8. Fourier Series of an even square wave.

Rotating Fourier Series

Chapter 8. Fourier Series of an even square wave.

Chapter 8.1 Introduction
So far, we have studied the function f(t), in which all harmonics were visible in the formula, e.g. f(t)=1.3+0.7cos(2t)+05.cos (4t). All this in order to check Fourier Series formulas. And when the harmonics are not seen in the function f(t)? As, for example, in a square wave f(t) with amplitude A=1, pulsation ω=1/sec, initial phase ϕ=0 and duty cycle w=50%.

Fig.8-1
Even square wave and Fourier Series formulas
Fig.8-1a
Even square wave f(t)  A=1, ω=1/sek, ϕ=0 i w=50%.
A few words of explanation. The parameters A, ω and ϕ are typical for trigonometric functions. It is difficult to assign them, for example, to a quadratic function. But a square wave does! The parameters A=1 and ω=1/sec are obvious. I will just add that ω=1/sec corresponds to the period T=2πsec. And ϕ=0 means that the wave starts in the “upper half of the rectangle”, just like ϕ=0 in the “upper half of the cosine”.
Fourier Series formulas
In the following points, we will analyze the F(njω0t) trajectories and the time plots of the square wave for n=0…8. As before, we did this with the trigonometric functions. But first, a few words how to read the nth harmonic hn(t) from the center of gravity of the nth trajectory F (njω0t).
Fig.8-1b
Formula for the nth center of gravity scn of the trajectory F(njω0t)
Here the centers of gravity scn agree with intuition and we will only assume they are true. In chapter 11 we will calculate them exactly with the WolframAlfa program.
Fig.8-1c
The formula for the nth complex Fourier Series coefficient cn This is double the center of gravity of the scn! Treat it as a complex amplitude for the nth harmonic.
Note
The formula is true for n=1,2,3 …
For n=0
c0=sc0=a0
This is the mean of f(t) over a period of 0…2π. Substitute n=0 is the creation of the integral as the formula Fig. 7-2c from chapter. 7
Fig.8-1d
cn  as a complex amplitude:
cn=an+jbn  algebraic version
cn=|cn|exp(jϕ) exponential version when:
-|cn| amplitude module as “an and bn ” Pythagoras
-ϕ phase as an/bn=tg(ϕ)
Fig.8-1e
Formula for the nth harmonic hn(t) with cosine/sine components
Fig.8-1f
The formula for the nth harmonic hn(t) as a cosine with a phase shift ϕ.
Note
The subject of the above formulas is discussed in p 7.2 of chapter. 7

Chapter 8.2 Even square wave f(t) and its trajectory F(njω0t) for n=0 i.e. F(0j1t)

Fig.8-2
Trajectory F(0j1t) of the function f(t) of an even square wave, i.e. with the following parameters:
n=0 Centrifuge doesn’t roytate!
A=1-amplitude
ω0=1/sec-pulsation corresponding to the period T=2πsec≈6.28sec.
ϕ=0
50% -duty cycle
a.Trajectory F(0j1t)
The stationary plane Z, where on the real axis Re Z the vector changes according to the function f(t) shown in Fig. 8-2b. In other words, they are “swinging” motions on the real axis Re Z between points (0,0) and (1,0).
b.Time plot f(t) of an even square wave.
Note that the run begins in the middle of the half period. Like the cosine function. Both animations in  a and b describe the same thing. Read parameters A, ω0, ϕ and duty cycle. Animation b is easier, but it is also possible with a. From the plot it is possible to calculate, even without integrals, the coefficient c0=a0=0.5 of the Fourier Series as the mean value f (t) for 0 … 2π. See Fig. 8-1c.
In the following subchapters, the Z plane will start to rotate at speeds of ω=-nω0, that is, from -1/sec, -2/sec … -8 /sec in the clockwise direction. The end of the vector will draw the trajectories F(njω0). From them we can read the harmonics for the pulsation nω0. As in the previous chapters, only instead of f(t) as a combination of sines and cosines, now there is f(t)=square wave.

Chapter 8.3 The first harmonic of an even square wave h1(t)=a1cos (1t).
Chapter 8.3.1 Trajectory F (njω0t) of an even 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. 8-3
Trajectory F (1j1t) of an even square wave
a.The radius R=1 as a vector (1,0) rotates at the speed ω=-1/sec around the point z=(0,0) during T=2π sec. The entire complex Z plane rotates with it. But be careful! The axes of the plane Re Z and Im Z stand still! The end of the vector would draw the entire circle as opposed to the semicircle in Fig. 8-3c.
b.Trajectory F(1j1t)=f (t)*exp(-1j1t) as a rotating vector modulated by the function f(t).
The function f(t) is a square wave in Fig. 8-2b. Look at a and imagine that the Z plane starts to rotate at ω=-1/sec. Remember that the Re Z and Im Z axes are stationary! This creates a rotating vector F(1j1t)=f (t)*exp(-1j1t). The Z plane will do 1 turn, and the R=1 radius will do 2 “quarter turns” with a 1π second pause in between. This is equivalent to one “half round“.
c.Trajectory F(1j1t) as a semicircle drawn by the end of the vector from b
Now imagine that during 1 turn of the Z plane, i.e. by the angle 0 … 2π, all vectors are added together. And their average is calculated. Some vector sc1 will be created. Will it be sc1=(0,0)? Probably not, but it will definitely be somewhere on the Re Z axis between (0,0) and (1,0). Rather closer to (0.0). It will be the vector sc1=(1/π, 0)! In other words, the center of gravity sc1=(1/π, 0) of the trajectory F(1j1t).
We will precisely calculate sc1 with the WolframAlfa program in chapter 11.

