electrical xe31eo2 circuits 2...

Fourier series

Electrical Circuits 2 – Lecture1

XE31EO2 - Pavel Máša - Fourier Series

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• Filtr

• RLC defibrillator

MOTIVATIONWHAT WE CAN'T EXPLAIN YET

Source voltage – rectangular waveformResistor voltage – sinusoidal waveform


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MOTIVATION ‐MATHEMATICALBASIC PHYSICAL DESCRIPTION, TRANSFORMATION

• Physical properties of basic elementary passive circuit elements in the electrical circuit are described by integral‐differential equations

– Versatile mathematical description– Quite complicated solution simplification by means of transformation

• What does it mean – transformation?– Complex mathematical operations (integrals and derivatives) replace

by more tricky

i(t) =1

L

Z t

0

u(¿ )d¿ + iL(0)i(t) =1

L

Z t

0

u(¿ )d¿ + iL(0)

u(t) =1

C

Z t

0

i(¿ )d¿ + uC(0)u(t) =1

C

Z t

0

i(¿ )d¿ + uC(0) i(t) = Cdu(t)

dti(t) = C

du(t)

dt

u(t) = Ldi(t)

dtu(t) = L

di(t)

dt

u(t) = Ri(t)u(t) = Ri(t) i(t) =u(t)

Ri(t) =

u(t)

R


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• What kind of transforms we know?– Stationary Steady State – DC analysis – special case, when both

current and voltage are constant, the derivative of constant is zero

• Capacitor:

with arbitrary voltage across capacitor passing current is still zero – we may replace it by open circuit

• Inductor:

with arbitrary passing current the voltage is still zero – inductor may be replaced by short circuit

– Sinusoidal Steady State – sinusoidal excitation

What determine the choice of transform?Waveform of voltage / current!

phasors


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What we are not able to solve?Periodical, but non‐sinusoidal waveformsSingle pulsesEvaluate waveforms after circuit is switched on / off

We have to introduce new mathematical tools and transforms into electrical circuit analysis

Fourier series (it is not true transform, but the Fourier transform will be derived from it; periodical non‐sinusoidal waveforms)

Fourier transform (only single pulses)

Laplace transform (general‐purpose, all kinds of possible waveforms, including phenomena after switching the circuit on / off)


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HARMONIC SYNTHESIS

If some circuit can change the rectangular waveform on sinusoidal, it is legitimate to suppose, the rectangular waveform has to contain such sinusoidal

First we try opposite procedure− What happens, when we add some sinusoidals together?

1. The same frequency

when we add 2 sinusoidals of the same frequency, the magnitude and phase changes, but still it is sinusoidal‐ in time domain

where

‐ or by means of phasors

0 0.5 1 1.5 2 2.5 3 3.5 4-50

0

50

0 0.5 1 1.5 2 2.5 3 3.5 4-80

-60

-40

-20

0

20

40

60

8050 sin(6.28 t) + 50 cos(6.28 t)


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2. Different frequencies

the sum is not sinusoidal, it may, but may not be periodical

it is valid → common period

example: T1 = 0.04 s-1, T2 = 0.06 s-1 → T = 3 · 0.04 = 2 · 0.06 = 0.12 s-1

special case:• We have distinct (fundamental) frequency ω0, all other frequencies are integral multiples, ω = kω0

(If we fulfill some conditions) the periodical function may be replaced by series of sinusoidal functions

0 2 4 6 8 10 12-50

0

50

0 2 4 6 8 10 12-100

-50

0

50

10050 sin(6.28 t) + 50 cos(1.9 t)

0 T/2 T 3T/2 2T 5T/2 3T-0.5

0

0.5

1

1.5

2

2.5k = 3

0 T/2 T 3T/2 2T 5T/2 3T

0

0.5

1

1.5

2

k = 3


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• Function expansion may be written

sine form

How to find Fourier series coefficients? It is not possible directly

general form

ω0 – fundamental (first) harmonics

kω0 – higher harmonics

It is mathematical mean value (DC component)

But how to compute ak, bk? ‐ excursion ‐ orthogonality

FOURIER SERIES

f(t) = B0 +

1Xk=1

Bmk sin(k!0t + Ãk)f(t) = B0 +

1Xk=1

Bmk sin(k!0t + Ãk)

f(t) =a0

2+

1Xk=1

(ak cos k!0t + bk sin k!0t)f(t) =a0

2+

1Xk=1

(ak cos k!0t + bk sin k!0t)

a0

2= B0 =

1

T

Z T

0

f(t)dta0

2= B0 =

1

T

Z T

0

f(t)dt


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Rectangularity – originally, in geometry perpendicularity, generalized on vector spaces(two vectors are orthogonal, when its inner product is zero), and functions.

