An Introduction to Integral Transforms
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An Introduction to Integral Transforms Tomáš Kozubek and Marek Lampart Ostrava
Transformation of the structure and content of higher education at VŠB-TUO NPO VŠB-TUO MSMT16605/2022 17. LISTOPADU 2175/15 708 00 OSTRAVA-PORUBA [email protected] www.vsb.cz
Contents Preface 4 1 Fourier series 5 1.1 Fourierseries ............................ 7 1.2 Fourier series in a complex field . . . . . . . . . . . . . . . . . . 9 1.3 Development of periodic function . . . . . . . . . . . . . . . . . 13 1.4 Sine and cosine series . . . . . . . . . . . . . . . . . . . . . . . . 20 1.5 Properties of Fourier series . . . . . . . . . . . . . . . . . . . . . 24 1.6 The space L2(a,b).......................... 28 1.7 Generalized Fourier sequence . . . . . . . . . . . . . . . . . . . . 32 1.8 Gibbsphenomenon......................... 38 1.9 Workedexample .......................... 40 1.10Appendix .............................. 45 2 Laplace transform 49 2.1 Properties of the Laplace transform . . . . . . . . . . . . . . . . . 53 2.2 Inverse Laplace transform . . . . . . . . . . . . . . . . . . . . . . 63 2.3 Applications of the Laplace transform . . . . . . . . . . . . . . . 67 References 77 Index 78 3
Preface Integral transforms as well as Fourier series are very interesting and powerful topics in the undergraduate mathematics course. Their importance to applications means that they can be studied both from a very pure perspective and a very applied perspective. This text book for students takes into account the varying needs and backgrounds of students in mathematics, science, and engineering. It covers two topics that feature in the course: •Fourier series, •Laplace transform. Since the topics of Fourier series and integral transform are not elementary subjects, there are some reasonable assumptions about what the reader knows. The reader should be confident with the relevant topics taught as standard in the area of real analysis of real functions of one and multiple variables, complex analysis, sequences, and series. This text is mostly a translation from the Czech original [8], and it is a natural continuation of the textbook An Introduction to Complex Analysis [2]. The authors are grateful to their colleagues for their comments that improved this text and to John Cawley who helped with the correction of many typos and English grammar. prof. RNDr. Marek Lampart, Ph.D. May 4, 2024 4
Chapter 1 Fourier series In a numerous technical problems one can find functions whose course is repeating. These functions describe periodic events in many physical processes (vibrations of constructions, steady movement of the piston of internal combustion engines, steady rotation movements) or mechanical oscillations (acoustic waves) and alternating electric current. Such functions are called periodic, that is if f(t)is a function of a real variable tand there is a positive real number Tsuch that for each tfrom the domain it holds that f(t)=f(t+T).(1.1) The number T, for which equation (1.1) holds, is called period, hence the function fis periodic with period T. If Tis a period of fthen nT is also a period for each nÀN. The smallest such T, for which equation (1.1) holds, is called the prime period. We point out that the prime period should not to exist. As an example, a constant function can be given that has as a period every positive real number, but does not have a prime period. Next, for simplicity we will use the notation of period, and from the context it will always be clear if the period is prime or not. For any ↵ÀRthe interval (↵,↵+T]is called the interval of periodicity; in particular the basic interval of periodicity is a special case where ↵=0or ↵=*T_2, that is the basic interval of periodicity takes the form (0,T]or (*T_2,T_2]. The next lemma shows that it is possible to restrict our attention to functions with the period 2⇡. Lemma 1 For every periodic function f(t)with a period of Tthere is a transformation of an argument1t=tr(x)such that the transformed function f(tr(x)) has a period of 2⇡. Proof: Let t=tr(x)= T 2⇡x. 1by a transformation of an argument we mean transformation of coordinates, such as in the case of transformation of Cartesian coordinates to polar, spherical of cylindrical ones. 5
6CHAPTER 1. FOURIER SERIES Then f(t)=f(tr(x)) = f⇠T 2⇡x⇡=g(x). The function g(x)is defined for every xand is periodic with a period of 2⇡: g(x+2⇡)=f⇠T 2⇡(x+2⇡)⇡=f⇠T 2⇡x+T⇡=f(t+T)=f(t)=g(x). ∏ An elementary example to consider is simple harmonic oscillation, which is given by a general sine function f(t)=Asin(!t +').(1.2) Here, the variable tis interpreted as time, Ais the amplitude indicating deviation from the equilibrium position, the argument !t+'is called the phase of oscillation, for t=0we get the initial phase and the constant !, and indicating the number of oscillations from 2⇡seconds is called the circular frequency (angular velocity). The time for one oscillation period is denoted by T, and in our case it is T=2⇡_!. In practice, however, we encounter more complex periodic functions (which we will show in the next sections) that can be written as the sum of an infinite series of simple harmonic oscillations, where the first term of this series has the same period as the given periodic function. The periods of the following oscillations are then a half, a third, etc., of the period of the first oscillation. This creates a periodic function expressing a compound harmonic oscillation, which is described by an infinite series with the terms un=Ansin(n!t +'n).(1.3) These can be equivalently written in the form un=ancos(n!t)+bnsin(n!t),(1.4) where for simplicity we put u0=a0 2.(1.5) The series ÿ … n=1 un=a0 2+ ÿ … n=1 (ancos(n!t)+bnsin(n!t)) (1.6) is called a trigonometric series. If the series converges, as will be shown later, then it converges to a function with a period2of T=2⇡_!, that is with the period of the element with index 1. The coefficients anand bnare called Fourier coefficients of the function f(t). 2Observe that the period does not have to be prime. If a1=b1=0then 2⇡_!is a period, but is not the smallest one. In addition, if there at least one of the coefficients a2or b2is non-zero, then the prime period is ⇡_!.
1.1. FOURIER SERIES 7 1.1 Fourier series In Section 1.5 will be shown that the trigonometric series a0 2+ ÿ … n=1 (ancos(nt)+bnsin(nt)) (1.7) is uniformly convergent in R, and its sum is a continuous periodic function f(t) with a period of the first term series2, that is T=2⇡(here, Lemma 1 was applied, hence T=2⇡and !=1). Consequently, f(t)=a0 2+ ÿ … n=1 (ancos(nt)+bnsin(nt)).(1.8) Now, the task is to find coefficients’ anand bnuniformly convergent trigonometric series (1.8) using the function f(t)which is its sum. To solve this problem the orthogonality of the systems function from Example 10 (page 31) on the interval [*⇡,⇡], which is an interval of a length 2⇡. The coefficient a0will be derived by integration of equation (1.8) from *⇡to ⇡. So, ⇡ *⇡ f(t)dt=⇡ *⇡Ha0 2+ ÿ … n=1 (ancos(nt)+bnsin(nt))Idt=⇡a0, a0=1 ⇡⇡ *⇡ f(t)dt,(1.9) here we used the fact that for each n,î⇡ *⇡cos(nt)dt=0and î⇡ *⇡sin(nt)dt=0. The coefficients anwill be derived from (1.8) multiplied by function cos(nt)and its integration on the same interval. Then using computations from Example 10 (on page 31) we get ⇡ *⇡ f(t) cos(nt)dt=an⇡ *⇡ cos2(nt)dt=an⇡, an=1 ⇡⇡ *⇡ f(t) cos(nt)dt.(1.10) The coefficients bnwill be found analogously, as anunder assumption (1.8) is multiplied by function sin(nt): ⇡ *⇡ f(t) sin(nt)dt=bn⇡ *⇡ sin2(nt)dt=bn⇡, bn=1 ⇡⇡ *⇡ f(t) sin(nt)dt.(1.11)
