Basic Laplace Transform Techniques

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Chapter 16 Basic Laplace Transform Techniques 16.1 Laplace transform The variation of the constant approach transforms a linear ¯rst-order ODE into an equivalent but simpler integration problem. In contrast, the Laplace transform transforms an ODE into an equivalent algebraic equation (without derivative). To do this the integration process must certainly be part of the way the Laplace trans- form is de¯ned (integration has to occur somewhere). 16.1.1 De¯nition De¯nition 16.1. The Laplace transform of f(t) is a function F(s) =Lff(t)g(s) de¯ned by Z 1 ¡st F(s)= f(t)e dt: (16.1) 0 The Laplace transform is an integral transform. Another notation for the Laplace transform of a function f(t) isLff(t)g(s), which makes it more clear as to whichfunctionf(t)wearetalkingabout,butisabitmorecumbersome. Sometimes we simply write Lff(t)g, but keep in mind that the Laplace transform of f(t) is a function of s only Wecanthinkofsasarealnumberwhichis¯xedwhenevaluatingtheintegral (16.1). Values of the Laplace transform for complex values of s can however be useful in practical applications. In particular, the Laplace transform becomes the Fourier transform when s = i with real (corresponding to a given frequency ). 2¼ Note that the Laplace transform only uses values of f(t) for t¸0. The Laplace transform is used extensively in Electrical Engineering and Con- trol Theory. The Fourier transform and other integral transforms are used in nu- merous applications (stability of algorithms, medical imaging,...) 6162 Chapter 16. Basic Laplace Transform Techniques 16.1.2 Laplace transform of standard functions Let's compute the Laplace transform of a few simple functions. Example 16.2 The Laplace transform of f(t)=1 is Z 1 ¡st Lf1g= 1¢e dt (16.2a) 0 · ¸ 1 ¡st e (assume s6=0) = (16.2b) ¡s 0 ¯ µ ¶ ¡st ¡st ¯ e e ¯ = lim ¡ (16.2c) ¯ t1 ¡s ¡s t=0 1 (assume s0) =0¡ (16.2d) ¡s 1 = : (16.2e) s Line (16.2a) uses the de¯nition (16.1) of the Laplace transform, substituting f(t)= 1. On Line (16.2b) the result of the integration is evaluated between t = 0 and t = 1 (understood in the limit sense). This yields (16.2c). Note that (16.2b) assumes s= 6 0 because of a division by 0 (if s=0 then (16.2a) diverges (equal1)). Someadditionalrestrictionmustbeplacedon sforthelimitappearingin(16.2c)to exist (be ¯nite), namely s0, in which case the limit vanishes. The result (16.2e) is recorded as entry 1 in Table 16.1. Example 16.3 The Laplace transform of f(t)=t (a given real number) is Z 1 ¡st Lftg= t¢e dt (16.3a) 0 · ¸ Z 1 1 ¡st ¡st e e (assume s6=0) = t¢ ¡ dt (16.3b) ¡s ¡s 0 0 µ ¶ ¯ Z 1 ¡st ¡st ¯ te te 1 ¡st ¯ = lim ¡ + e dt (16.3c) ¯ t1 ¡s ¡s s 0 t=0 1 (assume s0) =0¡0+ Lf1g (16.3d) s 1 1 = (16.3e) s s 1 = : (16.3f) 2 s Line (16.3a) uses the de¯nition (16.1) of the Laplace transform, substituting f(t)= t. Integration by parts yields (16.3b), which can be written as (16.3c). To obtain (16.3d) note that the limit vanishes provided s0, and the integral was computed inExample16.2astheLaplacetransformoff(t)=1. Theresult(16.3f)isrecorded as entry 2 in Table 16.1.16.1. Laplace transform 63 at Example 16.4 The Laplace transform of f(t)=e (a given real number) is Z 1 at at ¡st Lfe g= e ¢e dt (16.4a) 0 Z 1 (a¡s)t = e dt (16.4b) 0 · ¸ 1 (a¡s)t e = (16.4c) a¡s 0 ¯ µ ¶ (a¡s)t (a¡s)t ¯ e e ¯ = lim ¡ (16.4d) ¯ t1 a¡s a¡s t=0 1 =0¡ (16.4e) a¡s 1 = : (16.4f) s¡a Line (16.4a) uses the de¯nition (16.1) of the Laplace transform, substituting f(t)= at e . The property (1.1) of exponential functions yields (16.4b), from which (16.4c) and(16.4d)follow(assumings= 6 a). Thelimitexists(andvanishes)providedsa. The result (16.4f) is recorded as entry 3 in Table 16.1. Example 16.5 The Laplace transform of f(t)=sin(bt) is Z 1 ¡st Lfsin(bt)g= sin(bt)¢e dt: (16.5a) 0 From (??) we obtain (with a¡s) · ¸ 1 ¡st ¡ ¢ e = ¡bcos(bt)+(¡s)sin(bt) (16.5b) 2 2 s +b 0 µ ¶ ¡st ¡ ¢ e ¡b = lim ¡bcos(bt)+(¡s)sin(bt) ¡ (16.5c) 2 2 2 2 t1 s +b s +b ¡b =0¡ (16.5d) 2 2 s +b b = : (16.5e) 2 2 s +b at Notethatthelimitin(16.5c)existsaslongastheexponentialfactor e hasa¯nite limit, which is the case for s0. An alternate way to obtain (16.5e) utilizes Euler's formula (1.10) and recog-64 Chapter 16. Basic Laplace Transform Techniques ibt nizing sin(bt) as=e . From Example 16.4 we obtain, with a=ib, ibt Lfsin(bt)g=Lf=e g ibt ==Lfe g 1 == s¡ib s+ib == (s¡ib)(s+ib) s+ib == 2 2 s +b b = : 2 2 s +b The result (16.5e) is recorded as entry 4 in Table 16.1. The existence of the Laplace transform for certain values of s hinges upon the ¡st convergence of the integral (16.1), i.e., on the behavior of the integrand f(t)e as t0 and t1. 2 t Example 16.6 If f(t)=e then Z Z Z Z 1 1 1 1 2 2 t ¡st t ¡st st ¡st e ¢e dt¸ e ¢e dt¸ e ¢e dt= dt=1 0 s s s for any (¯xed) value of s. Thus the integral (16.1) diverges for all values of s, i.e., 2 t f(t)=e does NOT have a Laplace transform. 1 Example 16.7 If f(t)= then (assume s0) t Z Z 1 1 ¡s 1 1¡e ¡st ¡st 0· ¢e dt· ¢e dt= t s 1 1 converges but Z Z Z 1 1 1 1 1 dt ¡st ¡s ¡s ¡s 1 ¡s ¢e dt¸ ¢e dt=e =e lnt =e (0¡(¡1))=1 0 t t t 0 0 0 1 diverges for any s0. Thus f(t)= does NOT have a Laplace transform. t for Lff(t)g to exist the function f(t) must be well-behaved around t = 0 (for ex- ample ¯nite and continuous at t=0) and not grow faster than regular exponentials ®t e as t 1 (such functions are of exponential order). When