Note to Fig. 8-3b,c
The above animations suggest that the upper and lower parts of the trajectory are separated by a gap T=π sec. But this is only at the beginning at time t=0…2π sec! After that, there is no this break and the movement along the trajectory is continuous.

Chapter 8.3.2 The first harmonic with constant c0 on the background of the even square wave, i.e. c0+h1(t)
or the first approximation of the even square wave.
acc. Rys. 8-1c
The constant component c0 is the average over the period T=2πsec, i.e. c0=a0=0.5. You can even calculate it without calculus.
Also c1 is the complex amplitude of the first harmonic.
c1=2*sc1=(2/π,0) czyli a1=2/π i b1=0
acc. Fig. 8-1e
h1(t)=(2/π)*cos(1t)≈0.637cos(1t)

Fig.8-4
S1(t)=c0+h1(t) that is the first harmonic with a constant c0=0.5 on the background of a square wave.
It is also the first approximation to our 50% even square wave.

Chapter 8.4 The second harmonic of an even square wave, or actually its absence, because c2=0-> h2 (t)= a2cos(2t)=0.
Chapter 8.4.1 Trajectory F(njω0t) of an even 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.8-5
Trajectory F(2j1t) of an even square wave
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-2/sec around the z=(0,0) and will make 2 turns in the time T=2π sec.
b. Trajectory F(2j1t) as a rotating vector modulated by the function f(t).
In 2π sec, the Z plane will make 2 turns, but radius R=1 will only do 2 “half turns” with a 1π sec.  gap in between.
c. Trajectory F(2j1t) as a circle drawn by the end of the vector b.
Evidently sc2=0. Note that from the circle itself, the value sc2=0 is not obvious! After all, a circle could have been drawn 1.5 times, for example. Fortunately, this 1 full turn, albeit with a break in the middle, is shown in b.
Conclusion
sc2=0 and therefore the harmonic for ω=2 / sec doesn’t exist.

Chapter 8.5 The third harmonic of an even square wave h3(t)=a3cos (3t).
Chapter 8.5.1 Trajectory F (njω0t) of an even square wave for n=3 and ω0=1/sec  i.e. F (3j1t).
The Z plane rotates at a speed of ω=-1/sec

Fig.8-6
Trajectory F(3j1t) of an even square wave
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-3/sec around z=(0,0) and will make 3 turns in the time T=2π sec.
b. Trajectory F(3j1t)=f (t)exp(-3j1t) as a rotating vector modulated by the f (t). At T=2π sec, the Z plane will make 3 turns, but radius R=1 only 2 times “3/4 turns” with a 1π sec gap in between. It turns out that at the time T=2π sec, the radius R=1 stays longer on the left half-plane than on the right. Therefore its average value as a vector will be (-1/3π, 0)
c. Trajectory F(3j1t) as a circle drawn by the end of the vector b. The center of gravity sc3(- 1/3π,0) results from the summation of the vectors in b and their mean at T=2π sec when 3 rotations of the Z plane are made. Its value is negative, unlike in Fig. 8-3b, where the vector made 2 “quarter turns” only on the positive Z half-plane.
Note
In the next subchapters there will be no such precise description of the trajectory F (njω0t).
The reader will notice that:
– for even rotational speeds of the Z plane, i.e. for ω=-2/sec, -4/sec, -6/sec … scn parameters are zero, i.e. there are no harmonics for even pulsations.
– for odd rotational speeds of the Z plane, i.e. for ω=-1/sec, -3/sec, -5/sec scn decrease to zero when ω=-∞. In other words, the harmonic amplitudes decrease to zero as the frequency increases to infinity.

Chapter 8.5.2 The third harmonic with constant c0 on the background of the even square wave, i.e. c0+h3(t)
acc. Fig. 8-1c
c3 is the complex amplitude of the third harmonic
c3=2*sc3=(-2/3π,0) czyli a3=-2/3π i b3=0
acc. Fig. 8-1e
h3(t)=-(2/3π)*cos(3t)≈-0.212cos(3t)

Fig.8-7
c0+h3(t)=0.5-(2/3π)*cos(3t)+0.5
So the third harmonic with a constant component c0 against the background of the square wave.

Chapter 8.5.3 Third approximation of an even square wave or S3 (t)=c0+h1(t)+h3(t).