8CHAPTER 1. FOURIER SERIES Formulae for the determinationofcoefficients arecalled(Euler)-Fourier. The given trigonometric series is called the Fourier series of functions f(t)and coefficients anand bnthe Fourier coefficients of the functionf(t). Naturally, this begs the question whether Fourier series (1.8) is convergent and if its sum equals f(t)in the interval [*⇡,⇡]. The answer is given by the following theorem, which will be proved in Section 1.5: Theorem 1 (Dirichlet’s) If the function f(t)fulfills the so called Dirichlet conditions, then a Fourier series of the function f(t)is convergent at every tto the value 1 2(f(t+ 0) + f(t*0)) and it holds that 1 2(f(t+ 0) + f(t*0)) = a0 2+ ÿ … n=1 ancos(nt)+bnsin(nt). Moreover, in points twhere f(t)is continuous, it is 1 2(f(t+ 0) + f(t*0)) = f(t). In the foregoing theorem we use standard notation3 f(t+ 0) = lim t1ôt+f(t1)af(t*0) = lim t1ôt*f(t1). Dirichlet conditions are the following: 1. the function f(t)is periodic, 2. the function f(t)has on the interval of periodicity only finite numbers of discontinuities of the first type, 3. the function f(t)has on the interval of periodicity piecewise continuous derivation. Example 1 The following functions do not fulfill the Dirichlet conditions on the interval [*⇡,⇡]: f1(t)= 2 1*t,f2(t) = sin ⇠2 2*t⇡. Really, f1(t)has at the point t0=1discontinuity of the second kind and f2(t)has on the neighborhood of the point t0=2infinitely many extremes. 3sometimes a shortened version is used; f(t+) resp. f(t*)
1.2. FOURIER SERIES IN A COMPLEX FIELD 9 The relations (1.9) – (1.11) can be generalized for functions with the period T= 2l, hence also for functions with the interval of periodicity [*l,l]. Using Lemma 1, the transformation t=⇡ ltcan be done, and we get for nÀNthe formulae: a0=1 ll *l f(t)dt,(1.12) an=1 ll *l f(t) cos(n⇡ lt)dt,(1.13) bn=1 ll *l f(t) sin(n⇡ lt)dt(1.14) and the Fourier series takes the form f(t)=a0 2+ ÿ … n=1 (ancos(⇡ lnt)+bnsin(⇡ lnt)).(1.15) 1.2 Fourier series in a complex field In Section 1.1 the Fourier coefficients anand bnwere derived from a Fourier series of a periodic function with a period of 2⇡. These formulae take the form: f(t)=1 2a0+ ÿ … n=1 (ancos(nt)+bnsin(nt)),(1.16) a0=1 ⇡⇡ *⇡ f(t)dt,(1.17) an=1 ⇡⇡ *⇡ f(t) cos(nt)dt,(1.18) bn=1 ⇡⇡ *⇡ f(t) sin(nt)dt.(1.19) Now, let us write functions sin(nt)and cos(nt)in series (1.16) using the following exponential form: cos(nt)=1 2(eint +e*int),(1.20) sin(nt)= 1 2i(eint *e*int)=*i 2(eint *e*int).(1.21) Substituting (1.20) and (1.21) into the series (1.16) we get f(t)=1 2a0+ ÿ … n=1 0an(eint +e*int 2)*ibn(eint *e*int 2)1=(1.22) =1 2a0+ ÿ … n=1 ⇠1 2(an*ibn)eint +1 2(an+ibn)e*int)⇡.(1.23)
16 CHAPTER 1. FOURIER SERIES Let us now compose a one-sided and two-sided phase and amplitude spectrum, using formulas (1.37), (1.38), (1.39), (1.40) and (1.24), (1.25), (1.26): A0=ÛÛÛÛ a0 2ÛÛÛÛ=0, An=ta2 n+b2 n=u0+(*1)n+1 2 n=2 n, cn=1 2(an*ibn)=1 2⇠0*i(*1)n+1 2 n⇡= i(*1)n1 n, 'n=*arg cn=h n l n j *⇡_2for n=…,*5,*3,*1,2,4,6,…, ⇡_2for n=…,*6,*4,*2,1,3,5,…. The two-sided amplitude (resp. phase) spectrum is shown in Figure 1.7 (resp. 1.8). The values of the coefficients are given in table 1.1. n-3 -2 -1 0 1 2 3 an— — — 0 0 0 0 bn— — — — 2 -1 2/3 cni_3*i_2 i 0*i i_2*i_3 cn1_3 1_21 0 1 1_2 1_3 An— — — 0 2 1 2/3 'n*⇡_2⇡_2*⇡_2—⇡_2*⇡_2⇡_2 Table 1.1: Table of coefficients of harmonic analysis of a function (1.41). Example 4 Let us develop the periodic function in the Fourier series f(t)with a basic interval of periodicity (*⇡,⇡](see Figure 1.9) given by: f(t)=h n l n j tfor tÀ[0,⇡], *tfor tÀ(*⇡,0), (1.46) and do a spectral analysis. We will analyze the problem exactly as in the previous example. Firstly, we verify Dirichlet’s conditions: 1. the function is obviously periodic,
1.3. DEVELOPMENT OF PERIODIC FUNCTION 17 Figure 1.7: Two-sided amplitude spectrum of the function (1.41). Figure 1.8: Two-sided phase spectrum of the function (1.41). Figure 1.9: Graph of the function (1.46)
18 CHAPTER 1. FOURIER SERIES Figure 1.10: Graph of the sum s2(t)of the development of (1.46). 2. the function is continuous, 3. the function has on the interval (k⇡,⇡+k⇡)a derivative for kÀZ. Hence, we can apply (1.9), (1.10) and (1.11) to get Fourier coefficients: a0=1 ⇡⇡ *⇡ f(t)dt=1 ⇡H0 *⇡ *tdt+⇡ 0 tdtI=⇡, an=1 ⇡⇡ *⇡ f(t) cos(nt)dt=1 ⇡H0 *⇡ *tcos(nt)dt+⇡ 0 tcos(nt)dtI= =2 ⇡n2((*1)n*1), bn=1 ⇡⇡ *⇡ f(t) sin(nt)dt=1 ⇡H0 *⇡ *tsin(nt)dt+⇡ 0 tsin(nt)dtI=0. It is worth noting that if a developed function is even, and all coefficients bn are zero, then the Fourier series will have only cosine elements; it will be even. This is not a coincidence, as we will show in the next section. The sought-after development of our function is therefore f(t)=⇡ 2*4 ⇡ ÿ … n=1 cos(2n*1)t (2n*1)2, where we applied an=0for even n. Due to Dirichlet’s theorem 5, the sum of this series equal f(t)for tÀR; a graph of the sum is in Figure 1.9. Partial sums of the first members: s2(t)=⇡ 2*4 ⇡0cos(t)+cos(3t) 91,(1.47) s3(t)=⇡ 2*4 ⇡0cos(t)+cos(3t) 9+cos(5t) 25 1,(1.48) are given in Figures 1.10 and 1.11 respectively.
1.3. DEVELOPMENT OF PERIODIC FUNCTION 19 Figure 1.11: Graph of the sum s3(t)of the development of (1.46). Let us now compose a one-sided and two-sided phase and amplitude spectrum, using formulas (1.37), (1.38), (1.39), (1.40) and (1.24), (1.25), (1.26): A0=ÛÛÛÛ a0 2ÛÛÛÛ=⇡_2, An=ta2 n+b2 n=u2 ⇡n2((*1)n*1) + 0 = 2 ⇡n2(*1)n*1, cn=1 2(an*ibn)=1 2⇠2 ⇡n2((*1)n*1) *i0⇡=1 ⇡n2((*1)n*1), 'n=*arg cn=h n l n j 0for n= ±2,±4,±6,…, ⇡for n= ±1,±3,±5,…. The two-sided amplitude (resp. phase) spectrum is shown in Figure 1.12 (resp. 1.13). The values of the coefficients are given in table 1.2. n-3 -2 -1 0 1 2 3 an— — — ⇡*4_⇡0*4_(9⇡) bn— — — — 0 0 0 cn*2_(9⇡)0*2_⇡ ⇡_2*2_⇡0*2_(9⇡) cn2_(9⇡)02_⇡02_⇡02_(9⇡) An— — — ⇡_2 4_⇡04_(9⇡) 'n⇡0⇡—⇡0⇡ Table 1.2: Table of coefficients of harmonic analysis of a function (1.46).
20 CHAPTER 1. FOURIER SERIES Figure 1.12: Two-sided amplitude spectrum of the function (1.46). Figure 1.13: Two-sided phase spectrum of the function(1.46). 1.4 Sine and cosine series In Exercises 3 and 4 we observed the interesting property that when the given function is odd (resp. even), then the Fourier series contains only sine (resp. cosine) elements. The following theorems describe this observation Theorem 2 Let f(t)be an odd function with a period of 2⇡satisfying Dirichlet’s conditions. Then its Fourier expansion contains only sine terms f(t)= ÿ … n=1 bnsin(nt).(1.49) Proof: Firstly, by Dirichlet’s theorem 5 the series (1.49) converges. Now we will show that an=0for nÀN‰{0} and bnÀR. Since the function f(t)is even, that is f(*t)=*f(t)for each t, and also f(t) cos(nt)is odd, we get
1.4. SINE AND COSINE SERIES 21 for nÀN a0=1 ⇡⇡ *⇡ f(t)dt=1 ⇡H0 *⇡ *f(*t)dt+⇡ 0 f(t)dtI= =1 ⇡0*⇡ 0 f(t)dt+⇡ 0 f(t)dt1=0, an=1 ⇡⇡ *⇡ f(t) cos(nt)dt= =1 ⇡H0 *⇡ f(t) cos(nt)dt+⇡ 0 f(t) cos(nt)dtI= =1 ⇡0*⇡ 0 f(t) cos(nt)dt+⇡ 0 f(t) cos(nt)dt1=0, where we used linear substitutions to calculate the first integral in the penultimate equality t=*t. The function f(t) sin(nt)is obviously even, for nÀNwe have bn=1 ⇡⇡ *⇡ f(t) sin(nt)dt= =1 ⇡H0 *⇡ f(t) sin(nt)dt+⇡ 0 f(t) sin(nt)dtI= =1 ⇡0⇡ 0 f(t) sin(nt)dt+⇡ 0 f(t) sin(nt)dt1= =2 ⇡⇡ 0 f(t) sin(nt)dt. ∏ Theorem 3 Let f(t)be an even periodic function with a period of 2⇡satisfying Dirichlet’s conditions. Then its Fourier evolution contains only cosine terms f(t)=a0 2+ ÿ … n=1 ancos(nt).(1.50) Proof: We will lead the proof as before, now f(*t)=f(t). Hence, the function f(t) sin(nt)is obviously odd and for nÀNwe have bn=1 ⇡⇡ *⇡ f(t) sin(nt)dt= =1 ⇡H0 *⇡ f(t) sin(nt)dt+⇡ 0 f(t) sin(nt)dtI= =1 ⇡0*⇡ 0 f(t) sin(nt)dt+⇡ 0 f(t) sin(nt)dt1=0.
