Rotating Fourier Series

Chapter 9. Fourier series of odd square wave.

Chapter 9.1 Introduction
This is the wave from the previous chapter shifted to the right (ie delayed) by a quarter period (ie by 90° or π/2). And just like the first harmonic of the previous one was the cosine, this one will be sine. This is an example of an odd function.
Fig. 9-1
Odd square wave f (t)  A=1, ω=1/sec and ϕ=-π/2.
The parameter ϕ=-π/2 of the odd wave means that it is delayed by π/2 in relation to the even one from the previous chapter. Like the sine, it lags π/2 relative to the cosine. In the following points we will analyze the trajectories F(njω0t) and the time plots of this square wave for n=0…8.

Chapter 9.2 Odd square wave f(t) and its trajectory F(njω0t) for n=0  i.e. F(0j1t)

Fig.9-2
Trajectory F(0j1) of the function f(t) of an odd square wave, i.e. with the following parameters:
A=1-amplitude
ω0=1/sec-pulsation corresponding to the period T=2πsec≈6.28sec.
ϕ=-π/2
duty cycle-50%
a. Trajectory F(0j1t) corresponding to trajectory F(njω0t) for n=0 and ω0=1/sec.
The stationary plane Z, where on the real axis Re Z the vector changes according to the function f(t) shown in b.
b. Time plot of f(t) odd square wave.
Note that the run starts at the beginning of the period. Like the sine function. Both animations in a and b describe the same thing. Read parameters a, ω0, ϕ and duty cycle. In the following subsections, the Z plane will start to rotate at speeds ω=-nω0, that is, at speeds -1/sec, –2/sec …- 8/sec. The vector will draw the trajectories F (njω0). From them we can read the harmonics for the pulsation nω0.

Chapter 9.3. First harmonic of odd square wave h1(t)=b1sin (1t).
Chapter 9.3.1 Trajectory F(njω0t) of an odd square wave for n=1 and ω0=1/sec,  i.e.  F (1j1t).
The Z plane rotates at a speed of ω=-1/sec

Fig. 9-3
Trajectory F(1j1t) of odd square wave
a. Radius R=1 as vector (1,0) rotates at speed ω=-1/sec around  z=(0,0) during T=2π sec and makes 1 turn.
b. Trajectory F(1j1t)=f (t)exp(-1j1t) as a rotating vector modulated by the f(t). The function f (t) is a square wave in Fig. 9-2b. The Z plane in a will start to rotate at the ω=-1/sec. This creates a rotating vector F(1j1t)=f(t) exp(-1j1t). The Z plane will do 1 turn, but the radius R=1 will only do 1 “half turn”.
c. Trajectory F(1j1t) as a semicircle drawn by the vector b.
During 1 turn of the Z plane, i.e. by the angle 0 … 2π, all vectors are vectorially added and their average over the period T=2π sec is calculated. It will be a vector,  otherwise sc1=(0,-1/π)=0-1j/π=-1j/π.

Chapter 9.3.2 The first harmonic on the background of an odd square wave, i.e. c0+h1(t) or the first approximation of an odd square wave.
Acc. Fig. 8-1c
The constant c0 is the average over the period T=2π sec, i.e. c0=a0=0.5.
Also c1 is the complex amplitude of the first harmonic c1=2sc1=(0,-2/π)=0-j2/π=-j2/π
that is a1=0 and b1=-2/π
Acc. Fig. 8-1e
h1(t)=(2/π)sin(1t)≈0.637sin (1t)

Fig.9-4
S1(t)=0.5+h1 (t)= (2/π) sin (1t) that is the first harmonic with a constant c0 on the background of a square wave.
It is also the first approximation of our square wave.

Chapter 9.4 The second harmonic of an odd square wave, or rather its absence, because c2=0 ->h2(t)=b2sin(2t)=0.
Chapter 9.4.1 Trajectory F (njω0t) of an odd square wave for n=2 and ω0=1/sec, i.e. F(2j1t).
The Z plane rotates at a speed of ω=-2/sec

Fig. 9-5
Trajectory F(2j1t) of odd square wave
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-2/sec around the point z=(0,0) and will make 2 turns in the time T=2π sec.
b. Trajectory F(2j1t) = f(t) exp(-2j1t) as a rotating vector modulated by the function f(t). Will make 1 turn in the first half cycle T/2=1π/sec and 0 turns in the second half period. The parameter sc2=0 as an average value of T=2π sec is obvious.
c. Trajectory F(2j1t) as a circle drawn by the the vector b.
Note
sc2=0 and therefore the harmonic for ω=2/sec doesn’t exist.

