# Fourier Series Classic

**Chapter 3 Complex Fourier Series**

**Chapter 3 Complex Fourier Series**

**Chapter 3.1 The complex function exp(jωt) as a rotating vector**In automatics, electrical engineering and many other fields, the complex function

**exp(jωt)**appears all the time. This is a special case of a

**complex function**whose domain is a real number time

**t**, not any complex number

**z.**

So every real number

**t**is assigned a complex number

**z=exp(jωt)**–>

**Note**.

**Fig. 3-1**The complex function

**z=1exp(jωt)**as a spinning vector where

**ω=2.62/sec**. author

**Chetvorno**.

On the

**right**you will see a

**complex z**plane with coordinates

**x,y**which treat as the

**real**coordinate

**Re z**and the

**imaginary**coordinate

**Im z**. The point

**z**rotates with a period of

**T=2.4sec**, i.e. with an angular speed of

**ω=2.62/sec**. Check it with your watch, e.g. by measuring the time of

**10**revolutions. The

**position**of the rotating

**z**point at timet

**t**is precisely the point

**z=1exp(jωt)**. For example, for

**ωt=0 sec**, the position

**z=1+j0=1.**

You can also calculate the positions using

**Euler’s**formula:

**exp(jωt)=cos(ωt)+jsin(ωt)**

For example, for

**ωt=0**and

**ωt=π/2=90º**, these are the points in the figure marked with an arrow:

**z=exp(j0)=1+j0**

**z=exp(jπ/2)=0+j1**

The larger

**ω**, the

**faster**the black point

**rotates**.

**On the left**you will see the movement of the

**red**dot, which is the projection of the rotating

**black**point on the

**y-axis**. harmonic motion described by the function

**y=sin(ωt)**.

**Note**

In the general case, the domain of a

**complex**function is any

**complex**number

**z**, the

**two-dimensional**complex plane, and the

**anti-domain**is also the

**two-dimensional**complex plane. So we enter a more difficult to imagine

**four-dimensional**space. Fortunately, in engineering, the domain of a

**complex function**is most often

**one-dimensional**time

**t**.

More about

**complex numbers**and

**functions**in the course „

**Complex Numbers**”

**Chapter 3.2 Complex version**** of ****a*sin(ωt)+b*cos(ωt) **

**Complex numbers**make calculations with trigonometric functions easier.

**Examples**

**1.**The trigonometric formula for

**a*sin(ωt)+b*cos(ωt)**is very complicated. In the

**complex version**, it is simple

**addition**.

**2**. Differentiated and integrated in the complex version is only

**+/-90º**rotation.

**Chapter 3.2.1 The complex function c*exp(jωt) as a rotating vector**

**Fig. 3-2**

c*exp(jωt)as a rotating vector

c*exp(jωt)

