10. Fourier Series of an square wave with an offset of -30°

Rotating Fourier Series

Chapter 10. Fourier series of a square wave with a shift of -30°.

Chapter10.1 Introduction

Fig.10-1
Square wave f(t)
A=1, ω=1/sec and ϕ=-30° or -π/6.
The parameter ϕ=-30° of the wave means that it is shifted by -30° in relation to the even wave from chapter 8. So it is not an even or odd function. Such “bland”. Even wave had only cosine harmonics, i.e. components bn=0. In turn, odd only sinusoidal, i.e. components an=0. Now we expect cosine and sine components to occur.
Desciption on Fig 8-1 chap 8

Chapter 10.2 A square wave with an offset of -30 ° and its trajectory F (nj1t) for n = 0 i.e. F (0j1t)

Fig.10-2
Trajectory F(0j1) of the function f(t) of an square wave with an offset of -30°  i.e. with the following parameters:
A=1-amplitude
ω0=1/sec-pulsation corresponding to the period T=2πsec≈6.28sec.
ϕ=-π/6=-30°
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. The time plot of f(t) of a square wave with ϕ=-π/6=-30°
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).

Chapter 10.3 The first harmonic of a square wave with an offset of -30 °, i.e. h1(t)= |c1|cos(1t-30°).
Chapter 10.3.1 Trajectory F (nj1t) of a square wave with a shift of -30 ° for n=1 and ω0 =1/sec, i.e. F(1j1t).

Fig. 10-3
Trajectory F(1j1t) of a square wave with a shift of -30°
Compare the drawing with the corresponding Fig. 8-3 in Chapter. 8 for an even square wave. There the vector sc1 had a displacement of ϕ=0° and was presented as a complex number in the algebraic form sc1=(1/π,0) or otherwise 1/π+j0. It could well have been the exponential form sc1=1/πexp(j0°), where the amplitude A=1/π and the phase ϕ=0° are clearly visible. In this chapter we will use the exponential version and therefore the center of gravity of the trajectory in c is sc1= |1/π| exp(-j30°). Tu |1/π| is the so-called sc1 module You can easily switch from exponential to algebraic and vice versa:
sc1=(1/π) exp(-j30 °)= (1/ π) [cos (-30 °) + jsin (-30°)] ≈0.275-j0.159
The vector sc1 is the average value of the rotating vector b in the period T=2π sec, calculated from the formula Fig. 7-2b Chapter 7.
“Revenons à nos moutons” ie. let’s come back to the topic
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   vector modulated by f(t). The f(t) is a square wave in Figure 10-2b.
The Z plane in a will start to rotate at ω=-1/sec. A spinning vector F(1j1t)=f (t)exp (-1j1t) will arise. The Z plane will do 1 turn, but the trajectory as a spinning vector F(1j1t)=f (t)exp (-1j1t) only 1 “half turn”.
c. Trajectory F(1j1t) as a semicircle drawn by the vector b. During 1 turn of the Z plane by the angle 0 … 2π, all vectors are added vectorally and their average over the period T=2π sec is calculated from the formula Fig. 7-2b chapter. 7. This is the vector sc1=(1/ π)exp(-j30°). The trajectory is rotated with respect to that of Fig. 8-3c chapter 8 by the angle ϕ=-30 °.

Note to Fig. 10-3b,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 10.3.2 The first harmonic on a square wave with a shift of -30 ° or c0 + h1 (t) = 0.5 + 1 * cos (1t-30 °) or the first approximation of a square wave with a shift of -30 °.
The constant component c0=a0 is the average over the period T=2π, i.e. c0=a0=0.5.
Acc. Fig. 10-3c
sc1=|sc1|*exp(jϕ)=(1/π)*exp(-j30°)
c1=2sc1=(2/π)*exp(-j30°)  i.e. |c1|=2/π  ϕ=-30 °
Acc. Fig. 8-1f chapter 8
h1(t)=(2/ π)*cos(1t-30 °)≈0.637*cos (1t-30 °)

Fig. 10-4
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  square wave.

Chapter 10.4 The second harmonic of a square wave with a shift of -30 °, or actually its absence because c2=0 ->h2(t)=0.
Chapter 10.4.1 Trajectory F (njω0t) of an 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. 10-5
Trajectory F(2j1t) of a square wave with a shift of -30 °
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 f(t). In 2π sec, the Z plane will make 2 turns, but radius R=1 will do only 1 turn consisting of a first 240° turn and a second 120° turnn with a 1π sec gap between them.
c. Trajectory F(2j1t) as a circle drawn by vector from b.
Evidently sc2=0. Note that from the circle alone, 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 at b.

