**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**

Evensquare wave and

Even

**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**

Formulafor the

Formula

**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

c0=sc0=a0

**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)**.

**Time plot**

b.

b.

**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

**sine**s 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 circl**e 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**.