**Rotating Fourier Series**

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

**Chapter 4.1 Introduction
**Each

**periodic**function

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

**sine/cosine**, i.e.

**harmonics**with pulsations

**1ω0, 2ω0,3ω0 …**Creation

**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 the square wave

**f(t)**with pulsation

**1ω0=1/sec**,

i.e. with the period

**T≈6.28sec**? The

**two-dimensional**and

**complex**version of the function

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

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

**n**trajectories for

**n**different harmonics with pulsations

**ω=n*ω0**. We will extract subsequent harmonics from each trajectory

**F(njω0t)**. This is more intuitive than dereferencing from the real

**time**function

**f(t)**.

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 puzzle “It swims in the pond, the duck is called”. What does swim in the pond? What is this for? Well, to learn about the

**Rotating Plane Method**. You will see how for almost every pulsation

**ω**, nothing is

**centrifuging**. Or else, harmonic

**A*cos (nω0t)**with amplitude

**A=0**is centriufuged. Which is

**nothing**. But for one

**particular**centrifufe pulsation, ie

**ω=4/sec**, the

**4rd**harmonic ie.

**h4(t)=0.5*cos(4t)**is centrifuged. Note that the centrifuge

**rotation**and harmonic

**pulsation**are the same here!

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

**Z**plane.

**Chapter 4.2 F(1j1t)=1*exp(-1j1t) that is, rotating radius R = 1 at speed ω = 1 / sec
**The complex function itself

**F(njω0t)=f(t)*exp(-njω0t)**looks quite exotic and it is not known how to go about 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 simpler! So the general

**F(njω0t)=f(t)*exp(-njω0t)**becomes the special

**F(1j1t)=1*exp(-1j1t)**. We know from

**Chapter 2**that it is a radius

**R**of length

**1**rotating at

**ω0=-1/sec**(clockwise). Its initial state is the vector

**(1,0)**.

**Fig.4-1**

**a.
**

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

**rotating vector**.

**b**.

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

**rotating**

**trajectory**.

In the animation you can only see one revolution lasting

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

**T**, the trajectory

**F(1j1t)**, unlike the rotating vector, would be stationary! Here was the angular speed of the plane

**1ω0=-1/sec**

For

**F(2j1t)=1*exp(-2j1t)**it will be

**2**times the speed.

For

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

**3**times …

For

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

**7**times larger circle

**a**… here

**f(t)=7cos(1t)**

Nothing more nothing less.

For each 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

**non-zero harmonics**. You won’t get a Nobel Price for that, but at least you’ve gotten used to the special case of the trajectory

**F(njω0t)=f (t)*exp(-njω0t)**when

**f (t)=1**.

**c**.

**F(0j1t)=(1,0)**as a

**vector**initial state when

**t=0**.

Previously, we studied the complex function

**F (njω0t)**for

**n=1,2,3 …**And what about

**n=0?**

From pure mathematics it follows that

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

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

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

**F(0j1t)=(1,0)**. Same as

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

**n = 0**!

**.**

d

d

Function

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

**t=0**. Here the trajectory has degenerated to the point

**(1,0**) on the

**Re z**axis.

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

See animation

**Fig.4-1a**where radius

**R=1**is rotting at

**1ω0**. Such are the laws of

**complex functions**that every “creation” on the complex plane

**Z**multiplied by

**exp(-1jω0t)**will rotate “clockwise” by the angle

**α=-1ω0t**around

**z=(0,0)**. So the angle

**α**is rotating with the speed

**-1ω0**, therefore the “creation” will also be rotating “clockwise” with the speed

**-1ω0**. For example, in

**Fig. 4-1a**it was the radius

**R=1**drawing in

**Fig.4-1b**the trajectory

**F (1jω0t) = 1 * exp (-1j1t)**. Why didn’t I choose a trajectory, e.g.

**F(1j1t)=0.5cos(4t)*exp(-1j1t)**? Because it’s harder to imagine than for

**f (t)=0.75+0.5cos (4t)**, as you’ll find out about it. However, we will come back to it in

**Chapter 4.4**.

**Note**

In each experiment, the

**n-th**trajectory

**F(njω0t)**will appear rotating around the point

**scn**. Let’s call it the

**center of gravity**of the trajectory, although we calculate it differently from the

**center of gravity**of a plane figure in mechanics! Most often it will be

**scn=(0,0)**, which is the

**center**of the complex

**Z**plane. But sometimes at some spin speeds

**ω=n*ω0**the point

**scn**will be

**non-zero**. I wonder at what speeds

**ω**and what will be the coordinates

**scn=(a, b)**of this point?

The

**center of gravity**of a

**scn**trajectory will often be intuitive. Eg

**Fig. 4-4b**. But this is not always the case. Then you just have to believe me. In

**Chapter 7**, you will learn how to accurately calculate the center of gravity

**scn**for the

**n-th**trajectory

**F(njω0t)**, and how it directly relates to the harmonic of

**ω=njω0**.

**Chapter 4.3.2 Trajectory F(njω0t)=f(t)*exp(-njω0t) for n=0
ie. F(0j1t)=0.75+0.5cos(4t) i.e non-rotating R(t)=0.75+0.5cos(4t)
**Radius

