Rotating Fourier Series

Chapter 5. How to filter out the harmonics with 1.3+0.7*cos(2t)+0.5*cos(4t)?

Chapter 5.1 Introduction
In the previous chapter, we extracted the harmonic 0.5*cos(4t) from the function f(t)=0.5*cos(4t) and from the function f(t)=0.5*cos(4t-30°) using a rotating speed of ω=n*ω0 Z planes. Now we will do the same, but with the function f(t)=1.3+0.7*cos(2t)+0.5*cos(4t). The pulsations of both cosines are different! I wonder how this will affect the f(t) plot and the rotating trajectories F(jnω0t)?
You will find that:
– regardless of the nω0 pulsation, the trajectory period F(jnω0t) is T=6.28 sec
– for ω=0 the function does not rotate!
– for ω=1*ω0=1/sec…8*ω0=8 the radius R=1 will make 1…8 turns
Will it be possible to extract the constant component c0=1.3 and the harmonics h2(t)=0.7*cos(2t) and h4(t)=0.5*cos(4t)?

Chapter 5.2 Trajectory F(0j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-0j1t) i.e. no rotation
A problem similar to Fig. 4-2 of the previous chapter. The radius R does not rotate, but changes according to periodic function R(t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t). So the trajectory in the complex plane Z is
F(0j1t)=f(t)*exp(-j0)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t). By the way, you will see the time chart of the f(t) function.

Fig. 5-1
Function F(0j1t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t) in complex and real version
Fig. 5-1a
F(0j1t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t) in complex version
Fig. 5-1b
F(0j1t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t) in real version
The average value of f(t) over the period T=2π≈6.28sec (and also T=πsec≈3.14sec) is c0=1.3.
It is also the centroid sc0=(1.3,0) of the “rocking” trajectory in Fig.5-1a when the plane rotation speed ω=0. Another name for parameter c0=1.3 is the constant component of f(t).
In the Fourier Series of the periodic function f(t) the first element is the constant component – coefficient c0=a0. This is the average f(t) over period T.
We calculate it as an integral from the formula below.

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
General formula for f(t) with period T
Formula for f(t)=1.3+0.7*cos(2t)+0.5*cos(4t) and T=2π
Formula for any f(t) and T=2π

Chapter 5.3 Trajectory F(1j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-1j1t) i.e. with rotation 1ω0=-1/sec
The radius R=1 rotates with 1ω0=-1/sec.
See the animation in Fig. 5-1a. The end of the vector moves “back and forth” around c0=(+1.3, 0) by formula R(t)=f(t)=1.3+0.7*cos(2t)+0.5*cos(4t). What if the vector started rotatimg at 1ω0=-1/sec? So “clockwise” and with a period of T≈6.28sec. Then its motion in the complex plane Z is described by the complex function F(1j1t).

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

Chapter 5.4 Trajectory F(2j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-2j1t) i.e. with rotation 2ω0=-2/sec

The animation lasts T≈6.28sec.
Fig. 5-4a
Radius R=0.5 will make 2 revolutions.
The complex function F(2j1t) as a rotating vector.
The variable-length vector R(t) will make 2 rotations.
The complex function F(2j1t) as a trajectory.
Second rotation on the same track. The centroid of the trajectory is now sc2=(0.35,0). It is a vector and can also be written in the complex exponential form sc2=R*exp(jϕ)=0.35*exp(j0°). In Chapter 7 you will learn that from the centroid of the trajectory you can easily read the 2nd harmonic of f(t) as h2(t)=0.7cos(2t). Look at f(t). Correct!
Don’t go into how we calculated the centroid sc2=(0.35,0). Trust your intuition for now, although it seems like it should be a bit more to the right. Note, however, that the velocity of the left side of the trajectory is slightly slower. So the distances from 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 5.5 Trajectory f(3j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-3j1t) i.e. with spin 3ω0=-3/sec

The animation lasts T≈6.28sec.
Fig. 5-5a
The radius R=0.5 will make 3 revolutions.
The complex function F(3j1t) as a rotating vector
Complex function F(3j1t) as a trajectory.
The centroid of the trajectory F(3j1t) is sc3=(0,0)=0. So the function f(t) has no harmonic with pulsation 3ω0=3/sec.

Chapter 5.6 Trajectory F(4j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-4j1t) i.e. with spin 4ω0=-4/sec

The animation lasts T≈6.28sec.
Fig. 5-6a
The radius R=0.5 will make 4 turns.
The complex function F(4j1t) as a rotating vector.
The variable-length vector R(t) will make 4 rotations.
Complex function F(4j1t) as a trajectory.
3rd and 4th rotation along the same track. Non-zero centroid sc4=(0.25,0)!
From the centroid sc4, the 4th harmonic of the f(t) function can be read as 0.5cos(4t).

Chapter 5.7 Rotating radius 5ω0=5/sec, so the trajectory F(j5ω0t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-j5ω0t) for 5ω0=-5/sec.

The animation lasts T≈6.28sec.
Fig. 5-7a
The radius R=0.5 will make 5 revolutions.
The complex function F(5j1t) as a rotating vector
Complex function F(5j1t) as a trajectory.
The centroid of the trajectory F(5j1t) is sc5=(0,0)=0. So the function f(t) has no harmonic with a pulsation of 5ω0=5/sec.

Chapter 5.8 Trajectory F(6j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-6j1t) so with rotation 6ω0=-6/sec

The animation lasts T≈6.28sec.
The radius R=0.5 will make 6 turns.
The complex function F(6j1t) as a rotating vector
Complex function F(6j1t) as a trajectory. 3 revolutions along the same path, which is why it seemingly stopped.
The centroid of the F(6j1t) trajectory is sc6=(0,0)=0*. So the function f(t) has no harmonic with a pulsation of 6ω0=6/sec.
*sc6=(0,0) is not as obvious as, for example, for Fig.5-7c. It seems too far to the left. Note, however, that the trajectory velocities on the left are lower than those on the right.

Chapter 5.9 Trajectory F(7j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-7j1t) i.e. with spin 7ω0=-7/sec

Complex function F(j7ω0t) for 7ω0=7/sec.
The animation lasts T≈6.28sec.
Fig. 5-9a
The radius R=0.5 will make 7 revolutions.
The complex function F(7j1t) as a rotating vector
Complex function F(7j1t) as a trajectory.
The centroid of the trajectory F(7j1t) is sc7=(0,0)=0. So the function f(t) has no harmonic with a pulsation of 7ω0=7/sec.

Chapter 5.10 Trajectory F(8j1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-8j1t) i.e. with spin 8ω0=-8/sec

The animation lasts T≈6.28sec
Fig. 5-10a
The radius R=0.5 will make 8 turns.
The complex function F(8j1t) as a rotating vector.
Complex function F(8j1t) as a trajectory.
The centroid of the trajectory F(8j1t) is sc8=(0,0)=0. So the function f(t) has no harmonic with a pulsation of 8ω0=8/sec.

Scroll to Top