We already know that

**exp(jωt)**is a rotating vector of length

**1**. It is a

**blue**vector in the

**initial**state for

**ωt=0**and after time

**t**, here e.g. corresponding to

**ωt=π/3=60º**. It is obvious that

**2*exp(jωt)**will be a vector of length

**2**, and

**3*exp(jωt)**of length

**3**, etc… What about the expression

**c*exp(jωt)**, when

**c**is a complex number, e.g.

**c=1.8+0.4j**? Vector

**c*exp(jωt)**will also rotate. The

**initia**l state is just

**c**as

**c*exp(jωt)**for

**ωt=0**, and after a time corresponding to

**ωt=π/3=60º**, the vector will also rotate by

**60º**as shown in the figure.

**Chapter 3.2.2 Components a1cos(1ωt) and b1sin(1ωt) as rotating vectors**

Consider the components

**a1cos(1ωt)**and

**b1sin(1ωt)**in

**Fig. 3-3**

They will facilitate the transition from the

**Trigonometric Fourier Series**to the

**Complex Fourier Series**.

**Fig.3-3**

The component

**a0**is a constant and

**a1,b1,a2,b2**… are the

**cosine/sine**

**amplitudes**. The

**1ω**pulsation corresponds to the period

**T**of

**f(t)**. If the parameters

**a0,a1,b1,a2,b2…**are arbitrary then we have the

**formula**for the

**sum**of

**cosine/sine**waves with pulsations

**1ω, 2ω, 3ω…**and amplitudes

**a1,a2…b1,b2…**If the coefficients are given in

**Fig’ 2-12**in

**Chapter 2.2**, then

**f(t)**is a

**Trigonometric Fourier Series**. Let’s assume that

**a1=1**and

**b1=0.7**and the pulsation, e.g.

**ω=2.62/sec**, was previously measured with a stopwatch. So we are looking for

**rotating vectors**representing the time chart

**1cos(2.62t)+0.7sin(2.62t)**. Vectors can also be treated here as

**complex numbers**.

**Fig.3-4**

**a1cos(1ωt)+b1sin(1ωt)**as a projection of the sum of rotating vectors

**(a1-jb1)exp(jωt)**to the real

**axis**

**Re z.**

**Fig. 3-4a**

Rotating vectors in the initial state, i.e. for

**.**

**1ωt=0****see**

a1

-jb1

sum a1-jb1–>a1

-jb1

sum a1-jb1–>

***Note**

Fig. 3-4bFig. 3-4

Rotating vectors and their sum over time

**t**. Here t is such that 1ωt=π/3=60º.

After time t, the rotating vectors are:

**a1*exp(jωt)**

**-jb1*exp(1jωt)**

**(a1-jb1)*exp(j1ωt)**

**The most important conclusion**

The expression

**a1cos(1ωt)+b1sin(1ωt)**in

**Fig 3-1a**is the projection of a rotating vecto

**(a1-jb1)*exp(**on the real axis

**exp(j1ωt)****Re z**!

Or otherwise

**a1cos(1ωt)**+**b1sin**=

**(1ωt)****Re z{**

**(a1-jb1)***

**exp(j1ωt)}****The proof of this is the school trigonometry in**

**Fig. 3-4b**where the lengths of the rotating vectors

**blue**and

**green.**

**are**

**a1=**

**1**and

**b1=0.7.**

The following waveforms correspond to the rotating vectors.

**Fig. 3-5**

Time charts corresponding to the rotating vectors in

**Fig. 3-4**.

The time chart of

**0.7cos(1ωt)**is

**90°**behind

**1cos(1ωt)**and their sum

**1.22cos(1ωt-35°)**is behind

**35°**. Compare with the rotating vectors in

**Fig. 3-4**.

***Note**

This number can also be represented as a modulus

**|c1|≈1.221**and a phase

**φ≈-35º**or equivalently as Euler

**1.221*exp(-j35º)**.

Modulus

**1.221**is Pythagoras with

**1**and

**0.7**and

**φ1=arctg(-0.7/1)≈-35º**. Check with a calculator.

**Chapter 3.3 Complex Fourier Series for 3 harmonics****Chapter 3.3.1 Real axis projection version**.

We will start with a **Fourier Series** in which **f(t)** has only **3** harmonics. This will be easier:**f(t)=a1cos(1ωt)+b1sin(1ωt)+a2cos(2ωt)+b2sin(2ωt)+a3cos(3ωt)+b3sin(3ωt)**

Doubts may arise here.**Firstly**

Why break into a **Fourier Series** a function **f(t)** which is itself a** Fourier Series**? The coefficients **a0, a1,b1,a2,b2** and** a3,b3** can be seen in** f(t)** and you no longer need to use the formulas in **Fig. 2-12** from the previous chapter. Yes, but we do it only for teaching reasons. Anyway, if it bothers you, treat **f(t)** like any other function and plug it into the formula in **Fig. 2-12**. You get the same function **f(t)**!**Secondly**

Why are there **6** components here and not **3** as in the title? – Because each single harmonic has a **cosine** and** sine** component.

