## 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
.
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
f(t)=0.75+0.5cos (4t)
. What if the vector rotate at the speed 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 °).