**Rotating Fourier Series**

**Chapter 9. Fourier series**** of an odd square wave****.**

**Chapter 9.1 Introduction**This is the wave from the previous chapter shifted to the

**right**(i.e. delayed) by a

**quarter**of a period (i.e. by

**90°**or

**π/2**). And just as the

**first**harmonic of the

**previous**one was a

**cosine**, this one will be a

**sine**. This is an example of an

**odd**function.

**Fig. 9-1****Odd** square wave **f(t) A=1, ω=1/sec** and **ϕ=-π/2**.

The parameter **ϕ=-π/2** of the **odd** wave means that it is delayed by** π/2** compared to the **even** wave from the **previous** chapter. Like the** sine**, it lags by **π/2** with respect to the **cosine**. In the next points, we will analyze the trajectories of **F(n*jω0t)** and the time charts of this **square** wave for **n=0…8**.

**Chapter 9.2 Odd square wave f(t) and its trajectory F(njω0t) for n=0, i.e. F(0j1t)**

**Fig.9-2**

The trajectory **F(0j1)** of the function **f(t)** of an **odd** square **wave**, i.e. with the parameters:**A=1**-ampliltude**ω0=1/sec**-pulsation corresponding to the period **T=2πsec≈6.28sec**.**ϕ=-π/2****50%**-filling**Fig.9-2a**

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 acc. to. the function **f(t)** shown in **Fig.9-2b**.**Fig.9-2b**

Time chart of **f(t)** of an **odd** square **wave**.

Note that the time chart starts at the **beginning** of the **period**. Just like the **sine** function. Both animations in **Fig.9-2a** and **Fig.9-2b** describe the same thing. Read the parameters **A,ω0,ϕ** and **filling**.

In the next **subchapters**, the **Z** plane will start rotating at speeds **ω= -nω0**, i.e. at speeds** -1/sec,-2/sec…-8/sec**. The end of the vector will draw trajectories **F(njω0)**. From them we will read the harmonics for the **nω0** pulsation.

**Chapter 9.3 First harmonic of an odd square wave h1(t)=b1*sin(1t).Chapter 9.3.1 Trajectory F(njω0t) of an odd square wave for n=1 and ω0=1/sec, i.e. F(1j1t).**

**Fig. 9-3**

Trajectory **F(1j1t)** of an **odd** **square wave****Fig.9-3a**

The radius **R=1** as a vector **(1,0)** rotates with a speed **ω=-1/sec** around the point **(0,0)** in time **T=2π sec** and will complete** 1** revolution.**Fig.9-3b**

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.9-2b**. The **Z** plane in **Fig.9-2a** will start rotating at a speed of** ω=-1/sec**. This will create a rotating vector** F(1j1t)=f(t)*exp(-1j1t)**. The **Z** plane will make** 1** revolution, but the radius **R=1** will only make **1** “half-turn”.**Fig.9-3c**

The trajectory of **F(1j1t)** as a semicircle drawn by the end of the vector in **Fig. 8-3b**.

During **1** rotation of the **Z** plane, i.e. by an angle of **0…2π**, all vectors are vector summed and their average is calculated in the period **T=2π sec**. It will be a vector, otherwise the point **sc1=(0,-1/π)=0-1j/π=-1j/π**.**Chapter 9.3.2 The first harmonic against the background of an odd square wave, i.e. c0+h1(t) or in other words, the first approximation of an odd square wave.**

Acc. to **Fig. 8-1c**

The constant component **c0** is the average in the period **T=2π**, i.e. **c0=a0=0.5**.

Also, **c1** is the** complex** amplitude of the **first** harmonic**c1=2*sc1=(0,-2/π)=0-j2/π=-j2/π i.e. a1=0 and b1=-2/π**

Acc. to **Fig. 8-1e****h1(t)=(2/π)*sin(1t)≈0.637sin(1t)**

**Fig.9-4S1(t)=0.5+h1(t)=(2/π)*sin(1t)**,

i.e. the

**first**harmonic with a

**constant**component

**c0**against the background of a

**square wave**.

This is also a

**first**approximation to our

**square wave**.

**Chapter 9.4 Second harmonic of an odd square wave, or rather its absence because c2=0 –>h2(t)=b2*sin(2t)=0.Chapter 9.4.1 Trajectory F(njω0t) of an odd 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.9-5**

Trajectory** F(2j1t)** of an **odd square wave****Fig.9-5a**

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.9-5b**

Trajectory **F(2j1t)=f(t)*exp(-2j1t)** as a rotating vector modulated by the function** f(t)**. It will make **1** revolution in the **first** half-period **T/2=1π/sec** and **0** revolutions in the **second half-period**. The parameter **sc2=0** as the average value of **T=2π sec** is obvious.**Fig.9-5c**

The trajectory of **F(2j1t)** as a circle drawn by the end of the vector in **Fig. 9-5b**.**Note****sc2=0** and therefore the harmonic for** ω=2/sec** does not exist.