In a specific example, it looks like this:**f(t)=1cos(1ωt)+0.2sin(1ωt)+0.6cos(2ωt)+0.4sin(2ωt)+0.4cos(3ωt)+0.4sin(3ωt)**

That is**a1=1 b1=0.2****a2=0.6 b2=0.4****a3=0.4 b3=0.4**

Other coefficients **a0,a4,b4,a5,b5…=0**The basic pulsation

**ω**can be arbitrary, but if you like to be specific, assume

**ω=2.6.2/sec**as in the animation in

**Fig. 3-1**.

The coefficients

**c1, c2, c3**can be assigned vectors-complex numbers:

**c1=a1-jb1=1-0.2j**

**c2=a2-jb2=0.6-0.4j**

**c3=a3-jb3=0.4-0.4j**

And the function

**f(t)**rotating vectors-complex numbers

**c1*exp(j1ωt)**

**c2*exp(j2ωt)**

**c3*exp(j3ωt)**

**Fig. 3-4**shows:

**a1cos(1ωt)+b1sin(1ωt)=Re z {(a1-jb1)*exp(j1ωt)}**—>the real part of the rotating vector

**(a1-jb1)*exp(j1ωt)**

Similarly for

**2ωt**,

**3ωt**

**a2cos(2ωt)+b2sin(2ωt)=Re z {(a2-jb2)*exp(j2ωt)}**

**a3cos(3ωt)+b3sin(3ωt)=Re z {(a3-jb3)*exp(j3ωt)}**

So the time function

**f(t)**is the

**real part**of the sum of rotating vectors

**c1*exp(j1ωt)+c2*expj2ωt)+c3*exp(j3ωt)**

where

**c1=a1-jb1, c2=a2-jb2, c3=a3-jb3**

Fig. 3-6

Fig. 3-6

**Fig. 3-6a**Rotating vectors corresponding to the function

**f(t)**

**Fig. 3-6b**The function

**f(t)**as the result of the sum of harmonics

**c1, c2, c3**are the rotating vectors

**c1*exp(j1ωt), c2*exp(j2ωt) and c3*exp(j3ωt)**in the initial state

**t=0**, and after time

**t**corresponding to the phases

**1ωt=30º**,

**2ωt=60º and 3ωt =90º**. It can be seen that after the same time vector

**c3**, which was the most “lagged”, rotated the most. No wonder, since it has the highest rotational speed of

**3ωt**. Notice that the rotating vectors

**c1*exp(j1ωt), c2*exp(j2ωt)**and

**c3*exp(j3ωt)**at

**t=0**have the values

**c1=1-0.2j**,

**c2=0.6-0.4j**and

**c3=0.4-0.4j**. It is enough to substitute

**t=0**into the formulas.

The

**sum**of the projections of rotating vectors on the

**Re z axis**is just a function of

**f(t)**

That is:

**f(t)=Re {c1*exp(j1ωt)+c2*exp(j2ωt)+c3*exp(j3ωt)}**

Two notess

**1.**After period

**T**, the rotating vectors will return to their initial state

**c1,c2,c3**. Only vector

**blue**will do

**1**,

**green**will do

**2**and

**red**

**3**turns.

**2.**The

**c1*exp(j1ωt) + c2*exp(j2ωt) + c3*exp(j3ωt)**is a complex periodic function with period

**T=2π/ω**.

**3.**Rotating vectors resemble a solar system with the

**Sun**at

**(0,0)**and orbits:

**c1*exp(j1ωt)**–

**Mars**orbit-largest

**c2*exp(j2ωt)**–

**Earth**orbit

**c3*exp(j3ωt)**–

**Venus**-smallest

**Mars**has the

**lowest**angular

**velocity**and

**Venus**the

**highest**, and in this the

**rotating**vectors are similar to

**planets**.

**Fig. 3-6**shows the individual rotating vectors, but you can’t see their

**sum**!

The version from

**Fig. 3-7**, in which the sum of the rotating vectors is the

**end**of the vector

**c3*exp(j3ωt)**, does not have this disadvantage. This is a more convenient method of adding vectors than the previous “parallelogram” method!

