Laplace transform. XE31EO2 - Pavel Máša. EO2 Lecture 3. XE31EO2 - Pavel Máša - Lecture 3

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1 Laplace transform EO2 Lecture 3 Pavel Máša

2 INTRODUCTION We know, the Fourier transform due to strict conditions of existence does not exists for number of very common waveforms even sin function have to be dumped How to ensure, the transform will exist for many waveforms (almost all physically feasible waveforms)? We dump it ourselves we will multiply the waveform by function e ¾t

3 FROM FOURIER TRANSFORM TO LAPLACE TRANSFORM We know the Fourier transform To extend the set of integrable functions, we dump f (t) function by But dumping has effect just when t 0, negative time amplifies the function, we have to introduce condition t 0 In circuit analysis we study tasks what happens after..., so this condition is not limiting The history of the circuit is described by initial conditions (voltage across capacitor, current passing inductor at t = 0). F(¾; j!) = Z +1 Direct Laplace transform 0 f (t)e ¾t e j!t dt = Z +1 0 f (t)e (¾+j!)t dt = Z +1 0 f(t)e pt dt e ¾t When σ = 0 the Laplace transform come into Fourier transform

4 Note fonts Two different fonts are used for typesetting of Fourier and Laplace transform Consequently, we can meet with two different symbols for typing of Fourier transform Laplace transform Note Fff(t)g Lff(t)g Instead of p operator s is sometimes used p / s is sometimes called complex frequency

5 e ¾t REGION OF CONVERGENCE For dumping function we defined necessary condition t 0 But this condition is not sufficient to ensure the (Fourier) integral of the function f(t)e ¾t converges When σ < 0(or certain number), it does not dump, but amplifies, depending up to properties of f(t) function The Laplace transform of the function f(t) exists for all complex numbers such that ¾>¾ min The part of p plane satisfying this condition is called region of convergence (don t confuse with region of stability, with lays in left part of p plane, left from the region of convergence (and left from imaginary axis)!!! poles diverge) (BIBO) stability BIBO Bounded Input Bounded Output If a system is BIBO stable, then the output will be bounded for every input to the system that is bounded. Passive circuit is always stable, if it contains non zero resistivity, zero resistivity limit of stability Active circuit (containing some amplifier) need to have feedback

6 EXAMPLE 1 POLE IN P PLANE pole (tends to ) P (p) = R =100Ð C =1mF pole 1 1+pRC p p = 10 Fourier transform (frequency characteristic)

7 EXAMPLE ZEROES AND POLES IN P PLANE 3 D view of p plane, side view sigma omega poles P (p) = pc(pl + R) p 2 LC + prc +1 Zeroes: p 01 =0; p 02 = 100 Poles: p p1;2 = 50 86:6j R =10Ð L =0:1H C =1mF the upper view of the same p plane poles zeroes Fourier transform imaginary axis

8 SELECTED PROPERTIES OF LAPLACE TRANSFORM property time domain Laplace transform (frequency domain) Linearity Time shifting Differentiation integration convolution df(t) dt Z t 0 f( )d f(t) g(t) = Z t 0 f( )g(t )d pf (p) f(0 + ) 1 p F (p)

9 TABLE OF SELECTED LAPLACE TRANSFORMS Time domain Frequency domain Region of convergence Dirac delta function Unit step function (DC voltage connected at time t = 0) Exponential decay 1

10 Time domain Frequency domain Region of convergence Exponentially decaying sine wave Exponentially decaying cosine wave Phase shifted sine wave

11 INVERSE LAPLACE TRANSFORM We start from inverse Fourier transform f(t) = 1 2¼ Z +1 1 F(j!)e j!t d! Using inverse Fourier transform we will evaluate dumped function f(t)e ¾t = 1 2¼ Z +1 1 F(p)e j!t d! Moving dumping on right side of equation f(t) = 1 2¼ Z +1 1 F(p)e ¾t e j!t d! = Z +1 By substitution of variables and integral limits we got Laplace transform 1 ff (p)g = f(t) = 1 2¼j Z ¾+1 ¾ 1 So far as it is possible, we don t use the inverse Laplace transform integral directly, but we try to use transform properties and known transforms from the Table 1 F(p)e pt d! F(p)e pt dp

12 THE PROCEDURE OF INVERSE LAPLACE TRANSFORM EVALUATION The resulting network function (e.g. transfer function P(p)) / variable (voltage transform U(p), current transform I(p), ) will be rational function, ratio of two polynomial functions First, by possible polynomial division we have to ensure, the degree of polynomial P(p) in nominator is smaller than degree of polynomial Q(p) in denominator; at once we can factor out the highest order polynomial coefficient in denominator We will compute both polynomial roots in nominator (zeros) and denominator (poles) of the function Now, the partial fraction decomposition procedure depends on pole type 1. Simple real roots

13 2. Repeated real roots with root multiplicity α, β, γ 3. The pair of complex conjugate roots

14 Now we have left to find coefficients A, B, 1. General method (compare coefficients at same powers) 1. Function F (p) (after partial fraction decomposition) multiply by initial denominator 2. Compare coefficients with the same power of p in nominator of original function 1. 2.

15 2. Single pole method it is not universal tool, when the polynomial has roots with multiplicity, it can be used only with root of highest power, other coefficients must be calculated using other methods In the function F (p) we will replace variable p by value of the root p i. Bracket, containing root p i must be removed (it is zero). Mathematical description: Example:

16 OPERATIONAL CHARACTERISTICS OF TWO PORTS Transform of derivative Transform of an integral multiplication by p (and initial condition) division by p Kirchhoff's laws in operational form If (σ = 0) and without initial conditions Fourier, if moreover sine waveform source sinusoidal steady state

17 INDUCTOR AND CAPACITOR EQUIVALENT CIRCUIT DIAGRAMS initial condition voltage source operational impedance initial condition current source operational admittance initial condition current source operational impedance operational admittance initial condition voltage source

18 EQUIVALENT CIRCUIT AND TERMINALS OF REAL CIRCUIT ELEMENT The capacitor was charged at 10 V. Find the current passing the capacitor. series equivalent circuit Ohm s law parallel equivalent circuit current divider??? But the current passing capacitor should be the same??? This is actual capacitor!!!

19 In time t = 0 an inductor was passed by the current i L (0) = 2 A. Find the Laplace transform of current passing the inductor when t > 0 and voltage transform at t > 0. parallel equivalent circuit series equivalent circuit

20 When the initial conditions are zero, the operational characteristics will be analogical to those in Fourier transform or sinusoidal steady state Impedance and admittance, including the input one of two ports Transfer function (voltage, current, )

21 EXAMPLE THE SAME AS IN THE LAST LECTURE U m The integrating network in the figure is excited by rectangular pulse in second figure. Compute the waveform of output voltage. The capacitor has zero voltage at the time of connection of source (zero initial condition). To find the solution we will use table of Laplace transforms rectangular pulse is superposition of two unit step functions multiplied by U m P (p) = 1+pRC 3. U 2 (p) =U 1 (p) P (p) = U m 1 1 e pt 0 p 1+pRC Statement in the square bracket will be temporary omitted (it is information about time delay, transformed later) 0 t 0 t The transform of the square bracket are two unit step functions, the second is time shifted by t 0

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