Fourier series. XE31EO2 - Pavel Máša. Electrical Circuits 2 Lecture1. XE31EO2 - Pavel Máša - Fourier Series

Transcription

1 Fourier series Electrical Circuits Lecture - Fourier Series

2 Filtr RLC defibrillator MOTIVATION WHAT WE CAN'T EXPLAIN YET Source voltage rectangular waveform Resistor voltage sinusoidal waveform - Fourier Series

3 MOTIVATION MATHEMATICAL BASIC PHYSICAL DESCRIPTION, TRANSFORMATION Physical properties of basic elementary passive circuit elements in the electrical circuit are described by integral differential equations u(t) =Ri(t) u(t) =L di(t) dt u(t) = C Z t i( )d + u C () Versatile mathematical description i(t) = u(t) R i(t) = L Quite complicated solution simplification by means of transformation What does it mean transformation? Complex mathematical operations (integrals and derivatives) replace by more tricky Z t i(t) =C du(t) dt u( )d + i L () - Fourier Series

4 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 - Fourier Series

5 What we are not able to solve? Periodical, but non sinusoidal waveforms Single pulses Evaluate 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) - Fourier Series

6 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?. The same frequency when we add sinusoidals of the same frequency, the magnitude and phase changes, but still it is sinusoidal in time domain where sin(6.8 t) + 5 cos(6.8 t) or by means of phasors - Fourier Series

7 . Different frequencies 5 5 sin(6.8 t) + 5 cos(.9 t) the sum is not sinusoidal, it may, but may not be periodical it is valid common period example: T =.4 s -, T =.6 s - T = 3.4 =.6 =. s - special case: We have distinct (fundamental) frequency ω, all other frequencies are integral multiples, ω = kω k = T/ T 3T/ T 5T/ 3T T/ T 3T/ T 5T/ 3T (If we fulfill some conditions) the periodical function may be replaced by series of sinusoidal functions.5.5 k = 3 - Fourier Series

8 Function expansion may be written sine form How to find Fourier series coefficients? It is not possible directly general form ω fundamental (first) harmonics kω higher harmonics a = B = T Z T f(t)dt f(t) =B + FOURIER SERIES X B mk sin(k! t + Ã k ) k= f(t) = a + X (a k cos k! t + b k sin k! t) k= It is mathematical mean value (DC component) But how to compute a k, b k? excursion orthogonality - Fourier Series

9 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?. Multiplication by constant. sin Z T ORTHOGONALITY Z T a Z T a μ sin k ¼ T t dt = μ cos k ¼ T t dt = μ sin k ¼ μ T t sin l ¼ T t dt = jif k = lj = T = jif k 6= lj = hf;gi w =.5.5 Z b a f(x)g(x)w(x)dx =: 3. cos Z T μ cos k ¼ μ T t cos l ¼ T t dt = jif k = lj = T = jif k 6= lj = Fourier Series

10 4. Sin and cos Z T μ sin k ¼ μ T t cos l ¼ T t dt = jif k = lj = = jif k 6= lj = 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 lω t and integrate it within one period, then, thanks to orthogonal properties of sin and cos, all terms, but l th sin term will be zeroised, that is just b l coefficient remains. Accordingly, when we multiply series by cos lω t function and by integration within one period we obtain a l coefficients. Z T e jk!t e jl!t dt = jk = lj = T = jk 6= lj = : - Fourier Series

11 FOURIER SERIES COEFFICIENTS DC part Cos terms Sin terms Sine form Conditions of existence of the Fourier series Dirichlet s conditions: Normalization is necessary The result of integrating, see orthogonality, T je a k,orb T k. Function f(t) is bounded within a period. Function exhibits only a finite number of extremes and discontinuities of the. kind - Fourier Series

12 SIN, OR COS? Above was stated sin form, but some books use cos form What is the difference? It is based on different phasor definition Calculations are completely the same, seeing that cos is phase shifted sin but whereas - Fourier Series

13 Even function may contain only even coefficients: cos Odd function may contain only odd coefficients: sin Antiperiodic function contains only odd coefficients BASIC PROPERTIES COMPUTATION When is possible to divide period in several (, 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 different Gibbs 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..5 k = T/ T 3T/ T 5T/ 3T.5 k = 5 k = 5.5 T/ T 3T/ T 5T/ 3T.5 T/ T 3T/ T 5T/ 3T - Fourier Series

14 Even, or odd? Mathematical condition: Odd function Even function Antiperiodic function 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 In both cases series contains only odd harmonics, though functions does not exhibit condition of the antiperiodic function Before we determine, if the function is even or odd / antiperiodic, it is necessary remove DC part - Fourier Series