**R**changes according to periodic function

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

**f(t)=R(t)=0.75+0.5cos (4t)**and doesn’t rotate

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

**c0=+0.75**on the

**Re z**axis according to the function

**f (t)=R(t)**. This is best shown in the animation.

**Fig.4-2
**Trajectory

**F(0j1t)=0.75+0.5cos(4t)**in the

**complex**and

**real**version

Like the number

**x=1**it can be the

**complex**version

**x=1+0j**or the real

**axis**only

**x=1**

**a**

**F(0j1t)=0.75+0.5cos(4t)**in the complex version because on the

**Z**plane

Indeed, every time

**t**corresponds to a point on the complex plane

**Z**, here on the real axis

**Re z**. Remember that every

**real number**is a

**complex number**, but not every

**complex number**is

**real**! The function

**f(t**) is a vector that keeps “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 is not its base period

**T=πsek/2≈1.57sec**!

**b**

**F(0j1t)=0.75+0.5cos(4t)**in the

**real**version, i.e. the classic function

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

The

**average**value of

**f(t)**in 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)**, i.e. the coefficient

**c0=a0**. We calculate it as the integral of the formula below.

**Fig.4-3
**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**

**General**formula for any

**f(t)**with period

**T**

**b**

**Special**formula for

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

**Note.**

Here the base 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=a0=+0.75**is obvious without integral calculus. The areas above and below the

**green**line

**c0=+0.75**are equal. Another look at

**c0=a0=+0.75**. It is also the center of gravity

**sc0=(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 so F (1j1t) = [0.75 + 0.5cos (4t)] * exp (-1j1t)
that is the rotating radius R (t) = 0.75 + 0.5cos (4t)**

See the animation in

**Fig. 4-3.a**. The end of the vector moves “round-trip” around the point

**c0=(+0.75,0)**acc. by the formula

**. What if the vector rotate at the speed**

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

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

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

**T≈6.28sec**.

Then its movement on the complex plane can be described as in the title of the chapter.

**Fig. 4-4**

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

The animation shows **one** period **T=2π/ω≈6.28sec** of the complex function **F(1j1t)**.

**a**

**F(1j1t)** as modulated by the function** f(t)=0.75+0.5cos (4t)** radius **R(t)**, the rotation of which takes **T=2π/ω0≈6.28sec**

As the radius **R** rotates, its length changes according to the function **R(t)=f(t)=0.75+0.5cos (4t)**.

**b**

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

The center of gravity of the trajectory **F(njω0t)** for **n=1** is evidently **sc1=(0,0)**. Conclusion–> **f (t)=0.75+0.5cos(4t)** there is no harmonic with pulsation **1ω0=1/sec**! Directly from the formula. This is not America’s discovery. But we got it from **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 as above. with modulated **R** radius, can be more generally.

It is a combination of **2** moves

**–** **harmonic** along the **Re z** axis in **Fig. a** described by the equation **f(t)= 0.75+0.5cos (4t)**, i.e. with the speed **ω=4/sec**

**– a** **rotating** complex plane around** z=(0,0)** with a speed of **ω0=-1/sec**.

The whole plane rotates with its contents (also with modulated radius **R**!), But the **Rez/Imz** axes remain stationary!

**Chapter 4.4 How to extract harmonic 0.5cos (4t) from the function f (t) = 0.5cos (4t) using a plane rotating at speed ω=n*ω0? **

**Chapter 4.4.1 Introduction **

We have the function **f(t)=0.5cos (4t)**. We know it, but “Someone” only knows that it is a **cosine** function. It does not know the amplitude **A=0.5**, **pulsation ω=4/sec**. He only sees the **harmonic motion** in **Fig. 4-4a**. Roughly, this is some kind of** sine/cosine** wave, but that is not enough to determine the exact formula for **f (t)**. However, it has the ability to rotate the complex plane at any speed **ω**. But let’s make it easier for the “Who” to do and let this speed be **n*ω0** for **n=0…8 **and** ω0=1/sec**. He knows that the** f(t)** pulsation is some multiple of **ω0=1/sec**. We know that this multiple is** n=4**, “Someone” is not. What is the function of **f (t)**? More exactly, what are the parameters of the function** 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. without rotation**

“Someone” scratched his head and “put” **f(t)** on the complex plane **Z**. First, without rotating the **Z** plane, that is **n=0 **or** 0jω0=0**.