**Fig. 3-7**

Sumof rotating vectors

Sum

**c1*exp(j1ωt) + c2*exp(j2ωt) + c3*exp(j3ωt)**

The first

**blue**vector rotates around

**(0,0)**,

**green**around the

**blue**end, and

**red**around the

**green**end. Subsequent linear speeds are increasing. It’s like the end of a shooting whip breaking the sound barrier.

The lower vector is the initial state at

**t=0**, and the upper one is the state after time

**t**corresponding to the phase

**1ωt=30º**of the

**blue**vector or, which is the same for the phase

**2ωt=60º**

**green**and

**3ωt=90º**

**red**.

After period

**T**when:

**–blue**will make

**1**turn

**–green**

**2**turns

**–red 3**turns

will be the

**initial**state again!

Coming back to astronomy this time:

**-blue**is the

**Earth**orbiting the

**Sun(0,0)**

**-green**is the

**Moon**orbiting the

**Earth**

**-red**is an artificial

**satellite**orbiting the

**Moon**.

The

**rotating**vectors

**c1*exp(j1ωt), c2*exp(j2ωt)**and

**c3*exp(j3ωt)**in

**Fig. 3-7**are accompanied by their projections on the

**Re z**axis, that is:

**a1cos(1ωt)+b1sin(1ωt)**

**a2cos(2ωt)+b2sin(2ωt)**

**a3cos(3ωt)+b3sin(3ωt)**

Therefore, it is obvious that:

**f(t)=Re {(a1-jb1)*exp(1jωt)+(a2-jb2)*exp(2jωt)+(a3-jb3)*exp(3jωt)}**

That is

**f(t)=Re {c1*exp(1jωt)+c2*exp(2jωt)+c3*exp(3jωt)}**

**Summary**

Fig. 3-8

Fig. 3-8

Complex Fourier Series for

**f(t)**with

**3**harmonics

**Fig. 3-8a**

General formula when

**f(t)**has

**3**arbitrary harmonics

**Fig. 3-8b**

Special formula when

**f(t)**has

**3**specific harmonics

**Chapter 3.3.2 Counter-rotating vector version, otherwise “without Re z” version**

Look again at

**Fig. 3-7**. It represents the

**sum**of the rotating vectors

**c1*exp(j1ωt)+ c2*exp(j2ωt)+c3*exp(j3ωt)**and its projection on the

**Re z**axis, here for phase

**1ωt=30º**. So

**c1=1-0.2j**rotated

**30º**. This corresponds to the equation in

**Fig. 3-9a**.

**Fig. 3-9****2** equivalent formulas for** f(t)** with **3** harmonics.

If in the formula **Fig.3-9a**:**1.** We will replace the rotating** blue** vector**– half** of it–>(**0.5-0.1j)exp(j1ωt**)**– “conjugate half”**—> **(0.5+0.1j)exp(-j1ωt)** which rotates in the opposite direction because it is **-j1ωt**!**2.** We will do the same with the **green** and **red** vectors.

This gives us the equivalent formula **Fig. 3-9b****Note**, that although the new formula is longer, it immediately yields a** real number**!**Figure 3-9c** is a generalization of **Fig. 3-9b**. Just remember that **c1, c2,** and **c3** are the halves of **c1, c2,** and **c3 **in **Fig. 3-8a**!**I emphasize**

Pairs of vectors of the same color are** conjugate** vectors rotating in opposite directions!!!

The interpretation of the formulas **Fig.3-9b** and **Fig.3-9c** is **Fig.3-10** with **3** pairs of oppositely rotating **conjugate** vectors.

**Fig. 3-10****f(t)** as the sum of counter-rotating pairs of vectors**A** – general formula for any pairs of vectors** c(-3), c(-2), c(-1), c(1), c(2)** and **c(3)****B –** special formula for specific pairs of vectors

At first glance you can see that:**1. The bottom** rotating vectors are the **halves** of the **bottom** rotating vectors in **Fig.3-7** for **t=0**.**2. The upper** and** lower** rotating vectors are **conjugate** numbers (“mirror image”)**3. The upper** and **lower** vectors rotate in opposite directions and their sum is:**f(t)=a1cos(1ωt)+b1sin(1ωt)+a2cos(2ωt)+ b2sin(2ωt)+a3cos(3ωt)+ b3sin(3ωt)****Explanation** for **p.3**

Formula **B** is pairs of vectors rotating in **opposite** directions with complex coefficients** c(-1), c1, c(-2),c2** and** c(-3),c3** “photographed” in **Fig. 3-10** at time **t=0**.

