**Rotating Fourier Series**

**Chapter 10. Fourier Series of a square wave with a shift -30º**

**Chapter 10.1 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°** relative to the even wave from **Chapter 8**. Therefore, it is not an **even** or **odd** function. So bland. The **even** wave had only** cosine** harmonics **an**, i.e.** an≠0 i bn=0** components. In turn, the **odd** one has only **sinusoidal** components **an=0** i **bn≠0**, i.e. components **an=0**. We now expect** cosine** and **sine** components to occur i.e. **an≠0 **i **bn≠0**.

**Chapter 10.2 Constant component, i.e. c0=0.5.**The constant component, as an average value in the period

**T=2π sec**, is

**c0=0.5**. It does not depend on the shift phase

**ϕ**.

**Chapter 10.2.1 Trajectory F(nj1t) for n=0 and ω0=1/sec, i.e. F(0j1t)**.

**Fig.10-2**

The trajectory **F(0j1)** of the function **f(t)** of a **square wave** with a shift of **-30°**, i.e. with the parameters:**A=1-amplitude****ω0=1/sec**-pulsation corresponding to the period **T=2πsec≈6.28sec**.**ϕ=-30°=-π/6****50%**-filling**Fig.10-2a**

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.10-2b**.**Fig.10-2b**

Time chart of **f(t)** of a **square wave** with **ϕ=-30°=-π/6**

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 10.3 First harmonic i.e. h1(t).****Chapter 10.3.1 Trajectory F(nj1t) 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 analogous **Fig.8-3** from **chapter 8** for an **even square wave**. There, the vector **sc1** had a shift **ϕ=0°** and was presented as a **complex** number in the algebraic form **sc1=(1/π,0)** or in other words **1/π+j0**. It could just as well have been the exponential form** sc1=1/π*exp(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 **centroid** of the trajectory in **Fig. 10-3c** is **sc1=|1/π|*exp(-j30°)**. Here **|1/π|** this is the so-called **sc1** module

You can easily switch from the exponential version to the algebraic version and vice versa:**sc1=1/π*exp(-j30°)=1/π*[cos(-30°)+jsin(-30°)]≈0.275-j0.159**

The** sc1** vector is the average value of the rotating vector from **Fig. 10-3b** in the period **T=2π sec**, calculated from the formula **Fig. 7-2b**, **chapter 7**.

Revenons à nos moutons, so let’s get back to the topic.**Fig.10-3a**

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 makes **1** revolution.**Fig.10-3b**

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. 10-2b**.

The **Z** plane in **Fig. 10-2a** will start rotating at a speed of **ω=-1/sec**. A rotating vector **F(1j1t)=f(t)*exp(-1j1t)** will be created. The **Z** plane will make **1** revolution, but the trajectory as a rotating vector **F(1j1t)=f(t)*exp(-1j1t)** will only make **1** half-turn.**Fig.10-3c**

The trajectory **F(1j1t)** as a semicircle drawn by the end of the vector in **Fig. 10-3b**.

During **1** rotation of the **Z** plane by an angle of** 0…2π**, all vectors are vector summed and their average in the period **T=2π sec** is calculated according to the formula **Fig. 7-2b** **chapter 7**. This is the vector **sc1=(1/π)*exp(-j30°)**. The trajectory is rotated relative to that in** Fig. 8-3c** **chapter 8** by an angle **ϕ = -30°**.

Note to **Fig. 10-3b,c**

The above animations suggest that the **upper** and **lower** parts of the trajectories are separated by a gap **T=π sec**. 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 10.3.2 The first harmonic in the background of a square wave i.e. c0+h1(t) or in other words, the first approximation of a square wave.**

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

Acc. to **Fig. 10-3c****sc1=|sc1|*exp(jϕ)=(1/π)*exp(-j30°)****c1=2*sc1=(2/π)*exp(-j30°) i.e. |c1|=2/π ϕ=-30°**

Acc. to **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)**

i.e. the **first** harmonic with a constant component **c0=0.5** against the background of a **square wave**.

This is a **first** approximation to our **square wave**.

**Chapter 10.4 Second harmonic or rather its absence because c2=0 –>h2(t)=0.Section 10.4.1 Trajectory F(njω0t) 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°****Fig.10-5a**

The radius **R=1** as a vector **(1,0)** rotates with a speed **ω=-2/sec** around the point **(0,0)** and will make **2** revolutions in time** T=2π sec**.**Fig.10-5b**

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 make only **1** revolution, consisting of the **first** revolution of** 240°** and the second revolution of **120°** with a break of **1π sec** between them.**Fig.10-5c**

The trajectory of **F(2j1t)** as a circle drawn by the end of the vector in **Fig. 10-5b**.

