9.Fourier series of an odd square wave - How does it work?/Jak to dziala?

Rotating Fourier Series

Chapter 9. Fourier series of an odd square wave.

Chapter 9.1 Introduction
This is the wave from the previous chapter shifted to the right (i.e. delayed) by a quarter of a period (i.e. by 90° or π/2). And just as the first harmonic of the previous one was a cosine, this one will be a 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 compared to the even wave from the previous chapter. Like the sine, it lags by π/2 with respect to the cosine. In the next points, we will analyze the trajectories of F(n*jω0t) and the time charts 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
The trajectory F(0j1) of the function f(t) of an odd square wave, i.e. with the parameters:
A=1-ampliltude
ω0=1/sec-pulsation corresponding to the period T=2πsec≈6.28sec.
ϕ=-π/2
50%-filling
Fig.9-2a
The trajectory F(0j1t) corresponds to the trajectory F(njω0t) for n=0 and ω0=1/sec.
A stationary Z plane, where on the real axis Re Z the vector changes acc. to. the function f(t) shown in Fig.9-2b.
Fig.9-2b
Time chart of f(t) of an odd square wave.
Note that the time chart starts at the beginning of the period. Just like the sine function. Both animations in Fig.9-2a and Fig.9-2b describe the same thing. Read the parameters A,ω0,ϕ and filling.
In the next subchapters, the Z plane will start rotating at speeds ω= -nω0, i.e. at speeds -1/sec,-2/sec…-8/sec. The end of the vector will draw trajectories F(njω0). From them we will read the harmonics for the nω0 pulsation.

Chapter 9.3 First harmonic of an odd square wave h1(t)=b1*sin(1t).
Chapter 9.3.1 Trajectory F(njω0t) of an odd square wave for n=1 and ω0=1/sec, i.e. F(1j1t).

Fig. 9-3
Trajectory F(1j1t) of an odd square wave
Fig.9-3a
The radius R=1 as a vector (1,0) rotates with a speed ω=-1/sec around the point (0,0) in time T=2π sec and will complete 1 revolution.
Fig.9-3b
Trajectory F(1j1t)=f(t)*exp(-1j1t) as a rotating vector modulated by the function f(t). The function f(t) is the square wave of Fig.9-2b. The Z plane in Fig.9-2a will start rotating at a speed of ω=-1/sec. This will create a rotating vector F(1j1t)=f(t)*exp(-1j1t). The Z plane will make 1 revolution, but the radius R=1 will only make 1 “half-turn”.
Fig.9-3c
The trajectory of F(1j1t) as a semicircle drawn by the end of the vector in Fig. 8-3b.
During 1 rotation of the Z plane, i.e. by an angle of 0…2π, all vectors are vector summed and their average is calculated in the period T=2π sec. It will be a vector, otherwise the point sc1=(0,-1/π)=0-1j/π=-1j/π.
Chapter 9.3.2 The first harmonic against the background of an odd square wave, i.e. c0+h1(t) or in other words, the first approximation of an odd square wave.
Acc. to Fig. 8-1c
The constant component c0 is the average in the period T=2π, i.e. c0=a0=0.5.
Also, c1 is the complex amplitude of the first harmonic
c1=2*sc1=(0,-2/π)=0-j2/π=-j2/π i.e. a1=0 and b1=-2/π
Acc. to 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),
i.e. the first harmonic with a constant component c0 against the background of a square wave.
This is also a first approximation to our square wave.

Chapter 9.4 Second harmonic of an odd square wave, or rather its absence
because c2=0 –>h2(t)=b2*sin(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 an odd square wave
Fig.9-5a
The radius R=1 as a vector (1,0) rotates with a speed ω=-2/sec around the point (0,0) and will complete 2 revolutions in time T=2π sec.
Fig.9-5b
Trajectory F(2j1t)=f(t)*exp(-2j1t) as a rotating vector modulated by the function f(t). It will make 1 revolution in the first half-period T/2=1π/sec and 0 revolutions in the second half-period. The parameter sc2=0 as the average value of T=2π sec is obvious.
Fig.9-5c
The trajectory of F(2j1t) as a circle drawn by the end of the vector in Fig. 9-5b.
Note
sc2=0 and therefore the harmonic for ω=2/sec does not exist.

