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Appendix A - A.1 The £ Laplace Transform Laplace...

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Unformatted text preview: ;% A.1 The £_ Laplace Transform Laplace transforms can be used to study the complete response characteristics of feedback systems, including the transient response. This is in contrast to Fourier transforms, in which the steady-state response is the main concern. In many applications it is useful to define the Laplace transform of f (t), denoted by £_{ f (t)} = F (s), as a function of the complex variable 5 = a + ja), where 00 no é f(t)e_“ dt, (A1) 0, which uses 0” (that is, a value just before I = O) as the lower limit of integra- tion and is referred to as the unilateral (or one-sided) Laplace transform.1 A function f (1‘) will have a Laplace transform if it is of exponential order, which means there exists a real number a such that [1330 |f(t)e“”| = 0. (A2) 1 Bilateral (or two-sided) Laplace transforms and the so-called 56+ transforms, in which the lower value of integral is 0+, also arise elsewhere. 811 812 AppendixA Laplace Transforms TABLE A1 The decaying exponential term in the integrand in effect provides a built-in convergence factor. This means that even if f (I) does not vanish as t —> co, the integrand will vanish for sufficiently large values of a if f does not grow at a faster than exponential rate. For example, 01er1 ' is of exponential order, whereas e‘2 is not. If F (5) exists for some so = 00 + jwo, then it exists for all values of s such that Re(s) z 00. (A.3) The smallest value of 00 for which F(s) exists is called the abscissa of conver- gence, and the region to the right of Re(s) z 00 is called the region of conver- gence. Typically, two-sided Laplace transforms exist for a specified range, a < Re(s) < fl, (A4) which defines the strip of convergence. Table A2 gives some Laplace transform pairs. Each entry in the table follows from direct application of the transform definition. A.1.1 Properties of Laplace Transforms In this section we will address and prove each of the significant properties of the Laplace transform as discussed in Chapter 3 and listed in Table A.1. In addition we show how these properties can be used through examples. Properties of Laplace Transforms Number LaplaceTransform Time Function Comment — F (s) f (t) Transform pair 1 ozF1(s) + flF2(s) af1(t) + fif2(t) Superposition 2 F(s)e“"A f(t — A) Time delay (A 2 0) 3 I712‘I F (%) f(at) Time scaling 4 F(s + a) e‘“’f(t) Shift in frequency 5 smF(s) — sm'1f(0) “Sm—zflo) — '- - — f(’”‘1)(0) f(’")(t) Differentiation 6 %F(s) ff(§) d; Integration 7 F1(s)F2(s) f1(t) * f2(t) Convolution 8 £11010 SF (5) f (0*) Initial Value Theorem 9 i213)“: (s) [1320 f (1) Final Value Theorem 10 515 gig F1(§)F2(s — 0d; f1(z)f2(t) Time product 11 _ ail—Sn” tf(z) Multiplication by time TABLE A.2 Table of Laplace Transforms Section A.1 The 68L Laplace Transform 813 Number F(s) f(t), t 2 0 1 1 8(t) 2 1/5 1(t) 3 1/32 t 4 2i/s3 :2 5 3l/s4 [3 6 ml/s"1+1 t’” 1 —a 7 s + a e ’ 8 1 —llt (s + a)2 te 1 1 _ 9 (s + (1)3 2—!t2e at 10 1 1 m—l —a (s+a)'" (”z—1)" e ' a —a 11 S(s + a) 1 — e ’ 12 5—2“: a) gm: — 1 + e‘“’) b — a —a —b 13 —(s+a)(s+b) e ’—e t 14 (s {fay (1 — at)e“” a2 