**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)**.

What if

**harmonics**are not visible 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 harmonic formulas**Fig.8-1a**

Even square wave **f(t)** **A=1, ω=1/sec, ϕ=0** and **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 only 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** wave”.**Harmonic** formulas

In the next points we will analyze the trajectories **F(n*jω0t)** and time charts of the **square wave** for **n=0…8**. Similarly to what we did before with **trigonometric** functions. But first, a few words on how to read the **nth** harmonic **hn(t)** from the **centroid** of the nth trajectory **F(njω0t)**.**Fig.8-1b**

Formula for the** nth** centroid **scn** of the trajectory **F(n*jω0t)**

In this **chapter**, the centroids of **scn** agree with intuition and we will only assume that they are true. In **Chapter 11,** we will calculate them precisely with the **WolframAlfa** program.**Fig.8-1c**

Formula for the **nth** complex **coefficient** of a Fourier series **cn**

This is after **doubling** the centroid **scn**! Treat this as the **complex** amplitude for the **nth** harmonic.**Note**

The formula is true for **n=1,2,3…****For n=0****c0=sc0=a0** and this is the average of **f(t)** over the **period.**

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

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

**Fig.8-2**The trajectory

**F(0j1)**of the function

**f(t)**of an

**even square wave**, i.e. with the parameters:

**A=1-amplltude**

**ω0=1/sec**-pulsation corresponding to the period

**T=2πsec≈6.28sec**.

**ϕ=0**

**50%**-filling

**Fig.8-2a**

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 according to. the function

**f(t)**shown

in

**Fig.8-2b**. In other words, these are jumping movements “back and forth” on the real axis

**Re Z**between the points

**(0,0)**and

**(1,0)**.

**Fig.8-2b**

Time chart

**f(t)**of an

**even square wave**.

Notice that the waveform begins halfway through the

**half-cycle**. Just like the

**cosine**function. Both animations in

**Fig. a**and

**b**describe the same thing. Read the parameters

**A,ω0,ϕ**and

**filling**parameter. Animating

**b**is easier, but animation

**a**is also possible. From the time chart it is possible to calculate, even without integrals, the coefficient

**c0=a0=0.5**of the

**Fourier Series**as the average value of

**f(t)**for

**0…2π**. See

**Fig.8-1c**.

In the next

**subchapters**, the

**Z**plane will start rotating at speeds

**ω=-nω0**, i.e.

**-1/sec,-2/sec…-8/sec**in the “clockwise” direction. The end of the vector will draw trajectories

**F(njω0)**. From them we will read the harmonics for the

**nω0**pulsation. As in previous chapters, only instead of

**f(t)**=combination of

**sines**and

**cosines**, it will now be

**f(t)=square wave**

Chapter 8.3 First harmonic of an even square wave h1(t) =a1*cos(1t).

Chapter 8.3 First harmonic of an even square wave h1(t) =a1*cos(1t).

