# Fourier Series Classic

**Chapter 2 Trigonometric Fourier series**

**Chapter 2.1 Introduction**Let us decompose into a

**Fourier Series**the function

**f(t)**with period

**T=2π/sec**corresponding to the pulsation

**ω=1/sec**. For now, take these formulas on word of honor.

**Fig. 2-1**

Decomposition of the function

**f(t)**on a

**Fourier Series**where:

**a0**-constant component

**1ω**– pulsation of the

**first harmonic**equal to the pulsation of the function

**f(t)**here

**1ω=1/sec**.

**2ω, 3ω…**

**– pulsations**of successive

**harmonics**as

**multiples**of the

**first**

**a1, a2 … –**

**amplitude**s of successive

**cosine harmonics**

**b1, b2 … – amplitudes**of successive

**sine harmonics**

Parameters

**a0, a1, a2 …b1, b2 …an,bn**calculated with the above integrals are specific values

E.g. for a

**odd**square wave (Fig. 2-17)

**a0=a1=a2=…=0**

**b2=b4=b6=…=0**

**b1=4/π=1.273, b3=4/3π, b5=4/5π…etc**

For a this square wave, there are only

**odd**coefficients

**bn**in the figure.

**Even**

**bn**and all

**an**are

**zero**. So there are only

**sines**. This is intuitive because

**sine**and

**square**wave are

**odd**functions. If the

**graph**were shifted

**π2**(90º) to the left, the

**square wave**would be an

**even**function and there would be only

**cosines**.

Most often the

**Fourier Series**is

**infinite**, but not always.

**Example**of sum of

**sine waves**:

**f(t)=1.25sin(1t)+0.25sin(2t)**

This series has only

**2**coefficients

**b1=1.25**and

**b2=0.25**. The others are

**zero**.

The pulsation

**ω**can be of any value. They are often

**ω=1/sec**or

**ω=2π/sec**, and then the

**Fourier Series**has the following form:

**Fig. 2-2**

**Fig. 2-2a ω=1/sec**

**slower**waveform corresponding to period

**T=2π sec ≅6.28**

**sec**

**Fig. 2-2b ω=2π/sec faster**run

**–> T=1 sec**.

**Chapter 2.2 Fourier Series Analysis – First Approach**

I wonder what was on **Jean Baptiste Joseph Fourier’s** mind in **1806**? How did he figure out that any periodical function **f(t)** is a **sum** of **sine** and **cosine**? I don’t know, but it could have been.

The trick is to find the appropriate coefficients **ω,a0,a1,a2…b1,b2…**of the equation** f(t)=…** from **Fig. 2-1**. Let’s approach the matter from the other side. i.e. we know the function **f(t)** and it is the** sum** of **3 sinusoids**. To make it easier, there are no **cosines**! To draw it and then also to calculate it, we will use the amazing program from the Internet **WolframAlpha**. You don’t need to know him. Call only and enter in the appropriate fields what the picture below tells you. In this way, the function **f(t)** in the range** t=-2π…+2π** will be drawn. You will see how simple and useful this program is. I wrote more about the** WolframAlfa** program in the **Rotating Fourier Series** course in **chapter 11.2**. So click **https://www.wolframalpha.com **and do as the picture says. Anyway, you can be less ambitious and stop at just watching ready-made charts.**Fig. 2-3**

The **f(t)** function was calculated and plotted with the command “plot”.

The domain of the function is **t=- ∞…+ ∞** although the graph shows only** t=-2π…+2π**

It is a** sum** of **sinusoids** with **pulsations** and** amplitudes**:**First **harmonic–> **ω=1/sec** (i.e. T=2π sec) with amplitude **b1=1****Second** harmonic–>**2ω=2/sec** with amplitude **b2=0.25****Third harmonic**–>**3ω=3/sec** with amplitude **b3=0.5**

The **first harmonic** corresponds to the pulsation **ω=1/sec** of our studied function** f(t)**, i.e. the period **T=2π sec**.