15 m 1—e“"(l+at) (17 - a)s -1] —a 16 (s+a)(s+b) be t—ae ’ 17 S2 : a2 sinat 18 s2 i a2 cosat 19 (_g+sa—_;—2a.lrb7 e‘m COSbt b —a - 20 m e ’ s1nbt 21 a2 + ’72 s[(s + (1)2 + 172] 1 — e“‘t (cos bt + % sin b1) 814 Appendix A Laplace Transforms Section A.1 The QC- Laplace Transform 815 1_ Superposition EXAMPLE A.2 Delayed Sinusoidal Signal One of the more important properties of the Laplace transform is that it is l'near. We can rove this as follows: > 1 p Find the Laplace transform of f(t) = A sin(t _ Id). £{Otf1(t) + 13f2(t)} = A [O‘f1(t) + flf2(t)le_“ 611 (A5) Solution. The Laplace tratnsform of sin(t) is a] fi(t)e’”dt+fi/ f2(t)e_“dt £{sin(t)}= 1 . 0 0 s2+1 = OlF1(S) + ,3F2(S). Therefore, 11$ng Eq. (A.7) we obtain The scaling property is a special case of this; that is, £{af(t)} = ocF(s). (A6) F“) 2 32 +16 EXAMPLE A.1 Sinusoidal Signal 3. Time Scaling Find the Laplace transform of f (t) = 1 + 2 sin(wt). If the time t is scaled by a factor a, then the Laplace transform of the time— Solution. The Laplace transform of sin(wt) is ' Sealed Signal is . (U 00 £{s1n<wr>} = + wzt F1<s> = /0 fem-m. Therefore, using Eq. (A5) we obtain Again we define t’ : at. As before dt/ = a d; and 1 20) s2 + 2ws + (02 F = — + : _'___. fl, I (S) s s2 +w2 s3 +w2s F1(s) = foo f(t’)e Wu dt’ 2 ip (i) 0 lal 10' it ~ (A.8) 2_ Time Delay , EXAMPLE A.3 Sinusaid with Frequency a) Suppose a function f (t) is delayed by A > 0 units of time. Its Laplace transform is 00 Find the Laplace transform 0f f0) = A sin(wz)_ F1(s) =/ (t — A)e’“dt. . 0 f Solution. The Laplace transform of sin(t) is Let us define t/ = t — A. Then dt’ = dt because A is a constant and f(t) = 0 fOI' t < 0. Thus £{Sin(t)} = 1 ' 00 00 52 +1 F s = t’ e_‘(’/+Mdt' = f t, e_S(’/+A)dt’. . 1( ) Li f( ) 0 f( ) Therefore, us1ng Eq. (A.8) we obtain Because e‘” is independent of time, it can be taken out of the integrand, so 1 1 F(S) : — ‘ i °° , , / ii . Iwi<giz+i F1(S)=e_s / f(t )e‘S’dt =6 5 F(s). (A./) _ w 0 _ 52 + ‘02 From this result we see that a time delay of A corresponds to multiplication of M; as expected. the transform by e‘”. i 816 Appendix A Laplace Transforms EXAMPLE A.4 EXAMPLE A.5 4. Shift in Frequency Multiplication (modulation) of f (t) by an exponential expression in the time domain corresponds to a shift in frequency: F1(s) = /We_“’f(t)e_”dt = [00 f(t)e_(s+“)’dt = F(s + (1). (A9) 0 0 Exponentially Decaying Sinusoid Find the Laplace transform of f (t) = A sin(wt)e’”’ . Solution. The Laplace transform of sin(a)t) is £{sin(a)t)} 2 S2 :wZ. Therefore, using Eq. (A.9) we obtain F( ) Au) 5 = ——. (s + a)2 + a)2 5. Differentiation The transform of the derivative of a signal is related to its Laplace transform and its initial condition as follows: df oo df _g _S 00 00 _“ £{El2/ (5)6 ‘Idtze ’f(t)|0_+s 0_ f(t)e dt. (A.10) Because f (t) is assumed to have a Laplace transform, e‘“ f (t) —> 0 as t —> 00. Thus £[f] = —f(0‘) +sF(s). (AM) Another application of Eq. (A.11) leads to em = 521%) — sf(0‘) — for). (A12) Repeated application of Eq. (A.11) leads to £{f’"(t)} = 3mm) — smrlfm — sm—Zflow — - - - — f(’"_1)(0’), (A13) where f m (t) denotes the mth derivative of f (t) with respect to time. Derivative of Cosine Signal Find the Laplace transform of g(t) = fifo), where f(t) = cos(wt). EXAMPLE