For example, the pair **(0.5-j0.1)exp(j1ωt)** and **(0.5+j0.1)exp(-j1ωt)** gives the sum = the **blue** vector on the **Re z** axis. It is equal to the **blue** horizontal vector on the **Re z** axis in **Fig. 3-10** is **1cos(1ωt)+0.2sin(1ωt)**. This vector will move according to this formula only along the **Re z** axis.

It is analogous with the rotating pair of **green** and **red**.

The sum of** all** the rotating vectors will give us a** particular** pattern **B** and a **general** pattern **A.**

**Chapter 3.4 Complex Fourier Series – general formula with real part****Chapter 3.4.1 Basics****And now** the most important**1. **Take any periodic function **f(t)** with period** T****2.** Calculate the fundamental pulsation **ω=2π/T****3.** Calculate the coefficients **a0,a1,b1,a2,b2,…an,bn…** and then the **real** coefficient** c0** and the** complex** coefficient **c1,c2,c3…cn…Fig. 3-11**Formulas for Fourier coefficients

Now we can determine the final general formula for the

**Complex Fourier Series with a real part.**

We can now find the final general formula for a

**Complex Fourier Series with a real part.**

**Fig. 3-12**

The formula is a generalization of

**Fig. 3-8a**. The constant component

**c0**appeared. Not very visible because of the

**yellow**color. It accurately approximates the function

**f(t)**only for

**n=∞**, but practically a

**dozen**, maybe a little more, of the first components is enough.

Each component

**c1*exp(1ωt)**,

**c2*exp(2ωt)**,

**c3*exp(3ωt),c4*exp(4ωt)… c∞*exp(∞ωt),**is a rotating vector as in

**Fig. 3-13.**

**Fig.3-13**

The **yellow** vector **c0** is a constant component and is the only one that does not rotate. Or formally it rotates, but with a pulsation **ω=0**. The others rotate faster and faster “each around the previous end”, i.e. indirectly around the fixed end** c0**. Successive vectors are getting shorter and shorter, and the last **c∞** is usually of **zero** length and rotates infinitely fast! The closer to** c∞**, the polyline in **Fig.3-13** approaches the continuos line! The projection of the end of the last vector on the real axis is just our periodic function **f(t)**. Try to imagine these rotating vectors. Doesn’t it remind you of the end of the shooting whip?

In other words

The vectors are rotating and the projection of endpoint** c∞** on the real axis moves exactly like the function **f(t)**!

Just as there are no ideals in life, the last rotating vector is e.g. **c20*exp(j20ωt)** and not** c∞*exp(j∞t)**. This is quite enough to approximate the **f(t)** function as accurately as possible.**Note:**

The “last” rotating **c∞*exp(j∞t)** usually has a length of **zero**. But the sum of all rotating vectors, the point **c∞*exp(j∞t)** points to, has a **finite** value You will find out in a moment!**Chapter 3.4.2 Example with a square wave**

Nothing appeals to the imagination like a specific example, e.g. with a **square wave**. It is discussed in detail in **chapters 2.5** and **2.6** as a **trigonometric series**.

