Fourier Series

Chapter 2 Trigonometric Fourier Series
Chapter 2.1 Introduction

Fig. 2-1
More friendly Fourier Series version-without sigma sign when:
1ω- first harmonic pulsation
2ω, 3ω…-
the consecutive  harmonics pulsations
a1, a2 …- the consecutive 
cosine harmonics amplitudes 
b1, b2 …- the consecutive 
sine harmonics amplitudes 
The ω pulsation is any value but for simpler calculations  may be ω=1 or ω=2π and Fourier Series then:

Fig. 2-2
(T=2π) is a slower process and ω=2π (T=1) is a faster one.

Chapter 2.2 Preliminary Fourier Series Analysis
You need to find the appropriate ω,a0,a1,a2…b1,b2…  coeffitients  of the  f(t)=… see Fig 2-1
We assume that we know the f(t) function first:

Fig. 2-3
So the f(t) function is a Fourier Series with the  1ω=1b1=1b2=0.75 b3=3 coeffitients. The other that is a0, a1,a2..and b3,b4…are zeroes. Note that the coeffitients number is finite here!
Let’s draw f(t) by the program I recommend it by the way.

Fig. 2.4
f(t) with its 3 harmonics.
Assume that we know the f(t) diagram only, not the formula. You know that there are 3 sine harmonics but you don’t know the ω1, ω2, ω3 pulsations and the b1, b2 b3 amplitudes.
About a0 constant coeffitient–>a0=0 and this transparently.
The first harmonic period should be the same as f(t) period, that is T=2π! Why? Because the first harmonic is the first f(t) function approximation. Conclusion–>  ω1=ω=1. And what about ω2 and ω3. They should be ω multipiles ω2=2 and ω3=3. Otherwise the f(t=2π)≠0f(t=4π)≠0… !
And what about b1, b2 b3 amplitudes? Let’s draw many  sums with different  b1, b2 b3 combinations and choose this one with the best f(t) approximation.
This method is labour-intensive and primitive some. We use it as a last resort, when we don’t have  better ideas. 

Chapter 2.3 Orthogonal functions
Chapter 2.3.1 Introduction
The Orthogonal Function knowledge is necessary for  Fourier Series analysis.

Fig. 2-5
Orthogonal functions definition
An example for electricians
The A/C coil current and voltage  are the orthogonal functions because the average coil energy in the period is 0. The coil charges  energy during T/2 and gives it back during the next T/2.
The most popular orthogonal functions are sine type*. So we will test different combinations of the sin(nωt) and cos(kωt) pairs. The Wolframalpha program will be very useful here.
The other orthogonal functions (Walsh functions for example) aren’t so important in the science area. 

Chapter 2.3.2 The Wolframalpha program
Fourier Series needs good integration knowledge.  But this is not mathematics course. You should know only, that the definite integral is a surface under function diagram! This integral will compute Wolframalpha instead of you.  Your job is only to write in the special instruction. So you don’t need to be a Wolframalpha world champion!
Are  sin(t) and cos(t) orthogonal for example? 

Fig. 2-6
The sin(t) and cos(t) are orthogonal
It isn’t possible simpler!

Chapter 2.3.3 sin(nωt) and cos(kωt) pairs are orthogonal for any n,k.
Call  and test orthogonality for n=2 and k=3  that is  for f1(t)=sin(2t) and f2(t)=cos(3t).
Write in dialog window  (see Fig. 2-7) “sin 2t cos 3t” instead of “sin t cos t” (as in Fig. 2-5)
Woflfram reaction

Fig. 2-7
sin(nω) and cos(kω) are orthogonal always. For n=k too.
The generalisation for all n,k combinations is in many  mathematics books. 

Chapter 2.3.4 sin(nωt), sin(kω)  or cos(nωt),cos(kωt) pairs are orthogonal for n≠k.
Please test orthogonality sin(2t),sin(3t) and cos(2t),cos(3t):
Woflfram reaction

Fig. 2-8
and cos(nω),cos(kω) are orthogonal for n≠k.

Chapter 2.3.5 sin(nωt), sin(nωt)  or cos(nωt),cos(nωt) pairs aren’t orthogonal!
Please test orthogonality sin(2t)sin(2t) and cos(2t)cos(2t):
Woflfram reaction

Fig. 2-9
and cos(nω),cos(nω) aren’t orthogonal. The inegral=T/2.

Chapter 2.4 What are a0, a1, a2…an,,, b1,b2…bn Fourier Series Parameters?
Chapter 2.4.1 Introduction
Why do we study functions orthogonality so much? You will know in a while.

Chapter 2.4.2 a0 coefficient
This is f(t) function average value.