Chapter 9.5 Third harmonic of an odd square wave h3(t)=b3*sin(3t).
Chapter 9.5.1 Trajectory F(njω0t) of an odd square wave for n=3 and ω0=1/sec, i.e. F(3j1t).
The Z plane rotates at a speed of ω=-3/sec

Fig.9-6
Trajectory F(3j1t) of an odd square wave
Fig.9-6a
The radius R=1 as a vector (1,0) rotates with a speed ω=-3/sec around the point (0,0) and will complete 3 revolutions in time T=2π sec.
Fig.9-6b
Trajectory F(3j1t)=f(t)*exp(-3j1t) as a rotating vector modulated by the function f(t).
It will make 1.5 revolutions in the first half-period T/2=1π/sec and 0 revolutions in the second half-period. In the period T=2π sec, the radius R=1 stays longer in the lower half-plane than in the upper one. 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
Fig.9-6c
The trajectory of F(3j1t) as a circle drawn by the end of the vector in Fig. 9-6b.
The centroid sc3=(0,-1/3π) results from the summation of the vectors in Fig. 9-6b and their average at time T=2π sec when 3 revolutions of the Z plane are made. The lower semicircle is drawn twice, and the upper one only 1 time. As if the lower semicircle was as more “heavy”.
Chapter 9.5.2 Third harmonic against an odd square wave, i.e. c0+h3(t).
Acc. to Fig. 8-1c from chapter 8
c3 is the complex amplitude of the third harmonic
c3=2*sc3=(0,-2/3π) i.e. a3=0 and b3=-2/3π
Acc. to Fig. 8-1e
h3(t)=(2/3π)*sin(3t)≈0.212sin(3t)

Fig.9-7
co+h3(t)=0.5+(2/3π)*sin(3t),
i.e. the third harmonic with a constant component c0 against the background of a square wave.
Chapter 9.5.3 Third approximation of an 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 similar to a square wave than the first one in Fig.9-4

Chapter 9.6 The fourth harmonic of an odd square wave, or rather its absence
because c4=0 –>h4(t)=c4*sin(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
Fig.9-9a
The radius R=1 as a vector (1,0) rotates with a speed ω=-4/sec around the point (0,0) and will complete 4 revolutions in time T=2π sec.
Fig.9-9b
Trajectory F(4j1t)=f(t)*exp(-4j1t) as a rotating vector modulated by the function f(t). It will make 2 revolutions in the first half-period T/2=1π/sec and 0 revolutions in the second half-period. The parameter sc4=0 as the average value of T=2π sec is obvious.
Fig.9-9c
The trajectory of F(4j1t) as a circle drawn by the end of the vector in Fig. 9-9b.
sc4=0
Note
sc4=0 and therefore the harmonic for ω=4/sec does not exist.

Chapter 9.7 Fifth harmonic of a symmetrical odd square wave h5(t)=b5*sin(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
Fig.9-10a
The radius R=1 as a vector (1,0) rotates with a speed ω=-5/sec around the point (0,0) and will complete 5 revolutions in time T=2π sec.
Fig.9-10b
Trajectory F(5j1t)=f(t)*exp(-5j1t) as a rotating vector modulated by the function f(t).
It will make 2.5 revolutions in the first half-period T/2=1π/sec and 0 revolutions in the second half-period. In the period T=2π sec, the radius R=1 stays longer in the lower half-plane than in the upper one. Therefore, its average value as a vector will be sc5=(0,-1/5π)=-1j/5π.
Fig.9-10c
The trajectory of F(5j1t) as a circle drawn by the end of the vector in Fig. 9-10b. The centroid sc5=(0,-1/5π) results from the summation of the vectors in Fig.9-10b and their average in the period T=2π sec, when 5 revolutions of the Z plane are made. As if the lower semicircle was “heavier” than the upper one
Chapter 9.7.2 The fifth harmonic against the background of an odd square wave, i.e. c0+h5(t).
Acc. to Fig. 8-1c from chapter 8
c5 is the complex amplitude of the fifth harmonic
c5=2*sc5=(0,-2/5π) i.e. 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)
i.e. the fifth harmonic with a constant component c0 against the background of a square wave.
Chapter 9.7.3 Fifth approximation of an odd square wave, i.e. S5=c0+h1(t)+h3(t)+h5(t).