22 CHAPTER 1. FOURIER SERIES Next, the function f(t) cos(nt)is even and for nÀN‰{0} it is an=1 ⇡⇡ *⇡ f(t) cos(nt)dt= =1 ⇡H0 *⇡ f(t) cos(nt)dt+⇡ 0 f(t) cos(nt)dtI= =1 ⇡0⇡ 0 f(t) cos(nt)dt+⇡ 0 f(t) cos(nt)dt1= =2 ⇡⇡ 0 f(t) cos(nt)dt. ∏ The previous proofs are also instructive for how to calculate the relevant coefficients. If the function f(t)is even with the period T=2lwith the basic interval of periodicity (*l,l], all coefficients will be an=0and bn=2 ll 0 f(t) sin ⇠n⇡ lt⇡dt. If the function f(t)is odd with the period T=2lwith the basic interval of periodicity (*l,l], all coefficients will be bn=0and an=2 ll 0 f(t) cos ⇠n⇡ lt⇡dt. Letusassumethat on the interval(0,l]isgivenafunctionf(t)satisfying Dirichlet’s conditions and we would like develop it into a Fourier series. Such a task can be performed in various ways. The entered function can be extended to an interval (*l,l], which we can do by defining the interval (*l,0) so that the extension is even or odd. Definition 1 Let f(t)be a piecewise continuous function on the interval (0,l].Odd periodic extension of the function f(t)with the basic interval of periodicity (*l,l] is a function g(t)defined by g(t)=h n l n j f(t)for tÀ[0,l], *f(*t)for tÀ(*l,0). (1.51) Definition 2 Let f(t)be a piecewise continuous function on the interval (0,l]. Even periodic extension of the function f(t)with the basic interval of periodicity (*l,l]is a function g(t)defined by g(t)=h n l n j f(t)for tÀ(0,l], f(*t)for tÀ(*l,0). (1.52)
1.4. SINE AND COSINE SERIES 23 Figure 1.14: Graph of odd extension of the function (1.53). The series (1.49) is called a sine Fourier sequence and the sequence (1.50) cosine Fourier sequence. The attentive reader has noticed that we assume a basic interval (0,l], which may seem restrictive when the specified function is defined on an interval (a,b]. Then just perform a coordinate transformation ⌧=t*aand transform the interval (a,b]on (0,l], where l=b*a. Example 5 Let’s develop the following function in the sine and cosine Fourier series f(t)=tsin(t)for tÀ(0,⇡].(1.53) Sine Fourier sequence Firstly, we make an odd extension (see Figure 1.14). The developing function has a period of 2⇡and the basic interval of periodicity (*⇡,⇡]. according to Theorem 2, an=0and bn=2 lîl 0f(t) sin ⇠n⇡ lt⇡dt.So, for n=2,3,…it is bn=2 ⇡⇡ 0 tsin(t) sin (nt)dt=4n ⇡ (*1)n*1 (n*1)2(n+ 1)2. For n=1we get b1=2 ⇡⇡ 0 tsin2(t)dt=⇡ 2. The series gets the form f(t)=⇡ 2sin(t)+ ÿ … n=2 4n ⇡ (*1)n*1 (n*1)2(n+ 1)2sin(nt). Cosine Fourier sequence Firstly, we make an even extension (see Figure 1.15). The developing function has a period of 2⇡and the basic interval of periodicity (*⇡,⇡]. according to Theorem 3, bn=0and an=2 lîl 0f(t) cos ⇠n⇡ lt⇡dt.So, for n=0,2,3,…it is an=2 ⇡⇡ 0 tsin(t) cos (nt)dt=2 (*1)n+1 (n*1)(n+ 1).
24 CHAPTER 1. FOURIER SERIES Figure 1.15: Graph of even extension of the function (1.53). For n=1we get a1=2 ⇡⇡ 0 tsin(t) cos (t)dt=1 2. The series gets the form f(t)=1+1 2cos(t)+ ÿ … n=2 2(*1)n+1 (n*1)(n+ 1) cos(nt). 1.5 Properties of Fourier series Theorem 4 For every piecewise continuous function f(t)on the interval [a,b]it holds that lim nôÿb a f(t) sin(nt)dt=0,(1.54) lim nôÿb a f(t) cos(nt)dt=0.(1.55) Proof: If the interval [a,b]has the length 2⇡, then it is easy to see that formulae (1.54) and (1.55) hold. If the interval [a,b]has a length greater than 2⇡, then split it into k+1intervals [a,b]=Lk Õ i=1 [a+ 2(i*1)⇡,a+2i⇡]M‰[a+2k⇡,a+2k⇡ +b], where the length of the last one is less then 2⇡. Now, extend the function f(t)on the right from point bsuch that it will equal zero on the interval [b,a+ 2(k+ 1)⇡]. Then b a f(t) sin(nt)dt= k+1 … i=1 a+2i⇡ a+2(i*1)⇡ f(t) sin(nt)dt and each of the integrals on the right is for nôÿin a limit equal to zero. Analogously to (1.55), proving this theorem. ∏
1.5. PROPERTIES OF FOURIER SERIES 25 The following theorem, which we mentioned in section 1.1 and used in section 1.3, gives the answer to the question of what conditions the functions must meet f(t), so that the respective Fourier series converge. Theorem 5 (Dirichlet’s) If the function f(t)fulfills the so called Dirichlet conditions, then the Fourier series of the function f(t)is convergent at every tto the value 1 2(f(t+ 0) + f(t*0)) and it holds that 1 2(f(t+ 0) + f(t*0)) = a0 2+ ÿ … n=1 ancos(nt)+bnsin(nt). Moreover in points t, where f(t)is continuous, it is 1 2(f(t+ 0) + f(t*0)) = f(t). Proof of Dirichlet’s theorem: It is possible to show that (see [9]), for every nit holds that 1= 1 ⇡⇡ 0 sin ⇠⇠1 2+n⇡t⇡ sin ⇠t 2⇡dt.(1.56) The second relation will be multiplied by f(t+ 0) + f(t*0) 2(1.57) and we get f(t+ 0) + f(t*0) 2=1 ⇡⇡ 04f(t+ 0) + f(t*0) 25sin ⇠⇠1 2+n⇡u⇡ sin ⇠u 2⇡du. (1.58) Now, we introduce Rn(t)=sn(t)*f(t+ 0) + f(t*0) 2,(1.59) where sn(t)=1 2a0+ n … k=1 (akcos(kt)+bksin kt).(1.60) To prove this theorem, it is now sufficient to show that lim nôÿRn(t)=0.
32 CHAPTER 1. FOURIER SERIES for m=nit is h n n l n n j î⇡ *⇡cos(mt) sin(mt)dt=0, î⇡ *⇡cos(mt) cos(mt)dt=⇡, î⇡ *⇡sin(mt) sin(mt)dt=⇡. (1.69) By normalizing a given sequence, we obtain an orthonormal system of functions: 1 ˘2⇡ ,cos(t) ˘⇡ ,sin(t) ˘⇡ ,cos(2t) ˘⇡ ,sin(2t) ˘⇡ ,… Example 11 The system of functions {eint}ÿ n=*ÿ is not on the interval [0,⇡]orthogonal. In really: (fm,fn)=⇡ 0 eimte*int dt=(*1)m*n*1 m*në0for m*nodd. Let a given sequence of functions be {fn}ÿ n=1 fromL2(a,b). Ifthere is a function fÀL2(a,b)such that lim nôÿÒfn*fÒ=0,(1.70) then the sequence is called {fn}ÿ n=1 convergent to fin the norm L2(a,b). Sometimes the designation convergence in a diameter or convergence in terms of standard deviation is also used. The sequence of functions {fn}ÿ n=1 from L2(a,b)converges on Mto a function funiformly, if for any ✏>0there is n0such that for every n>n0and every zÀM is fn(z)*f(z)<✏. Note that if we know the sequence {fn}ÿ n=1 is uniformly convergent, it is also convergent. The following example shows that this may not be the case. Example 12 The geometric sequence {tn}ÿ n=1 is convergent but not uniformly on the interval [0,1). It is easy to see that a given sequence converges to 0on the interval [0,1). Next, it holds that sup tn=1for tÀ[0,1), so 0is not a limit, hence a given sequence is not convergent uniformly. 1.7 Generalized Fourier sequence As we have already described in the previous paragraphs, the whole theory of Fourier series arose from the need to develop a given periodic function into a periodic function formed by trigonometric functions; see Example 10. Most of the theorems valid for Fourier series remain valid if we replace the trigonometric functions in the original considerations with a system of functions that are orthogonal or orthonormal (see Section 1.6).