A.6 Section A1 The £_ Laplace Transform 817 Solution. The Laplace transform of cos(wt) is F(s) = £{cos(wt)} : s2 in‘ Using Eq. (A.1l) with f(0‘) : 1,we have (1)2 5 0(3) — c.C{g(t)} — s - 52 +w2 1— 32 +w2' 6. Integration Let us assume we wish to determine the Laplace transform of the integral of a time function—that is, to find Firs) = at {/0 nods} = [000 megkwt Employing integration by parts, where t V a 2/ f(§)d§ and dv = e‘S’dt, 0 we get 00 I 00 1 ' 1 F1(S) = [—iefl" (/0 f(E)dS>] —/0 —;e_”f(t)dt = ;F(S)- (A14) 0 Time Integral 0f Sinusoidal Signal Find the Laplace transform of f (z) = fol sin wt d 1. Solution. The Laplace transform of sin(wt) is £ ' n} — w {s1n(a) 52 + w2 Therefore, using Eq. (A14) then F ‘ Cl) (s) _ s3 + (025 818 Appendix A Laplace Transforms EXAMPLE A.7 7. Convolution Convolution in the time domain corresponds to multiplication in the frequency domain. Assume that £{f1(t)} = F1(s) and £{f2(t)} = F2(s). Then 00 00 t £{f1(t)*f2(t)}= f0 f1(t)*f2(t)e“‘dt= f0 [f0 f1(r)f2(t—I)dr:le““dt. Reversing the order of integration and changing the limits of integration yield £{f1(t)*f2(t)}=f0 / f1(r)f2(t—r)e_s’dtdt. Multiplying by e‘srest results in 00 00 £{f1(t) * f2(t)} =f f1(T)e_Stl:/ f2(l‘ — T)€_‘Y(’_I)dt] dt. 0 I If t’ét — t, then £{f1(’)* 150)} = [0 f1(T)e_“dt f0 f2(r’)e-sr'dt, £{f1(t) * f2(t)} = F1(S)F2(s), This implies that £—1{F1<s)F2<s)} = flu) * f2(t). (A15) Ramp Response of a First-Order System Find the ramp response of a first order system with a pole at +a. Solution. Let f1 (t) = t be the ramp input and f2 (t) = e‘” be the impulse response of the first order system. Then, using Eq. (A.15) we find 1 1 £‘1{ } = flu) * f2(t) szs—a = /0 f1(r)f2(t-r)dr = / re“(’_r)dr 0 1 = —2(e‘" —at — 1). a Section A1 The at- Laplace Transform 819 8. Time Product Multiplication in the time domain corresponds to convolution in the frequency domain: _ ac +1 00 1 £{fi(t)f2(t)} = E F1(§)F2(s — 861$- ac—joo To see this, consider the relation ammo» = [0 f1(t)f2(t)e_“dt. Substituting the expression for f1(t) given by Eq. (3.14) yields 00 ac+joo £{f1(t)f2(t)}= /0 [571?] f F1(5)e€’ds]f2(t)e-stdt. c_j0° Changing the order of integration results in ae+j°° °c{f1(t)f2(t)} = i F1(£)/0 f2(t)e_(s_§)tdtdé. 2”] oc—joo Using Eq. (A9), we get ac+j00 — 277:]. ac—joo 1 £{f1(t)f2(t)} F1(€)F2(s — EMS = EH“) * F20). (A16) 9. Multiplication by Time Multiplication by time corresponds to differentiation in the frequency domain. Let us consider d d 00 —st EEF(S)=$/(; e f(t)dt = foo —te‘“f(t)dt 0 — [00 e‘“[tf(t)]dt 0 = -£{tf(t)}- Then d £{tf(t)} = _d_§F(s)’ (A17) which is the desired result. 820 AppendixA Laplace Transforms EXAMPLE A.8 Time Product of Sinusoidal Signal Find the Laplace transform of f (t) : tsin wt. Solution. The Laplace transform of sin wt is . a) £{s1n(a)t)} = s2+w2. Hence using Eq. (A.17), we obtain F(s)— d a) _ 2ws ‘ _ ds s2+w2 —(s2+w2)2' A.1.2 Inverse Laplace Transform by Partial- Fraction Expansion As we saw in Chapter 3, the easiest way to find f (I) from its Laplace transform F (s), if F(s) is rational, is to expand F (s) as a sum of simpler terms that can be found in the tables via partial fraction expansion. We have already discussed this method in connection with simple roots in Section 3.1.5. In this