The next **ω** pulsations are **multiples** of the **first** harmonic. Other parameters, i.e.** b4,b5…a0, a1,a2..**are **zero**. This is an example of a **Fourier Series** with a finite number of **n=3** elements that I mentioned in **Chapter 1**.

Now imagine you only know the graph of **f(t)** but you don’t know the formula. All you know is that there are **3** **sinusoidal** components with unknown **ω1, ω2, ω3** pulsations and unknown** b1, b2, b3** amplitudes .

As for the constant component **a0,** it is evident that **a0=0**. The fields **“over”** and **“under f(t)”** are the same!

The period **T** of the first harmonic** ω1** should be the same as that of **f(t)**, i.e. **T=2π** sec! After all, the **first** harmonic is the first approximation of this **f(t)** function. So **ω1=ω=1/sec**.

What is **ω2** and **ω3**? They should be multiples of** ω=1/se**c, i.e.** ω2=2/sec** and **ω3=3/sec**. Otherwise, the harmonics would not meet at **t=-2π, t=0, t=+2π…** And they must meet for **f(t)=0** at these points. Thanks to this, **f(t)** repeats every **T=2π sec**.

Now let’s move on to the other **parameters**, i.e. to the **amplitudes** of the **b1, b2 b3 ** **sine**? Why not study their various combinations and choose such **b1, b2 **and** b3** at which the **sum** of** sine** waves is closest to the **f(t)** function? After many, maybe even** 100** attempts to draw the graph, we will get something closest to **f(t)** in **Fig. 2-3**. The combination will be, for example, **b1=0.98,** **b2=0.2** and **b3=0.48**. Almost a success. These values are close to the formula. Let me remind you that we only know the graph **f(t)** but without the formula! A primitive, laborious and guesswork-like method. However, it is often used when the creative vema fails. It’s always better than waving a white flag.

**Chapter 2.3 Fourier Series Analysis – Second approach with harmonic extraction****Mr. Fourie**r realized that the “guess-guess” in the previous chapter was only a prelude to solving the problem. How to cleverly extract individual **3** unknown **harmonics**? In fact, only **3** amplitudes **b1, b2** and **b3** because we already know the **1ω**, **2ω** and **3ω** pulsations. And maybe so?**Fig. 2-4**How to extract the

**constant**component,

**first, second**and

**third harmonics**?

**Fig. 2-4a**

Napoleon, who knew mathematics, asked Fourier*.

Dear Jean Baptiste – I thought of some function

**f(t)**consisting of

**3**sine waves.

What’s the function? For simplicity, I will say that its graph is

**Fig. 2-3**and its successive pulsations are

**1ω=1/sec, 2ω=2/sec and 3ω=3/sec**. I emphasize. You don’t know the formula of the

**f(t)**function in

**Fig. 2-3**, you only know its values in the range

**-2π…+2π**, i.e. its

**graph**.

**Fig. 2-4b**

Fourier wrote down this function with

**3**unknown coefficients

**b1, b2**and

**b3**and started thinking.

First let’s find

**b1**.

**Fig. 2-4c**

I’m looking for

**b1**, so let’s multiply both sides of the equation by

**sin(1t)**! Where did it come from? I guess he didn’t know himself. Probably intuition. Looking ahead, if you were looking for

**b2**, for example, multiply by

**sin(2t)**.

**Fig. 2-4d**

Or maybe we could

**integrate**both sides in the period

**T=2π**. Where did

**Fourier’s**idea come from? Probably intuition.

**Fig. 2-4e**

The integral broken down into sums.