1.7. GENERALIZED FOURIER SEQUENCE 33 Therefore, the question arises whether the given function f(t), which is integrable with the square on the interval [a,b], develops into an infinite series ÿ … n=0 ↵n'n(t) using an orthonormized system of functions {'n}ÿ n=0,'nÀL2(a,b)and determines the coefficients ↵n. 1. Approximation We will approximate the function f(t)by a polynomial Tnbased on the smallest standard deviation, provided that {'n(t)}ÿ n=0 is an orthonormal system of functions. So the question is how to choose the coefficients ↵nin the polynomial Tk(t)=↵0'0(t)+↵1'1(t)+5+↵k'k(t), to make the value of the integral Ik=1 b*ab a [f(t)*Tk(t)]2dt minimal. Let’s modify the given integral Ik=1 b*a0b a [f(t)]2dt*2b a [f(t)Tk(t)] d t+b a [Tk(t)]2dt1. Then it is b a f(t)Tk(t)dt= k … n=0 ↵nb a f(t)'n(t)dt. Denote an=îb af(t)'n(t)dtand let’s call this number the Fourier coefficient of the function f(t)with respect to a given set of functions {'n(t)}ÿ n=0. Then we have b a f(t)Tk(t)dt= ÿ … n=0 ↵nan. Now let’s compute b a [Tk(t)]2dt=b a [ k … n=0 ↵n'n]2dt= =b a`rrrp k … n=0 ↵2 n'2 n+2 k … m,n=0 n<m ↵n↵m'n'masssq dt= = k … n=0 ↵2 n,
34 CHAPTER 1. FOURIER SERIES due to the orthonormality of the system{'n(t)}ÿ n=0. The integral therefore has the form Ik=1 b*aHb a f(t)2dt+ k … n=0 (↵2 n*2↵nan+a2 n*a2 n)I=(1.71) =1 b*aHb a f(t)2dt+ k … n=0 (↵n*an)2* k … n=0 a2 nI.(1.72) This equation applies to any choice of coefficients ↵n. Integral Iktherefore, has a minimum value when selected ↵n=an. Denote the polynomial Tkwith coefficients ↵n=anas Pkand relevant integrals Jk. Then Jk=1 b*ab a [f(t)*Pk(t)]2dt=1 b*aHb a [f(t)]2dt* k … n=0 a2 nI.(1.73) Because for every k Jkg0, it is k … n=0 a2 nfb a [f(t)]2dt.(1.74) the inequality (1.74) is called Bessel. Because the formulae (1.73) and (1.74) hold for any k, infinite sequence ÿ … n=0 a2 n is convergent, since all partial sums are smaller than the given fixed positive number according to (1.74) îb a[f(t)]2dt. 2. Closedness Let us now address the natural question of whether lim kôÿJk=0. Orthonormal systems that meet this property are called closed. Therefore, the following applies to them: lim kôÿb a [f(t)* k … n=0 an'n(t)]2dt=b a [f(t)]2dt* ÿ … n=0 a2 n=0, so b a [f(t)]2dt= ÿ … k=0 a2 k,(1.75)
1.7. GENERALIZED FOURIER SEQUENCE 35 equality (1.75) is called Parseval. We have already proved that the sequence of trigonometric functions is in the interval [0,2⇡]orthonormal, see Example 10. Closedness can be proved by Fejér theorems. 3. Orthonormality In the previous sections, we have already shown that the orthonormality of a system of functions significantly simplifies the development of a given function using these functions. Unfortunately, a power sequence of functions 'k(t)=tkwhere k=0,1,…(1.76) it is not orthonormal or normalized, despite the frequent task of developing a function f(t)into a power series. The way to compile the coefficients of such a power series is known; it is enough to develop the given function f(t)into a Taylor series. This method is theoretically good, but in practice we come across a problem that may not have a solution, namely determining the values of all orders of derivation of a given function. This problem can be eliminated by not developing the function with the system (1.76), but by a system composed of polynomials, which forms an orthonormed set of functions on a given interval. Theorem 7 (Schmidt’s) Let {'n(t)}ÿ n=0 (1.77) be a sequence of continuous and nonzero functions on the interval [a,b]such that each finite block '0(t),'1(t),…,'k(t)denotes k+1 linearly independent functions. Then a sequence of functions can be created from this sequence { n(t)}ÿ n=0 (1.78) continuous on this interval [a,b]such that 1. each of its finite blocks 0(t), 1(t),…, k*1(t)denotes klinearly independent functions, 2. eachfunction k(t)isalinearcombinationoffunctions'0(t),'1(t),…,'k*1(t), 3. the sequence (1.78) forms an orthonormal sequence. The proof of this theorem has two parts. The first part is constructive; the appropriate orthonormal system is constructed. In the second, the validity of the properties described by the sentence is verified. Now let’s do the construction, as the verification of the given properties is easy to do and has been left for the reader as an exercise. Construction of the sequence(1.78): 1. Put 0(t)='0(t) c0 ,(1.79)
36 CHAPTER 1. FOURIER SERIES where c2 0=îb a'2 0(t)dt. The function 0(t)is obviously normalized. Hence, b a 2 0(t)dt=1_c2 0b a '2 0(t)dt=1. 2. Now, introduce 1(t)='1(t)*a10 0(t),(1.80) where a10 is picked in such a way that the function 1(t)is orthogonal to the function 0(t), so it will be b a 1(t) 0(t)dt=b a ['1(t)*a10 0(t)] 0(t)dt=0. From here we have a10 =b a a10 2 0(t)dt=b a '1(t) 0(t)dt. Denote c2 1=b a 2 1(t)dt, then the function 1(t)=1(t) c1 (1.81) is on the interval [a,b]orthonormal to the function 0(t). Similarly, we introduce a function 2(t)='2(t)*a20 0(t)*a21 1(t),(1.82) where we choose a20 and a21 such that 2(t)is orthogonal to the functions 0(t) and 1(t). We can easily deduce that a20 =b a '2(t) 0(t)dt,(1.83) a21 =b a '2(t) 1(t)dt.(1.84) Then the function 2(t)=2(t) c2 where c2 2=b a 2 2(t)dt,(1.85) is normalized and orthogonal to 0(t)and 1(t). 3. To specify additional members of the sequence (1.78) it is enough to proceed analogously with the help of mathematical induction. Note the numbers ckare generally different from zero with respect to the linear independence of each finite section of the sequence (1.77). From the construction (1.78), the linear independence of each finite block then follows.
1.7. GENERALIZED FOURIER SEQUENCE 37 Example 13 Apply Theorem 7 to the power sequence of functions (1.76) on the interval [*1,1]. Using the process described above, we derive polynomials which, except for constant factors, are so-called Legender polynomials. Let’s construct the first three members of the sought orthonormed system. Firstly, '0(t)=1and c2 0=1 *1 '2 0(t)dt=2, so c0=˘2 and 0(t)= 1 ˘2 . Next, put 1(t)='1(t)*a10 0(t)=t*a10 1 ˘2 , a10 =1 *1 '1(t) 0(t)dt=1 *1 t ˘2 =0, hence 1(t)=t and c2 1=1 *1 2 1(t)dt=1 *1 t2dt=2 3. So, we get 1(t)=1(t) c1 =t ˘2_3 =u3 2t. Now, let 2(t)='2(t)*a20 0(t)*a21 1(t)=t2*1 3, c2 2=1 *1 2 2dt=8 45 and 2(t)=2(t) c2 =u5 2⇠3 2t2*1 2⇡. The first three members of the search sequence are: 0(t)= 1 ˘2 , 1(t)=u3 2t, 2(t)=u5 2⇠3 2t2*1 2⇡.