**Section 8.3.1 Trajectory of F(njω0t) of an even square wave for n=1 and ω0=1/sec, i.e. F(1j1t).**

**Fig.8-3**Trajectory

**F(1j1t)**of an

**even square wave**

**Fig.8-3a**

The radius

**R=1**as a vector

**(1,0)**rotates with speed

**ω=-1/sec**around the point

**(0,0)**in time

**T=2π sec**. The entire complex plane

**Z**rotates with it. But be careful! The axes of the

**Re Z**and

**Im Z**planes don’t rotate! The

**end**of the

**vector**would draw an entire circle unlike the

**semi-circle**in

**Fig.8-3c**. The average value of the sum of rotating vectors during

**1**revolution, i.e. during

**2π sec**, is the zero vector

**z=(0,0)**. The

**z**parameter is the so-called centroid of the trajectory, i.e. circle. Here z coincides with the centroid

**scn**known from physics.

**Fig.8-3b**

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.8-2b**. Look at

**Fig.8-2a**and imagine that the

**Z**plane starts

**rotating**at a speed of

**ω=-1/sec**. As a reminder, the

**Re Z**and

**Im Z**axes are stationary! This will create a rotating vector

**F(1j1t)=f(t)*exp(-1j1t)**. The

**Z**plane will make

**1**revolution, and the radius

**R=1**will make

**2**“quarter turns” with a break of

**1π sec**in between. This is equivalent to one “half-turn”.

**Fig.8-3c**

The trajectory of

**F(1j1t)**as a semicircle drawn by the

**end**of the vector in

**Fig. 8-3b**.

Now imagine that during

**1**rotation of the

**Z**plane, i.e. by an angle of

**0…2π**, all vectors are summed (vectorically!) and their

**average**is calculated. Some

**sc1**vector will be created. Will it be

**sc1=(0,0)**as in

**Fig.8-3a**? Probably not, but it will definitely lie somewhere on the

**Re Z**axis between

**(0,0)**and

**(1,0)**. Rather closer to

**(0,0)**than

**(1,0)**. It will be the vector

**sc1=(1/π, 0)**! Otherwise, the centroid

**sc1=(1/π, 0)**of the trajectory

**F(1j1t)**. We will precisely calculate the

**sc1**parameter of the trajectory using the

**WolframAlfa**program in

**Chapter 11**.

**Note**to

**Fig. 8-3b,c**

The above animations suggest that the upper and lower “quadrants” of the trajectories are separated by a gap

**T=πsec**. But this is only the case at the beginning at time

**t=0…2π sec**! Later, there is no break and the movement along the trajectory is continuous.

**Chapter 8.3.2 First harmonic against the background of an even square wave, i.e. c0+h1(t) or in other words, the first approximation of an even square wave.**

Acc. to

**Fig. 8-1c**the constant component

**c0**is the average over the period

**T=2π**that is,

**c0=a0=0.5**. You can calculate it even without integral calculus.

**c1**is the

**complex**amplitude of the

**first**harmonic

**c1=2*sc1=(2/π,0) i.e. a1=2/π**and

**b1=0**

Acc. to

**Fig. 8-1e**

**h1(t)=(2/π)*cos(1t)≈0.637cos(1t)**

**Fig.8-4S1(t)=c0+h1(t) **i.e. the

**first**harmonic with a

**constant**component

**c0=0.5**against the background of a

**square wave**. This is also a

**first**approximation to our even

**50%**square wave.

Chapter 8.4 Second harmonic of an even square wave, or rather its absence because

c2=0 –>h2(t)=a2*cos(2t)=0.

Chapter 8.4 Second harmonic of an even square wave, or rather its absence because

c2=0 –>h2(t)=a2*cos(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****Fig.8-5a**

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.8-5b**

Trajectory **F(2j1t)=f(t)*exp(-2j1t)** as a rotating vector modulated by the function** f(t)**. In **2π sec**, the **Z** plane will make **2** revolutions, but the radius **R=1** will only make **2** “half-turns” with a break of **1π sec** in between.**Fig.8-5c**

The trajectory of **F(2j1t)** as a circle drawn by the end of the vector in** Fig. 8-5b**.

Obviously** sc2=0**. Note that from the circle itself the value **sc2=0** is not obvious! After all, the** circle** could have been drawn, for example, **1.5** times. Fortunately, this **1** complete revolution, although with a break in the middle, is visible in **Fig. 8-5b****Note****sc2=0** and therefore the harmonic **for ω=2/sec** does not exist.

Chapter 8.5 Third harmonic of an even symmetric square wave h3(t) =a3*cos(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=-3/sec**

**Fig.8-6**

Trajectory **F(3j1t)** of an** even square wave****Fig.8-6a**

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.8-6b**

Trajectory **F(3j1t)=f(t)*exp(-3j1t)** as a rotating vector modulated by the function **f(t)**.

In time **T=2π sec**, the **Z** plane will make** 3** revolutions, but the radius **R=1** will only make **2 **times “**3/4 revolutions**” with a break of **1π** sec in between. It will turn out that during **T=2π sec**, the radius **R=1** stays longer in the **left** half-plane than in the **right** one. Therefore, its average value as a vector will be **(-1/3π, 0)****Fig.8-6c**

The trajectory of **F(3j1t)** as a circle drawn by the end of the vector in **Fig. 8-6b**.