An animation similar to **Fig. 4-2** will result, but without the constant component **c0**.

**Fig.4-5**

Trajectory **F(0j1t)=f(t)=0.5cos (4t)** in the** complex** and real **version**

**a**

**F(0j1t)=f (t)=0.5+cos(4t)** in the **complex** version

**b**

**F(0j1t)=f(t)=0.5+cos(4t)** in the **real** version

It’s just a function of **f(t)**! Its average value of **f(t)** in the period **T=2π≈6.28sec** (and also **T=πsek/2≈1.57sec**) is **c0=0**. I guess everyone can see it, even without integration. Another look at **c0=a0=0**. It is also the center of gravity **sc0=(0.0)** of the “swinging” **trajectory** in **Fig. a**, when the plane rotation speed **ω=0**.

**Conclusion**.

We already know the first parameter of **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 rotation 1ω0=-1/sec**

In this and the following subsections, a pulsating ratdius **R(t)=0.5cos(4t)** will start to rotate. We start with the lowest speed **1ω0=-1/sec**. Here the animation will be **T=2π/ω0≈6.28sec** and the radius** R = 0.5** from **Fig. 4-6a** will make **1** revolution. In the following speeds, i.e. **2ω0=-2/sec, 3ω0=-3/sec …8ω0=-8/sec**. The time of each animation will be the same **T≈6.28sec**. It means the radius **R=0.5** will make **2.3 … 8** turns. Or else, the complex plane **Z** will perform **2.3 … 8** turns.

**Fig.4-6
**

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

The animation lasts

**T=2π/ω0≈6.28sec**. You’ll admit that animation, especially

**Fig. c**, gives more information than a naked drawing.

**a**

The radius

**R=0.5**with

**1ω0=-1/sec**will make

**1**revolution.

**b**

During rotation, the length of the

**radius**changes according to the function

**R(t)=f(t)=0.5cos (4t)**. So the complex function is realized as a rotating vector, i.e. the trajectory

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

**c**

Complex function

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

**center of gravity**of the trajectory

**F(1j1t)**is evidently

**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**. ie from

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

**R**becomes “negative” like every

**cosine**. Therefore, earlier in

**Chapter 4.3.3**we examined the “easier” radius

**R**, which is still “positive”.

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

**Fig.4-7**

**F(2j1t)=0.5cos(4t)]*exp (-j2t)**

The animation lasts **T≈6.28sec**.

**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. A second rotation on the same track and therefore seemingly stopped. The center of gravity of the trajectory **F(2j1t**) is **sc2 = (0,0)**. So the 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 rotation 3ω0=-3/sec
**

**Fig.4-8 **

**F(3j1t)=0.5cos(4t)]*exp (-j3t)**

The animation lasts T≈6.28sec.

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

**Note**:

The faint **red** horizontal line will be useful later for comparison with **Fig. 4-16c**

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

**Fig. 4-9 **

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

The animation lasts **T≈6.28sec**.

**a**

Radius **R = 0.5** will make **4** turns.

**b**

Complex function **F(4j1t)** as a rotating vector. A vector **R(t)** of variable length will make **8** revolutions in a circle. So **2** times more than **R** in **a**!

**c**

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

You only see the** first** of the **eight** turns of the **trajectory**. Something interesting happens with the **center of gravity** of the **sc4** trajectory. It is not as before (and later too!) **zero**, but **sc4=(0.25,0)**. It is a vector and can also be written as exponential **sc4=R *exp (jϕ)=0.25*exp (j0 °)**. In **Chapter 7** you will learn that from the center of gravity of the trajectory for a **4*ω0** pulsation as **sc4=(0.25.0)** or **sc4=0.25*exp(j0 °)**, you can easily read the **4th** harmonic of **f (t)** as **0.5cos (4t)**. You will also learn how to calculate this center of gravity** sc4**.

**Note**

Also note that the vector** b** and the trajectory **c** have **2** times more pulsation than the rotating vector **R=0.5** in **a**.

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

**Fig. 4-10 **

**F(5j1t)=0.5cos (4t)]*exp (-j5t)**

The animation lasts **T≈6.28sec**.