Two functions are orthogonal, if the following condition is fulfilled:

We are interesting in sin and cos functions – are they orthogonal?

1. Multiplication by constant

2. sin

3. cos

ORTHOGONALITY

hf; giw =

Z b

a

f(x)g(x)w(x) dx = 0:hf; giw =

Z b

a

f(x)g(x)w(x) dx = 0:

0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1

0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1

Z T

0

a0

2¢ sin

μk2¼

Tt

¶dt = 0

Z T

0

a0

2¢ sin

μk2¼

Tt

¶dt = 0

Z T

0

a0

2¢ cos

μk2¼

Tt

¶dt = 0

Z T

0

a0

2¢ cos

μk2¼

Tt

¶dt = 0

Z T

0

sin

μk2¼

Tt

¶¢ sin

μl2¼

Tt

¶dt = jif k = lj =

T

2

= jif k 6= lj = 0

Z T

0

sin

μk2¼

Tt

¶¢ sin

μl2¼

Tt

¶dt = jif k = lj =

T

2

= jif k 6= lj = 0

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1

Z T

0

cos

μk2¼

Tt

¶¢ cos

μl2¼

Tt

¶dt = jif k = lj =

T

2

= jif k 6= lj = 0

Z T

0

cos

μk2¼

Tt

¶¢ cos

μl2¼

Tt

¶dt = jif k = lj =

T

2

= jif k 6= lj = 0

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1


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4. Sin and cos

5. Complex exponential (phasor)

Now we know what orthogonality is – but how it helps us to find Fourier series coefficients?

Periodic function is expanded by series

If we multiply this series by function sin lω0t and integrate it within one period, then, thanks to orthogonal properties of sin and cos, all terms, but lth

sin term will be „zeroised“, that is just bl coefficient remains.

Accordingly, when we multiply series by cos lω0t function and by integration within one period we obtain al coefficients.

Z T

0

sin

μk2¼

Tt

¶¢ cos

μl2¼

Tt

¶dt = jif k = lj = 0

= jif k 6= lj = 0

Z T

0

sin

μk2¼

Tt

¶¢ cos

μl2¼

Tt

¶dt = jif k = lj = 0

= jif k 6= lj = 0

0 0.2 0.4 0.6 0.8 1

−0.4

−0.2

0

0.2

0.4

0.6

0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1

Z T

0

ejk!0t ¢ e¡jl!0t dt = jk = lj = T

= jk 6= lj = 0:

Z T

0

ejk!0t ¢ e¡jl!0t dt = jk = lj = T

= jk 6= lj = 0:


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• DC part

• Cos terms

• Sin terms

• Sine form

Conditions of existence of the Fourier seriesDirichlet’s conditions:

1. Function f(t) is bounded within a period2. Function exhibits only a finite number of extremes and discontinuities of the 1. kind

FOURIER SERIES COEFFICIENTS

Normalization is necessary‐ The result of integrating, see orthogonality,

je ak

T

2, or bk

T

2ak

T

2, or bk

T

2


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• Above was stated sin form,

• but some books use cos form

– What is the difference?

It is based on different phasor definitionCalculations are completely the same, seeing that cos is phase shifted sin

SIN, OR COS?

but whereas


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• Even function may contain only even coefficients: cos• Odd functionmay contain only odd coefficients: sin• Antiperiodic function contains only odd coefficients

When is possible to divide period in several (2, 4) equivalent parts and the area under the function waveform (except of sign) is also the same, it is possible to compute coefficients just in one part of period (of course, normalizing have to be changed) – rectangular, triangular, ... waveform

Fourier series is just approximation, original waveform and its approximation may be differentGibbs phenomenon – if the function has jump discontinuity, then Fourier series exhibits oscillations near the jump. Even if the number of terms of the Fourier series is increasing to the infinity, the overshot will be still about 8.95 % greater than original function. With increasing number of terms the width of the first overshot decreases, but its height never converges to the original function, but noted value.