Obviously **sc2=0**. Note that from just the circle, the value of** sc2=0** is not obvious! After all, the circle could have been drawn, for example,** 1.5 times**. Fortunately, this **1** full revolution, although with a break in the middle, is visible at **10-5b**.**Note****sc2=0** and therefore the harmonic for **ω=2/sec** does not exist.

**Chapter 10.5 Third harmonic i.e. h3(t).****Chapter 10.5.1 Trajectory F(njω0t) for n=3 and ω0=1/sec i.e. F(3j1t).**

The **Z** plane rotates at a speed of **ω=-3/sec**

**Fig.10-6**

Trajectory **F(3j1t)** of a square wave with a shift of **-30°****Fig.10-6a**

The radius **R=1** as a vector **(1,0)** rotates at a speed of **ω=-3/sec** around the point **(0,0)** and makes **3** revolutions in time **T=2π sec**.**Fig.10-6b**

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

In time **2π sec**, the **Z** plane will make** 3** revolutions, but the radius **R=1** will make only **1.5** revolutions, consisting of a first **360°** revolution and a second **180°** revolution with a break of **1π sec** between them. Radius **R=1** stays longer in the **upper** half-plane than in the **lower** one. Therefore, the **average** value of the vector is **sc3=(0.1/3π)=1j/3π****Fig.10-6c**

The trajectory of** F(3j1t)** as a circle drawn through the end of the vector in **Fig. 10-6b**.

Centroid **sc3=( 0.1/3π)=1j/3π**.**Chapter 10.5.2 Third harmonic against a square wave i.e. c0+h3(t).**

Acc. to** Fig. 10-3c****sc3=|sc3|*exp(jϕ)=(1/3π)*exp(j90°)****c1=2*sc1=(2/3π)*exp(j90°) i.e. |c1|=2/3π ϕ=90°**

Acc. to** Fig. 8-1f, chapter 8****h3(t)=(2/3π)*cos(3t+90°)≈-0.212*sin(3t)**

**Fig.10-7****S3(t)=c0+h3(t)**

i.e. the third harmonic with a constant component **c0=a0=0.5** against the background of a **square wave**.**Chapter 10.5.3 Third wave approximation i.e. S3=c0+h1(t)+h3(t).**

**Fig.10-8****S3(t)=c0+h1(t)+h3(t)**

The **third** approximation is more similar to a **square wave** than the **first** one in **Fig. 10-4**

**Chapter 10.6 The fourth harmonic or rather its absence because c4=0 –>h4(t)=0.**

**Chapter 10.6.1 Trajectory F(njω0t) 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°****Fig.10-9a**

The radius** R=1** as a vector **(1,0)** rotates at a speed of **ω=-4/sec** around the point **(0,0)** and will make** 4** revolutions in time **T=2π sec**.**Fig.10-9b**

Trajectory **F(4j1t)=f(t)*exp(-4j1t)** as a rotating vector modulated by the function **f(t)**.

In **2π sec** the** Z** plane will make 4 revolutions, but the radius** R=1** will make **2** full revolutions consisting of the** first** **480°** and the **second 240°** with a break of **1π sec** between them.**Fig.10-9c**

The trajectory of **F(2j1t)** as a circle drawn by the end of the vector in **Fig. 10-9b**.

Obviously **sc4=0**.**Note****sc4=0** and therefore the harmonic for** ω=2/sec** does not exist.

**Chapter 10.7 Fifth harmonic i.e. h5(t).****Chapter 10.7.1 Trajectory F(njω0t) for n=5 and ω0=1/sec, i.e. F(5j1t).**

The **Z** plane rotates at a speed of** ω=-5/sec**

**Fig. 10-10**

Trajectory **F(5j1t)** of a **square wave** with a shift of **-30°****Fig.10-10a**

The radius** R=1** as a vector **(1,0)** rotates at a speed of **ω=-5/sec** around the point **(0,0)** and will make** 5** revolutions in time **T=2π sec**.**Fig.10-10b**

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

In** 2π sec**, the **Z** plane will make **5** revolutions, but the radius** R=1** will make only **2.5** revolutions (**900°**), including the first **585°** and the second **315°** with a break of** 1π sec** 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°)****Fig.10-10c**

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

Centroid **sc5=(1/5π)*exp(-j150°)**.**Chapter 10.7.2 Fifth harmonic against a square wave i.e. c0+h5(t).**Acc. to