**Fig. 2-4f**

Exclusion of

**b1, b2**and

**b3**before the integral.

**Fig. 2-4g**

And now the most important. Two

**integrals**are zeroed and the

**first**one is

**π**!

Explanation in

**chapter 2.6**.

**Fig. 2-4h**

Formula for

**b1**for pulsation

**ω=1**, more precisely

**ω=1/sec**. So for

**T=2π/ω=2π**more precisely

**T=2πsec**

**Fig. 2-4i**

Formula for

**b1**for arbitrary pulsation

**ω**. So for

**T=2π/ω**.

For

**ω=1**, the formula transforms into

**Fig. 2-4h**.

**Fig. 2-4j**

The final formula for

**a0, b1, b2**and

**b3**for

**f(t)**from the plot in

**Fig. 2-3**.

**-a0**is the constant component

**f(t)**which is the arithmetic mean. This is the mean of a function with period

**T**.

**-b1**is described in

**Fig. 2-4a…i**

**-b2**If in

**Fig. 2-4c**we substitute

**sin(1t)**for

**sin(2t)**, we get

**b2**

**-b3**If in

**Fig. 2-4c**we substitute

**sin(1t)**for

**sin(3t)**, we get

**b3**

***Note**

I don’t know if it was exactly like that, but the fact is that the gentlemen knew each other. Napoleon himself was even a good mathematician and is said to be the author of the theorem about the so-called Napoleon Points. Some, probably not fond of the chief, question it.

**Chapter 2.4 Checking the formula****Chapter 2.4.1 Introduction**

After applying the formula **Fig. 2-4j** for the function **f(t)** from **Fig. 2-4a** we get?**a0=0****b1=1****b2=0.25****b3=0.5**

A bit like the riddle “The duck floats on the water, the duck’s name is”. After all, you can see it in **Fig. 2-4a**! Yes, but these coefficients are only for calculating the value of **f(t)**! Only to obtain, I hope, the correct coefficients from the integration formulas **Fig. 2-4j**. In other words, as if we had only **f(t)** graphs at our disposal.**Chapter 2.4.2 Constant component-a0**

Enter the formula for **a0** from Fig. 2-4j when **T=2π** to wolframAlpha.

Call up **https://www.wolframalpha.com** and do what the picture says.**Fig. 2-5**

Calculation of **a0**–constant component

As we expected **a0=0**. This can be seen in **Fig. 2-3**. The “areas above and below” are equal.**Chapter 2.4.3 First Harmonic Amplitude-b1**

Call up **https://www.wolframalpha.com** and do what the picture says.**Fig. 2-6**

Calculation of **b1** – first harmonic amplitude

That’s right, **b1=1**, plus the integral is represented as an area**Chapter 2.4.4 Second Harmonic Amplitude-b2**

Call up **https://www.wolframalpha.com** and do what the picture says.**Fig. 2-7**

Calculation of **b2**– second harmonic amplitude

That’s right, **b2=0.25**. Note that the “top” and “bottom” surfaces subtract.**Chapter 2.4.5 Third Harmonic Amplitude-b3**

Call up **https://www.wolframalpha.com** and do what the picture says.

**Fig. 2-8**

Calculation of **b3**– **third** harmonic amplitude

That’s right, **b3=0.5****Conclusions**

Calculations from **chapters 2.4.2…5** confirmed the truth of the formula **Fig. 2-4j**

**Chapter 2.5 Orthogonal Functions****Chapter 2.5.1**

Go back to** Fig. 2-4g** in **Chapter 2.3** for a moment. There, **two** integrals were **zero** and one had the value of** π**. Now you will find out why?

Because in the** zeros** there were products of **orthogonal functions**, and in the one with **π** there were **products** of **non-orthogonal** functions.**Fig. 2-9**Definition of orthogonal functions

For an

**electrician**, the

**sinusoidal current**and

**voltage**across the

**coil**are

**orthogonal**functions because the energy in period

**T**is

**zero**! For one half of the period, the coil receives energy and for the other half it gives it back.

We will study the

**orthogonality**of various combinations of pairs of

**sin(nωt)**and

**cos(kωt)**functions.

It turns out that:

**Only**the combinations

**sin(nt)*sin(nt)**and

**cos(nt)*cos(nt)**are

**not orthogonal**and their

**integral=π**

**The other**combinations

**sin(nt)*cos(kt)**are

**orthogonal**and their

**integral=0**.

**Chapter 2.5.2 Are sin(t) and cos(t) orthogonal?**

**The Fourier Series**requires the ability to

**integrate**. In order not to go too deep into mathematics, into some antiderivatives and other integration by parts, we will again use the wonderful program

**Wolframalpha**From the

**integration**itself, it is enough to know that the

**integral**is defined by the

**area**

**under the function**.

For example, we will check whether

**sin(t)**and

**cos(t)**are

**orthogonal**.

Click

**https://www.wolframalpha.com**and do as the picture says.

**Fig. 2-10**

How did

**WolframAlpha**check the

**orthogonality**of the functions

**f1(t)=sin(t)**and

**f2(t)=cos(t)**?

It couldn’t be simpler and more intuitive!

In the following sections, we will check the

**orthogonality**of various combinations of

**sin(nωt)**and

**cos(kωt)**.

You will find that the functions

**sin(nωt)**and

**cos(kωt)**are almost always orthogonal except for the pairs

**sin(nωt)**and

**sin(nωt)**or

**cos(nωt)**and

**cos(nωt)**whose integrals are

**T/2**.

**Chapter 2.5.3 The pairs sin(nωt) and cos(kωt) are orthogonal**

Call

**www.wolframalpha.com**and check orthogonality e.g. for

**ω=1, n=2**and

**k=3**i.e. for

**f1(t)=sin(2t)**and

**f2(t)=cos(3t)**.

In the dialog box, instead of

**“sin t cos t”**enter

**“sin 2t cos 3t”**

**Wolfram Allfa**will calculate something like this

**Fig. 2-11**

The pairs

**sin(nω)**and

**cos(kω)**are always orthogonal. Also for

**n=k**.

I leave the generalization to arbitrary pairs of

**sin(nωt)**and

**cos(kωt)**to mathematicians. This remark also applies to the next examined pairs of functions.

**Note:**

Although the periods of

**sin(2t)=2π/ω=2π/2=π**and

**sin(3t)=2π/3**are different, the

**period**of the product

**sin(2t)sin(3t)**is equal to the period of

**sin(t)**, which is

**2π ≈6.28**. This can be seen in

**Fig. 2-11**! Try to prove it.

In general, the period

**T**of

**sin(nωt)*cos(kωt)**is equal to the period

**sin(ωt)**so

**T=2π/ω**.

**Chapter 2.5.4 The pairs sin(nωt) and sin(kωt) and cos(nωt) and cos(kωt) for n≠k are orthogonal.**

I suggest you check the orthogonality of the function pairs

**sin(2t)sin(3t)**and

**cos(2t)cos(3t)**yourself with

**Wolfram Alpha**.

The result should be as follows:

**Fig.2-12**

Orthogonalityof pairs of functions

Orthogonality

**sin(2t)**and

**sin(3t)**as well as

**cos(2t)**and

**cos(3t)**and their generalization.

**Chapter 2.5.5 The pairs sin(nωt) and sin(nωt) and cos(nωt) and cos(nωt) are not orthogonal!!!**

I suggest you check the

**orthogonality**of the function pairs

**sin(2t)sin(2t)**and

**cos(2t)cos(2t)**yourself with

**WolframAlpha**.

The result should be as follows:

**Fig.2-13**

Orthogonality of pairs of functions

**sin(2t)**and

**sin(2t)**as well as

**cos(2t)**and

**cos(2t)**and their generalization.

The functions are

**not**

**orthogonal**and their

**integral = π**for

**ω=1**and generally

**integral = T/2**(half period) for any

**ω**pulsation.

**Chapter 2.6 What are the parameters a0, a1, a2…an,,, b1,b2…bn Fourier series?****Chapter 2.6.1 Introduction**

You’ll see in a moment how **orthogonality** makes it easier to find a **Fourier Series**. By the way. **Orthogonal** or **perpendicular** can be **vectors**, but **functions**? In **Hilbert spaces** they are supposedly the same as in our “human” **spaces**. Here the functions **sin(ωt)** and **cos(ωt)** are associated with **rotating** **vectors** with speed **ω**. The vector **cos(ωt)** leads the vector **sin(ωt)** by 90º and both **vectors** are **perpendicular** to each other. Otherwise, **sin(ωt)** and **cos(ωt)** are** orthogonal**.

