5. How to filter harmonic from f(t)=1.3+0.7cos(2t)+0.5cos(4t)?

Rotating Fourier Series

Chapter 5. How to extract harmonics from  f(t)=1.3+0.7cos(2t)+0.5cos(4t)?

Chapter 5.1 Introduction
In the previous chapter, we extracted the harmonic 0.5cos(4t) from the functions f(t)=0.5cos(4t) and f(t)=0.5cos(4t-30 °) using the
rotating Z plane method. Now we will do the same, but with the function f (t)=1.3+0.7cos (2t)+0.5cos (4t). The pulsations of both cosines are different and the constant co=1.3 component was added! I wonder how this will affect the graph of f(t) and the rotating trajectories of F(jnω0t)?

Chapter 5.2 How to extract harmonics from the function f (t)=1.3+0.7cos (2t)+0.5 cos (4t)
using a plane rotating at speed ω=nω0?
Chapter 5.2.1 Introduction
You will find that regardless of the pulsation of nω0, the period of the trajectory F(jnω0t) is (also)*, T=6.28sec
and for:
– ω = 0ω0=0/sec the trajectory doesn’t rotate
– ω=1*ω0=1/sec …8 ω0 =8/sec the radius R=1 will make 1…8 turns
Will it be possible to extract the constant c0=1.3 and harmonics h2(t)=0.7cos (2t) and h4(t)=0.5cos (4t)?
*Period T may be less. Hence “also”.

Chapter 5.2.2 Trajectory F (0j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-0j1t), i.e. without rotation
The problem is similar to Fig. 4-2 of the previous chapter. Radius R doen’t rotate, but varies according to periodic function
R(t)=f(t)=1.3+0.7cos(2t)+0.5cos(4t). So the trajectory on the complex plane Z is F (0j1t) = f (t) * exp (-0j1t)=1.3+0.7cos(2t)+0.5cos(4t).
By the way, you’ll see a time graph of the  f(t).

Fig. 5-1
F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t) in the complex and real version
a
F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t) in the complex version
b
F(0j1t)=f (t)=1.3+0.7cos (2t)+0.5cos (4t) in the real version
Conclusion.
The average value of f(t) in the period T=2π≈6.28sec (and also T= πsek≈3.14sec) is c0=1.3. It is also the center of gravity sc0=(1.3,0) of the “swinging” trajectory in Fig.5-1a, when the plane rotattion speed ω=0. Another name for the parameter c0=1.3 is the constant component of the function f (t).
Note.
In the Fourier Series of the periodic function f(t), the first element is the constant component-the coefficient c0. This is the mean of f(t) over the period of T. We calculate it as the integral of the formula below.

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

Chapter 5.2.3 Trajectory F (1j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-1j1t) i.e. with rotation 1ω0=-1/sec
Radius R=1 spins at 1ω0=-1/sec.
See the animation in Fig. 5-1a. The end of the vector moves “there and back again” around c0=(+1. 3, 0) acc. to
formula R(t)=f(t)=1.3+0.7 cos (2t) + 0.5 cos (4t). What if the vector started to spin at 1ω0 =-1/sec? So “clockwise” and
with the period T≈6.28sec. Then its motion on the complex plane Z is described by the complex function F(1j1t).

Fig.5-3
F(1j1t)=[1.3+0.7cos (2t)+0.5cos (4t)]*exp(-j1t)
The animation lasts T=2π/ω0≈6.28sec.
a
The radius R = 0.5 with 1ω0=-1/sec will make 1 revolution.
Notice that we chose R=0.5 as the reference for the constant radius. Also in the next animations
b
During rotation, the length of the radius changes according to the function R(t)= f(t)=1.3+0.7cos (2t)+0.5cos (4t).
So the complex function is realized as a rotating vector F(1j1t)=[1.3+0.7cos (2t)+0.5cos (4t)]*exp(-j1t)
c
Complex function F(1j1t) as a trajectory.
It is drawn by the vector from b
The center of gravity of the trajectory F(1j1t) is sc1 = (0,0)= 0 . So the f(t) has no harmonic with pulsation 1ω0=1/sec.