**Fig. 10-5c**

**sc5=|sc5|*exp(jϕ)=(1/5π)*exp(-j150°).**

**c5=2*sc5=(2/5π)*exp(-j150°) i.e. |c1|=2/5π ϕ=-150°**

Acc. to

**Fig. 8-1f, chapter 8**

**h5(t)=(2/5π)*cos(5t-150°)≈-0.127*cos(5t+30°)**

**Fig. 10-11****S5(t)=c0+h5(t)**

i.e. the **fifth** harmonic with a constant component **c0=a0=0.5** against the background of a **square wave**.**Chapter 10.7.3 Fifth wave approximation ****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 similar to a **square wave** than the** third** in **Fig. 10-8**

**Chapter 10.8 The sixth harmonic or rather its absence because c6=0 –>h6(t)=0.****Section 10.8.1 Trajectory F(njω0t) 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°****Fig.10-13a**

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.10-13b**

Trajectory **F(6j1t)=f(t)*exp(-6j1t)** as a rotating vector modulated by the function **f(t)**.

In a time of** 2π sec**, the **Z** plane will make** 6** revolutions, but the radius** R=1** will make a full **3** revolutions consisting of the first** 720°** and the second** 360°** with a break of** 1π sec** between them.**Fig.10-13c**

The trajectory of **F(6j1t)** as a circle drawn by the end of the vector in **Fig. 10-13b**.

Obviously** sc6=0**.**Note****sc6=0** and therefore the harmonic for** ω=6/sec** does not exist.

**Chapter 10.9 The seventh harmonic i.e. h7(t).****Section 10.9.1 Trajectory F(njω0t) 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 it.

**Fig. 10-14**

Trajectory **F(7j1t)** of a square wave with a shift of **-30°****Fig.10-14a**

The radius **R=1** as a vector **(1,0)** rotates at a speed of** ω=-7/sec** around the point **(0,0)** and completes **7** revolutions in time **T=2π sec**.**Fig.10-14b**

Trajectory **F(7j1t)=f(t)*exp(-7j1t)** as a rotating vector modulated by the function** f(t)**.

In time** 2π sec**, the **Z** plane will make** 7** revolutions **(2520°)**, but the radius **R=1** will make only **3.5** revolutions (1260°)**,** including** 2** and **1/3** revolutions (840°) in the** first** cycle and** 1** and **1/6** in the** second** cycle rotation (420°) with a **1π sec** break between them. 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°)****Fig.10-10c**

Trajectory **F(7j1t)** as a circle drawn by the end of the vector in **Fig. 14b**.

Centroid **sc7=(1/7π)*exp(-j30°)**.**Chapter 10.9.2 The seventh harmonic against a square wave i.e. c0+h7(t).**

Acc. to **Fig. 10-14c****sc7=|sc7|*exp(jϕ)=(1/7π)*exp(-j30°).****c7=2*sc7=(2/7π)*exp(-j30°) i.e. |c7|=2/7π ϕ=-30°**

Acc. to** Fig. 8-1f**, **chapter 8****h7(t)=(2/7π)*cos(7t-30°)≈-0.127*cos(7t-30°)**

**Fig.10-15****S7(t)=c0+h7(t)**

i.e. the **seventh** harmonic with a constant component **c0=a0=0.5** against the background of** a** square wave.**Chapter 10.9.3 Seventh wave approximation i.e. S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)**

**Fig. 10-16****S7(t)=c0+h1(t)+h3(t)+h5(t)+h7(t)**

The **seventh** approximation is more similar to a **square wave** than the **fifth** one in **Fig. 10-12**

**Chapter 10.10 Eighth harmonic of a square wave or rather its absence because c8=0 –>h8(t)=0.**

**Chapter 10.10.1 Trajectory F(njω0t) for n=8 and ω0=1/sec, i.e. F(8j1t).**

**Fig.10-17**

Trajectory **F(8j1t)** of a square wave with a shift of **-30°****Fig.10-17a**

The radius **R=1** as a vector **(1,0)** rotates at a speed **ω=-8/sec** around the point **(0,0)** and will complete **8** revolutions in time **T=2π sec**.**Fig.10-17b**

Trajectory **F(8j1t)=f(t)*exp(-8j1t)** as a rotating vector modulated by the function **f(t)**.

In time **2π sec**, the **Z** plane will make **8** revolutions, but the radius** R=1** will make the full **4** revolutions only in the **first** stage.**Fig.10-17c**

The trajectory of **F(8j1t)** as a circle drawn by the end of the vector in **Fig. 10-17b**.

Obviously **sc8=0**.**Note****sc8=0** and therefore the harmonic for **ω=8/sec** does not exist.