38 CHAPTER 1. FOURIER SERIES Consequently, we can develop the function f(t), which is continuous on the interval [*1,1], into a series made up of these polynomials. 1.8 Gibbs phenomenon In the previous sections we dealt with the development of functions from L2(a,b) into the Fourier series. We already know that the Fourier series of functions fÀ L2(0,2⇡)converges in a norm of the space L2(0,2⇡)to the function f. In addition, if other conditions are met (see Theorem 2.11 from [7]), then the Fourier series converges uniformly. A simple example of a function that does not meet these conditions is f0(t)=⇡*t 2pro tÀ(0,2⇡), which is periodically extended to the whole R; see Figure 1.18. It can be verified (see Chapter 2 from [12]), that f0(t)= ÿ … k=1 sin(kt) k,(1.86) here equality applies in the sense of convergence in L2(l,l+2⇡),lÀR, and also in terms of the uniform convergence on each interval with extreme points 2l⇡ +✏and 2(l+1)⇡*✏, where ✏À(0,⇡)(see Theorem 2.4 from [7]). The problem is therefore the points of discontinuity of the function f0; for example, we do not know what is happening at the point t=0. At this point the sum of the series (1.86) equals 0, thus, the average limit on the right and left is zero. 1 2⌅f0(0 + 0) + f0(0 *0)⇧=0. Let us now focus on the error of the partial sum of the series (1.86) Rn(t)=sn(t)*f0(t)= n … k=1 sin(kt) k*⇡*t 2for tÀ(0,2⇡). We can easily verify that R® n(t)=1 2+ n … k=1 cos(kt)= sin ⇠⇠n+1 2⇡t⇡ 2 sin ⇠1 2t⇡, Rn(0) = *⇡ 2 and hence Rn(t)=*⇡ 2+t 0 sin ⇠⇠n+1 2⇡x⇡ 2 sin ⇠1 2x⇡dx.
1.8. GIBBS PHENOMENON 39 The function sn(t)is increasing in a neighborhood of zero. Now let’s look for such a point tn>0, in which the function Rn(t)has a local extreme and is closest the zero. From the equation R® n(t)=0we get sin ⇠⇠n+1 2⇡t⇡=0 and so xn=⇡⇠n+1 2⇡*1 . If we denote n+1_2=pand apply the substitution px =swe get Rn(tn)=⇡_p 0 sin(px) 2 sin ⇠1 2x⇡dx*⇡ 2=⇡ 0 sin(s) 2psin 0s 2p1*⇡ 2. For a large enough p(that is large enough for n) it is 2psin 0s 2p1g0, for sÀ(0,⇡)and lim pôÿ2psin 0s 2p1=s. Now we find the integral majorant and we get lim nôÿRn(tn)=⇡ 0 sin(s) sds*⇡ 2 . =0,18⇡ and for a large nit is sn(tn). =1,18⇡ 2. Each partial sum sn(t)has a maximum that exceeds by about 18% the maximum of the function f0, see Figure 1.18. This phenomenon is called Gibbs phenomenon. The maximum with increasing nis still significantly different to the maximum of the function f0, just the point tnis tending to zero, in which the maximum is reached. Therefore, we can never achieve a partial sum series (1.86) that approximates the function f0uniformly. It is possible to demonstrate (see Chapter 6 in [7]) that each function f, which has a finite number of points of discontinuity of the first kind, can be written in the form f(t)=g(t)+h(t), whereg(t)isa function thatmeetsouradditionalconditionsimposedon the function f0and whose Fourier series converges uniformly, and where h(t)= m … i=1 cif0(t*ti)
40 CHAPTER 1. FOURIER SERIES f(t) t ⇡_2 *⇡_2 2⇡*2⇡ f0 sn - 6 HHHHHHHHHHHH HHHHHHHHHHHH Figure 1.18: Gibbs phenomenon is a function that captures the jumps of a function f. From the properties of the function f0we know that the Fourier series of a function fconverges at every point tto the value 1 2[f(t+ 0) + f(t*0)] and that in every neighborhood of the function f, the Gibbs phenomenon manifests itself. Thus, a partial sum of the Fourier series of a function fwill be in a neighborhood of each point of discontinuity tiand acquire, except for a negligible deviation, a value 1 2[f(ti+ 0) + f(ti*0)] ± 1 21,18[f(ti+ 0) + f(ti*0)], therefore, the partial sums will not converge uniformly around the point of discontinuity. 1.9 Worked example In this section we will solve one example in which we will look for the Fourier series of a periodic signal. As follows from the previous sections, the Fourier series of a periodic signal is a mathematical notation of the statement that a periodic signal f(t)with a repetition frequency of 1_Tcan be composed of a constant signal and
1.9. WORKED EXAMPLE 41 harmonic signals of frequencies 1_kT where k=1,2,3,…. Hence, f(t)=A0+A1cos(!t +'1) ´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨ the first (basic) harmonic + +A2cos(2!t +'2)+A3cos(3!t +'3)+ +5+Akcos(k!t +'k) ´≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠¨ k-th harmonic (higher) +5 =A0 ´Ø¨ direct component (mean value) + ÿ … k=1 Akcos(k!t +'k), ´≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠¨ alternating component (1.87) where Akis the amplitude of k-th harmonic component, k! is the circular repetition frequency k-th harmonic component and 'kis the initial phase of k-th harmonic element. From the above formula (1.87) it is obvious that each periodic signal has an alternating and direct element. The direct component is equal to the mean value of the signal over the repetition period. The alternating component consists of harmonic signals with zero mean values, so it is the original signal devoid of the direct component. The alternating component contains the so called first harmonic of a frequency, which is the same as the repetition frequency of the periodic signal, and from the higher harmonic, of which there is generally an infinite number and whose frequency is an integer multiple of the frequency of the first harmonic. In decomposition of the above periodic signal (1.87) the sub-components are unambiguous and every two different periodic signals of the repetition frequency !are unambiguously represented by different pairs of sets {A0,A1,…Ak,…}and {'0,'1,…'k,…}; see the section Fourier series in the complex plane 1.2. The graphical representation of these sets in the form of spectral lines on the frequency axis is called the spectrum of the signal. If a signal passes through an electrical circuit, we can understand it as the passage of a set of its harmonic components. Due to the different transmission capabilities of the circuit at different frequencies, the individual harmonic components at the output of the circuit will be differently attenuated and phase shifted, so that the output signal will also be periodic, but will be distorted compared to the input signal. The signal spectrum, resp. the distribution of its spectral lines on the frequency axis, together with the frequency characteristics of the circuit, provides a useful and illustrative tool for understanding the phenomena associated with signalcircuit interactions. Because the harmonic signal can be written in other forms than those shown in the discussed formula (1.87) (specifically in the decomposition into sine and cosine
48 CHAPTER 1. FOURIER SERIES The task for any initial data (1.88) is uniquely solvable and the appropriate solution is given by formula (1.95). The conditions (1.96) and (1.97) are very restrictive and difficult to guarantee. For this reason, we are looking for a solution to problem (1.88) in the form of an infinite sum, and we express it in the form of a Fourier series u(x,t)= ÿ … n=1 ⇠Ancos ⇠n⇡ct l⇡+Bnsin ⇠n⇡ct l⇡⇡sin ⇠n⇡x l⇡,nÀN.(1.98) The constants Anand Bnare then given as Fourier coefficients of sine developments of functions '(x)and (x), hence '(x)= ÿ … n=1 Ansin ⇠n⇡x l⇡, (x)= ÿ … n=1 n⇡c lBnsin ⇠n⇡x l⇡. In other words, solving problem (1.88) for the wave equation, it is at all times texpressed in the form of a Fourier sine series in the variable x, if the initial conditions of '(x)and (x)can be expressed. It turns out that for a sufficiently wide class of functions, such a decomposition is possible, and the respective series converge. In this case, we calculate the coefficients using the formulae An=2 ll 0 '(x) sin ⇠n⇡x l⇡dx, Bn=2 n⇡c l 0 (x) sin ⇠n⇡x l⇡dx. Finally, it should be noted that convergence needs to be discussed for correctness. Convergence considerations need to be applied to functions 'and ;, this have been done in the previous sections.
