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

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).

Fig.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 real 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

Fig.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

Fig.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

Fig.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 nω0, 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.4-16
F(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 of sc4 you can read the 4th harmonic of f(t) as 0.5cos(4t-30°).