Chapter 10.5 The third harmonic of a square wave with an offset of-30 °, i.e. h3(t)=|c1|cos(1t-30°).
Chapter 10.5.1 Trajectory F (nj1t) of a square wave with a shift of -30 ° for n=3 and ω0 =1/sec, i.e. F(3j1t).
The Z plane rotates at a speed of ω=-3/sec

Fig. 10-5
Trajectory F(3j1t) of a square wave with a shift of -30 °
a. The radius R=1 as a vector (1,0) rotates at the speed ω=3/sec around the point z=(0,0) and will make 3 turns in the time T=2π sec.
b. Trajectory F(2j1t)=f (t)exp(-2j1t) as a rotating vector modulated by f(t). In 2π sec, the Z plane will make 3 turns, but radius R=1 will do only 1.5 turns consisting of a first 360° turn and a second 180° turn with a 1π sec gap between them. Radius R=1 stays longer in the upper half-plane than in the lower half-plane. Therefore, the average value of the vector is sc3 = (0.1 / 3π) = 1j / 3π
c. Trajectory F(3j1t) as a circle drawn by the vector b.
Center of gravity sc3=(0.1/3π)=1j/3π.

Chapter 10.5.2 The third harmonic on a square wave background with an offset of -30 °, i.e. c0+h3 (t).
Acc. Fig. 10-3c
sc3=|sc3|exp(jϕ)=(1/3π)exp(j90°)
c1=2sc1=(2/3π)exp(j90°) that is | c1 |=(2/3π) ϕ=90°
Acc. Fig. 8-1f chapter 8
h3 (t)=(2/3π)cos(3t+90 °)≈-0.212sin (3t)

Fig.10-7
S3(t)=c0+h3(t) that is the third harmonic with a constant c0=0.5 on the background of a square wave.

Chapter 10.5.3 Third approximation of the square wave with an offset of -30 °, i.e. S3=c0+h1(t)+h3(t).

Fig.10-8
S3(t)=c0+h1(t)+h3(t)
The third approximation is more like a square wave than the first one in Fig. 10-4

Chapter 10.6 The fourth harmonic of a square wave with a shift of -30 °, or actually its absence because c4=0 ->h4(t)=0.
Chapter 10.6.1 Trajectory F(njω0t) of an 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.10-9
Trajectory F(4j1t) of a square wave with a shift of -30°
a. The radius R=1 as a vector (1,0) rotates at the speed ω=-4/sec around 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 f(t).
In 2π sec, the Z plane will make 4 turns, but the radius R=1 will make 2 turns consisting of the first 480° and the second 240° with a 1π sec gap between them.
c. Trajectory F(4j1t) as a circle drawn by the end of the vector from b.
Evidently sc4=0.
Conclusion
sc4=0 and therefore the harmonic for ω=4/sec is doesn’t exist.

Chapter 10.7 The fifth harmonic of a square wave with an offset of -30 °, i.e. h5(t)=|c5|cos(5t-30°).
Chapter 10.7.1 Trajectory F (nj1t) of a square wave with a shift of -30 ° for n=5 and ω0 =5/sec, i.e. F(5j1t).
The Z plane rotates at a speed of ω=-3/sec

Fig. 10-10
Trajectory F(5j1t) of a square wave with a shift of -30 °
a. The radius R=1 as a vector (1,0) rotates at the speed ω=5/sec around the point z=(0,0) and will make 5 turns in the time T=2π sec.
b. Trajectory F(5j1t)=f (t)exp(-5j1t) as a rotating vector modulated by f(t)
 In 2π sec, the Z plane will make 5 turns, but radius R=1 will only do 2.5 turns (900°) including the first 585° and the
second 315° with a 1π sec gap between them. Radius R=1 stays longer in the lower left quadrant than in the others.
Therefore the average value of the vector is sc5=(1/5π)exp(-j150 °)
c. Trajectory F(5j1t) as a circle drawn the vector b.
Center of gravity sc5=(1/5π)exp(-j150 °).

Chapter 10.5.2 The fifth harmonic on a square wave background with an offset of -30 °, i.e. c0+h5 (t).
Acc. Fig. 10-10c
sc5=|sc5|exp(jϕ)=(1/5π)exp(-j150°)
c5=2sc5=(2/5π)exp(-j150°) that is |c5|=(2/5π) and ϕ=-150°
Acc. Fig. 8-1f chapter 8
h5 (t)=(2/5π)cos(5t+150 °)≈-0.127cos(5t+30°)

Fig. 10-11
S5(t)=c0+h5(t) that is the fifth harmonic with a constant component c0=a0=0.5 against the square wave.

