## Rotating Fourier Series

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

**Chapter 9.1 Introduction
**This is the wave from the previous

**chapter**shifted to the

**right**(ie delayed) by a

**quarter**period (ie by

**90°**or

**π/2**). And just like the

**first**harmonic of the

**previous**one was the

**cosine**, this one will be

**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**in relation to the

**even**one from the previous

**chapter**. Like the

**sine**, it lags

**π/2**relative to the

**cosine**. In the following points we will analyze the trajectories

**F(njω0t)**and the

**time plots**of this

**square wave**for

**n=0…8**.

Description on

**Fig- 8-1 chap. 8**

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

**Fig.9-2
**Trajectory

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

**f(t)**of an

**odd**square

**wave**, i.e. with the following parameters:

**A=1**-amplitude

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

**T=2πsec≈6.28sec**.

**ϕ=-π/2**

**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.**Time plot of

**f(t)**odd square

**wave**.

Note that the run starts at the beginning of the period. Like the

**sine**function. Both animations in

**a**and

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

**a**,

**ω0**,

**ϕ**and

**duty cycle**. 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)**. From them we can read the

**harmonics**for the pulsation

**nω0**.

**Chapter 9.3. First harmonic of odd square wave h1(t)=b1sin (1t).
Chapter 9.3.1 Trajectory F(njω0t) of an odd 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. 9-3**

Trajectory **F(1j1t)** of odd square **wave**

**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 **rotating** vector **modulated** by the **f(t)**. The function **f (t)** is a square **wave** in **Fig. 9-2b**. The** Z** plane in **a** will start to **rotate** at the **ω=-1/sec**. This creates a rotating vector **F(1j1t)=f(t) exp(-1j1t)**. The **Z** plane will do **1** turn, but the radius **R=1** will only do **1** “half turn”.

**c.** Trajectory **F(1j1t)** as a **semicircle** drawn by the vector **b**.

During **1** turn of the **Z plane**, i.e. by the **angle 0 … 2π**, all vectors are vectorially added and their average over the period **T=2π sec** is calculated. It will be a **vector**, otherwise **sc1=(0,-1/π)=0-1j/π=-1j/π**.

**Chapter 9.3.2 The first harmonic on the background of an odd square wave, i.e. c0+h1(t) or the first approximation of an odd square wave.**

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

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

Also **c1** is the complex amplitude of the first harmonic **c1=2sc1=(0,-2/π)=0-j2/π=-j2/π**

that is **a1=0** and **b1=-2/π**

Acc. **Fig. 8-1e**

**h1(t)=(2/π)sin(1t)≈0.637sin (1t)
**

**Fig.9-4
**

**S1(t)=0.5+h1 (t)= (2/π) sin (1t)**that is the

**first harmonic**with a constant

**c0**on the background of a square wave.

It is also the first approximation of our square wave.

**Chapter 9.4 The second harmonic of an odd square wave, or rather its absence, because c2=0 ->h2(t)=b2sin(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 odd **square** wave

**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 the function** f(t)**. Will make** 1** turn in the **first half** cycle **T/2=1π/sec** and **0** turns in the **second half** period. The parameter **sc2=0** as an **average** value of **T=2π sec** is obvious.

**c.** Trajectory **F(2j1t)** as a circle drawn by the the vector** b**.

**Note**

**sc2=0** and therefore the harmonic for** ω=2/sec** doesn’t exist.

**Chapter 9.5.The third harmonic of odd square wave h3(t)=b3sin(3t).
Chapter 9.5.1 Trajectory F(njω0t) of an odd square wave for n=3 and ω0=1/sec, i.e. F (5j1t).
**The

**Z**plane rotates at a speed of

**ω=-3/sec**

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

**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(3j1t)=f(t)exp(-3j1t)** as a **rotating** vector **modulated** by the **f(t)**. Will make **1.5** turns in the **first** half period **T/2=1π sec** and **0** turns in the **second** half period. In the period **T=2π sec**, the radius **R=1** stays longer on the** lower** half-plane than on the **upper** half-plane. 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
**

**c.**Trajectory

**F(3j1t)**as a

**semicircle**drawn by vector

**b**.

The center of gravity

**sc3=(0,-1/3π)**results from the summation of the vectors

**b**and their mean at time

**T=2π sec**, when

**3**rotations of the

**Z**plane are made. The lower semicircle is drawn

**2**times, and the upper one is only

**1**time. As if the bottom semicircle is

**twice**as “heavy”.

**Chapter 9.5.2 The third harmonic on an odd square wave background, i.e. c0+h3(t).
** Acc.

**Fig. 8-1c chapter. 8**

**c3**is the

**complex**amplitude of the

**third**harmonic

**c3 = 2sc3=(0,-2/3π)**so

**a3=0**and

**b3=-2/3π**

Acc.