a Laplace trans- form exists the limit at1 such as the ones appearing in (16.2a), (16.2a) or (16.2a) typically vanishes provided suitable restrictions are placed on s. 16.2 Basic properties Two basic properties make the Laplace transform well-suited for solving linear ODEs.16.2. Basic properties 65 16.2.1 Linearity The Laplace transform of a linear combination of two (or more) functions is equal to the linear combination of the respective Laplace transforms. Mathematically, Lf®f(t)+¯g(t)g=®Lff(t))g+¯Lfg(t)g=®F(s)+¯G(s): (16.6) This linear property easily follows from the linearity property for integrals. Example 16.8 2 3 Lf2t¡3sintg=2Lftg¡3Lfsintg= ¡ 2 2 s s +1 using Table 16.1, entries 2 and 4 (with b=1). 16.2.2 Derivative formula 0 The derivative formula relates the Laplace transform of the derivative f (t) of a function f(t) to the Laplace transform F(s) of the function f(t) itself, namely 0 Lff (t)g=sLff(t)g¡f(0)=sF(s)¡f(0): (16.7) The derivative formula (16.7) is recorded in Table 16.1 as entry 15. Example 16.9 Lf1g=sLftg¡0 is easily veri¯ed using Table 16.1, entries 1 and 2 (f(t)=t). 0 Example 16.10 For f(t)=sin(bt) we have f (t)=bcos(bt) and f(0)=0, so that b Lfbcos(bt)g=sLfsin(bt)g¡0=s 2 2 s +b using Table 16.1, entry 4. Using the linear property (16.6) we then obtain b bLfcos(bt)g=s ; 2 2 s +b i.e., s Lfcos(bt)g= ; 2 2 s +b which is recorded as entry 5 in Table 16.1. at 0 at Example 16.11 For f(t)=e we have f (t)=ae and f(0)=1, so that at at Lfae g=sLfe g¡1: (16.8) Using the linear property (16.6) and Table 16.1, entries 2 we verify that 1 1 a =s ¡1 s¡a s¡a66 Chapter 16. Basic Laplace Transform Techniques indeed. at Note that we could have used (16.8) to ¯nd Lfe g, had we not known its expression from a previous calculation: 1 at at at at aLfe g=Lfae g=sLfe g¡1 ) Lfe g= : s¡a We include here a proof of the derivative formula (16.7) because it is short and simple: Z 1 0 0 ¡st (de¯nition) Lff (t)g= f (t)e dt 0 Z 1 ¡ ¢ d ¡st 1 ¡st (integration by parts) =f(t)e ¡ f(t) e dt 0 dt 0 Z ³ ´ 1 ¡st ¡st = lim f(t)e ¡f(0)+s f(t)e dt (16.9) t1 0 =¡f(0)+sLff(t)g: ®t The limit in (16.9) is guaranteed to exist ifjf(t)j·Ce for all t¸0 (exponential order). The derivative formula then holds for s®. The derivative formula can be applied recursively to yield formulas for the Laplace transform of higher order derivatives of f(t). In particular, 00 0 0 Lff (t)g=Lf(f )(t)g 0 0 =sLff (t)g¡f (0) 0 =s(sLff(t)g¡f(0))¡f (0) 2 0 =s Lff(t)g¡sf(0)¡f (0) 2 0 =s F(s)¡sf(0)¡f (0): (16.10) This formula is recorded in Table 16.1, entry 16. 