Chapter 5.2.4 Trajectory F (2j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-2j1t) i.e. with rotation 2ω0=-2/sec

Fig5-4
F(2j1t)=[1.3+0.7cos (2t)+0.5cos (4t)]*exp(-2j1t)
The animation lasts T=2π/ω0≈6.28sec.
a
The radius R = 0.5 with 1ω0=-1/sec will make 2 rotations.
b
Complex function F(2j1t) as a rotating vector.
The vector R(t) of variable length will make 2 rotations
c
Complex function F(2j1t) as a trajectory.
Second rotation on the same track. The center of gravity of the trajectory is now sc2=(0.35.0). It is a vector and can also be written as a complex exponential sc2=R*exp(jϕ) = 0.35*exp (j0 °). In Chapter 7, you will learn that from the center of gravity of the trajectory, you can easily read the 2nd harmonic of f(t) as h2 (t)=0.7cos (2t). Look at f(t). That’s right!
Note.
Don’t go into detail as we found the center of gravity sc2=(0.35.0). Trust your gut for now, although it seems like it should be a bit more to the right. Note, however, that the speed on the left side of the trajectory is slightly slower. So the distances from the 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 5.2.5 Trajectory F 3j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-3j1t) i.e. with rotation 3ω0=-3/sec

Fig.5-5
F(3j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-3j1t)
The animation lasts T≈6.28sek.
a
The radius R=0.5 with 1ω0=-1/sec will make 3 rotations.
b
Complex function F(3j1t) as a rotating vector.
The vector R(t) of variable length will make 3 rotations
c
Complex function F(3j1t) as a trajectory.
The center of gravity of the trajectory F(3j1t) is sc3=(0,0).So the f(t) has no harmonic with pulsation 3ω0=3/sec.

Chapter 5.2.6 Trajectory F 4j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-4j1t) i.e. with rotation 4ω0=-4/sec

Rys.5-6
F(4j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-4j1t)
The animation lasts T≈6.28sek.
a
The radius R=0.5 with 4ω0=-4/sec will make 4 rotations.
b
Complex function F(4j1t) as a rotating vector.
The vector R(t) of variable length will make 4 rotations
c
Complex function F(4j1t) as a trajectory.
3rd and 4th rotation along the same track. Non-zero center of gravity sc4 = (0.25.0)! From the center of gravity, the 4th harmonic of the function f(t) can be read as 0.5cos (4t).

Chapter 5.2.7 Trajectory F 5j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-5j1t) i.e. with rotation 5ω0=-5/sec

Fig.5-7
F(5j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-5j1t)
The animation lasts T≈6.28sek.
a
The radius R=0.5 with 5ω0=-5/sec will make 5 rotations.
b
Complex function F(5j1t) as a rotating vector.
The vector R(t) of variable length will make 5 rotations
c
Complex function F(5j1t) as a trajectory.
The center of gravity of the trajectory F(5j1t) is sc5=(0,0).So the f(t) has no harmonic with pulsation 5ω0=5/sec.

Chapter 5.2.8 Trajectory F 6j1t) = [1.3+0.7cos(2t)+0.5cos(4t)]exp (-6j1t) i.e. with rotation 6ω0=-6/sec

Fig.5-8
F(6j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-6j1t)
The animation lasts T≈6.28sek.
a
The radius R=0.5 with 6ω0=-6/sec will make 6 rotations.
b
Complex function F(6j1t) as a rotating vector.
The vector R(t) of variable length will make 6 rotations
c
Complex function F(5j1t) as a trajectory.
The center of gravity of the trajectory F(6j1t) is sc6=(0,0). So the f(t) has no harmonic with pulsation 6ω0=6/sec.

Chapter 5.2.9 Trajectory F 7j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-7j1t) i.e. with rotation 7ω0=-7/sec

Fig.5-9
F(7j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-7j1t)
The animation lasts T≈6.28sek.
a
The radius R=0.5 with 6ω0=-7/sec will make 7 rotations.
b
Complex function F(7j1t) as a rotating vector.
The vector R(t) of variable length will make 7 rotations
c
Complex function F(7j1t) as a trajectory.
The center of gravity of the trajectory F(7j1t) is sc7=(0,0).So the f(t) has no harmonic with pulsation 7ω0=7/sec.

Chapter 5.2.10 Trajectory F 8j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-8j1t) i.e. with rotation 8ω0=-8/sec

Fig.5-10
F(8j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-8j1t)
The animation lasts T≈6.28sek.
a
The radius R=0.5 with 8ω0=-8/sec will make 8 rotations.
b
Complex function F(8j1t) as a rotating vector.
The vector R(t) of variable length will make 8 rotations
c
Complex function F(8j1t) as a trajectory.
The center of gravity of the trajectory F(8j1t) is sc8=(0,0).So the f(t) has no harmonic with pulsation 8ω0=8/sec.