## 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 c

**omplex 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 **7**turns 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°)

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 **firs**t 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.