0 0 Example 16.12 Take f(t) = sin(bt), so that f(0) = 0, f (t) = bcos(bt), f (0) = b 00 2 and f =¡b sin(bt). Then the formula (16.10) reduces to 2 2 Lf¡b sin(bt)g=s Lfsin(bt)g¡s¢0¡b; (16.11) i.e., by linearity and Table 16.1, entries 4, b b 2 2 ¡b =s ¡b; 2 2 2 2 s +b s +b which is easily veri¯ed. Note that (16.11) could be used to determineLfsin(bt)g: b 2 2 ¡b Lfsin(bt)g=s Lfsin(bt)g¡s¢0¡b ) Lfsin(bt)g= : 2 2 s +b16.3. Solution of linear IPVs 67 16.3 Solution of linear IPVs Examples 16.11 and 16.12 show that the Laplace transform of certain functions can easilybeobtainedbyusingthelinearityproperty(16.6)andthederivativeformulas (16.7), (16.10). Here we formalize this idea by showing that the Laplace transform of the solution of the second-order constant coe±cients linear IVP 8 00 0 au +bu +cu=g(t); u(0)=®; : 0 u(0)=¯ (if a= 6 0) can be obtained this way. 16.3.1 Obtaining the Laplace transform of the solution Apply the Laplace transform to both sides of the ODE and use linearity (14) and the derivative formulas (15, 16): 00 0 au + bu + cu = g(t) L 00 0 aLfu g +bLfug +cLfug = Lfg(t)g ? ? ? 2 0 =U(s) =G(s) =s Lfug¡su(0)¡u(0) 2 =s U(s)¡®s¡¯ ? =sLfug¡u(0) =sU(s)¡® Note how the ICs are used right away in the determination of U(s). Collecting terms yields 2 (as +bs+c)U(s)¡a®s¡(a¯+b®)=G(s); i.e., G(s) a®s+a¯+b® U(s)= + : 2 2 as +bs+c as +bs+c (16.12) U(s) with trivial ICs U(s) with no forcing ®=0; ¯ =0 g(t)=0; G(s)=0 (start at rest) (HODE) 1 The quantity is called a transfer function. 2 as +bs+c Example 16.13 Find U(s) if u(t) is the solution of the IVP ½ 0 u =2u; u(0)=1:68 Chapter 16. Basic Laplace Transform Techniques 0 u ¡ 2u = 0 L 0 Lfug¡2Lfug = Lf0g ? ? ? =0 =U(s) =sLfug¡u(0) =sU(s)¡1 Thus 1 sU(s)¡1¡2U(s)=0 ) U(s)= : s¡2 2t This expression is of course the Laplace transform of u(t)=e , according to entry 3 of Table 16.1. Example 16.14 Find U(s) if u(t) is the solution of the IVP ½ 0 at u ¡2u=e ; (16.13) u(0)=1; where a is a (constant) parameter. 0 2t u ¡ 2u = e L 0 2t Lfug¡2Lfug = Lfe g ? ? ? 1 =U(s) =sLfug¡u(0) = (3) s¡2 =sU(s)¡1 Thus 1 1 1 sU(s)¡1¡2U(s)= ) U(s)= + : 2 s¡2 (s¡2) s¡2 This expression can be recognized as 1 1 2t 2t 2t 2t Lfte +e g=Lfte g+Lfe g= + 2 (s¡2) s¡2 according to entries 8 and 3 of Table 16.1, respectively. A direct veri¯cation t shows that u(t)=(t+1)e is indeed the solution of the IVP (16.13). Example 16.15 Find U(s) if u(t) is the solution of the IVP 8 00 0 2u +3u +u=10cost; u(0)=1; (16.14) : 0 u(0)=¡1:16.3. Solution of linear IPVs 69 00 0 2u + 3u + u = 10cost L 00 0 2Lfu g +3Lfug +Lfug = 10Lfcostg ? ? ? 2 0 1 =U(s) =s Lfug¡su(0)¡u(0) = 2 2 =s U(s)¡s+1 s +1 ? =sLfug¡u(0) =sU(s)¡1 Thus 10 2 2(s U(s)¡s+1))+3(sU(s)¡1)+U(s)= ; 2 s +1 i.e., 10 2 (2s +3s+1)U(s)¡2s¡1= ; 2 s +1 or 10 2s+1 U(s)= + : (16.15) 2 2 2 (s +1)(2s +3s+1) 2s +3s+1 The second term in (16.15) can be simpli¯ed if we factorize the denominator: 2s+1 2s+1 1 ¡t = = =Lfe g; 2 2s +3s+1 (2s+1)(s+1) s+1 according to entry 3 of Table 16.1. Unfortunately the ¯rst term in (16.15) cannot be (as) easily written as the Laplace transform of a recognizable function in Table 16.1. 