Fourier transform. XE31EO2 - Pavel Máša. EO2 Lecture 2. XE31EO2 - Pavel Máša - Fourier Transform

Transcription

1 Fourier transform EO2 Lecture 2 Pavel Máša - Fourier Transform

2 INTRODUCTION We already know complex form of Fourier series f(t) = 1X k= 1 A k e jk! t A k = 1 T 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? Z T f(t)e jk! t dt - Fourier Transform

3 What if we will extend period of rectangular pulse? Or narrow rectangular pulse? FOURIER SERIES PERIOD EXTENSION - Fourier Transform

4 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 t is, the more spectral terms falls in one period of the function. Narrowing constant T, lesser t, k represents still the same frequency (frequency) distance of neighbouring terms is the same function extends across towards Widening constant t, T extends towards, k represents still less frequencies (frequency) of neighbouring terms falls up to zero function stay at place, but magnitude plunge! - Fourier Transform

5 FROM FOURIER SERIES TO FOURIER TRANSFORM Period extension towards Initially discrete frequency becomes continuous Magnitude of frequency spectra (primarily distinct harmonics) tends to Coefficients (harmonics) have to be multiplied by period T (or!) Direct Fourier transform - Fourier Transform

6 The waveform is qualified by series INVERSE TRANSFORM If T, then Inverse Fourier transform So far is possible, we don t use definition integral directly, but we try to use transform properties and known transforms - Fourier Transform

7 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 discrete systems (sound and image digital processing CD, SACD and DVD players, home cinemas, ) - Fourier Transform

8 The transform is a analysis tool of frequency spectra of waveforms the maxima of F(jω) functions denotes harmonics of frequency spectra Dumping is essential to satisfy conditions of existence of the transform ju(j!)j Fourier transform Function is stretched it has limited frequency resolution - Fourier Transform

9 BASIC PROPERTIES property time domain frequency domain linearity Time shifting Frequency shifting (modulation) Differentiation Integration Scaling Exponential pulse transform f(t) =af 1 (t)+bf 2 (t) F(j!)=aF 1 (j!)+bf 2 (j!) f(t) =f 1 (t t ) F(j!)=F 1 (j!)e j!t f(t) =f 1 (t)e j! t F(j!)=F 1 j(!! ) f(t) = dn f 1 (t) dt n f(t) =( t) n f 1 (t) Z t F(j!) =(j!) n F 1 (j!) F(j!) = dn F 1 (j!) d(j!) n f(t) = f 1 (t)dt F(j!) = F 1(j!) 1 j! f(t) =f 1 (at) F(j!)= 1 μ j! jaj F 1 a f(t) =e at F(j!)= 1 j! + a - Fourier Transform

10 EXAMPLE Find the Fourier series of rectangular waveform on the figure. U m = 2 V, T =.1 s, t =.25 s Fourier Transform

11 Find the Fourier transform of the rectangular pulse on the figure. Compare result with Fourier series above. U m = 2 V, t =.25 s transform series Magnitude of Fourier transform changes with the length of the pulse! - Fourier Transform

12 EXAMPLE 2 U 2 U 1 Find the Fourier transform of the pulse on the figure. U 1 = 1 V, U 2 = 2V, t =.5 s The waveform may be considered as superposition of two distinct waveforms rectangular, U m = 1 V saw tooth, U m = 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 t and it is shifted in time applied properties are scaling, a =.5 and time shifting within -.25 s magnitude scaling (time) scaling time shifting t - Fourier Transform

13 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 analysis APPLICATION IN ELECTRICAL CIRCUIT ANALYSIS 1. 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 circuit 1. Find Fourier transform of input waveform 2. Find Fourier transform of output waveform 3. Find transfer function of the circuit - Fourier Transform

14 EXAMPLE U m Integrating network in the figure is excited by rectangular pulse on the figure. Find waveform of the output voltage U 1 (j!)= Z t e j!t U m e j!t dt = U m j! Now, we should just find inverse transform ½Z t ¾ 1 1. We know, is transform of the integral j! u 2( )d = 1 j! U 1 e j!t m 1+j!RC first, we find functionu u 2 (t) and then we will integrate it t 1 P(j!)= 1+j!RC 1 U 2 (j!)= 1+j!RC U m e j!t 1 j! 2. The transform 1 e j!t U m is superposition of two different functions, e j!t is the time delay t 1+j!RC ½ ¾ = 1 1+j!RC RC e 1 RC t ) u 2(t) = U ³ m e 1 RC t 1(t) e 1 RC (t t) 1(t t ) Z RC t 4. u 2( )d = U μz t Z t h i m e 1 RC d RC ( t) d = U m (1 e t RC )1(t) (1 e t t RC )1(t t ) RC t = U m e j!t 1 j! e 1 t Note. it is possible find it more easily partial fractions, Laplace transform... t - Fourier Transform

15 To the input of the circuit we connected the waveform On the output we measured the waveform EXAMPLE u 1 (t) =1e 5t V u 2 (t) =6:¹6(e 5t e 2t )V Find transfer function of the circuit. Find suitable circuit diagram. 1 U 1 (j!) = j! +5 6:¹6 U 2 (j!) = j! + 5 6: ¹6 j! +2 = 6: ¹6 2 6:¹6 5 (j! + 5)(j! + 2) = 1 (j! +5)(j! + 2) (j!+5)(j!+2) P(j!) = U 1 2(j!) U 1 (j!) = 1 j!+5 = 1 j! Fourier Transform