**Chapter 9.5 Third harmonic of an odd square wave h3(t)=b3*sin(3t).Chapter 9.5.1 Trajectory F(njω0t) of an odd 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.9-6**

Trajectory **F(3j1t)** of an **odd square wave****Fig.9-6a**

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.9-6b**

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

It will make **1.5** revolutions in the first half-period **T/2=1π/sec** and **0** revolutions in the **second half-period**. In the period **T=2π sec**, the radius **R=1** stays longer in the **lower half-plane** than in the **upper one**. Therefore, its **average** value as a vector will be **sc3=(0,-1/3π)=-1j/3π** and not **sc3=(0,0)** as e.g. in **Fig.9-5c****Fig.9-6c**

The trajectory of **F(3j1t)** as a circle drawn by the** end** of the vector in **Fig. 9-6b**.

The centroid **sc3=(0,-1/3π)** results from the summation of the vectors in **Fig. 9-6b** and their average at time **T=2π sec** when **3** revolutions of the **Z** plane are made. The **lower** semicircle is drawn **twice**, and the upper one only **1** time. As if the lower semicircle was as more “heavy”.**Chapter 9.5.2 Third harmonic against an odd square wave, i.e. c0+h3(t).**

Acc. to **Fig. 8-1c** from** chapter 8****c3** is the **complex** amplitude of the **third** harmonic**c3=2*sc3=(0,-2/3π) **i.e.** a3=0 and b3=-2/3π****Acc.** to **Fig. 8-1e****h3(t)=(2/3π)*sin(3t)≈0.212sin(3t)**

**Fig.9-7****co+h3(t)=0.5+(2/3π)*sin(3t)**,

i.e. the** third** harmonic with a constant component **c0** against the background of a** square wave**.**Chapter 9.5.3 Third approximation of an odd square wave, i.e. S3(t)=c0+h1(t)+h3(t).**

**Fig.9-8**

**S3(t)=c0+h1(t)+h3(t)=0.5+(2/π)*sin(1t)+(2/3π)*sin(3t)**

The

**third**approximation is more

**similar**to a

**square wave**than the

**first**one in

**Fig.9-4**

**Chapter 9.6 The fourth harmonic of an odd square wave, or rather its absence because c4=0 –>h4(t)=c4*sin(4t)=0.Chapter 9.6.1 Trajectory F(njω0t) of an odd 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.9-9**

Trajectory **F(4j1t)** of an **even square wave****Fig.9-9a**

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.9-9b**

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

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

**Chapter 9.7 Fifth harmonic of a symmetrical odd square wave h5(t)=b5*sin(5t).Chapter 9.7.1 Trajectory F(njω0t) of an odd 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.9-10**

Trajectory **F(5j1t)** of an **even square wave****Fig.9-10a**

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.9-10b**

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

It will make **2.5** revolutions in the **first half-**period **T/2=1π/sec** and **0** revolutions in the **second half-period**. In the period **T=2π sec**, the radius **R=1** stays longer in the **lower half-plane** than in the **upper one**. Therefore, its **average** value as a vector will be **sc5=(0,-1/5π)=-1j/5π**.**Fig.9-10c**

The trajectory of **F(5j1t)** as a circle drawn by the end of the vector in **Fig. 9-10b**. The centroid **sc5=(0,-1/5π)** results from the summation of the vectors in **Fig.9-10b** and their **average** in the period **T=2π sec**, when **5** revolutions of the **Z** plane are made. As if the lower semicircle was “heavier” than the upper one**Chapter 9.7.2 The fifth harmonic against the background of an odd square wave, i.e. c0+h5(t).**

Acc. to **Fig. 8-1c** from **chapter 8****c5** is the **complex** amplitude of the** fifth** harmonic**c5=2*sc5=(0,-2/5π) **i.e. **a5=0 and b5=-2/5π****Acc.** to **Fig. 8-1e****h5(t)=(2/5π)*sin(5t)≈0.127sin(5t)**

**Fig. 9-11****co+h5(t)=0.5+(2/5π)*sin(5t)**i.e. the

**fifth**harmonic with a constant component

**c0**against the background of

**a square wave**.

**Chapter 9.7.3 Fifth approximation of an odd square wave, i.e. S5=c0+h1(t)+h3(t)+h5(t).**

**Fig.9-12****S5(t)=c0+h1(t)+h3(t)+h5(t)=0.5+(2/π)*sin(1t)+(2/3π)*sin(3t)+(2/5π) )*sin(5t)**

The **fifth** approximation is more similar to a **square wave** than the **third** one in **Fig.9-8**.

