xe31eo2 eo2 pavel –lecture máša...

Fourier transform

EO2 – Lecture 2

Pavel Máša

XE31EO2 - Pavel Máša - Fourier Transform

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• We already know complex form of Fourier series

Series frequency spectra is discrete

Circuits in the sinusoidal steady state must be solved step by step (very laborious)

We can expand only periodic waveforms

? What if we connect to the circuit single pulse source?

? How to find output voltage by one computation, not laborious step by step calculation, term by term?

INTRODUCTION

f (t) =

1Xk=¡1

Akejk!0tf (t) =

1Xk=¡1

Akejk!0t

Ak =1

T

Z T

0

f (t)e¡jk!0t dtAk =1

T

Z T

0

f (t)e¡jk!0t dt


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What if we will extend period of rectangular pulse?

Or narrow rectangular pulse?

FOURIER SERIES – PERIOD EXTENSION


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• Both when we narrow the pulse, or when we extend the period, the envelope curve is function

• The larger the ratio of period T and pulse width t0 is, the more spectral terms falls in one period of the function.

Narrowing – constant T, lesser t0, k represents still the same frequency(frequency) distance of neighbouring terms is the same – function extends across towards ∞

Widening – constant t0, T extends towards ∞, k represents still less frequencies(frequency) of neighbouring terms falls up to zero – function stay at place, but magnitude plunge!


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Period extension towards ∞

Initially discrete frequency becomes continuous

Magnitude of frequency spectra (primarily distinct harmonics) tends to 0

Coefficients (harmonics) have to be multiplied by period T (or → 0 !)

Direct Fourier transform

FROM FOURIER SERIES TO FOURIER TRANSFORM


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• The waveform is qualified by series

• If T→ ∞, thenInverse Fourier transform

So far is possible, we don’t use definition integral directly, but we try to use transform properties and known transforms

INVERSE TRANSFORM


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Function has to satisfy Dirichlet’s conditions (Fourier series !)

Function has to be absolutely integrable

considerably restrictive condition, when transform is not applicable to so common waveform, such as unit step function (DC voltage / current)

It satisfies condition

NECESSARY CONDITIONS

Transform is essential tool of description of frequency properties of discretesystems (sound and image digital processing – CD, SACD and DVDplayers, home cinemas, …)


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The transform is a analysis tool of frequency spectra of waveforms– the maxima of F(jω) functions denotes harmonics of frequency spectra

Fourier transform

Dumping is essential to satisfy conditions of existence of the transform

jU(j!)jjU(j!)j

Function is „stretched“it has limited frequency resolution


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BASIC PROPERTIES

property time domain frequency domain

linearity

Time shifting

Frequency shifting (modulation)

Differentiation

Integration

Scaling

Exponential pulse transform

f(t) = af1(t) + bf2(t) F(j!) = aF1(j!) + bF2(j!)

f(t) = f1(t¡ t0) F(j!) = F1(j!)e¡j!t0

f (t) = f1(t)ej!0t F(j!) = F1

¡j(! ¡ !0)

¢f (t) =

dnf1(t)

dtnF(j!) = (j!)n F1(j!)

F(j!) =dnF1(j!)

d(j!)nf (t) = (¡t)nf1(t)

f (t) =

Z t

¡1f1(t) dt F(j!) =

F1(j!)

j!

f (t) = f1(at) F(j!) =1

jaj F1

μj!

a

¶f (t) = e¡at

F(j!) =1

j! + aXE31EO2 - Pavel Máša - Fourier Transform

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EXAMPLE

Find the Fourier series of rectangular waveform on the figure. Um = 2 V, T = 0.1 s, t0 = 0.025 s

-15 -10 -5 0 5 10 15-0.2

0

0.2

0.4


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Find the Fourier transform of the rectangular pulse on the figure. Compare result with Fourier series above.Um = 2 V, t0 = 0.025 s

-1000 -800 -600 -400 -200 0 200 400 600 800 1000-0.02

0

0.02

0.04

-15 -10 -5 0 5 10 15-0.2

0

0.2

0.4

Magnitude of Fourier transform changes with the length of the pulse!

series

transform


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EXAMPLE 2

Find the Fourier transform of the pulse on the figure.U1 = 1 V, U2 = 2V, t0 = 0.05 s

U1U1

U2U2

t0t0

• The waveform may be considered as superposition of two distinct waveforms– rectangular, Um = 1 V– saw tooth, Um = 1 V