section, we discuss partial-fraction expansion for cases of complex and repeated roots. Complex Poles In the case of quadratic factors in the denominator, the numerator of the quadratic factor is chosen to be first-order as shown in Ex- ample 3.8. Whenever there exists a complex conjugate pair of poles such as C C 1 + 2*, 5—191 S—p1 F(s) = we can show that C2 = Cf (see Problem 3.1), and that f(t) = C15plt + Cfepi" = 2Re(C1e1’1’). Assuming that p1 = a + jfl, we may rewrite f (t) in a more compact form as N) = 2Re{C1eP1'} = 2Re11C1 lei arg(Cl)e‘“+J‘/3>’} (A18) = ZlClleo" cos[,3t + arg(C1)]. EXAMPLE A.9 Partial-fraction Expansion: Distinct Complex Roots Find the function f (t) for which the Laplace transform is 1 HS) 2 5(32 +s +1)" Section A.1 The £_ Laplace Transform 821 Solution. We rewrite F (s) as C1 C23 +C3 F(S)=T+s2+s+1' Using the cover-up method, we find C1 to be C1 = SF(S)|s=0 = 1. Setting C1 = 1 and then equating the numerators in the partial fraction expansion relation, we obtain (s2 +s + 1) + (C25 + C3)s = 1. After solving for C2 and C3, we find that C2 = —1 and C3 = —1. To make it more suitable for using the Laplace transform tables, we rewrite the partial fraction as 1+1 5+5 2 1 2 + 3 ' (S + Z 4 From the tables we have 3 1 _, . 3 f(t)= <1—e"/2cos\/;t—fie /2s1n\/;t)1(z) = (1 — %e"/2 cos (gt — %>) 1(1). Alternatively, we may write F(s) as C1 C2 C; (A19) =— + *, . F(s) s+s—p1 s—p1 where p1: —% + j?. C1 = 1 as before, and now 1 . 1 C2 = (s — P1)F(S)ls=p1= —§ ivm C, 1 .1 2— 2 Jzfi and f(t) 2 (1+ 2|Czle‘“ cos[;6t + arg(C2)])1(t) 3 5 = (1+ %e”/2 cos [£1 + %]) 1(1) 2 (1— if“ cos [£1 — 1]) 1(1). fl 2 6 822 Appendix A Laplace Transforms Section A.1 The £_ Laplace Transform 823 The latter partial fraction expansion can be readily computed using MATLAB, EXAMPLE A_1O Partial-Fraction Expanszon: Repeated Real R00“ num = 1' % form numerator . ’ ' th t has the La lace transform den = conv([1 0],[1 1 1]); % form denominator F1nd the functlon f(t) a P [r,p,k] = residue(num,den) % compute residues s+3 F(S) : —_—(s + 1)“ + 2)2 . which yields the following result: r : [—0'5000 + 0'2887i _ 0'5000 _ 0288-” “3000],; Solution. We write the partial fraction as p = [—0.5000 + 0.8660i — 0.5000 * 0.8660i 0]’; k = ; (:1 Q C; + ‘ . s+l+s+2 (s+2)2 F(s) = and agrees with the above hand calculations. Note that if we are using the tables, the first method is preferable, while the second method is preferable for checking MATLAB results. Then s + 3 _ 2 Repeated Poles For the case where F (s) has repeated roots, the procedure to C1 = (S + 0“” = (S + 32 21 — ’ compute the partial-fraction expansion must be modified. If [71 is repeated “:71 S— . . . . d three times, we wrlte the partial fract1on as C2 : _ [(5 + 2)2F(s)] |S=_2 : _27 C1 C2 C3 C4 C” s F(S)= + 2+ 3+ +~~+ ~ s+3 S — 191 (s — m) (s - Pl)‘ s — p4 s — M C, = (s + 2)2F(s) = S + 1 = —1- 5:42 S=*2 We determine the constants C4 through Cn as discussed previously. If we multiply both sides of the above equation by (s — p1)3, The function f(t) is N) = (2e” — 262’ —te’2')1(z). C (S — P03 3 — — 2 — . . ' n . . 