Fig. 2-10
 a0 as  an  average value

Chapter 2.4.3 a2 coefficient
Let’s compute a2 for example.

Rys. 2-11
How do compute a2 coeffitient?
Fig. 2-11a Write Fourier Series formula. You know a0 coeffitient only now.
Fig. 2-11b Multiply 2 formula sides by cos(2ωt) because we are looking for a2
Fig. 2-11c  Integrate 
2 formula sides in the 0…T range 
Rys. 2-11d Definite integral are computed acc. to the appropriate figures
Rys. 2-11e Included zeroes in Fig. 2-11d
Rys. 2-11f Final a2 Fourier Series Formula

Chapter 2.5 Fourier Series formula
The remaining a1, a3, … an,… are computed analogously.
The remaining b1, b2,…, bn… are computed analogously but there is sin(nωt) instead of cos(nωt).

Fig. 2-12
Fourier Series formula

Chapter 2.6 Fourier Series for square wave

Fig. 2-13
Square wave period T=1 corresponding to ω=2π
Wolfram Alpha (see chapter 2.3.2) will be used for integrals calculation.

Fig. 2-14
a0, a1
and b1 calculation for Fourier Series of the Fig. 2-13 square wave
Please write given instruction in dialog window. There is a little calculation correction. Other say a little and innocent fraud. The Wolfram Alfa calculated a0, a1 as a very, very small number. I “improved” it for number 0. b1 was “improved” likewise that is b1=1.27324 was corrected for b1=4/π as almost the same number.
The other a2, a3,a4,a5,a7 and b2,b3,b4,b5,b7 coeffitients are calculated similarly by Wolfram Alfa. The number “in red border” should be replaced apparently by 4,6,8,10,12 i 14.
The undermentioned table is a effect of the Wolfram calculations

Fig. 2-15
a0,a2, a3,a4,a5,a7 and b1,b2,b3,b4,b5,b7 coeffitients of the Fig. 2-13 square wave.
By the by. All a1, a2, a3… are 0. It’s typical for odd functions. The even functions have b1, b2, b3… =0. The intuition confirms it.
Let’s use a program and draw a diagram of the first 4 square wave harmonics.

Fig. 2-16
4 harmonics of the square wave.
The first harmonic pulsation is ω=2π corresponding to T=1. The sine amplitudes b1, b3, b5 and b7 are from Fig. 2-15. The 5 (and more…) harmonics diagram is more similar to the ideal  square wave of course. 

Chapter 2.8 Square Wave Fourier Series calculated by Wolfram Alpha special instruction
Wolfram Alpha is equipped with the special Fourier Series instructions. You don’ t need to calculate integrals as in the previous chapter.
Let’s test it for square wave from Fig. 2-13. The result should be the same.
The program will calculate the first square wave harmonics. 
Write the given instruction in the dialog window.

Fig. 2-17
The instruction effect
The program will show many different things. There are more crucial here.
Don’t interest yourself “Complex Fourier Series” so long. We will return to this subject in the Chapter 4. The more important are “Diagram” and “First 7 harmonics”.
The conclusion:
The effect is exactly as in the Chapter 2.7. But You are less tired!
Let’s test this marvel for harmonics yet.

Fig. 2-18
9 square wave harmonics
This square wave approximation is better.
Go baldheaded and try with 17 harmonics now!

Fig. 2-19
square wave harmonics
This is the best approximation. Unfortunately, the “First 17 harmonics” window does’nt exist here. It is just as well that you can read adquate data from the “Complex Fourier Series” 
You will fail with 19 square wave harmonics. Why? I don’t know. Maybe we use the free-nonpayable Wolfram Alpha version?

Chapter 2.9 Fourier Series for other periodic functions
Chapter 2.9.1 Function sin(t)
It’s very a simple function. What does Wolfram Alpha say?

Fig. 2-20
harmonics of the sin(t)
The instruction in the dialog window concerns harmonics of the periodic function Fourier Series when  f(t=sin(t). The additional Wolfram Alpha  assumption–> T=2π  i.e. ω=1. It concerns the next subchapters too.
The function has only harmonic–>sin(t)

Chapter 2.9.2 “Saw” that is function t
I remind you that  functionis periodical type with T=2π  i.e. ω=1.

Rys. 2-21
5 and harmonics of the  “Saw” i.e. function t. The harmonics  “Saw” approximation is better of course.

Chapter 2.9.3 Parabola that is function t^2

Fig. 2-22
and 9 harmonics of the Parabola 
Note that harmonics  approximation in the range -π…+π is almost as ideal Parabola.
By the by. There is a0≠0.

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