**Fourier ****Transform**

**Chapter 3 Fourier Transform of a single square wave pulse part 1
**

**Chapter.3.1 Introduction**

It is difficult to find a simpler

**non-periodic**function

**f(t)**than a single rectangular pulse in

**Fig. 3-1a**. It is an

**even**function with amplitude

**A=1**and duration

**Tp=1sec**. The

**Fourier Transform**is generally a

**complex**function. But thanks to the

**even**of

**f(t)**, its Fourier Transform

**F(ω)**is also a

**real**function. And this one is easier to analyze than the

**complex**function. Let me remind you that every

**real**function and

**real**number is

**complex**, but not vice versa! Due to the length, the topic has been divided into

**chapters 3**and

**4**.

**Chapter 3.2 A single square wave pulse as a signal with an infinite period To=∞.**

**Fig. 3-1
**

**a.**Single rectangular pulse

– amplitude

**A=1**

– duration

**Tp=1sec**

The impulse is an

**even**function.

**b.**Periodic function as a sequence of pulses with period

**To**, duration

**Tp=1sec**and amplitude

**A=1**.

A single rectangular pulse as

**a**, can be treated as a

**periodic**function

**b**with infinite period

**To=∞**.

**Chapter 3.3 Square waves with increasing period To.
**We will examine several square waves of different

**To**but with the same amplitude

**A=1**and duration

**Tp=sec**.

**Fig. 3-2
**

**Periodic**functions with pulse

**A=1**,

**Tp=1sec**and increasing period

**To**.

**a. To=2 sec**–>

**ωo=1π/sec**

b. To=4 sec–>

b. To=4 sec

**ωo=1π/2sec**

c. To=8 sec–>

c. To=8 sec

**ωo=1π/4sec**

**d. To=16 sec**–>

**ωo=1π/8sec**

Waves

**2, 4, 8, 16 sec**are increasingly

**similar**to a

**single pulse**. And the most is

**To=16sec**. So we can calculate the

**Fourier**coefficients

**a(n)**for this wave from the known formulas and say that we have “almost” decomposed the “single”

**f(t)**from

**Fig.6-1a**into

**harmonics**.

**Note:**

The pulses should be synchronized. They are “almost” in the animation. Sorry.

**Rozdz.3.4 What formula will we use for bar diagrams?
**The

**bar diagrams**applies only to

**even**functions and is the most intuitive version of the

**Fourier Series**. We will examine the effect of the increasing period

**To**of

**Fig. 3-2**on the

**bar diagrams**. This will make it easier for us to understand the idea of the

**Fourier Transform**as a “descendant” of the

**Fourier Series**. We will start with the general trigonometric formula.

**Fig. 3-3
**

**Trigonometric Fourier**formula for

**any**function

**f(t)**with period

**To**.

In the complex version

**c(n)=a(n)-jb(n)**.

**Note:**

The interval of integration in formulas

**c,d,e**is

**-To/2…+To/2**. It can also be different, e.g.

**0…To**or

**-To/4…+To3/4**,

provided that its length is

**To**. For the

**even**function

**f(t)**there are no

**b(n)**components and the formula will simplify.

**Fig. 3-4**

**Trigonometric**

**Fourier Series**formula for an

**even**periodic function

**f(t)**.

All bar diagrams, animations in this and the next chapter will be related to the above formula. We will specialize it even more for a square wave

**f(t)**with duration

**Tp=1sec**and amplitude

**A=1**. It pays off because there will be a lot of repetitive calculations. The period of the wave

**To**increases from

**2 sec**to

**∞**with a

**constant**pulse duration

**Tp=1sec**. So the

**duty cycle**decreases from

**50%**to

**0%**.

**Fig. 3-5**

**Fourier Series**formulas for the

**square wave**in

**Fig.**

**3-1b**for different

**To**, that is different

**ω0=2π/To**.

Derivation of the formulas in

**Fig. 3-11**.

They apply to square waves in which:

**–**rectangular pulse has a constant value

**A=1**and

**Tp=1sec**.

**–**only the period of occurrence of these pulses

**To**increases.

Thanks to them, we can easily determine the coefficients

**a(n)**necessary for the animation of bar diagrams.

**a.**Definition of the

**sinc(x)**function.

The function

**f(x)=sin(x)/x**appears so often in

**signal theory**that a special name

**sinc(x)**was created for it. Note that although every period

**T=πsec**the function resets to

**0**, it is not

**periodic**! It tends to

**0**when

**x–>+/-∞**and

**sinc(0)=1**is its maximum. It’s a bit weird at first because we’re dividing by

**0**! But there is a

**“0/0”**–>

**de l’Hospital’s**rule and everything is lege artis.

**b.**The

**sinc(x)**function graph

Characteristic points

**x=0,π/2**,

**π**and

**2π**are shown.

The chart shows, for example, that:

**–**

**sinc(0)=1**

**–**

**sinc(π/2)=2/π≈0.637**

**– sinc(1π)=sinc(2π)=…sinc(nπ)=0**

**– sinc(+∞)= sinc(-∞)=0**

**c.**Formula for

**a(n)**as a function of

**ω0**pulsation. There is the amplitude of the

**nth**harmonic for the

**n*ω0**pulsation, where

**ω0**is the pulsation of the periodic function

**f(t)**.

**d.**The formula for

**a(n)**as a function of period

**To**. It is also the amplitude of the

**nth**

**harmonic**for the period

**To/n**, where

**To**is the period of the periodic function

**f(t)**. We will use this formula most often. Recall that

**ω0=2π/To**.

For example, let’s calculate

**a1(ω0=1π/sec)=a1(To=2sec)**. That is, the

**first harmonic**for the square wave in

**Fig. 3-2a**.

We will use the

**3-5d**formula

**a(1)=(1/2)*sinc(1*π/2)=1/π≈0.318**

We used the plot of

**Fig. 3-5b**, where

**sinc(π/2)=2/π≈0.637**

**Note**

**1.**Formulas are valid for

**n=-∞…0…+∞**. For

**n=0**, the harmonic is the constant component

**a(0)**.

**2.**They come from the

**integral**formula

**Fig.3-4d**.

For doubters – derivation in

**chapter 3.5.7**

**Chapter 3.5 Bar diagrams for periodic functions
**More precisely for

**periodic**functions

**a, b, c**and

**d**from

**Fig. 3-2**.

**Chapter 3.5.1 Introduction**

Recall the

**4**animations from

**Fig. 3-2**.

Each of them is:

**-even**function of time f(t)

**-a square**wave consisting of square pulses

**Tp=1sec**repeating every period

**To**.

**-the first**pulse begins during half of its duration

**Tp=1sec**and therefore

**f(t)**is an

**even**function.

We will examine the effect of increasing period

**To**on the

**bar diagrams**. We will use the formula

**Fig. 3-5d**. It was assumed that

**A=1**and

**Tp=1sec**. Therefore, the only variables in the formula

**Fig. 3-5d**are

**To**as the period of the

**first harmonic**of the function and

**n**as the

**nth harmonic**.

**Chapter 3.5.2 Bar diagram for wave A=1, Tp=1sec and To=2sec**

For the first wave

**f(t)**, i.e. for

**Fig. 3-2a**, we have already made a bar diagram in

**subchapter 2.4 chapter 2**. This is a typical square wave with amplitude

**A=1**and period

**To=2sec**. Its duty cycle is

**1/2**. In the formula of

**Fig.3-5d**, the only variable is

**n**because the period of the wave is constant

**To=2sec**. It corresponds to the pulsation

**ω0=1π/sec**.

Let’s calculate

**a(-n)=a(+n)**for

**n=0…3**. and

**To=2sec**according to the formula

**Fig. 3-5d**.

**n=0–>ω=0*ω0=0**

a(0)=(1/2)sinc*(0*π/2)=1/2

**=0.5**bo sinc(0)=1 acc.

**Fig. 3-5b**

n=1–>ω=1*ω0=1π/sek

a(-1)=a(+1)=(1/2*)sinc(1*π/2)=1/(2π)

n=1–>ω=1*ω0=1π/sek

**≈0.318**

n=2–>ω=2*ω0=2π/sek

a(-2)=a(+2)=(1/2)*sinc(2*π/2)

n=2–>ω=2*ω0=2π/sek

**=0**

n=3–>ω=3*ω0=3π/sek

a(-3)=a(+3)=(1/2)*sinc(3*π/2)=-1/(3*2π)