Chapter 9.5.The third harmonic of odd square wave h3(t)=b3sin(3t).
Chapter 9.5.1 Trajectory F(njω0t) of an odd square wave for n=3 and ω0=1/sec,  i.e.  F (5j1t).
The Z plane rotates at a speed of ω=-3/sec

Fig. 9-6 Trajectory F(3j1t) of odd square wave
a. Radius R=1 as vector (1,0) rotates at speed ω=-1/sec around  z=(0,0) during T=2π sec and makes 1 turn.
b. Trajectory F(3j1t)=f(t)exp(-3j1t) as a rotating vector modulated by the f(t). Will make 1.5 turns in the first half period T/2=1π sec and 0 turns in the second half period. In the period T=2π sec, the radius R=1 stays longer on the lower half-plane than on the upper half-plane. Therefore its average value as a vector will be sc3=(0,-1/3π)=- 1j/3π and not sc3=(0,0) as e.g. in Fig. 9-5c
c. Trajectory F(3j1t) as a semicircle drawn by vector b.
The center of gravity sc3=(0,-1/3π) results from the summation of the vectors b and their mean at time T=2π sec, when 3 rotations of the Z plane are made. The lower semicircle is drawn 2 times, and the upper one is only 1 time. As if the bottom semicircle is twice as “heavy”.

Chapter 9.5.2 The third harmonic on an odd square wave background, i.e. c0+h3(t).
Acc. Fig. 8-1c chapter. 8 c3 is the complex amplitude of the third harmonic
c3 = 2sc3=(0,-2/3π) so a3=0 and b3=-2/3π
Acc. Fig. 8-1e chapter. 8
h3(t)= (2/3π)sin (3t)≈0.212sin (3t)

Fig. 9-7
co+h3(t)=0.5+(2/3π)*sin(3t) that is, the third harmonic with a constant c0 on the background of the square wave.

Chapter 9.5.3 The third approximation of the odd square wave, i.e. S3(t)=c0+h1(t)+h3(t).

Fig.9-8
S3(t)=c0+h1(t)+h3(t)=0.5+(2/π)sin(1t)+(2/3π)sin(3t)
The third approximation is more like a square wave than the first one in Fig. 9-4

Chapter 9.6 The fourth harmonic of an odd square wave, or rather the lack of it, because c4=0 ->h4 (t)=c4sin(4t)=0.
Chapter 9.6.1 Trajectory F(njω0t) of an odd square wave for n=4 and ω0=1/sec, i.e. F(4j1t).

The Z plane rotates at a speed of ω=-4/sec

Fig. 9-9
Trajectory F(4j1t) of an even square wave
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-4/sec around the z=(0,0) and will make 4 turns in the time T=2π sec.
b. Trajectory F(4j1t)=f (t)exp(-4j1t) as a rotating vector modulated by the f(t). Will make 2 turns in the first half period T/2=1π/sec and 0 turns in the second half period. The parameter sc4=0 as an average value of T=2π sec is obvious.
c. Trajectory F(4j1t) as a circle drawn by the the vector b.
sc4=0
Note
sc4=0 and therefore the harmonic for ω=4/sec doesn’t.

Chapter 9.7 The fifth harmonic of odd square wave h5(t)=b5sin (5t).
Chapter 9.7.1 Trajectory F(njω0t) of an odd square wave for n=5 and ω0=1/sec,  i.e.  F (5j1t).
The Z plane rotates at a speed of ω=-5/sec

Fig. 9-10
Trajectory F(5j1t) of an even square wave
a. The radius R=1 as a vector (1.0) rotates at the speed ω=-5/sec around the z=(0.0) and makes 5 turns in the time T=2π sec.
b. Trajectory F(5j1t)=f(t)exp(-5j1t) as a rotating vector modulated by f(t). Will make 2.5 turns in the first half cycle T/2=1π/sec and 0 turns in the second half cycle. In the period T= 2π sec, the radius R=1 stays longer on the lower half-plane than on the upper half-plane. Therefore its average value as a vector will be sc5=(0,-1/5π)=-1j/5π and not sc5=(0,0) as e.g. in Fig. 9-9c
c. Trajectory F(5j1t) as a circle drawn by the vector b. The center of gravity sc5=(0,-1/5π) results from the summation of the vectors in b and their mean over the period T=2π sec, when 5 turns of the Z plane are made.  As if the bottom semicircle is “heavier” than the top one.
if the bottom semicircle is “heavier” than the top one.

Chapter 9.7.2. The fifth harmonic on an odd square wave background, i.e. c0+h5(t).
Acc. to Fig. 8-1c chapter. 8
c5 is the complex amplitude of the fifth harmonic
c5=2sc5=(0,-2/5π) so a5=0 and b5=-2/5π
Acc. to Fig. 8-1e
h5(t)=(2/5π) sin(5t)≈0.127sin(5t)

Fig. 9-11
co+h5(t)=0.5+(2/5π)sin(5t) that is, the fifth harmonic with a constant c0 on the background of the square wave.