Chapter 2 Laplace transform The Laplace transform is an effective method for solving various practical tasks in the fields of mathematical physics, electrical engineering, and control systems. This chapter is motivated by the works [5] - [13] and is organized as follows. We will first introduce the necessary terms, then in section 2.1 we will pay attention to the properties of the Laplace transform. In section 2.2 we will learn to perform the inverse Laplace transformation, and finally in section 2.3 we will look at the following application examples: solution of differential equations (and systems), examples from areas of electrical engineering and control systems. We will consider complex functions fof the real variable tÀ(*ÿ,ÿ), that is f:RôC, and complex variable p=x+iyÀC. Let us assume that the improper integral ÿ 0 f(t)e*pt dt(2.1) exists and has a finite value for at least one p. Then the integral (2.1) is called the Laplace integral of the function f. Example 14 Let’s calculate the Laplace integral of a function f(t)=1. By (2.1) we have ÿ 0 f(t)e*pt dt=ÿ 0 e*pt dt= lim ↵ôÿ↵ 0 e*pt dt= lim ↵ôÿ01 p*1 pe*p↵1. Since ↵ÀR, for p=x+iyit holds that e*p↵=e*x↵. So, for Re p>0it holds that lim↵ôÿe*p↵ =0, and the Laplace integral of the function f(t)=1for Re p>0converges and equals the function 1_p. For Re pf0the Laplace integral does not exist. Example 15 Let’s calculate the Laplace integral of a function f(t)=eat where 49
50 CHAPTER 2. LAPLACE TRANSFORM aÀC. By (2.1) we have ÿ 0 f(t)e*pt dt=ÿ 0 eate*pt dt= lim ↵ôÿ↵ 0 e(a*p)tdt= = lim ↵ôÿ01 a*pe(a*p)↵*1 a*p1=1 p*a for Re(p*a)>0. So, the Laplace integral f(t)=eat converges for Re p>Re ato the function 1_(p*a)and otherwise diverges. Definition 3 Let fbe a complex function of a real variable tÀ(*ÿ,ÿ). Let MœCbe a set of all p, for which the Laplace integral (2.1) converges. Then the complex function Fdefined by F(p)=ÿ 0 f(t)e*pt dt(pÀM)(2.2) is called the Laplace image of the function f. A given map that assigns a function fto its Laplace image F, is called a Laplace transform, and is denoted by L(f(t)) = F(p). Definition 4 The function fis called a subject (sometimes also preimage or original), if the following conditions are fulfilled: 1. fis on the interval [0,ÿ)piecewise continuous, 2. f(t)=0for each t<0, 3. there is a real number M>0and ↵such that for each tÀ[0,ÿ)it holds that f(t)fMe↵t.(2.3) Definition 5 Let ↵0= inf{↵ÀR:↵satisfies (2.3)}. The number ↵0is called the growth index of the function f. An important example of the subject is the Heaviside function shown in Figure 2.1, defined by ⌘(t)=h n l n j 0,for t<0, 1,for tg0. (2.4) Theorem 8 (on the existence of the Laplace image) Letfbeapreimagewiththe growth index ↵0. Then the Laplace integral F(p)=ÿ 0 f(t)e*pt dt converges in the half-plane Re p>↵0(see Figure 2.1) absolutely and defines Laplace image L(f(t)) = F(p), which is in this half-plane analytical function.
51 ⌘(t) t0 1 - 6 s c Figure 2.1: Heaviside function. Re p>↵0 ↵0x0 y - 6 Figure 2.2: Half-plane Re p>↵0. Proof: First, let’s prove the absolute convergence of the integral in the half-plane Re p>↵0. The existence of the integral î⌧ 0f(t)e*pt dt, for every ⌧>0, follows from the fact that f(t)is on [0,ÿ)piecewise continuous. If p=x+iyand Re p> ↵0, then e*pt=e*xt. From the third condition to the subject, then for each ↵such that x>↵>↵0it follows that f(t)e*ptfMe↵te*xt =Me(↵*x)t. So, ÛÛÛÛÿ 0 f(t)e*pt dtÛÛÛÛfÿ 0f(t)e*ptdtfMÿ 0 e(↵*x)tdt=4Me(↵*x)t ↵*x5ÿ 0 . Since ↵*x<0, it is limtôÿe(↵*x)t=0and ÛÛÛÛÿ 0 f(t)e*pt dtÛÛÛÛfM x*↵.(2.5) Thus we have proved that the Laplace integral converges absolutely in the half-plane Re p>↵0. Now, we will show that Fis analytic in the half-plane Re p>↵0. Let ↵1be such that xg↵1>↵>↵0. Then from the third condition we have the subject f(t)e*ptfMe(↵*↵1)t.
52 CHAPTER 2. LAPLACE TRANSFORM Theexpressionofthe right side ofinequalitydoesnotdependonpandsince↵*↵1< 0, integral ÿ 0 Me(↵*↵1)tdt converges. Thus the Laplace integral converges in the half-plane Re pg↵1absolutely. Moreover, it holds that ) )p[f(t)e*pt]=*tf(t)e*pt and *tf(t)e*ptfMte(↵*↵1)t, where ÿ 0 Mte(↵*↵1)tdt=M (↵*↵1)2. That means, the integral ÿ 0 *tf(t)e*pt dt converges in the half-plane Re pg↵1. Overall, therefore, the Laplace integral of a function fcan be integrable according to the parameter p. That is, the function Fis analytical in the half-plane Re pg↵1>↵0. Since ↵1was arbitrarily taken, it follows that Fis analytical in the half-plane Re p>↵0.∏ Corollary 1 Let L(f(t)) = F(p)and x=Rep. Then limxôÿF(p)=0. Proof: The statement follows from (2.5) and the proof of Theorem 8. ∏ Example 16 Let’s find a function f(t)such that its Laplace image equals the function ˘p. Firstly, limxôÿ˘x+iyë0. That is, according to the previous Corollary 1 the function ˘pcannot be a Laplace image of any function f(t). Proof of the following statement, which gives us complete information about the behavior of the function Fin the neighborhood of ÿ, exceeds the scope of this text and can be found, for example, in [7]. Theorem 9 (the first limit) Let fbe a preimage with a growth rate ↵0a↵>↵0. Then for the Laplace image Fof the function fit holds that lim pôÿ Re pg↵ F(p)=0.
2.1. PROPERTIES OF THE LAPLACE TRANSFORM 53 2.1 Properties of the Laplace transform In this section we will deal with the basic properties of Laplace transformation. We will formulate a theorem, which will be the corner stone for the operator calculus, and give a number of examples to illustrate it. These examples will give us a large number of images of the functions that are important and used in practice. We close the section with the second and third limit theorem and Duhamel’s formula. Theorem 10 (rules of the operator calculus) Letfkbe preimages, L(fk(t)) = Fk(p) and ckÀCfor k=1,2,…n. Then I. linearity L( n … k=1 ckfk(t)) = n … k=1 ckFk(p), II. time scaling L(f(t)) = 1 F⇠p ⇡,>0, III. Laplace domain shifting L(eatf(t)) = F(p*a), IV. derivative by a parameter )f(t,) ) =)F(p,) ) ,kde L(f(t,)) = F(p,), V. time shifting for every ⌧>0it holds that L(f(t*⌧)⌘(t*⌧)) = e*⌧pF(p), VI. time domain derivative L(f(n)(t)) = pnF(p)*pn*1f(0+)*5*f(n*1)(0+), where fand its derivatives up to the order n*1are continuous and f(i)(0+)= limtô0+f(i)(t), VII. Laplace domain derivative L(*tf(t)) = F®(p),
54 CHAPTER 2. LAPLACE TRANSFORM VIII. time domain integration L0t 0 f(⌧)d⌧1=F(p) p, IX. Laplace domain integration L0f(t) t1=ÿ p F(z)dz= lim Re qôÿq p F(z)dz, where f(t)_tis a preimage with a growth rate ↵0,îÿ pF(z)dzexists, and graph of the integrating curveîÿ pF(z)dzis a subset of Re p>↵0. Proof: I. linearity From the linearity of the integral we have L( n … k=1 ckfk(t)) = ÿ 0Hn … k=1 ckfk(t)Ie*pt dt= = n … k=1 ckÿ 0 fk(t)e*pt dt= n … k=1 ckFk(p), where îÿ 0fk(t)e*pt dt=Fk(p). Note that the integral ÿ 0Hn … k=1 ckfk(t)Ie*pt dt converges in half-plane Re p>↵0where ↵0= maxi=1,2,…n{↵i 0}and ↵i 0is the growth index fifor every i. II. time scaling By (2.2) we have L(f(t)) = ÿ 0 f(t)e*pt dt. Let’s substitute on the right side u=t, (that is dt=1_du), then L(f(t)) = 1 ÿ 0 f(u)e*pu_du=1 F⇠p ⇡. Obviously, if ↵0is a growth index of fdepending on t, then ↵0is a growth index of fdepending on u. III. Laplace domain shifting By (2.2) we have L(eatf(t)) = ÿ 0 eatf(t)e*pt dt=ÿ 0 f(t)e*(p*a)tdt=F(p*a),
2.1. PROPERTIES OF THE LAPLACE TRANSFORM 55 where îÿ 0f(t)e*(p*a)tdtconverges to Re(p*a)>↵0,↵0is a growth index of f. IV. derivative by a parameter The proof of this part goes beyond the complexity of this text, and can be found in [7]. V. time shifting By (2.2) we have L(f(t*⌧)⌘(t*⌧)) = ÿ 0 f(t*⌧)⌘(t*⌧)e*pt dt. Moreover,f(t*⌧)⌘(t*⌧)=0for every tÀ(0,⌧), so L(f(t*⌧)⌘(t*⌧)) = ÿ ⌧ f(t*⌧)⌘(t*⌧)e*pt dt. If we introduce substitutiont*⌧=u, then we get L(f(t*⌧)⌘(t*⌧)) = ÿ 0 f(u)⌘(u)e*p(u+⌧)du=e*p⌧ ÿ 0 f(u)e*pu du, consequently, L(f(t*⌧)⌘(t*⌧)) = e*p⌧ ÿ 0 f(t)e*pt dt=e*p⌧F(p). VI. time domain derivative Firstly, let’s prove the case for i=1, i.e. L(f®(t)) = pF (p)*f(0+). By (2.2) we have L(f®(t)) = ÿ 0 f®(t)e*pt dt, for Re p>↵0. We note that if ↵0is the growth index of the function f®, then it is also the growth index of f. We now calculate the integral on the right hand using the method of integration by parts, i.e. for tÀ(0,ÿ)putting u(t)=f(t),v(t)=e*pt, u®(t)=f®(t),v®(t)=*pe*pt. Then L(f®(t)) = [f(t)e*pt]ÿ 0+pÿ 0 f(t)e*pt dt.(2.6) Since f(t)fMe↵t (↵>↵0) for every psuch that Re p>↵>↵0then it holds that f(t)e*ptfMe(↵*Re p)t.
