0°)**.

In

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

**centroid**of the

**trajectory**for the pulsation

**4*ω0 as sc4=(0.25,0)**you can read the

**4th**harmonic of the function

**f(t)**as

**0.5cos(4t)**.

**Note**

Also note that the vector

**Fig.4-9b**and the trajectory

**Fig.4-9c**have

**2**times more

**pulsation**than the rotating vector

**R=0.5**from

**Fig.4-9a**.

**Chapter 4.4.7 Trajectory F(5j1t)=0.5cos(4t)*exp(-5j1t) i.e. with rotating -5ω0=-5/sec**

**F****ig.4-10**

**F(5j1t)=0.5cos(4t)]*exp(-5j1t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-10a**

The radius

**R=0.5**will make

**5**revolutions.

**Fig.4-10b**

The complex function

**F(5j1t)**as a

**rotating vector**.

**Fig.4-10c**

**Complex**function

**F(5j1t)**as a

**trajectory**.

The centroid of the trajectory

**F(5j1t)**is

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

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

**5ω0=5/sec**.

**Chapter 4.4.8 Trajectory F(6j1t)=0.5cos(4t)*exp(-6j1t) i.e. with rotating -6ω0=-6/sec**

**Fig****.4-11**

**F(6j1t)=0.5cos(4t)]*exp(-6j1t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-11a**

The radius

**R=0.5**will make

**6**turns.

**Fig.4-11b**

The complex function

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

**Fig.4-11c**

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 4.4.9 Trajectory F(7j1t)=0.5cos(4t)*exp(-7j1t) i.e. with rotating -7ω0=-7/sec**

**Fig.4-12**

**F(j7t)=0.5cos(4t)]*exp(-7j1t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-12a**

Radius

**R=0.5**with

**ω0=7/sec**will make

**7**revolutions.

**Fig.4-12b**

The complex function

**F(j7t)**as a rotating vector.

Fig.4-12c

Complex function F(j7t) as a trajectory.

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

**Chapter 4.4.10 Trajectory F(8j1t)=0.5cos(4t)*exp(-8j1t) i.e. with rotating -8ω0=-8/sec**

**F****ig.4-13**

**F(8j1t)=0.5cos(4t)]*exp(-8j1t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-13a**

The radius

**R=0.5**will make

**8**turns.

**Fig.4-13b**

The complex function

**F(8j1t)**as a

**rotating vector**.

**Fig.4-13c**

Complex function

**F(8j1t)**as a

**trajectory**. The centroid of the trajectory

**F(8j1t)**is sc8=(0,0). So the function

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

**8ω0=8/sec**.

**Chapter 4.4.11 Summary of trajectories F(njω0t)=0.5cos(4t)*exp(-jω0t) for different rotatings nω0**

The most important

**3**conclusions from the above trajectories

**1.**When the pulsations

**ω**of the periodic function

**f(t)**and the rotating of the

**Z plane**are the same (ω=-4ω0), then the trajectory rotates around the

**non-zero**center of gravity

**s4c=(0.25,0)**. This means that for this

**pulsation**there is a harmonic

**f(t)=2*0.25cos(4t)=0.5cos(4t)**. Why exactly “

**2*0.25=0.5**″, you will learn in

**Chapter 7**.

**2.**For the remaining

**non-zero**pulsations

**ω**, the trajectories rotatate around

**scn=(0,0)**. This means that all other

**ω**pulsations do not contain harmonics.

**3.**From the

**non-rotating**trajectory (ω=0) the constant component can be read, here

**c0=0**. In

**Fig.4-2a**, the constant component is

**c0=0.75**

Let’s combine the previous animations into one.

**Fig.4-14**Trajectories

**F(jnω0t)=0.5cos(4t)*exp(-jnω0t)**for different

**nω0**pulsations.

**1.**The trajectory for

**ω=0**pulses around

**sc0=c0=0**. It is the

**zero**constant component

**c0**of the function

**f(t)=2*0.25cos(4t)=0.5cos(4t)**and at the same time the centroid

**sc0**of the trajectory

**F(jnω0t)**for

**n=0**and

**ω0=1/sec**.

**2.**The trajectory for

**ω=-4ω0=-4/sec**pulses around

**sc4=(0.25,0)**. From it you can read the

**4th**(and only!)

**harmonic**of the function

**f(t)=0.5cos(4t)**.

**3.**For the remaining rotating speeds n

**ω0**, the trajectories rotate around

**zero**centroids

**sc1=sc2=sc3=sc5=sc6=sc7=sc8=(0,0)**. This means that for these

**nω0**pulsations, the function

**f(t)=0.5cos(4t)**has no harmonics. This is obvious, but we got it by the

**z-plane**rotating method.

**Chapter 4.5 Using a plane rotating at ω=n*ω0, how do you extract the 0.5cos(4t-30°) harmonic from f(t)=0.5cos(4t-30°)?Chapter 4.5.1 Introduction**In