Go back there for a while. A square wave with a pulsation of** ω=2π** (i.e. **T=1sec**) is**Fig. 3-14**Trigonometric Series of a square wave with

**ω=2π/sec.**

**Note**, that the constant component

**c0=0**, which is obvious and

**c2=c4=…=c2n=0**because the square wave

**f(t)**in

**Fig. 2-13**is an

**odd**function. Also the “cosine” coefficients are zero b1=b2=b3=0….

From the sine/cosine coefficients in

**Fig. 3-14**and the formulas in

**Fig. 3-11**, it follows directly that:

**c1=-4j/π, c3=-4j/3π, c5=-4j/5π, c7=-4j/7π**

and The

**Fourier Trigonometric Series**of a

**square wave**can also be represented as a

**Complex Fourier Series.**

Notice that you didn’t even have to calculate the

**integrals**. But if you insisted, because you like

**2**mushrooms in

**borscht**, then substituting

**f(t)**from

**Fig. 3-14**into the formula of

**Fig. 3-11**, the result would be identical.

**Fig. 3-15**

Complex Fourier Series of a

**square wave**

The initial state, i.e. for

**t=0**, are successive vertical vectors on the lower imaginary semi-axis

**Re z**. Only the first

**4**are visible, as

**c1**,

**c3**,

**c5**and

**c7**. The

**“even”**vectors

**c0,c2,c4,c6**…as stated earlier, are

**zero**. The last

**c∞**, although its length is

**zero**, lies at infinity on the negative “end” of the imaginary axis Im z. Here read the concluding remark of the previous section.

I also showed the state of the vectors when

**c1**rotated by

**1ωt=60º**,

**c3**by

**3ωt=180º**,

**c5**by

**5ωt=300º**, and

**c7**by

**7ωt=420º**. Evidently, the vector

**c7exp(7jωt)**was the fastest to make the largest rotation (

**420º**i.e.

**full rotation + 60º**), especially compared to the slowest

**c1exp(1jωt)**. And how to read

**f(t)**as a sum of rotating vectors? Here you go, for

**t=0**it is a projection on the

**Re z**axis, i.e.

**f(0)=0**, and for

**t**corresponding to a rotation of

**1ωt=60º=π/6**it is also

**approximately**the projection of the

**end**of the vector

**c7exp(7jωt)**on

**Re z**.

Why

**roughly**? Because I haven’t yet considered the remaining vectors

**c9exp(9jωt)…c∞exp(∞jωt)**ending at the finite point

**c∞**. This point is somewhere near the end of the vector

**c7exp(7jωt)**and you have to visualize it somehow. Only the projection of

**this**vector produces a perfect

**square wave**.

Let’s consider the graph in

**Fig. 3-16**again. For simplicity, let’s assume that our function

**f(t)**has only

**3 first**harmonics, i.e. only

**3**vectors

**c1, c3, c5**will rotate with velocities

**1ωt, 3ωt**and

**5ωt**.

**Fig. 3-16**

**Fig. 3-16a**Initial state for phase

**1ωt=0**.

**Fig. 3-16b**State for

**1ωt**phase. The vector

**c5exp(j5ωt)**rotates around the

**c3exp(j3ωt)**end, which in turn spins around the

**c1exp(j1ωt)**end. This resembles a system with the

**Sun**at

**(0,0)**, the

**Earth**revolving around the

**Sun**, the

**Moon**around the

**Earth**, and the

**Satellite**around the

**Moon**. After a time corresponding to

**1ωt=2π**, the initial state will again be as in

**Fig. 3-16a**, except that “

**Earth**” will make

**1**, “

**Moon**”

**2**and “

**Satellite**”

**3**rotations.

The projection of the “

**Satellite**” on the

**Re z**axis follows the formula

**{**

**c1exp(j1ωt)+c3exp(j3ωt)+c5exp(j5ωt)}**.

These are the

**first 3**“non-zero” harmonics of the decomposition of a

**square wave**into a

**Trigonometric Fourier Series**.

The expression

**c1exp(j1ωt)+c3exp(j3ωt)+c5exp(j5ωt)**alone are the

**first 3**components of the

**Complex Fourier Series**of a

**square wave**.

Note that the path of the

**Satellite**around the

**Sun**(and not around the

**Moon**) is no longer a circle but some strange contortion! It turns out that with the right combination of complex coefficients