**Fig. 8-1e chapter. 8**

h3(t)= (2/3π)sin (3t)≈0.212sin (3t)

h3(t)= (2/3π)sin (3t)≈0.212sin (3t)

**Fig. 9-7
**

**co+h3(t)=0.5+(2/3π)*sin(3t)**that is, the

**third**harmonic with a constant c0 on the background of the

**square wave**.

**Chapter 9.5.3 The third approximation of the 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 like a

**square wave**than the

**first**one in

**Fig. 9-4**

**Chapter 9.6 The fourth harmonic of an odd square wave, or rather the lack of it, because c4=0 ->h4 (t)=c4sin(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

**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)**. Will make **2** turns in the **first **half period **T/2=1π/sec **and **0** turns in the** second** half period. The parameter **sc4=0** as an **average** value of **T=2π sec** is obvious.

**c. **Trajectory **F(4j1t)** as a circle drawn by the the vector **b**.

**sc4=0**

**Note**

**sc4=0** and therefore the harmonic for** ω=4/sec** doesn’t.

**Chapter 9.7 The fifth harmonic of odd square wave h5(t)=b5sin (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

**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 **f(t)**. Will make **2.5** turns in the **first** half cycle **T/2=1π/sec** and **0** turns in the** second** half cycle. In the period **T= 2π sec**, the radius **R=1** stays **longer** on the** lower** half-plane than on the** upper** half-plane. Therefore its average value as a vector will be **sc5=(0,-1/5π)=-1j/5π** and not **sc5=(0,0)** as e.g. in **Fig. 9-9c**

**c.** Trajectory** F(5j1t)** as a circle drawn by the vector **b**. The center of gravity **sc5=(0,-1/5π)** results from the summation of the vectors in** b** and their **mean** over the period **T=2π sec**, when **5** turns of the **Z** plane are made. As if the bottom semicircle is “heavier” than the top one.if the bottom semicircle is “heavier” than the top one.

**Chapter 9.7.2. The fifth harmonic on an odd square wave background, i.e. c0+h5(t).**

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

**c5** is the **complex** amplitude of the **fifth harmonic**

**c5=2sc5=(0,-2/5π) so 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)**that is, the

**fifth**harmonic with a constant

**c0**on the background of the

**square wave**.

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

**Fig.9-12
**

**S5(t)=c0+h1(t)+h3(t)=0.5+(2/π)sin(1t)+(2/3π)sin(3t)+(2/5π)sin(5t)**

The

**fifth**approximation is more like a

**square wave**than the

**third**one in

**Fig. 9-8**

**Chapter 9.8 The sixth harmonic of an odd square wave, or rather the lack of it, because c6=0 ->h6(t)=c6sin(6t)=0.
Chapter 9.8.1 Trajectory F(njω0t) of an odd square wave for n=6 and ω0=1/sec, i.e. F6j1t).**

**The**

**Z**plane rotates at a speed of

**ω=-6/sec**

**Fig. 9-13**

Trajectory **F(6j1t)** of an **odd** 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)**.

Will make **3** turns in the first half cycle **T/2=1π/sec** and **0** turns in the **second** half cycle. The parameter **sc=0** as an average value of **T=2π sec** is obvious.

**c.** Trajectory **F(6j1t)** as a circle drawn by the vector **b**.

**sc6= 0
Conclusion**

**sc6=0**and therefore the harmonic for

**ω=1/6sec**doesn’t exist.

**Chapter 9.9 The seventh harmonic of odd square wave h7(t)=b7sin (5t).
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(7j1t)** of an** odd** 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 **f(t)**.

At the 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**.

**c. **Trajectory** F(7j1t)** as a circle drawn by vector from **b**.

Notice the word “**3.5 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 9.9.2. The seventh harmonic on an odd square wave background, i.e. c0 + h7 (t).**

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

**c7** is the **complex** amplitude of the **seventh **harmonic

**c7=2sc7=(0,-2/7π) so 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)**that is, the

**seventh**harmonic with a constant

**c0**on the background of the

**square wave**.

**Chapter 9.9.5 The seventh approximation of the odd 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 like a

**square wave**than the

**fifth**one in

**Fig. 9-12**

**Chapter 9.10 The eighth harmonic of an odd square wave, or rather the lack of it, because c8=0 ->h8(t)=c8sin(8t)=0.
Chapter 9.10.1 Trajectory F(njω0t) of an odd 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. 9-17**

Trajectory **F(8j1t)** of an **odd** 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)**.

Will make **4** turns in the first half cycle **T/2=1π/sec** and **0** turns in the **second** half cycle. The parameter **sc8=0** as an average value of **T=2π sec** is obvious.

**c.** Trajectory **F(8j1t)** as a circle drawn by the vector **b**.

**sc8= 0
Conclusion**

**sc8=0**and therefore the harmonic for

**ω=1/8sec**doesn’t exist.

**Chapter 9.11 Remaining harmonics of an odd 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 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 **ideal** square wave** f(t)** in **Fig. 9-1**.