Fig.8-8
S3(t)=c0+h1(t)+h3(t)
The third approximation is more like a square wave than the first one in Fig. 8-4.
What about the second approximation? So with S2(t)=c0+h1(t)+h2(t)? It is the same as the first one, ie S1(t)=c0+h1(t) because h2(t)=0 and we omit it. It is similar with the fourth, sixth approximation … because h4(t)=h6(t)= …=0.

Chapter 8.6 The fourth harmonic of an even square wave, or actually its absence, because c4=0->h4(t)= a4cos(4t)=0.
Chapter 8.6.1 Trajectory F(njω0t) of an even 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. 8-9
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). It makes 2 rotations during T=2π sec with a 1π sec gap between. The parameter sc4=0 as an average value of T=2π sec is obvious.
c. Trajectory F(4j1t) as a circle drawn by the end of the vector b.
sc4=0
Note
sc4=0 and therefore the harmonic for ω = 4 / sec is zero. The harmonic doesn’t exist

Chapter 8.7 The fifth harmonic of an even square wave h5(t)=a5cos(5t).
Chapter 8.7.1 Trajectory F(njω0t) of an even 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. 8-10
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 the f(t). At the time T=2π sec, the radius R = 1 stays longer on the right half-plane. Therefore, its average value as a vector will be sc5 =(1/5π, 0).
c. Trajectory F(5j1t) as a circle drawn by the end of the vector from b. Notice the word “5/4 turn” that appears. This should convince you that the trajectory F (5j1t) stays longer on the right half-plane Z. Otherwise, this part of the trajectory is “heavier”.

Chapter 8.7.2 The fifth harmonic with constant c0 on the background of the even square wave, i.e. c0+h5(t)
acc. Fig. 8-1c
c5 is the complex amplitude of the fifth harmonic
c5=2sc5=(-2/5π,0) czyli a5=-2/5π i b5=0
acc. Fig. 8-1e
h5(t)=-(2/5π)cos(5t)≈-0.127cos(5t)

Fig. 8-11
c0+h5(t)=0.5+(2/5π)*cos(5t)
So the fifth harmonic with a constant component c0 against the background of the square wave.

Chapter 8.7.3 Fifth approximation of an even square wave or S5(t)=c0+h1(t)+h3(t)+h5(t).

Fig. 8-12
S5(t)=c0+h1(t)+h3(t)+h5(t)
The fifth approximation is more like a square wave than the third one in Fig. 8-8.

Chapter 8.8 The sixth harmonic of an even square wave, or actually its absence, because c6=0->h6(t)= a6cos(6t)=0.
Chapter 8.8.1 Trajectory F(njω0t) of an even square wave for n=6 and ω0=1/sec, i.e. F(6j1t).

Fig. 8-9
Trajectory F(6j1t) of an even 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). It makes 2 rotations during T=2π sec with a 1π sec gap between. The parameter sc6=0 as an average value of T=2π sec is obvious.
c. Trajectory F(6j1t) as a circle drawn by the end of the vector b.
sc6=0
Note
sc6=0 and therefore the harmonic for ω=6/sec is zero. The harmonic doesn’t exist

Chapter 8.9 The seventh harmonic of an even square wave h7(t)=a7cos(5t).
Chapter 8.9.1 Trajectory F(njω0t) of an even 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. 8-14
Trajectory F(7j1t) of an even 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 the f(t). At the time T=2π sec, the radius R = 1 stays longer on the right half-plane. Therefore, its average value as a vector will be sc7 =(1/7π, 0).
c. Trajectory F(7j1t) as a circle drawn by the end of the vector from b. Notice the word “7/4 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 8.9.2 The seventh harmonic with constant c0 on the background of the even square wave, i.e. c0+h5(t)
acc. Fig. 8-1c
c7 is the complex amplitude of the seventh harmonic
c7=2sc7=(-2/7π,0) czyli a7=-2/7π i b7=0
acc. Fig. 8-1e
h7(t)=-(2/7π)cos(5t)≈-0.091cos(7t)

Fig. 8-15
c0+h7(t)=0.5+(2/7π)*cos(7t)
So the seventh harmonic with a constant component c0 against the background of the square wave.

Chapter 8.9.3 Seventh approximation of an even square wave or S5(t)=c0+h1(t)+h3(t)+h5(t)+h7(t).

Rys.8-16
S5(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)
The seventh approximation is more like a square wave than the fifth one in Fig. 8-12.

Chapter 8.10 The eighth harmonic of an even square wave, or actually its absence, because c8=0-> h8(t)= a8cos(8t)=0.
Chapter 8.10.1 Trajectory F(njω0t) of an even 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. 8-17
Trajectory F(8j1t) of an even 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). It makes 8 turns during T=2π sec with a 1π sec gap between. The parameter sc8=0 as an average value of T=2π sec is obvious.
c. Trajectory F(8j1t) as a circle drawn by the end of the vector b.
sc8=0
Note
sc8=0 and therefore the harmonic for ω=8/sec is zero. The harmonic doesn’t exist

Chapter 8.11 Remaining harmonics of an even 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 8-16 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. 8-1.