Chapter 10.7.3 Fifth approximation of a square wave with an offset of -30 °, i.e. S5(t)=c0+h1(t)+h3(t)+h5 (t).

Fig.10-12
S5(t)=c0+h1(t)+h3(t)+h5(t)
The fifth approximation is more like a square wave than the third one in Fig. 10-8

Chapter 10.8 The sixth harmonic of a square wave with a shift of -30 °, or actually its absence because c6=0 ->h6(t)=0.
Chapter 10.8.1 Trajectory F(njω0t) of a square wave with a shift of -30 ° for n=6 and ω0=1/sec, i.e.F(6j1t).

Fig. 10-13
Trajectory F(6j1t) of a square wave with a shift of -30°
a. The radius R=1 as a vector (1.0) rotates at the speed ω=-6/sec around the point z=(0.0) and makes 6 turns in the time T=2π sec.

b. Trajectory F(6j1t)=f (t exp(-6j1t) as a rotating vector modulated by f(t).
In 2π sec, the Z plane will make 6 turns, but radius R=1 will do a full 3 turns consisting of the first 2 turns and the second 1 turn with a 1π sec gap between them.
c. Trajectory F(6j1t) as a circle drawn by the the vector b.
Evidently sc6=0.
Conclusion
sc6=0 and therefore the harmonic for ω=6/sec is zero.

Chapter 10.9 The seventh harmonic of a square wave with a shift of -30 ° or  h7(t)=c7*exp(jωt).
Chapter 10.9.1 Trajectory F(njω0t) of a square wave with a shift of -30 ° for n=7 and ω0=1/sec, i.e. F(7j1t).
The Z plane rotates at a speed of ω=-7/sec
The sc7 vector is so small you can barely see it. You have to imagine him.

Fig. 10-14
Trajectory F(7j1t) of a square wave with a shift of -30°
a. The radius R=1 as a vector (1.0) rotates at the speed ω=-7/sec around the point z=(0.0) and will make 7turns in the time T=2π sec.
b. Trajectory F(7j1t)=f(t)exp(-7j1t) as a rotating vector modulated by f(t).
In 2π sec, the Z plane will make 7 turns (2520°), but the radius R=1 will only do 3.5 turns (1260°) including 2 and 1/3 turns-obrotu (840 °) in the first cycle and 1 and 1/6 turns-obrotu in the second cycle rotation (420 °) with a 1π sec gap between them. The radius R=1 stays a little longer in the lower right quadrant than in the others. Therefore the average value of the vector is sc7=(1/ 7π)exp(-j30 °)
c. Trajectory F(7j1t) as a circle drawn by the vector b.
Center of gravity sc7=(1/7π)exp (-j30°)..

Chapter 10.9.2 The seventh harmonic against the square wave with an offset of -30 °, i.e. c0+h7(t).
Acc. Fig. 10-14c
sc7=|sc7|*exp(jϕ)=(1/7π)*exp(-j30°).
c7=2*sc7=(2/7π)*exp(-j30°) czyli |c7|=2/7π  ϕ=-30°
Acc. Fig. 8-1f chapter 8
h7(t)=(2/7π)*cos(7t-30°)≈-0.127*cos(7t-30°)

Fig. 10-15
S7(t)=c0+h7(t) that is the seventh harmonic with a constant component c0=a0=0.5 against the square wave.

Chapter 10.9.3 Seventh approximation of the square wave with an offset of -30 °, i.e. S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t).

Figs. 10-16
S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)
The seventh approximation is more like a square wave than the fifth in Fig. 10-12

Chapter 10.10 The eighth harmonic of a square wave with a shift of -30 °, or actually its absence because c8=0–>h8(t)=0.
Chapter 10.10.1 Trajectory F (njω0t) of a square wave with a shift of -30 ° for n = 8 and ω0 = 1 / sec, i.e. F (8j1t).

Fig. 10-17
Trajectory F(86j1t) of a square wave with a shift of -30°
a. The radius R=1 as a vector (1.0) rotates at the speed ω=-8/sec around the point z=(0.0) and makes 8 turns in the time T=2π sec.
b. Trajectory F(8j1t)=f (t) exp(-8j1t) as a rotating vector modulated by f(t).
In 2π sec, the Z plane will make 8 turns, but radius R=1 will do full 4 turns in the first stage only.
c. Trajectory F(8j1t) as a circle drawn by the the vector b.
Evidently sc8=0.
Conclusion
sc8=0 and therefore the harmonic for ω=8/sec is doesn’t exist.