56 CHAPTER 2. LAPLACE TRANSFORM Moreover limtôÿMe(↵*Re p)t=0and lim tôÿf(t)e*pt =0.(2.7) If we substitute (2.7) into (2.6) we get L(f®(t)) = pF (p)*f(0+). For arbitrary nthe proof can be done by mathematical induction. VII. Laplace domain derivative From the proof of Theorem 8 it follows that ÿ 0 f(t)e*pt dt and ÿ 0 *tf(t)e*pt dt and they are uniformly convergent in the half-plane Re pg↵>↵0, where ↵0 is growth index f. The Laplace integral of the function fcan be differentiated according to the parameter p, so, if L(f(t)) = ÿ 0 f(t)e*pt dt=F(p), then F®(p)=ÿ 0 ) )p(f(t)e*pt)dt=ÿ 0 *tf(t)e*pt dt. Finally, by (2.2) we have L(*tf(t)) = F®(p). VIII. time domain integration Denote L(g(t)) = L0t 0 f(⌧)d⌧1=G(p). Apparently it is true that g®(t)=f(t),g(0) = 0. It follows from the property of deriving an object that L(g®(t)) = pG(p)*g(0+). Hence, L(f(t)) = pG(p). Consequently, G(p)=F(p) p.
2.1. PROPERTIES OF THE LAPLACE TRANSFORM 57 IX. Laplace domain integration Let L(f(t)_t)=G(p). Then by Theorem 8 the function Gis analytical in the half-plane Re p>↵0. Since G(p)fÿ 0ÛÛÛÛ f(t) te*ptÛÛÛÛdtfMÿ 0 e*(*↵0)tdt=M *↵0 , where Mis a positive constant and Re p=>↵0, we have lim ôÿG(p)=0.(2.8) From the property of image derivation we have L(*f(t)) = G®(p) and F(p)=*G®(p). Then for integrals of functions Fa*G®we have G(p)*G(q)=q p F(z)dz, here we integrate along the curve with endpoints pand qfulfilling the condition Re q>Re p>↵0. Now just go to the limit in the previous formula Re q=ôÿ; we apply (2.8) and get G(p)=ÿ p F(z)dz. ∏ Examples 17 to 24 demonstrate how to use properties of Theorem 10. Example 17 Find the Laplace image of a function f(t) = sin(!t). According to Euler’s formulae, it holds that sin(!t)=ei!t *e*i!t 2i . If we put in Exercise 15 parameter a= ±i!and apply I. from Theorem 10, then for Re p>Re(±i!)=Im !we get L(sin(!t)) = 1 2i 01 p*i!*1 p+i!1=! p2+!2. Example 18 Find the Laplace image of a function f(t)=eat sin(!t). We will use the result of Example 17 for the calculation and property III. from Theorem 10. So, for Re p>Re a+Im !we have L(eat sin(!t)) = ! (p*a)2+!2.
64 CHAPTER 2. LAPLACE TRANSFORM Theorem 17 Let Fbe analytic in Cup to finitely many singular points aiÀC (i=1,2,…n). Let for every aÀRsuch that for a>maxi=1,2,…,n{ai}it holds that: 1. there is a sequence of circle lines kiwith a center at 0and radii Ri, for which it holds that a<R1<5<Rn<…alimnôÿRn=ÿsuch that lim nôÿmax pÀkn {F(p)}=0, 2. the integral îa+iÿ a*iÿF(p)dphas a finite value. Then on Rthere is the continuous preimage fthat is given by f(t)=h n l n j≥n i=1 Res[F(p)ept]p=aifor t>0, 0 for tf0. (2.13) Remark 5 Note that the residues in formula (2.13) are calculated in the singularities of the function F(p)ept. The functions ept are analytic in C, therefore, they are crucial for calculating the residua function F(p). If, for example, aiis a simple pole, then according to the rules for counting residues Res[F(p)ept]p=ai=eaitRes[F(p)]p=ai. If aiis a pole of the second order we get Res[F(p)ept]p=ai= lim pôai d dp[(p*ai)2F(p)ept]= = Res[F(p)]p=aieait+ lim pôai [(p*ai)2F(p)]teait. We notice that images play a significant role, F, which has a rational form, see Examples 17 - 24. Let us now consider the inverse Laplace transform of the functions F(p)=P(p) Q(p),(2.14) where P(p)and Q(p)are polynomials over a complex field. Theorem 18 The function (2.14) is a Laplace image of some preimage if and only if deg(P(p)) <deg(Q(p)). Proof: Firstly, in the half-plane Re pf↵it holds that lim pôÿF(p) = lim pôÿ P(p) Q(p)=0.
2.2. INVERSE LAPLACE TRANSFORM 65 This only happens in cases where deg(P(p)) <deg(Q(p)) (deg denotes degree of the polynomial). On the other hand, let deg(P(p)) <deg(Q(p)). This means that there is decomposition of the right hand (2.14) on partial fractions over the field of complex numbers F(p)=P(p) Q(p)= m … k=1 rk … l=1 Pkl (p*ak)l,(2.15) where Pkl ÀC,rkis a multiple of the root akof polynomial Q(p)and mis the number of different zero points of the polynomial Q(p). From Example 19 we know that L0tl*1 (l*1)!eakt 1=1 (p*ak)l. Due to the linearity of the Laplace transform and (2.15) we have for t>0 F(p)=P(p) Q(p)=LHm … k=1 rk … l=1 Pkltl*1 (l*1)!eaktI. Then from Theorem 16 it follows that f(t)=h n l n j≥m k=1 ≥rk l=1 Pkltl*1 (l*1)! eaktfor t>0, 0for t<0. ∏ From the proof of the foregoing theorem it follows that Theorem 19 (the second theorem on decomposition) The Laplace imageFof the function fis a rational function if and only if if for t>0we can describe the formula fas a linear combination of functions in the form tneat, where nÀN0and aÀC. Example 28 Find out the preimage of the function F(p)= p+1 p2*p. To do this we use the notes in Remark 5. The function Fhas two single poles, 0and 1. So, we have Res[F(p)ept]p=0 =*1,Res[F(p)ept]p=1 =2et. Then by Equation (2.13) of Theorem 17 we have for t>0 f(t)=*1+2et.
66 CHAPTER 2. LAPLACE TRANSFORM Example 29 Findoutthepreimageof thefunctionF(p)= 1 (p+ 1)(p*1)3(p2+ 1). The computations will be done analogously to the previous example, hence using Remark 5. The function Fhas three single poles *1,±i and one pole of the third order. Hence, Res[F(p)ept]p=*1=*1_16 e*t,Res[F(p)ept]p=i =1_8eit Res[F(p)ept]p=*i=*1_8e*it,Res[F(p)ept]p=1 =2t2*6t+5 2et. Then by Equation (2.13) Theorem 17 we have for t>0 f(t)=*1_16 e*t+1_8eit*1_8e*it+2t2*6t+5 2et. Table 2.1: Table of Laplace transforms of common functions Time domain Laplace domain Region of convergence 11 pRe(p)>0 eat 1 p*aRe(p)>Re(a) sin(!t)! p2+!2Re(p)>0 cos(!t)p p2+!2Re(p)>0 sinh(!t)! p2*!2Re(p)>! cosh(!t)p p2*!2Re(p)>! eat sin(!t)! (p*a)2+!2Re(p)>a
2.3. APPLICATIONS OF THE LAPLACE TRANSFORM 67 Table 2.1: Table of Laplace transforms of common functions Time domain Laplace domain Region of convergence eat cos(!t)p*a (p*a)2+!2Re(p)>a tn,nÀNn! pn+1 Re(p)>0 tneat,nÀNn! (p*a)n+1 Re(p)>Re(a) tsin(!t)2p! (p2+!2)2Re(p)>0 tcos(!t)p2*!2 (p2+!2)2Re(p)>0 2.3 Applications of the Laplace transform Solution of ordinary differential equations Consider the Cauchy problem for a linear differential equation with constant coefficients ai(i=1,2,…,n) and the initial conditions x(n)+a1x(n*1) +5+a(n*1)x®+anx=f,(2.16) x(t0)=x0,x®(t0)=x® 0,…,x(n*1)(t0)=x(n*1) 0.(2.17) Next, suppose the right side of the equation fand a solution xincluding their derivatives up to the order of nare preimages. Under these conditions, we can solve the given problem by the Laplace transform. Without loss to generality, we can assume that the initial conditions are given in point t0=0, hence x(0+)=x0,x®(0+)=x® 0,…,x(n*1)(0+)=x(n*1) 0.(2.18)
68 CHAPTER 2. LAPLACE TRANSFORM Denote L(x(t)) = X(p)and L(f(t)) = F(p). Then the equation (2.16) can be written as (Theorem 10) [pnX(p)*pn*1x0*pn*2x® 05*x(n*1) 0]+ a1[pn*1X(p)*pn*2x0*pn*3x® 05*x(n*2) 0]+ 4 a(n*1)[pX(p)*x0]+ anX(p)=F(p). (2.19) After the adjustments we get X(p)=F(p)*P(p) Q(p),(2.20) where Q(p)=pn+a1pn*1+5+an*1p+anis a characteristic polynomial of Equation (2.16) and the degree of the polynomial Pis at most (n*1). Now simply find for the function Xits preimage x. According to the uniqueness of such an inverse Laplace transform, such an object is then (Theorem 16) a solution of differential equation (2.16) on the interval (0,ÿ). Remark 6 1. The procedure described above is called an operator method. 2. The Equation (2.19) is called an operator. 3. The advantage of the operator method is the simplicity of the solution operations. 4. With the solution we get a straight particular solution (if the initial conditions are not known, we get a general solution). Example 30 Let us solve the differential equation h n l n j x®® *2x®+x=4, x(0+)=0,x®(0+)=1. (2.21) We will proceed with the operator method described above. So, put L(x(t)) = X(p), then L(x®(t)) = pX(p), L(x®®(t)) = p2X(p)*1.