The centroid **sc3=(-1/3π, 0)** results from the summation of the vectors in **Fig. 8-6b** and their average at time **T=2π sec** when **3** revolutions of the **Z** plane are made. Its value is **negative**, unlike in **Fig.8-3b**, where the vector has made **2** “quarter turns” only on the positive **Z** half-plane.**Note**

In the next **subchapters** there will be no such detailed description of the trajectory **F(njω0t)**.**The** reader will notice that:

for **even** rotational speeds of the **Z** plane, i.e. **for ω=-4/sec, -6/sec, -8/sec…****scn** parameters are **zero**, i.e. there are **no** harmonics for even **ω**.

for **odd** rotational speeds of the **Z** plane, i.e. for** ω=-5/sec, -7/sec…****scn** parameters are **non-zero** and decrease to** zero** when **ω=-∞.** In other words, the **harmonic** amplitudes decrease to **zero** as the **frequency** increases to **infinity**.**Chapter 8.5.2 Third harmonic against the background of an even square wave, i.e. c0+h3(t).**

Acc. to **Fig. 8-1c****c3** is the** complex** amplitude **of** the** third** harmonic**c3=2*sc3=(-2/3π,0) i.e. a3=-2/3π** and **b3=0**

Acc. to **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**

That is, the

**third**harmonic with a constant component

**c0**against the background of a

**square wave**.

**Chapter 8.5.3 Third approximation of an even square wave, i.e. S3(t)=c0+h1(t)+h3(t).**

**Fig.8-8**

**S3(t)=c0+h1(t)+h3(t)**

The

**third**approximation is more similar to a

**square wave**than the

**first**one in

**Fig.8-4**

What about the

**second**one? So from

**S2(t)=c0+h1(t)+h2(t)**? It is the same as the

**first one**, i.e.

**S1(t)=c0+h1(t)**because

**h2(t)=0**and we ignore it. It is similar with the

**fourth, sixth**approximation…, because

**h4(t)=h6(t)=…=0.**

Chapter 8.6 Fourth harmonic of an even symmetric square wave, or rather its lack because

c4=0 –>h4(t)=a4*cos(4t)=0.

Chapter 8.6 Fourth harmonic of an even symmetric square wave, or rather its lack because

c4=0 –>h4(t)=a4*cos(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****Fig.8-9a**

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.8-9b**

Trajectory** F(4j1t)=f(t)*exp(-4j1t)** as a rotating vector modulated by the function **f(t)**. It will make **2** revolutions in a period of **T=2π sec** with a break of **1π sec** in between. The parameter **sc4=0** as an average value for **T=2π sec** is obvious.**Fig.8-9c**

The trajectory of **F(4j1t)** as a circle drawn by the end of the vector in **Fig. 8-9b**.**sc4=0****Note****sc4=0** and therefore the harmonic for** ω=4/sec** does not exist.

**Chapter 8.7 Fifth harmonic of a symmetrical even square wave h5(t)=a5*cos(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****Fig.8-10a**

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.8-10b**

Trajectory **F(5j1t)=f(t)*exp(-5j1t)** as a rotating vector modulated by the function **f(t)**.

At 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)**.**Fig.8-10c**

The trajectory of **F(5j1t)** as a circle drawn by the end of the vector in **Fig. 8-10b**.