BASIC PROPERTIES – COMPUTATION

0 T/2 T 3T/2 2T 5T/2 3T-0.5

0

0.5

1

1.5

2

2.5k = 25

0 T/2 T 3T/2 2T 5T/2 3T

0

0.5

1

1.5

2

k = 25

0 T/2 T 3T/2 2T 5T/2 3T

0

0.5

1

1.5

2

k = 15


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Mathematical condition:

– Odd function

– Even function

– Antiperiodic function

In both cases series contains only odd harmonics, though functions does not exhibit condition of the antiperiodic function

Even, or odd?

00

00

In this case the condition u(t) = u(-t) is valid ⇒ function is even, series has only cos terms and DC part

Here, u(t) ≠ -u(-t) ⇒ according to the definition, the function is not even nor odd, but, despite it, series has only sin terms and DC part

Before we determine, if the function is even or odd / antiperiodic, it is necessary remove DC part


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• It is not another form of Fourier series, it is just modification of sin form

PHASOR REPRESENTATION

Bmk sin(k!0t + Ãk) = ImhBmke

jÃkejk!0ti

= ImhBmkejk!0t

iBmk sin(k!0t + Ãk) = Im

hBmke

jÃkejk!0ti

= ImhBmkejk!0t

i

f(t) = B0 +

1Xk=1

ImhBmke

jk!0ti

f(t) = B0 +

1Xk=1

ImhBmke

jk!0ti

Using single phasors Bmk we can analyze electrical circuits – separately, one by one

• Relation to the basic form

Bmk = BmkejÃk = Bmk(cosÃk + j sinÃk) = bk + jakBmk = BmkejÃk = Bmk(cosÃk + j sinÃk) = bk + jak


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• It is still rather laborious to find Fourier series in phasor notation – it is possible to simplify the procedure?

• It is possible to write function of a real variable by the means of complex functions?

• Euler formula

cos function may be written using two half‐sized phasors turning in opposite direction

COMPLEX FORM OF FOURIER SERIES

ej!0t = cos !0t + j sin!0tej!0t = cos !0t + j sin!0t

ej!0t + e¡j!0t = cos !0t + j sin !0t + cos(¡!0t)| {z }cos!0t

+ j sin(¡!0t)| {z }¡j sin !0t

= 2 cos!0tej!0t + e¡j!0t = cos !0t + j sin !0t + cos(¡!0t)| {z }cos!0t

+ j sin(¡!0t)| {z }¡j sin !0t

= 2 cos!0t

Line spectra


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sin function may be also written in the meaning of two half sized rotating phasors, but they are shifted by 90 with respect to the cos function

ej!0t ¡ e¡j!0t = cos !0t + j sin !0t ¡ cos(¡!0t)| {z }cos !0t

¡ j sin(¡!0t)| {z }¡j sin !0t

= 2j sin !0tej!0t ¡ e¡j!0t = cos !0t + j sin !0t ¡ cos(¡!0t)| {z }cos !0t

¡ j sin(¡!0t)| {z }¡j sin !0t

= 2j sin !0t

But now, it is not function of a real variable?

sin !0t = ¡1

2j¡ej!0t ¡ e¡j!0t

¢sin !0t = ¡1

2j¡ej!0t ¡ e¡j!0t

¢Phasor, part of coefficients of complex form of Fourier series

It represents the shift of the sin function with respect to the cosfunction


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• Using the sin function, expressed as two rotating phasors, we may evaluate sin form of the Fourier series