2.3. APPLICATIONS OF THE LAPLACE TRANSFORM 69 Next L(4) = 4_p,Re p>0. The corresponding operator equation has the form p2X(p)*1*2pX(p)+X(p)=4_p. We express X(p) X(p)= p+4 p(p*1)2,Re p>1. After decomposition into partial fractions we get X(p)=4 p*4 p*1+5 (p*1)2. The inverse Laplace transform gives for tg0the solution x(t)=4*4et+5tet. In the procedure described above, decomposition into partial fractions can be avoided. Notice that the function X(p)= p+4 p(p*1)2 has at the point 0a single pole and at the point 1a pole of the second order, and uses the algorithm explained in Chapter 2.2 (see Theorem 17 and Example 29). Then according to the known formulas for calculating residues, we get Res[X(p)ept]p=0 =4, Res[X(p)ept]p=1 =5tet*4et. Based on the inverse Laplace transform, we get the solution x(t)=4*4et+5tet. Example 31 Let us solve the differential equation h n l n j x®® +4x= 2 cos(2t), x(0+)=0,x®(0+)=4. (2.22) We will proceed with the operator method described above. So, put L(x(t)) = X(p), then L(x®(t)) = pX(p), L(x®®(t)) = p2X(p)*4. Next L(2 cos(2t)) = 2p_(p2+4). The corresponding operator equation has the form (p2+ 4)X(p)*4= 2p p2+4.
70 CHAPTER 2. LAPLACE TRANSFORM We express X(p): X(p)= 4 p2+4+2p (p2+ 4)2. The inverse Laplace transform gives the solution x(t)=1_2 (4 + t) sin(2t). Example 32 (discontinuous right-hand side I) Let us solve the differential equation h n l n j x®® +x=f(t), x(0+)=1,x®(0+)=*1, (2.23) where f(t)=h n l n j 1,for 0 ftf1 0,for t>1. (2.24) Let’s proceed similarly to the previous examples. Put L(x(t)) = X(p), then L(x®(t)) = pX(p)*1, L(x®®(t)) = p2X(p)*p+1. Next L(f(t)) can be calculated directly from definition (2.2): L(f(t)) = ÿ 0 f(t)e*pt dt=1 0 e*pt dt=1 p(1 *e*p).(2.25) Or, notice that f(t)=⌘(t)*⌘(t*1) and by property V. of Theorem 10 again we get (2.25). The corresponding operator equation has the form (p2+ 1)X(p)*p+1= 1 p(1 *e*p). We express X(p)after decomposition to partial fractions X(p)=1 p*1 (p2+ 1) *01 p*p p2+11e*p. The inverse Laplace transform gives the solution x(t) = (1 *sin(t))⌘(t)*(1 *cos(t))⌘(t*1) or without using ⌘(t) x(t)=h n l n j 1*sin(t),tÀ[0,1), cos(t)*sin(t),tg1.
2.3. APPLICATIONS OF THE LAPLACE TRANSFORM 71 Example 33 (discontinuous right-hand side II) Let us solve the differential equation h n l n j x®® +x=f(t), x(0+)=1,x®(0+)=0, (2.26) where f(t)=h n l n j 1,for 0 ftf3 2,for t>3. (2.27) Let’s proceed analogously to the previous example. Put L(x(t)) = X(p), then L(x®®(t)) = p2X(p)*p. Next, it is enough to observe that f(t)=⌘(t)+⌘(t*3), and by property V. of Theorem 10 we get L(f(t)) = 1 p(1 + e*3p).(2.28) The corresponding operator equation has the form (p2X(p)*p)+X(p)=1 p(1 + e*3p). We express X(p): X(p)= p (p2+ 1) +1 p(p2+ 1)(1 + e*3p). The inverse Laplace transform gives the solution x(t) = 1 + (1 *cos(t*3))⌘(t*3) or without using ⌘(t) x(t)=h n l n j 1,tÀ[0,3), 2*cos(t*3),tg3. Example 34 (shifted initial conditions) Let us solve the differential equation h n l n j x®® +3x®+2x=et, x(1+)=1,x®(1+)=1. (2.29)
72 CHAPTER 2. LAPLACE TRANSFORM Because the initial conditions are not given in point t0=0, we have to do the substitution t=⌧+1ax(t)=x(⌧+ 1) = y(⌧). Then the equation takes the form h n l n j y®® +3y®+2y=e⌧+1, y(0+)=1,y®(0+)=1. (2.30) Hence, put L(y(t)) = Y(p), then L(y®(⌧)) = pY (p)*1, L(y®®(t)) = p2Y(p)*p*1. Next L(e⌧+1)=eL(e⌧)=e1 p*1. We express Y(p)after the decomposition to partial fractions Y(p)= e_6 p*1+3*e_2 p+1 +e_3*2 p+2 . We get from the inverse Laplace transform for ⌧g0, the solution y(⌧)=e_6e⌧+ (3 *e_2) e*⌧+(e_3*2) e*2⌧. By the inverse substitution ⌧=t*1and y(⌧)=x(t)we have for tg1the solution x(t)=e_6et*1+ (3 *e_2) e1*t+(e_3*2) e2*2t. Example 35 Let us solve the system of differential equations h n n l n n j x®*x+y=2, x*y®*y=et, x(0+)=1,y(0+)=1. (2.31) Let L(x(t)) = X(p)and L(y(t)) = Y(p), then L(x®(t)) = pX(p)*1and L(y®(t)) = pY (p)*1. The corresponding operator system has the form h n l n j (p*1)X(p)+Y(p)=p+2 p, X(p)*(p+ 1)Y(p)=*p p*1. (2.32) After treatment and decomposition into partial fractions h n l n j X(p)= 2 p3+1 p2+1 p*1, Y(p)= 2 p3*1 p2+1 p. (2.33)
2.3. APPLICATIONS OF THE LAPLACE TRANSFORM 73 The inverse Laplace transform gives the solution h n l n j x(t)=t2+t+et, y(t)=t2*t+1. (2.34) Tasks in electrical engineering Consider first the simple oscillation circuit shown in Figure 2.3 described by an integral-differential equation Ldi(t) dt+Ri(t)+ 1 Ct 0 i(⌧)d⌧=u(t),(2.35) where L,R, and Care the induction, resistance and capacitance constants. Next, u is the electromotive voltage and iis the current. Without prejudice to generality, we can assume that no current flows through the entire circuit at the beginning. This initial condition corresponds to a switching situation. Thus, no current a passes through the circuit i(0+)=0. Then the last member left parties (2.35) represents the voltage on the capacitor plates, which at the beginning is zero. Denote L(i(t)) = I(p)and L(u(t)) = U(p); the functions I(p)and U(p)are called operator current resp. operator voltage. Then from property VI. from Theorem 10 we have L0di(t) dt1=pI(p), from VIII. of Theorem 10 we have L0t 0 i(⌧)d⌧1=I(p) p. We rewrite the equation (2.35) into an operator form LpI(p)+RI(p)+I(p) Cp =U(p), after adjustment I(p)= U(p) Lp +R+1 Cp =U(p) Z(p),(2.36) where Z(p)is the operator impedance of the circuit. Formula (2.36) is called the operator form of Ohm’s law. To the end, the inverse Laplace transform then from (2.36) determines the circuit current i.
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