Fig.9-12
S5(t)=c0+h1(t)+h3(t)+h5(t)=0.5+(2/π)*sin(1t)+(2/3π)*sin(3t)+(2/5π) )*sin(5t)
The fifth approximation is more similar to a square wave than the third one in Fig.9-8.

Chapter 9.8 The sixth harmonic of an odd square wave, or rather its lack
because c6=0 –>h6(t)=b6*sin(6t)=0.
Chapter 9.8.1 Trajectory F(njω0t) of an odd square wave for n=6 and ω0=1/sec, i.e. F(6j1t).

Fig.9-13
Trajectory F(6j1t) of an odd square wave
Fig.9-13a
The radius R=1 as a vector (1,0) rotates with a speed ω=-6/sec around the point (0,0) and will complete 6 revolutions in time T=2π sec.
Fig.9-13b
Trajectory F(6j1t)=f(t)*exp(-6j1t) as a rotating vector modulated by the function f(t). It will make 3 revolutions in the first half-period T/2=1π/sec and 0 revolutions in the second half-period. The parameter sc6=0 as the average value of T=2π sec is obvious.
Fig.9-13c
The trajectory of F(6j1t) as a circle drawn by the end of the vector in Fig. 9-13b.
sc6=0.
Note
sc6=0 and therefore the harmonic for ω=1/6sec does not exist.

Chapter 9.9 Seventh harmonic of an odd square wave h7(t) =b7*sin(7t).
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(5j1t) of an odd square wave
Fig.9-14a
The radius R=1 as a vector (1,0) rotates with a speed ω=-7/sec around the point (0,0) and will complete 7 revolutions in time T=2π sec.
Fig.8-14b
Trajectory F(7j1t)=f(t)*exp(-7j1t) as a rotating vector modulated by the function f(t).
At 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.
Fig.9-14c
The trajectory of F(7j1t) as a circle drawn by the end of the vector in Fig. 9-17b.
Notice the words “3.5 turns” appear. This should convince you that the F(7j1t) trajectory stays longer in the lower half-plane Z. Otherwise, this part of the trajectory is “heavier”.
Chapter 9.9.2 The seventh harmonic against the background of an even square wave, i.e. c0+h7(t)
Acc. to Fig. 8-1c from chapter 8
c7 is the complex amplitude of the seventh harmonic
c7=2*sc5=(0,-2/7π) i.e. 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),
i.e. the seventh harmonic with a constant component c0 against the background of a square wave.
Chapter 9.9.3 The seventh approximation of an even 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 similar to a square wave than the fifth in Fig.9-12

Chapter 9.10 The eighth harmonic of an even square wave, or rather its lack
because c8=0 –>h8(t)=b8*cos(8t)=0.
Chapter 9.10.1 Trajectory F(njω0t) of an even square wave for n=8 and ω0=1/sec, i.e. F(8j1t).

Fig.9-17
Trajectory F(8j1t) of an even square wave
Fig.9-17a
The radius R=1 as a vector (1,0) rotates at a speed ω=-8/sec around the point (0,0) and will make 8 revolutions in time T=2π sec.
Fig.9-17b
Trajectory F(8j1t)=f(t)*exp(-8j1t) as a rotating vector modulated by the function f(t). It will make 4 revolutions in the first half-period T/2=1π/sec and 0 revolutions in the second half-period. The parameter sc8=0 as the average value of T=2π sec is obvious.
Fig.9-17c
The trajectory of F(8j1t) as a circle drawn by the end of the vector in Fig. 9-17b.
sc8=0
Note
sc8=0 and therefore the harmonic for ω=1/8sec does not exist.

Chapter 9.11 Remaining harmonics of an even square wave, i.e. for n=9,10,11…∞
We noticed that the centroids scn of the trajectories approach scn=(0,0) as ω increases. In addition, the scn of even harmonics is zero. This means that the harmonics decrease with increasing frequency and for an infinitely high frequency ω, the harmonic amplitudes are zero, i.e. they disappear.
The animation of 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 square wave f(t) as in Fig. 9-1.