15 PHASOR REPRESENTATION It is not another form of Fourier series, it is just modification of sin form h i B mk sin(k! t + à k )=Im B mk e jã k e jk! t =Im hb i mk e jk! t f(t) =B + X Im hb i mk e jk! t k= Using single phasors B mk we can analyze electrical circuits separately, one by one Relation to the basic form B mk = B mk e jã k = B mk (cos à k + j sin à k )=b k + ja k - Fourier Series

16 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 COMPLEX FORM OF FOURIER SERIES e j! t =cos! t + j sin! t e j! t + e j! t =cos! t + j sin! t +cos(! t) + j sin(! {z } t) =cos! {z } t cos! t j sin! t cos function may be written using two half sized phasors turning in opposite direction Line spectra - Fourier Series

17 sin function may be also written in the meaning of two half sized rotating phasors, but they are shifted by 9 with respect to the cos function e j! t e j! t =cos! t + j sin! t cos(! t) j sin(! {z } t) =jsin! {z } t sin! t = j e j! t e j! t cos! t j sin! t But now, it is not function of a real variable? Phasor, part of coefficients of complex form of Fourier series It represents the shift of the sin function with respect to the cos function - Fourier Series

18 Using the sin function, expressed as two rotating phasors, we may evaluate sin form of the Fourier series B mk sin(k! t + Ã k ) = j B mk e jk! t B mk e jk! t = = j (b k + ja k ) e jk! t (b k ja k ) e jk! t = = a k jb k e jk! t + a k + jb k e jk! t = = A k e jk! t + A k e jk! t = ja k = A kj = = A k e jk! t + A k e jk! t Here is defined relationship between coefficients in complex form and basic form f(t) = X k= A k e jk! t A k = T Z T f(t)e jk! t dt Beware of summation limits Coefficients are phasors, A is DC component - Fourier Series

19 u(t) [V] DC component a 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 3 = B = T b k = T = 4 k¼ t [s] Z T T u(t)dt = : EXAMPLE Find the Fourier series of the rectangular waveform in the figure. The period is T =. s. Z :5 3dt + : Z : :5 = 3 :5 3 + : :5 = Z T k =; 3; 5;::: sin k! t dt = 4 T k odd = k even u(t) =+ cos k! t X k= k! T n 3t :5 dt = + t o : = :5 = = ¼! T = 4 T 4 (k )¼ sin(k )k! t k ¼ T μ cos k ¼ T T +cos - Fourier Series

20 LINE SPECTRA a k b k 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 a is DC component We discuss discrete spectra harmonics take just certain values - Fourier Series

21 Complex form of rectangular waveform in our example is A k = T = j k¼ Z T Ã DC term e jk! t dt = T cos k¼ {z } j sin {z k¼ } A = B = a = e jk! t Complex terms, evaluated using basic form coefficients A k = a k jb k Fourier series u(t) =+ X k= = j 4¼ k¼ T jk!! = j k¼ = j k¼ = j (k )¼ ej(k )! t Tjk ¼ T k odd k odd ³ e jk¼ T e = j k¼ e jk¼ = - Fourier Series

22 TIME SHIFT a k.7.6 A k A k = T Z T Z T +t f(t t )e jk! t dt = = t t dt =d = = f( )e jk! ( +t ) d = A k e jk! t T t X f(t t )= A k e jk! (t t ) k= time shift means turn of each phasor about different angle different turning speed!!! b k arg(a k ) A k Fourier Series 3 33

23 PROPERTIES Linearity Time reversion Time shift Change of time scale Derivation Integral modulation X a i x i (t) i X a i A k;i x( t) A k x(t t ) A k e jk! t x(at) A k change of period T a dx dt Z t x( )d < x(t)e jk! t i jk! A k A k ; if A = jk! A k K - Fourier Series

24 Now we are able to explain first motivation example Odd rectangular waveform, U m = V, T =.5 ms We know the rectangular waveform may be approximated by series or u(t) = u(t) = X k= X k= The voltage across resistor in sinusoidal steady state is We have to calculate voltage of each term of the series separately!!! k = 5.7 less k = less k = 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 4U m (k )¼ sin(k )k! t ju m (k )¼ ej(k )! t R U k = U k (jk! ) 3 L L C +(jk! ) L C + jk! (L + L )+R 4U m U m(k ) = (k )¼ 8 8 :(k ) + j(k )(4:3 8:(k ) ) - Fourier Series

25 Transient Waveform detail Zoomed in y axis!!! - Fourier Series

26 Filter frequency response - Fourier Series