**Chapter 2.6.2 Coefficient a0**

This is the average value of the function **f(t)** ie**Fig. 2-14**Coefficient

**a0**as the

**average**value of the function

**f(t)**in period

**T**

**Chapter 2.6.3 Coefficient a2**

This is a bit like

**chapter 2.3**on extracting harmonics Let’s calculate, for example, the coefficient

**a2**.

**Fig. 2-15**

How to calculate

**a2**coefficient?

**Fig. 2-15a**

Write the formula for the

**Fourier Series**with the coefficients not yet known. We already know the coefficient

**a0**.

**Fig. 2-15b**

Multiply both sides of the equation by

**cos(2ωt)**because we are looking for

**a2**

**Fig. 2-15c**

Calculate the

**definite integral**of both sides of the equation in the range

**0…T**

**Fig. 2-15d**

The values of the integrals have been calculated with reference to the relevant figures. The integral at

**a0**is

**zero**, as with

**cosines**.

**Fig. 2-15e**

After including the

**zeros**in

**Fig. 2-15d**

**Fig. 2-15f**

Final

**a2 formula**

**Chapter 2.7 Determination of the Fourier series formula**

We have just determined the coefficient **a2** of the **Fourier Series**. Similarly, we compute any **an** by multiplying both sides of the equation in **Fig. 2-15b** by **cos(nωt)** instead of **cos(2ωt)**. However, if we multiply by **sin(nωt)** we get the coefficient **bn**. Taking into account the formula for **a0** in **Fig. 2-14**, the final formula for the coefficients** a0,an** and **bn** of the **Fourier Series** is as follows:**Fig. 2-16**

**Fourier Series**formula

Note the pulsation

**ω**of the periodic function

**f(t)**and its relation to the period

**T**.

**Chapter 2.8 Determination of the Fourier Series for a Square WaveChapter2.8.1 Fourier series of a square wave directly from the formula**That is, for the time course presented in the time chart.