**Chapter 4.4**there was a function

**f(t)=0.5cos(4t)**, now

**f(t)=0.5cos(4t-30°)**. How will this affect the trajectories

**F(jnω0t)=f(t)*exp(-jnω0t)**? We expect that for

**n≠4**the trajectories of

**F(jnω0t)**will also rotate around

**scn=(0,0)**. And around “what” will rotate for

**n=4**, i.e. what will be

**sc4**?

Will we also read from this “what” harmonic

**0.5cos(4t-30°)**?

**Chapter 4.5.2 Trajectory f(0j1t)=0.5cos(4t-30°)*exp(-0j1t) i.e. no rotation**

**Fig.4-15**Function

**0.5cos(4t-30°)**in

**complex**version

**f(t)**and

**classic**

**f(t)**

**Fig.4-15a**

**f(t)= 0.5cos(4t-30°)**in the

**complex**version

The constant component

**c0=0**, i.e.

**f(t)**“swings” around

**sc0=(0,0)**. Almost as in

**Fig.4-5a**. Find a slight difference visible at the initial moment in relation to

**Fig.4-5a**.

**Fig.4-15b**

The

**classic**version

**0.5cos(4t-30°)**i.e. the

**time chart.**

**Chapter 4.5.3 Trajectory f(3j1t)=0.5cos(4t-30°)*exp(-3j1t) i.e. with rotation 3ω0=-3/sec**

Previously, for

**f(t)=0.5cos(4t)**, we studied

**8**trajectory rotation speeds, i.e. for

**n*ω0**when:

**n=4**then there was a

**non-zero**

**centroid**of the trajectory

**sc4=(0.25,0)**

**n≠4**then there was a

**zero**

**centroid**

**scn**of the trajectory

**scn=(0,0)**

For

**f(t)=0.5cos(4t-30°)**it will be similar. Therefore, we will check only for one “zero” rotation, e.g. for

**n=3**, i.e. for

**n*ω0=3/sec**

**Fig.4-16F(3j1t)=0.5cos(4t–30°)*exp(-j3t)≈(0.433-j0.25)*cos(4t)*exp(-j3t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-16a**

The radius

**R=0.5**z will make

**3**turns.

**Fig.4-16b**

The complex function

**F(3j1t)**as a

**rotating vector**.

**Fig.4-16c**

**Complex**function

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

It is

**rotated**by some angle (

**-30°**) relative to the trajectory

**f(3j1t)=0.5cos(4t)*exp(-j3t)**from

**Fig.4-8c**. This can be seen by the

**red**line in both drawings. The

**centroid**for rotation

**ω=-3/sec**is

**sc3=(0,0)**. So the function has no harmonic for

**ω=3/sec**.

**Chapter 4.5.4 Trajectory F4j1t)=0.5cos(4t-30°)*exp(-4j1t) i.e. with rotation -4ω0=-4/sec**

**Fig.4-17**

**F(4j1t)=0.5cos(4t-30°)*exp(-j4t)**

The animation lasts

**T≈6.28sec**.

**Fig.4-17a**

The radius

**R=0.5**will make

**4**turns.

**Fig.4-17b**

The complex function

**F(4j1t)**as a

**rotating**vector

The

**variable-length**vector

**R(t)**will make

**8**revolutions in a

**circle**. That’s why you only see the

**first**turn.

**Fig.4-17c**

Complex function

**F(4j1t)**as a

**trajectory**.

The vector from

**Fig. 4-17b**will make

**8**revolutions around the

**circle**. The centroid is not

**zero**as before, but

**sc4=0.25*exp(-j30°)**or as

**sc4=(a,b)≈(0.433,-0.25)**. In

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

**f sc4**you can read the

**4th**harmonic of

**f(t)**as

**0.5cos(4t-30°)**.