**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 **center of gravity** 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 rotation 6ω0=-6/sec
**

**Fig.4-11
**

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

The animation lasts

**T≈6.28sec**.

**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

**center of gravity**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 rotation 7ω0=-7/sec
**

**Fig.4-12
**

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

The animation lasts

**T≈6.28sec**.

**a**

Radius

**R=0.5**with

**ω0=7/sec**will make

**7**turns.

**b**

Complex function

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

**c**

Complex function

**F(j7t)**as a trajectory.

The

**center of gravity**of the trajectory

**F(7j1t)**is

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

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

**7ω0=7/sec**.

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

**Fig. 4-13**

**F (8j1t)=0.5cos (4t)*exp (-j8t)**

The animation lasts **T≈6.28sec**.

**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 center of gravity** of the trajectory **F(7j1t)** is **sc8=(0,0)**. So the function **f(t)** has no harmonic with a pulsation of **8ω0=8/sec**.

** 4.4.11 Summary of trajectory F (njω0t)=0.5cos(4t)*exp(-jω0t) for different nω0 rotations.**

The most important **3** conclusions from the above trajectories

**1.** When the **pulsations ω** of the periodic function **f(t)** and **rotation** are the same, the trajectory rotates around the **non-zero** **center of gravity sc4=(0.25.0)**. This means that for this **pulsation** there is a harmonic **f(t)=2*0.25cos(4t)=0.5cos(4t)**. You will learn why** “2*0.25= 0.5”** in **Chapter 7**.

**2.** For the remaining **non-zero** pulsations **ω0**, the trajectories rotate around **sc4=(0,0)**. This means that all other **ω0** pulsations do not contain harmonics.

**3.** The **constant** component can be read from the **non-rotating** trajectory **(ω=0*ω0=0)**, 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*ω0=0/sec**rotates 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 center of gravity

**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**harmonic (and the only one!) Harmonic of the function

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

**3.**For the remaining speeds

**nω0**, the trajectories rotate around the

**zero**centers of gravity

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

**ω**the function

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

**Z**plane method.

**Chapter 4.5 How to extract harmonic 0.5cos (4t-30°) from the function f (t)=0.5cos (4t-30°) using a plane rotating at speed ω=n*ω0? **

**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**trajectories

**F(jnω0t**

**)**will also rotate around

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

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

**sc4**be?

**Chapter 4.5.2 Trajectory F(njω0t)=f(t)*exp(-njω0t) for n=0
ie. F(0j1t)=0.5cos(4t-30°) i.e non-rotating R(t)=0.5cos(4t-30°)
**

**Fig.4-15
**Trajectory

**F(0j0t)=0.5cos (4t-30 °)**in the

**complex**version and

**real**version

**Fig. 4-15a**

Trajectory

**F(0j0t)=0.5cos (4t-30 °)**in the

**complex**version

The

**constant**component

**c0 =0**, ie

**f(t)**“swings” around

**sc0=(0,0)**. Almost like

**Fig.4-5a**. Find the slight difference you can see at the beginning of

**Fig. 4-5a**.

**Fig. 4-15b**

Trajectory

**F(0j0t)=0.5cos (4t-30 °)**in the

**real**version, i.e. the time plot.

**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 tested

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

**n*ω0**when:

**n=4**then there was a

**non-zero**trajectory center of gravity

**sc4=(0.25.0)**

**n≠4**then was the

**zero**center of gravity sc of the trajectory

**scn=(0.0)**

For

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

**n=3**that is for

**n*ω0=3/sec**.

**Fig.4-16
F(3j1t)=0.5cos(4t**

**–**

**30°**

**)*exp(-j3t)≈(0.433-j0.25)*cos(4t)*exp(-j3t)**

The animation lasts

**T≈6.28sec**.

**a**

Radius

**R=0.5**z will make

**3**turns.

**b**

Complex function

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

**c**

Complex function

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

It is rotated by some angle(

**-30°**?) with respect to the trajectory

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

**Fig. 4-8c**. You can see it by the

**red**line in both pictures. The center of gravity for revolutions

**ω=3/sec**is

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

**for ω=3/sec**.

**Chapter 4.5.4 Trajectory F(4j1t)=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**.

**a**

Radius

**R = 0.5**will make

**4**turns.

**b**

Complex function

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

A vector

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

**8**revolutions in a

**circle**. Therefore, you only see the first turn.

**c**

Complex function

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

The vector

**b**will make

**8**turns in a circle. It 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 center of gravity

**sc4**, the

**4th**harmonic of

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

**0.5cos (4t-30 °)**.