Bmk sin(k!0t + Ãk) = ¡1

2j¡Bmke

jk!0t ¡B¤mke

¡jk!0t¢

=

= ¡1

2j£(bk + jak) ejk!0t ¡ (bk ¡ jak) e¡jk!0t

¤=

=ak ¡ jbk

2ejk!0t +

ak + jbk

2e¡jk!0t =

= Akejk!0t + A¤

ke¡jk!0t = jA¤

k = A¡kj =

= Akejk!0t + A¡ke

¡jk!0t

Bmk sin(k!0t + Ãk) = ¡1

2j¡Bmke

jk!0t ¡B¤mke

¡jk!0t¢

=

= ¡1

2j£(bk + jak) ejk!0t ¡ (bk ¡ jak) e¡jk!0t

¤=

=ak ¡ jbk

2ejk!0t +

ak + jbk

2e¡jk!0t =

= Akejk!0t + A¤

ke¡jk!0t = jA¤

k = A¡kj =

= Akejk!0t + A¡ke

¡jk!0t

Here is defined relationship between coefficients in complex form and basic form

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

Beware of summation limits Coefficients are phasors, A0 is DC component


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• DC component

• After subtraction of DC component from the waveform we find, the function is– Odd ⇒

– Antiperiodic ⇒

• Sin terms – the area between a function and t axis is the same under and over the t axis ⇒ it is enough calculate coefficients just in first half‐period

• Resulting series

EXAMPLE

00

u(t)

[V]

1

3

t [s] T

Find the Fourier series of the rectangular waveform in the figure. The period is T = 0.1 s.

a0

2= B0 =

1

T

Z T

0

u(t) dt =1

0:1

Z 0:05

0

3 dt +1

0:1

Z 0:1

0:05

1 dt = 10n£

3t¤0:05

0+

£t¤0:1

0:05

o=

= 30 ¢ 0:05 ¡ 30 ¢ 0 + 10 ¢ 0:1 ¡ 10 ¢ 0:05 = 2

a0

2= B0 =

1

T

Z T

0

u(t) dt =1

0:1

Z 0:05

0

3 dt +1

0:1

Z 0:1

0:05

1 dt = 10n£

3t¤0:05

0+

£t¤0:1

0:05

o=

= 30 ¢ 0:05 ¡ 30 ¢ 0 + 10 ¢ 0:1 ¡ 10 ¢ 0:05 = 2

k = 1; 3; 5; : : :k = 1; 3; 5; : : :

bk =2T2

Z T2

0

1 ¢ sin k!0tdt =4

T

·¡ cos k!0t

k!0

¸T2

0

=

¯̄̄̄!0 =

2¼

T

¯̄̄̄=

4

T

1

k 2¼T

μ¡ cos k

2¼

T

T

2+ cos 0

¶=

4

k¼k odd

= 0 k even

bk =2T2

Z T2

0

1 ¢ sin k!0tdt =4

T

·¡ cos k!0t

k!0

¸T2

0

=

¯̄̄̄!0 =

2¼

T

¯̄̄̄=

4

T

1

k 2¼T

μ¡ cos k

2¼

T

T

2+ cos 0

¶=

4

k¼k odd

= 0 k even

u(t) = 2 +

1Xk=1

4

(2k ¡ 1)¼sin(2k ¡ 1)k!0tu(t) = 2 +

1Xk=1

4

(2k ¡ 1)¼sin(2k ¡ 1)k!0t


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LINE SPECTRA

• The coefficients are represented as points, emphasized by vertical line– On the x axis lies k indexes

– The y axis is the value of distinct terms of the series

– (zero) term a0 is DC component –

– We discuss discrete spectra – harmonics take just certain values

0 2 4 6 8 10 12 14 16 18 20

0

1

2

ak

0 2 4 6 8 10 12 14 16 18 20-1

0

1

bk


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• Complex form of rectangular waveform in our example is