We know the transform of rectangular waveform from previous example, but we have to modify it – the magnitude is different, as well as time t0 and it is shifted in time

applied properties are scaling, a = 0.5 and time shifting within -0.025 s

magnitude scaling (time) scaling time shifting


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In contrast to Fourier series (where we have to compute with each harmonic separately), we may find waveforms of circuit variables with non‐sinusoidal excitation likewise using sinusoidal steady state analysis1. Find Fourier transform of excitation pulse

2. Using transfer function (same, as in sinusoidal steady state), find transform of output voltage

3. Using inverse transform we find output voltage waveform

When we know input and output voltage waveforms, we can find frequency response of the circuit1. Find Fourier transform of input waveform

2. Find Fourier transform of output waveform

3. Find transfer function of the circuit

APPLICATION IN ELECTRICAL CIRCUIT ANALYSIS


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Integrating network in the figure is excited by rectangular pulse on the figure. Find waveform of the output voltage.

1.

2.

3.

Now, we should „just“ find inverse transform…

1. We know, is transform of the integral

first, we find function and then we will integrate it

2. The transform is superposition of two different functions, is the time delay t0

3.

4.

t0

0

Um

t

EXAMPLE

U1(j!) =

Z t0

0

Ume¡j!t dt = Um

·e¡j!t

¡j!

¸t0

0

=Um

¡j!

£e¡j!t0 ¡ 1

¤U1(j!) =

Z t0

0

Ume¡j!t dt = Um

·e¡j!t

¡j!

¸t0

0

=Um

¡j!

£e¡j!t0 ¡ 1

¤P(j!) =

1

1 + j!RCP(j!) =

1

1 + j!RC

U2(j!) =1

1 + j!RC¢ Um

¡j!

£e¡j!t0 ¡ 1

¤U2(j!) =

1

1 + j!RC¢ Um

¡j!

£e¡j!t0 ¡ 1

¤

1

j!

1

j!

½Z t

0

u0

2(¿)d¿

¾=

1

j!¢ Um

1¡ e¡j!t0

1 + j!RC

½Z t

0

u0

2(¿)d¿

¾=

1

j!¢ Um

1¡ e¡j!t0

1 + j!RCu

0

2(t)u0

2(t)

Um1¡ e¡j!t0

1 + j!RCUm

1¡ e¡j!t0

1 + j!RCe¡j!t0e¡j!t0

¡1

½1

1 + j!RC

¾=

1

RCe¡

1RC t ) u

0

2(t) =Um

RC

³e¡

1RC t ¢ 1(t) ¡ e¡

1RC (t¡t0) ¢ 1(t¡ t0)

´¡1

½1

1 + j!RC

¾=

1

RCe¡

1RC t ) u

0

2(t) =Um

RC

³e¡

1RC t ¢ 1(t) ¡ e¡

1RC (t¡t0) ¢ 1(t¡ t0)

´Z t

0

u0

2(¿)d¿ =Um

RC

μZ t

0

e¡1

RC ¿d¿ ¡Z t

t0

e¡1

RC (¿¡t0)d¿

¶= Um

h(1¡ e

tRC )1(t)¡ (1¡ e¡

t¡t0RC )1(t¡ t0)

iZ t

0

u0

2(¿)d¿ =Um

RC

μZ t

0

e¡1

RC ¿d¿ ¡Z t

t0

e¡1

RC (¿¡t0)d¿

¶= Um

h(1¡ e

tRC )1(t)¡ (1¡ e¡

t¡t0RC )1(t¡ t0)

iNote. – it is possible find it more easily – partial fractions, Laplace transform... XE31EO2 - Pavel Máša - Fourier Transform

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To the input of the circuit we connected the waveform

On the output we measured the waveform

Find transfer function of the circuit. Find suitable circuit diagram.

EXAMPLE

u1(t) = 10e¡500t Vu1(t) = 10e¡500t V

u2(t) = 6:¹6(e¡500t ¡ e¡2000t) Vu2(t) = 6:¹6(e¡500t ¡ e¡2000t) V

U1(j!) =10

j! + 500U1(j!) =

10

j! + 500

U2(j!) =6:¹6

j! + 500¡ 6:¹6

j! + 2000=

6:¹6 ¢ 2000¡ 6:¹6 ¢ 500

(j! + 500)(j! + 2000)=

10000

(j! + 500)(j! + 2000)U2(j!) =

6:¹6

j! + 500¡ 6:¹6

j! + 2000=

6:¹6 ¢ 2000¡ 6:¹6 ¢ 500

(j! + 500)(j! + 2000)=

10000

(j! + 500)(j! + 2000)

P(j!) =U2(j!)

U1(j!)=

10000(j!+500)(j!+2000)

10j!+500

=1000

j! + 2000P(j!) =

U2(j!)

U1(j!)=

10000(j!+500)(j!+2000)

10j!+500

=1000

j! + 2000


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