n=3–>ω=3*ω0=3π/sek

**≈-0.106**

**Note**

You can check the above data with

**WolframAlfa**. How? See

**Chapter 11.2**

**WolframAlpha program**of the article

**“Rotating Fourier Series”**from the top tab.

E.g. To calculate a(-3)=a(+3) call

**WolframAlfa**

Click

**https://www.wolframalpha.com**

Then type or paste into thewindow

**0.5*sinc(3*pi/2)**and do what the picture says.

**Fig. 3-6**

The program correctly calculated

**a(-3)=a(+3)=(1/2)sinc(3*π/2)=-1/3π≈-0.106103**.

**Fig. 3-7
**Square wave bar diagram

**A=1, Tp=1sec, To=2sec**from

**Fig. 3-2a**

That is, for the

**first**wave

**f(t)**in

**Fig. 3-2a**with

**1/2**duty cycle.

The state of

**harmonics**before the animation, i.e. in the initial state, i.e. for

**t=0**. What does the picture show?

**1.**Non-pulsating

**DC**component

**a0=0.5**

**2.**

**First**harmonic pulsations

**1ω0=+1π/sec**and

**-1ω0=-1π/sec**

**3.**There are only

**odd**harmonics

**+-1*π/sec, +-3*π/sec, +-5*π/sec…**

**4.**Coefficients

**a(+1), a(-1), a(+3), a(-3)**are the amplitudes of

**+0.318, -0.106**of the pulsating bars. For example, negative

**-0.106**means that this was the value of the

**3rd**harmonic in the initial state for

**t=0**.

**5.**Only

**4**harmonics and component

**a0=+0.5**are shown. Other i.e. for

**ω=-∞…-5π/sec**and

**ω=+5π/sec…+∞**are invisible.

**6.**Invisible harmonics, however, are included in the pulsating bar

**f(t)**. It is the

**sum**of all harmonics, invisible too! Note that this is a perfect square wave. No “waves”, i.e. without the

**Gibbs**effect.

**7.**

**A**

**square wave**as a pulsating line

**f(t)**is in phase with the

**first**harmonic.

**8.**

**f(t)=…-(1/3π)*cos(3π*t)+(1/1π)*cos(1π*t)+0.5++(1/1π)*cos(1π* t)-(1/3π)*cos(3π*t)…**

The bar diagram animation and the above equation represent exactly the same thing. The

**f(t)**function is a

**vertical line**appearing every

**To= 2sec**, i.e. with a pulsation of

**1ω0=+1π/sec**. You see the harmonics in equation

**8**as pulsating bars with

**ω=-3π/sec, ω=-1π/sec, ω=0, ω=+1π/sec, ω=+3π/sec**. The remaining harmonics are invisible, but included in the total on

**f(t)**. That’s why you see the

**f(t)**band as perfect jumps every

**To=2sec**.

**Chapter 3.5.3 Bar diagram for wave A=1, Tp=1sec and To=4sec**

That is, for the

**second**wave

**f(t)**in

**Fig. 3-2b**with

**1/4**duty cycle.

**Fig. 3-8
**Square wave bar diagram

**A=1, Tp=1sec, To=4sec–>ω0=π/2sec**from

**Fig. 3-2b**

Let’s calculate

**a(-n)=a(+n)**for

**n=0…6**according to formula

**Fig. 3-5d**.

**n=0**–>

**ω=0*ω0=0**

a(0)=(1/4*)sinc(0*π/4)=(1/4)*1=1/4

**=0.25**

n=1–>

n=1

**ω=1*ω0=1*π/2sek**

a(-1)=a(+1)=(1/4)sinc(1*π/4)

**≈0.225**

**n=2–>ω=2*ω0=2*π/2sek=1*π/sek**

a(-2)=a(+2)=(1/4)*sinc(2*π/4)≈0.159

**n=3**–>

**ω=3*ω0=3*π/2sek**

a(-3)=a(+3)=(1/4)*sinc(3*π/4)

**≈0.159**

n=4–>

n=4

**ω=4*ω0=4*π/2sek=2*π/sek**

a(-4)=a(+4)=(1/4)*sinc(4*π/4)=0

**n=5**–>

**ω=5*ω0=5*π/2sek**

a(-5)=a(+5)=(1/4)*sinc(5*π/4)

**≈-0.045**

n=6–>ω=5*ω0=6*π/2sek=3*π/sek

a(-6)=a(+6)=(1/4)*sinc(6*π/4)

n=6–>ω=5*ω0=6*π/2sek=3*π/sek

**≈-0.053**

**Conclusions**

1.The

1.

**DC**component

**a0**and the harmonic amplitudes are

**2**times

**smaller**than for the previous square wave in

**Fig. 3-2b**. This agrees with intuition, because a rectangular pulse is

**2**times less frequent.

**2.**The envelope of the pulsating bars is also

**2**times

**smaller**vertically. This follows from the formula

**Fig. 3-5d**. This means that the zero points of

**ω=1π, 2π, 3π…**are the same. This, of course, applies to all square waves in

**Fig. 3-2**.

**3.**The bars are

**twice**as densely spaced. This also follows from the formula

**Fig. 3-5d**, but I will try to support it with intuition. It’s probably harder to choose the

**cosines**so that they add up to a pulse that occurs less often! The job needs to be more subtle. i.e. harmonics are not only

**2**times smaller, but there are

**2**times more of them. Just like with more Lego bricks, but smaller and more complicated, you can build a more complicated car. This “rarer impulse” is a more complicated car.