**Fig 2-17**

Square wave with period

**T=1 sec**corresponding to pulsation

**ω=2π/sec**

Let’s find

**a0, a1,a2,…a7**and

**b1,b2,…b7**. Let’s use the general formula

**Fig. 2-16**. We will use the

**Wolfram Alfa**program to calculate the integrals.

**Fig. 2-18**is the result of calculating the coefficients

**a0, a1**and

**b1**. I will admit to a slight

**correction-fraud**. The program calculated

**a1**as an unimaginably small number, but still different from

**0**, and

**b1**as

**1.27324**, and not as mathematics says

**b1=4/π**. I corrected this imperfection, also for the other coefficients.

**Fig**.

**2-18**

Calculation of

**a0, a1**and

**b1**for a square wave

**Fourier Series**from

**Fig. 2-17**

The remaining coefficients

**a2, a3,a4,a5,a7**and

**b2,b3,b4,b5,b7**are also calculated with WolframAlfa. All you have to do is enter

**4, 6, 8, 10, 12**and

**14**in the dialog box, respectively, in the place of “

**2**in the

**red**border”. The following table should appear.

**Fig. 2-19**

Coefficients

**a0,a2, a3,a4,a5,a7**and

**b1,b2,b3,b4,b5,b7**of the

**Square Wave Fourier Series**of Figure

**2-17**

By the way. The function is

**odd**and therefore all coefficients

**a1, a2, a3**… are

**zero**. If it was

**even**(e.g. shifted to the

**left**by

**0.25**) then the coefficients

**b1, b2, b3**… would be

**zero**while

**a1, a2, a3**…

**non-zero**. This is intuitive.

Let’s make a graph using the program

**http://pl.easima.com**

**Fig. 2-20**

**Square wave**decomposition into

**4 harmonics**.

Actually, the

**first 7**, because

**b2=b4=b6=0**. The pulsation of the first harmonic is

**ω=2π**. The period

**T=1**corresponds to it. The more harmonics, the closer their sum is to a

**square wave**.

**Chapter 2.8.2 Fourier series of a square wave using the special functions of Wolfram Alfa**

Previously, we determined the

**7**harmonics of a

**square wave**using the general

**integration**instructions of the

**WolframAlfa**program. The program also has

**specialized**instructions for determining the

**Fourier Series**. All you have to do is

**enter**them into the

**dialog box**and it will calculate it and make a graph! You don’t even need to know

**integral**s.

Let’s check

**WolframAlfa**for the same

**square wave**. The effect should be

**identical**.

Let’s not go into the details of the

**instruction**typed into the

**dialog box**. Let it suffice to know that it computes

**Fourier Series**parameters for a

**square wave**with the parameters of

**Fig. 2-17**. It will count the first

**7**harmonics. If you wanted to count them, e.g.

**9**, instead of

**7**, just enter

**9**.

**Fig. 2-21**

**Square wave**distribution from

**Fig. 2-17**into

**7 harmonics**

The program will show us a lot of different things. Only the most important ones are shown in

**Fig. 2-21**. The

**Complex Form of Fourier Series**will be covered in

**Chapter 4**. The important thing is that the graph and coefficients are exactly the same as in

**Fig. 2-20**. And how much less work!

Then let’s check this miracle to determine the first

**9**harmonics.

**Fig. 2-22**

Square

**wave distribution**from

**Fig. 2-13**into

**9 harmonics**.

This is a more accurate approximation of a square wave.

Is it, for example,

**17 harmonics**? There should be an even more accurate approximation.

**Fig. 2-23**

Square

**wave distribution**from

**Fig. 2-17**into

**17**

**harmonics**

The chart more accurately represents a

**square wave**. Unfortunately, he did not give us the trigonometric version of the

**Fourier Series**as before. This is not a major drawback, because the relevant parameters can be read from the

**complex form**. When trying to determine

**19**and more harmonics, the

**WolframAlfa**program does not work anymore. I suspect it has some special gates though. Maybe the paid

**WolframAlfa Pro**will do it?

The overshoot at the beginning of each pulse is called the

**Gibs**effect. Appears when

**n**has a finite value and disappears when

**n=∞**.

**Chapter 2.9 Determination of Fourier Series for other periodic functions****Chapter 2.9.1 Function 1sin(t)**

I wonder how **WolframAlfa** will behave with such a trivial function?**Fig. 2-24**

Distribution of** sin(t)** into **5 harmonics**.

The above instruction concerns the decomposition of the periodic function (here **sin(t)**) into **5** harmonics, with the additional assumption that **T=2π**, i.e. **ω=1**.

It is obvious that the** first** harmonic is the same function, i.e. **sin(t)** and the other harmonics do not exist, or otherwise their amplitudes are **zero**. The tacit assumption **T=2π**, i.e. **ω=1** also applies to the following examples.**Chapter 2.9.2 “Ramp;” or function t**Let me remind you that this is a periodic function with

**T=2π**

**Fig. 2-25**

The

**“Ramp”**decomposition, i.e. the function

**t**into the

**5th**and

**9th**harmonics.

**Chapter 2.9.3 The quadratic function, the function t^2Fig. 2-26**

Decomposition of

**a quadratic function**into

**5th**and

**9th**

**harmonics**. Note that in the range

**-π…+π**, the

**9**-harmonic approximation visually does not differ from the

**quadratic**function. By the way, we learned that there is such a thing as a constant component

**a0**. It wasn’t there in the previous examples.