**Chapter 9.8 The sixth harmonic of an odd square wave, or rather its lack because c6=0 –>h6(t)=b6*sin(6t)=0.**

**Chapter 9.8.1 Trajectory F(njω0t) of an odd square wave for n=6 and ω0=1/sec, i.e. F(6j1t).**

**Fig.9-13**

Trajectory **F(6j1t)** of an **odd square wave****Fig.9-13a**

The radius **R=1** as a vector **(1,0)** rotates with a speed **ω=-6/sec** around the point **(0,0)** and will complete **6** revolutions in time **T=2π sec**.**Fig.9-13b**

Trajectory **F(6j1t)=f(t)*exp(-6j1t)** as a rotating vector modulated by the function** f(t)**. It will make **3** revolutions in the **first half**-period **T/2=1π/sec** and **0** revolutions in the** second half**-period. The parameter **sc6=0** as the average value of** T=2π sec** is obvious.**Fig.9-13c**

The trajectory of **F(6j1t)** as a circle drawn by the end of the vector in **Fig. 9-13b**.**sc6=0**.**Note****sc6=0** and therefore the harmonic for **ω=1/6sec** does not exist.

**Chapter 9.9 Seventh harmonic of an odd square wave h7(t) =b7*sin(7t).Chapter 9.9.1 Trajectory F(njω0t) of an odd 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.9-14**

Trajectory **F(5j1t)** of an **odd square wave****Fig.9-14a**

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 in the **lower half-plane**. Therefore, its average value as a vector will be **(-1/7π, 0)=-j1/7π.****Fig.9-14c**

The trajectory of **F(7j1t)** as a circle drawn by the end of the vector in **Fig. 9-17b**.

Notice the words “**3.5 turns**” appear. This should convince you that the **F(7j1t)** trajectory stays longer in the **lower half-**plane **Z**. Otherwise, this part of the trajectory is “heavier”.**Chapter 9.9.2 The seventh harmonic against the background of an even square wave, i.e. c0+h7(t)**

Acc. to **Fig. 8-1c** from **chapter 8****c7** is the** complex** amplitude of the **seventh** harmonic**c7=2*sc5=(0,-2/7π) i.e. a7=0 and b7=-2/7π**

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

**Fig. 9-15****co+h7(t)=0.5+(2/7π)*sin(7t)**,

i.e. the **seventh** harmonic with a constant component **c0** against the background of** a square wave**.**Chapter 9.9.3 The seventh approximation of an even square wave, i.e. S7=c0+h1(t)+h3(t)+h5(t)+h7(t).**

**Fig.9-16****S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)=0.5+(2/π)*sin(1t)+(2/3π)*sin(3t) +(2/5π)*sin(5t)+(2/7π)*sin(7t)**

The** seventh** approximation is more similar to a square wave than the** fifth** in **Fig.9-12**

**Chapter 9.10 The eighth harmonic of an even square wave, or rather its lack because c8=0 –>h8(t)=b8*cos(8t)=0.**

**Chapter 9.10.1 Trajectory F(njω0t) of an even square wave for n=8 and ω0=1/sec, i.e. F(8j1t).**

**Fig.9-17**

Trajectory **F(8j1t)** of an **even square wave****Fig.9-17a**

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.9-17b**

Trajectory **F(8j1t)=f(t)*exp(-8j1t)** as a rotating vector modulated by the function **f(t)**. It will make **4** revolutions in the** first half**-period **T/2=1π/sec** and **0** revolutions in the **second half-period**. The parameter** sc8=0** as the average value of** T=2π sec** is obvious.**Fig.9-17c**

The trajectory of **F(8j1t)** as a circle drawn by the end of the vector in **Fig. 9-17b**.**sc8=0****Note****sc8=0** and therefore the harmonic for **ω=1/8sec** does not exist.

**Chapter 9.11 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, the **scn** of** even** harmonics is **zero**. This means that the harmonics **decrease** with **increasing** frequency and for an** infinitely** high frequency **ω**, the harmonic amplitudes are** zero**, i.e. they disappear.

The animation of** Fig.9-16** 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 **square** wave **f(t)** as in **Fig. 9-1**.