• DC term

• Complex terms, evaluated using basic form coefficients

• Fourier series

Ak =1T2

Z T2

0

1 ¢ e¡jk!0t dt =2

T

·e¡jk!0t

¡jk!0

¸T2

0

=2

¡Tjk2¼T

³e¡jk 2¼

2T2 ¡ e0

´=

j

k¼

¡e¡jk¼ ¡ 1

¢=

=j

k¼

Ãcos k¼| {z }

§1

¡ j sin k¼| {z }0

¡1

!=¡2j

k¼k odd

A0 = B0 =a0

2= 2A0 = B0 =

a0

2= 2

Ak =ak ¡ jbk

2=

0 ¡ j 4¼k¼

2=¡2j

k¼k oddAk =

ak ¡ jbk

2=

0 ¡ j 4¼k¼

2=¡2j

k¼k odd

u(t) = 2 +

1Xk=¡1

¡2j

(2k ¡ 1)¼ej(2k¡1)!0tu(t) = 2 +

1Xk=¡1

¡2j

(2k ¡ 1)¼ej(2k¡1)!0t


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TIME SHIFT

A0

k =1

T

Z T

0

f(t ¡ t0)e¡jk!0t dt =

¯̄̄̄¿ = t¡ t0

dt = d¿

¯̄̄̄=

=1

T

Z T+t0

t0

f(¿ )e¡jk!0(¿+t0) d¿ = Ake¡jk!0t0

A0

k =1

T

Z T

0

f(t ¡ t0)e¡jk!0t dt =

¯̄̄̄¿ = t¡ t0

dt = d¿

¯̄̄̄=

=1

T

Z T+t0

t0

f(¿ )e¡jk!0(¿+t0) d¿ = Ake¡jk!0t0

f(t ¡ t0) =

1Xk=¡1

Akejk!0(t¡t0)f(t ¡ t0) =

1Xk=¡1

Akejk!0(t¡t0)

time shift means turn of each phasor about different angle – different turning speed!!!

0.2

0.4

0.6

0.8

30

210

60

240

90

270

120

300

150

330

180 0

Ak

-2 -1 0 1 2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20

-1

-0.5

0

0.5

1

ak

0 5 10 15 20

-1

-0.5

0

0.5

1

bk

-20 -10 0 10 200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

⏐Ak⏐

-20 -10 0 10 20-4

-2

0

2

4

arg(Ak)


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Linearity

Time reversion

Time shift

Change of time scale

Derivation

Integral

modulation

PROPERTIES

Xi

aixi(t)X

i

aixi(t)X

i

aiAk;i

Xi

aiAk;i

x(¡t)x(¡t) A¡kA¡k

x(t ¡ t0)x(t ¡ t0) Akejk!0t0Akejk!0t0

x(at)x(at) Ak change of periodT

aAk change of period

T

a

dx

dt

dx

dtjk!0Akjk!0AkZ t

¡1x(¿ )d¿ < 1

Z t

¡1x(¿ )d¿ < 1 Ak

jk!0; if A0 = 0

Ak

jk!0; if A0 = 0

x(t)ejK!0tx(t)ejK!0t Ak¡KAk¡K


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Now we are able to explain first motivation example

Odd rectangular waveform, Um = 1V, T = 1.25 ms

• We know the rectangular waveform may be approximated by series

or

• The voltage across resistor in sinusoidal steady state is

• We have to calculate voltage of each term of the series separately!!!– k = 1 5.7 × less

– k = 3 225.5 × less

– k = 5 1175.1 × less

• The magnitude of voltage across resistor is decreasing quickly – important is only first term – What happens if we decrease / extend period?

BACK TO MOTIVATION EXAMPLE

u(t) =

1Xk=1

4Um

(2k ¡ 1)¼sin(2k ¡ 1)k!0tu(t) =

1Xk=1

4Um

(2k ¡ 1)¼sin(2k ¡ 1)k!0t

u(t) =

1Xk=¡1

¡2jUm

(2k ¡ 1)¼ej(2k¡1)!0tu(t) =

1Xk=¡1

¡2jUm

(2k ¡ 1)¼ej(2k¡1)!0t

U2k = U1kR

(jk!0)3L1L2C + (jk!0)2L1C + jk!0(L1 + L2) + RU2k = U1k

R

(jk!0)3L1L2C + (jk!0)2L1C + jk!0(L1 + L2) + R

U2m(2k¡1) =4U1m

(2k ¡ 1)¼¢ 8

8 ¡ 20:02(2k ¡ 1)2 + j(2k ¡ 1)(124:13 ¡ 80:12(2k ¡ 1)2)U2m(2k¡1) =

4U1m

(2k ¡ 1)¼¢ 8

8 ¡ 20:02(2k ¡ 1)2 + j(2k ¡ 1)(124:13 ¡ 80:12(2k ¡ 1)2)


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Waveform detail

Zoomed in y axis!!!Transient


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Filter frequency response


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