9 . (S _ p1) F“) — CI (5 p1) + C2 (S p1) + C3 + + S _ p " 7 (A20) The partial fraction computation can also be carried out usmg MATLAB s resndue . . _ f t' , and set s = P1, then all the factors on the r1ght SIde of Eq. (A20) w111 go to ““0 Ion zero except C3, which is num = [1 3]; % form numerator den = conv([1 1],[1 4 4]); % form denominator C3 = (s — p1)3F(S)|S:”1 [r,p,k] = residue(num,den) % compute resrdues as before. To determine the other factors, we differentiate Eq. (A20) with WhiCh yields the fOHOWing r6511“ respect to the Laplace variable 3: r = [_2 _1 21’; p = [—2 —2 —1]’; and k = fl; and agrees with the hand calculations. d Cn(S — [71)3 ds d d—[(S—P1)3F(S)]=2C1(s—p1)+C2+...+— ] (A21) S S _ Pr: Again, if we set s = p], we have A.1.3 The Initial Value Theorem We discussed the Final Value Theorem in Chapter 3. A second valuable Laplace transform theorem is the Initial Value Theorem, which states that 1t 1s always possible to determine the initial value of the time function f (z) from 1ts Laplace d C2 = d—s[(S — p1)3F(S)ls=Pl' Similarly, if we differentiate Eq. (A21) again and set s 2 171 a second time, we get 1 d2 transform. We may also state the theorem in this way: . 3 C1 = idszlfi " P1) F(S)]s=pl- In general, we may compute Ci for a factor with multiplicity k as The Initial Value Theorem For any Laplace transform pair, 1 di ' lim sF(s) = f(0+) (A22) Cr—i=_ ..— kF. , '=0,...,k—1. _‘ H00 A i! ids’ [(g p1) (”iii-q” l l 824 Appendix A EXAMPLE A.11 Laplace Transforms We may show this as follows. Using Eq. (A.11), we obtain df _ °° _ df — = F — 2 St — . . £{dll s (s) f(0) /_ e dt dt (A23) Let us consider the case where s —> 00 and rewrite the integral as °O mew) _ f” _s,df(t) [0+ _S,df(t) ffie _—dt dt—_ 0+e —dt dt+ _ e —dt dt. Taking the limit of Eq. (A23) as s ——> 00, we get 0+ 00 lim [SF(S) — f(0‘)] = 11m [/ god—{1:2 dl +/ e—stw dt] 5—)00 0 s—->oo + dl The second term on the right side of the above equation approaches zero as s —> 00 because e‘“ —> 0. Hence Slingotsns) — now] = Slingolflm) — f(0‘)l = f<0+> — f(0‘), OI‘ lim sF(s) = f(0+). 3%00 In contrast with the Final Value Theorem, the Initial Value Theorem can be applied to any function F (s). Initial Value Theorem Find the initial value of the signal in Example 3.10. Solution. From the Initial Value Theorem, we get y(0+) = lim sY(s) = lim s 3 — 0, s—>oo xeoc 5(5 — 2) _ which checks with the expression for y(t) computed in Example 3.10. For a thorough study of Laplace transforms and extensive tables, see Churchill (1972) and Campbell and Foster (1948); for the two-sided transform, see Van der P01 and Bremmer (1955). A Review of Complex Variables ' ' ' lex variables theory This a endix 1s a brief summary of some results on comp . with eiiiphasis on the facts needed in control theory. For a comprehensxve study of basic complex variables theory, see standard textbooks such as Brown an Churchill (1996) or Marsden and Hoffman (1998). 8.1 Definition of a Complex Number The complex numbers are distinguished from purely real numbers in that they also contain the imaginary operator, which we shall denote ]. By definition, j2 —1 0r j=\/—1. (13-1) A complex number may be defined as A = a + ja), (B2) where or is the real part and a) is the imaginary part, denoted respectively as a = Re(A), a) = Im(A). (B3) Note that the imaginary part of A is itself a real number. 825 ...
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