Notice the **“5/4 turn” text** that appears. This should convince you that the **F(5j1t)** trajectory stays longer on the **right** half-plane of **Z**. Otherwise, this part of the trajectory is **“heavier”**.**Chapter 8.7.2 Fifth harmonic against the background of an even square wave, i.e. c0+h5(t).**

Acc. to **Fig. 8-1c****c5** is the** complex** amplitude of the **fifth** harmonic

c5=2*sc5=(2/5π,0) i.e. a5=2/5π and b5=0

Acc. to** Fig. 8-1e****h5(t)=(2/5π)*cos(5t)≈0.127cos(5t)**

**Fig. 8-11****c0+h5(t)=0.5+(2/5π)*cos(3t)**

That is, the **fifth** harmonic with a constant component **c0** against the background of a** square wave**.**Chapter 8.7.3 Fifth approximation of an even square wave, i.e. S5=c0+h1(t)+h3(t)+h5(t).**

**Fig.8-12**

**S5(t)=c0+h1(t)+h3(t)+h5(t)**

The

**fifth**approximation is more similar to a

**square wave**than the

**third**one in

**Fig.8-8**

**Chapter 8.8 Sixth harmonic of an even symmetric square wave, or rather its lack because c6=0 –>h6(t)=b6*cos(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.8-13**

Trajectory **F(6j1t)** of an **even square wave****Fig.8-13a**

The radius **R=1** as a vector **(1,0)** rotates at a speed** ω=-6/sec** around the point **(0,0)** and will complete 6 revolutions in time **T=2π sec**.**Fig.8-13b**

Trajectory **F(6j1t)=f(t)*exp(-6j1t)** as a rotating vector modulated by the function** f(t)**. It will make** 2×1.5 revolutions** in a period of **T=2π sec** with a break of **1π sec** in between.**Fig.8-13c**

The trajectory of **F(6j1t)** as a **circle** drawn by the **end** of the vector in** Fig. 8-13b**.**sc6=0****Note****sc6=0** and therefore the harmonic for **ω=6/sec** does not exist.

**Chapter 8.9 Seventh harmonic of an even square wave h7(t)=a7*cos(7t).****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(5j1t)** of an** even square wave****Fig.8-14a**

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 on the** left** half-plane. Therefore, its average value will be **sc7=(-1/7π, 0)**.**Fig.8-14c**

The trajectory of **F(7j1t)** as a circle drawn by the end of the vector in **Fig. 8-17b**.

Notice the words **“1 and 3/4 turns”** that appear. This should convince you that the **F(7j1t)** trajectory stays longer on the** left** half-plane** Z**. Otherwise, this part of the trajectory is “heavier”.**Chapter 8.9.2 The seventh harmonic against the background of an even square wave, i.e. c0+h7(t).**

Acc. to **Fig. 8-1c****c3** is the **complex** amplitude of the **third** harmonic**c7=2*sc7=(-2/7π,0)** i.e. **a7=-2/7π** and **b7=0**

Acc. to **Fig. 8-1e****h7(t)=-(2/7π)*cos(7t)≈-0.091cos(7t)**

**Fig. 8-15****h7(t)+0.5**, i.e. the** seventh** harmonic with a constant component** c0** against the background of a **square wave**.**Chapter 8.9.3 The seventh approximation of an even square wave, i.e. S7=c0+h1(t)+h3(t)+h5(t)+h7(t).**

**Fig.8-16**

**S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)**

The

**seventh**approximation is more similar to a

**square wave**than the

**fifth**in

**Fig.8-12**

**Chapter 8.10 The eighth harmonic of an even square wave, or rather its lack because c8=0 –>h8(t)=b8*cos(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).**

**Fig.8-17**

Trajectory** F(8j1t)** of an** even square wave****Fig.8-17a**

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.8-17b**

Trajectory **F(8j1t)=f(t)*exp(-8j1t)** as a rotating vector modulated by the function **f(t)**. It will make **4** revolutions in a period of **T=2π sec** with a break of **1π sec** in between.**Fig.8-17c**

The trajectory of **F(8j1t)** as a circle drawn by the end of the vector in **Fig. 8-17b**.**sc8=0****Conclusion****sc8=0** and therefore the harmonic for **ω=8**/sec does not exist.

**Chapter 8.11 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, **scn** of **even** harmonics are** zero**. This means that the harmonics **decrease** with **increasing** frequency and for an **infinitely** high pulsation **ω**, the harmonic amplitudes are** zero**, i.e. they **disappear**.

The animation of** 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**.