**cn**and a large

**n**, the

**satellite**will rotate along any path. Even

**square**! So, given the appropriate

**cn**, we can draw any images. This is what the theory of image processing has been dealing with recently.

**Fourier**must be turning over in his grave.

**Note**

The

**circle**in

**Fig. 3-1**is a function that assigns to each time

**t**a corresponding point

**z=1exp(jωt)**in the plane of complex numbers

**z**.

By analogy

The

**path**of the sum of

**3**vectors in

**Fig. 3-16**is a function that assigns to each time

**t**a corresponding point

**z=c1exp(j1ωt)+c3exp(j3ωt)+c5exp(j5ωt)**in the plane of complex numbers

**z**.

**Chapter 3.4.3 Animation (from youtube)**Click on YouTube’s „play” triangle. GLV author’s nickname.

**Fig. 3-17**Animation of a Complex Fourier Series of a square wave

The spinning upper

**4**vectors are the

**4 components**of the formula

**f(t)=Re{c0+c1exp(j1ωt)+c3exp(j3ωt)+c5exp(j5ωt)+c7exp(j7ωt)}**Successive harmonics with decreasing amplitudes and increasing pulsations of

**1ωt, 3ωt, 5ωt**and

**7ωt**are perfectly presented. They are vector sums and sine sums. Of course, both give the same result, which is a function of time

**f(t)**close to a

**square wave**.

**Chapter 3.5 Complex Fourier Series – general formula without real part****And now** the most important thing, the second tadam!!!**1.** If you take any periodic function** f(t)** with period **T****2.** Calculate the fundamental pulsation **ω=2π/T****3.** Calculate the coefficients **a0,a1,b1,a2,b2,…an,bn…** and then the complex coefficients**…c(-n)…c(-3),c(-2),c(-1),c0,c1,c2,c3…cn…** acc. formulas** Fig. 3-18****then**

you will get the formula** Complex Fourier Series**–>**Fig.3-19****Note** for the above designations. e.g. **c(-3)** is **c** with negative subscript** -3**. I can’t save the subscript in **WordPress**.**Fig. 3-18a**The formula for the

**complex**coefficients

**c(n)**of the

**Complex Fourier Series**“without real part”

**Fig. 3-18b**

The same formula for c(n) after taking into account

**Euler’s**formula

**exp(jωt)=cos(ωt)+jsin(ωt)**

Compared to formula

**Fig. 3-11**

**1. c0**, i.e. the constant component coefficient is identical

**2.**The other coefficients cn are the complex numbers and halves of

**Fig. 3-11**(because

**an**and

**bn**are the halves of

**an**and

**bn**of

**Fig. 3-11**)

**3.**The pairs

**c(n)**and

**c(-n)**are conjugate numbers. So their sum

**c(n)+c(-n)=2a(n)**is a

**real**number.

**Fig. 3-19**

Formula for

**Complex Fourier Series**

**Negative**pulsations and coefficients with

**negative**indexes appeared. There are now

**2**times more* of them with

**2**times smaller

**modulus**, but they all add up to the same real function

**f(t)**!

*compared to

**Fig. 3-11**

**Fig. 3-20****Complex Fourier Series** as a sum of counter-rotating **vectors**

More precisely, “Sum of counter-rotating vectors + constant component **c0**“

The justification is very simple:**Each** pair of **oppositely** rotating vectors is, for example:**c(3)exp(j3ωt)+c(-3)exp(-j3ωt)=a(3)cos(3ωt)+ b(3)sin(3ωt)**. This is explained in detail in **Chapter 3.3.2**.**So** the sum of all **pairs** + the constant component** c0=a0**

will give us the formula **Fig. 3-3****That** is the formula **Fig. 3-19**

In an elegant and condensed form, it looks like this**Fig. 3-21Complex Fourier Series**

The “no real part” formula is more common than the equivalent “real part” version.

Therefore, it is simply the formula for the

**Complex Fourier Series**.

By the way, you learned that there is such a thing as

**positive**and

**negative**frequency/pulsation.

**Positive**have sinusoids whose vectors rotate counter-clockwise,

**negative**– clockwise. It used to intrigue me. One can somehow imagine that the

**zero**frequency is

**DC**(direct current). But

**negative**? Something more “solid than solid”? And how to live here? Now I know.