**4.**Pay attention to the less frequent impulse

**A=1 Tp=1sec**.

**Chapter 3.5.4 Bar diagram for wave A=1, Tp=1sec and To=8sec**

That is, for the

**third**wave

**f(t)**in

**Fig. 3-2c**with

**1/8**duty cycle.

**Fig. 3-9
**Square wave bar diagram

**A=1, Tp=1sec, To=4sec–>ω0=π/2sec**from

**Fig. 3-2c**

Let’s calculate

**a(-n)=a(+n)**for

**n=0…12**according to formula

**Fig. 3-5d**.

**n=0**–>

**ω=0*ω0=0**

a(0)=(1/8)*sinc(0*π/8)=1/8

**=0.125**

.

.

**n=12**–>

**ω=12*ω0=3*π**

a(12)=(1/8)*sinc(12*π/8)

**≈-0.026**

I gave only

**2**calculations, you can calculate the rest with

**WolframAlfa**program.

The pulse

**f(t)**occurs every

**To=8 sec**.

**Chapter 3.5.5 Bar diagram for wave A=1, Tp=1sec and To=16sec**

That is, for the

**fourth**wave

**f(t)**in

**Fig. 3-2d**with

**1/16**duty cycle.

**Fig. 3-10
**Square wave bar diagram

**A=1, Tp=1sec, To=16sec–>ω0=π/8sec**from

**Fig. 3-2d**

Let’s calculate

**a(-n)=a(+n)**for

**n=0…24**according to formula

**Fig. 3-5d**.

**n=0**–>

**ω=0*ω0=0**

a(0)=(1/16)* sinc(0*π/16)=1/16

**=0.0625**

…

…

**n=24**–>

**ω=24*ω0=3*π**

a(24)=(1/16)*sinc(24*π/16)

**≈-0.013**

The pulse

**f(t)**occurs every

**To=16 sec**.

**I emphasize**again.

The

**bar diagram**, and especially its animation, is the most intuitive representation of a

**Fourier Series**. You can see what you can’t see in the formula

**Fig.3-4**. i.e. mutual correlation between successive harmonics as in

**Fig. 3-10.**And animation is already a cosmos! You can see the

**f(t)**-left side of the

**Fourier Series**as a sum of swinging

**bars**. The function

**f(t)**is an impulse

**A=1,**

**Tp=1sec**appearing as a vertical line every

**To=16sec**.

**Chapter 3.5.6 Effect of increasing To period on the bar diagram.**

In

**Fig. 3-2**, the pulse has a constant amplitude

**A=1**and duration

**Tp=1sec**. The period

**To**of the square wave increases. The impulse becomes more and more solitary, gradually resembling a

**single impulse**. On ordinary, read

**static**, diagrams, this is not visible, but on

**animations**it is.

Let’s get back to the conclusions, so what happens to the bars when the period

**To**increases? Recall that the

**nth**bar is the

**nth**harmonic. You see only the most important initial bars in the range. The others, invisible and tending to zero, you have to imagine.

**Conclusions**

**1.**The bars are more and more densely located

**2.**The envelope of the bars flattens out.

**3.**Zeros

**ω=…-3π,-2π,-1π,0,+1π,+2π,+3π…**are constant.

**4.**Max

**1/2.1/4.1/8.1/16**correspond to

**ω=0 bars**, i.e. the constant component

**a(0)**of the

**Fourier Series**of square waves from

**Fig. 3-2a,b,c,d**.

**5.**The sum of all bars-harmonics is a function of

**f(t)**and appears every

**2, 4, 8 or 16 seconds**as a vertical line – a perfect square wave.

**Chapter 3.5.7 Derivation of the formula for volunteers**

**Fig. 3-11
**Derivation of the

**Fourier Series formula**for the wave from

**Fig. 3-1b**

This is a highly specialized function

**f(t)**and applies only to a sequence of pulses with amplitude

**A=1**and duration

**Tp=1sec**. Only the period of the function

**To**(or the pulsation

**ωo**) and

**n**changes. The

**ωo**pulsation concerns the

**first**harmonic and is also the distance

**Δω=ω0**between the bars. Basic knowledge of calculus is enough. The formula is also valid for the constant component–>

**n=0**.