Chapter 9.7.3 The fifth approximation of the odd square wave, i.e. S5=c0+h1(t)+h3(t)+h5(t)

Fig.9-12
S5(t)=c0+h1(t)+h3(t)=0.5+(2/π)sin(1t)+(2/3π)sin(3t)+(2/5π)sin(5t)
The fifth approximation is more like a square wave than the third one in Fig. 9-8

Chapter 9.8 The sixth harmonic of an odd square wave, or rather the lack of it, because c6=0 ->h6(t)=c6sin(6t)=0.
Chapter 9.8.1 Trajectory F(njω0t) of an odd square wave for n=6 and ω0=1/sec, i.e. F6j1t).

The Z plane rotates at a speed of ω=-6/sec

Fig. 9-13
Trajectory F(6j1t) of an odd square wave
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-6/sec around the z=(0,0) and will make 6 turns in the time T=2π sec.
b. Trajectory F(6j1t)=f (t)exp(-6j1t) as a rotating vector modulated by the f(t).
Will make 3 turns in the first half cycle T/2=1π/sec and 0 turns in the second half cycle. The parameter sc=0 as an average value of T=2π sec is obvious.
c. Trajectory F(6j1t) as a circle drawn by the  vector b.
sc6= 0
Conclusion

sc6=0 and therefore the harmonic for ω=1/6sec doesn’t exist.

Chapter 9.9 The seventh harmonic of odd square wave h7(t)=b7sin (5t).
Chapter 9.9.1 Trajectory F(njω0t) of an odd square wave for n=7 and ω0=1/sec,  i.e.  F (7j1t).
The Z plane rotates at a speed of ω=-7/sec

Fig. 9-14
Trajectory F(7j1t) of an odd square wave
a. The radius R=1 as a vector (1.0) rotates at the speed ω=-7/sec around the z=(0.0) and makes 7 turns in the time T=2π sec.
b. Trajectory F(7j1t)=f(t)exp(-7j1t) as a rotating vector modulated by f(t).
At the time T=2π sec, the radius R=1 stays longer in the lower half-plane.
Therefore, its average value as a vector will be (-1/7π,0)=-j1/7π.

c. Trajectory F(7j1t) as a circle drawn by vector from b.
Notice the word “3.5 turns” that appears. This should convince you that the trajectory F (7j1t) stays longer on the left half-plane Z.  Otherwise, this part of the trajectory is “heavier”.

Chapter 9.9.2. The seventh harmonic on an odd square wave background, i.e. c0 + h7 (t).
Acc. to Fig. 8-1c chapter. 8
c7 is the complex amplitude of the seventh harmonic
c7=2sc7=(0,-2/7π) so a7=0 and b7=-2/7π
Acc. to Fig. 8-1e
h7(t)=(2/7π) sin(7t)≈-0.091sin(7t)

Fig. 9-15
co+h7(t)=0.5+(2/7π)*sin(7t) that is, the seventh harmonic with a constant c0 on the background of the square wave.

Chapter 9.9.5 The seventh approximation of the odd square wave, i.e. S7=c0+h1(t)+h3(t)+h5(t)+h7(t).

Fig.9-16
S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)=0.5+(2/π)*sin(1t)+(2/3π)*sin(3t)+(2/5π)*sin(5t)+(2/7π)*sin(7t)
The seventh approximation is more like a square wave than the fifth one in Fig. 9-12

Chapter 9.10 The eighth harmonic of an odd square wave, or rather the lack of it, because c8=0 ->h8(t)=c8sin(8t)=0.
Chapter 9.10.1 Trajectory F(njω0t) of an odd square wave for n=8 and ω0=1/sec, i.e. F(8j1t).

The Z plane rotates at a speed of ω=-8/sec

Fig. 9-17
Trajectory F(8j1t) of an odd square wave
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-8/sec around the z=(0,0) and will make 8 turns in the time T=2π sec.
b. Trajectory F(8j1t)=f (t)exp(-8j1t) as a rotating vector modulated by the f(t).
Will make 4 turns in the first half cycle T/2=1π/sec and 0 turns in the second half cycle. The parameter sc8=0 as an average value of T=2π sec is obvious.
c. Trajectory F(8j1t) as a circle drawn by the  vector b.
sc8= 0
Conclusion

sc8=0 and therefore the harmonic for ω=1/8sec doesn’t exist.

Chapter 9.11 Remaining harmonics of an odd square wave, i.e. for n = 9,10,11 … ∞
We noticed that the center of gravity scn of the trajectory approaches scn=(0,0) as ω increases. In addition, the scn of the even harmonics are zero. This means that the harmonics decrease with increasing frequency, and for the infinitely high ω, the harmonic amplitudes are zero, i.e. they disappear.
The animation Fig 9-16 shows that for n=7, the sum of S7(t)=c0+h1 (t)+h3(t)+h5(t)+h7(t).
And when the number of harmonics is infinitely large, i.e. n=∞?
Then the sum of these harmonics S∞ (t)=c0+h1(t)+h3(t)+h5(t)+h7(t)+h9 (t)+ … + h∞ (t) is the ideal square wave f(t) in Fig. 9-1.