16.3.2 Inverse transform Inordertodeterminethesolutionu(t)inExample(16.15)weneedto¯ndafunction of t whose Laplace transform is equal to the ¯rst term on the right-hand side of (16.15). The good news is that the Laplace transform is a one-to-one map, i.e., two distinct functions will have two distinct Laplace transforms. This permits us to de¯ne an inverse transform: L ¡1 ¡ u(t) U(s); Lfu(t)g=U(s) , u(t)=L fU(s)g: á ¡1 L The formula for the inverse transform of a function F(s) looks deceptively similar to the formula (16.1) for the Laplace transform: Z 1 ¡1 st L fF(s)g= F(s)e ds: (16.16) 2i¼70 Chapter 16. Basic Laplace Transform Techniques 1 Exceptfortheconstantfactor ,thisformulalooksvery\symmetric"from(16.1): 2i¼ exchange the roles of t and s, change the sign in the exponential. The bad news is that the integration must be carried out along a path in the complex plane, rather than over a range of real values (integration \limits" were left out in (16.16)). As a result, this formula is not practical in this course, even for functions as simple as 1 F(s)= =Lf1g. s The alternative to formula (16.16) is to use a table of transforms with su±- ciently many formulas to provide a way to determine an inverse transform. For- tunately, many Laplace transform functions F(s) are rational functions (see Table 16.1) which can be written as linear combinations of simpler rational functions us- ing the partial fraction decomposition (PFD). It is advisable at this point to review how a PFD can be determined by consulting Section 3.4 page 23. ½ ¾ 1 ¡1 Example 16.16 Determine f(t)=L . 2 s +3s+2 1 1 1 1. PFD: = ¡ (see Example 3.6) 2 s +3s+2 s+1 s+2 2. Inverse transform each term in the PFD of F(s) using Table 16.1: 1 1 F(s)= ¡ s+1 s+2 ¡1 ¡1 L L ¡t ¡2t ¡1 ¡t ¡2t e e )f(t)=L fF(s)g=e ¡e : (3) (3) ½ ¾ s+7 ¡1 Example 16.17 Determine f(t)=L . 2 s +2s+10 1. Factor denominator: cannot be done with real coe±cients since 2 2 2 s +2s+10=(s+1) ¡1+10=(s+1) +9: Thus s+7 s+7 = : 2 2 s +2s+10 (s+1) +9 2. PFD setup: already in PFD form. 3. Inverse transform each term in the PFD of F(s) using Table 16.1: express the numerator in the form s+7=(s+1)+2(3). Then s+1 3 F(s)= + 2 2 2 (s+1) +9 (s+1) +9 ¡1 ¡1 L L ¡t ¡t ¡t e cos(3t) e sin(3t) )f(t)=e (cos(3t)+2sin(3t)): (12) (11)16.3. Solution of linear IPVs 71 Example 16.18 We revisit Example 16.15 and consider 10 F(s)= : 2 2 (s +1)(2s +3s+1) 1. Factor denominator (real coe±cients): 10 10 = : 2 2 2 (s +1)(2s +3s+1) (s +1)(2s+1)(s+1) 2. PFD setup: 10 as+b c d = + + : (16.17) 2 2 (s +1)(2s+1)(s+1) s +1 2s+1 s+1 3. Determine unknown coe±cients: (i) Multiply by common denominator and simplify factors: 2 2 10=(as+b)(2s+1)(s+1)+c(s +1)(s+1)+d(s +1)(2s+1): (16.18) (ii) Expand and compare powers of s: 3 2 3 2 3 2 10=(2as +(3a+2b)s +(a+3b)s+b)+c(s +s +s+1)+d(2s +s +2s+1): 3 s : 0 = 2a+c+2d; 2 s : 0 = 3a+2b+c+d; (16.19) s: 0 = a+3b+c+2d; 1: 10 = b+c+d: This system is not trivial. We can set it up in matrix form 2 32 3 2 3 2 0 1 2 a 0 6 76 7 6 7 3 2 1 1 b 0 6 76 7 6 7 = 4 54 5 4 5 1 3 1 2 c 0 0 1 1 1 d 10 and use the row reduction method of Section ?? page ??:72 Chapter 16. Basic Laplace Transform Techniques 2 3 2 3 1 2 0 1 2 0 1 0 1 0 2 1 6 7 6 7 3 2 1 1 0 r r 3 2 1 1 0 1 1 6 7 2 6 7 4 5 4 5 1 3 1 2 0 ¡ 1 3 1 2 0 0 1 1 1 10 0 1 1 1 10 2 3 2 3 1 1 1 0 1 0 1 0 1 0 2 2 r r ¡3r 2 2 1 1 1 1 6 7 6 7 0 2 ¡ ¡2 0 r r 0 1 ¡ ¡1 0 2 2 2 2 4 6 7 6 7 r r ¡r 3 3 1 1 1 4 5 4 5 0 3 1 0 ¡ 0 3 1 0 2 2 ¡ 0 1 1 1 10 0 1 1 1 10 2 3 2 3 1 1 1 0 1 0 1 0 1 0 2 2 r r ¡3r 3 3 2 1 4 1 6 7 6 7 0 1 ¡ ¡1 0 r r 0 1 ¡ ¡1 0 3 3 4 5 4 6 7 6 7 r r ¡r 4 4 2 5 16 4 5 4 5 0 0 4 0 ¡ 0 0 1 0 4 5 ¡ 5 5 0 0 2 10 0 0 2 10 4 4 2 3 2 3 1 1 1 0 1 0 1 0 1 0 2 2 5 1 1 1 6 7 6 7 r r ¡ r 0 1 ¡ ¡1 0 r ¡ r 0 1 ¡ ¡1 0 4 4 3 4 4 4 6 4 7 2 6 4 7 16 16 4 5 4 5 ¡ 0 0 1 0 ¡ 0 0 1 0 5 5 0 0 0 ¡2 10 0 0 0 1 ¡5 2 3 2 3 1 r r ¡r 1 0 0 5 1 0 0 0 ¡3 1 1 4 1 2 r r ¡ r 1 1 3 1 2 6 7 6 7 r r +r 0 1 ¡ 0 ¡5 0 1 0 0 ¡1 2 2 4 1 6 4 7 6 7 r r + r 2 2 3 16 4 4 5 4 5 r r ¡ r 0 0 1 0 16 0 0 1 0 16 3 3 4 5 ¡ ¡ 0 0 0 1 ¡5 0 0 0 1 ¡5 Therefore a=¡3; b=¡1; c=16; d=¡5; (ii') Thealternateapproachofpickingsuitablevaluesofsin(16.18)canright away deliver some of the coe±cients: s=¡1: 10=¡2d ) d=¡5; 1 5 s=¡ : 10= c ) c=16: 2 8 2 Obtaining a and b (which correspond to a quadratic denominator s + 2 1) would require substituting s = i (so that s + 1 = 0) and lead to calculations with complex numbers. Alternately we can substitute other interesting values: s=0: 10=b+c+d ) b=¡1; s=1: 10=6(a+b)+4c+6d ) a=¡3: 4. Inverse transform each term in the PFD of F(s) using Table 16.1: ¡3s¡1 16 ¡5 F(s)= + + 2 s +1 2s+1 s+1 s 1 1 1 =¡3 ¡ + 8 ¡ 5 1 2 2 s +1 s +1 s+ s+1 2 ¡1 ¡1 ¡1 ¡1 L L L L 1 ¡ t ¡t 2 cost sint e e (5) (4) (3) (3) 1 ¡1 ¡ t ¡t 2 )f(t)=L fF(s)g=¡3cost¡sint+8e ¡5e :16.3. Solution of linear IPVs 73 The solution of the IVP (16.14) of Example 16.15 thus becomes 1 ¡t ¡ t ¡t 2 u(t)=f(t)+e =¡3cost¡sint+8e ¡4e : (16.20) 16.3.3 Application to linear systems Short table of Laplace transforms Table 16.1 includes the Laplace transform of standard functions. Conversely, the table can be used to identify the inverse transform of functions of s listed in the column labelled \U(s)". Typical mistakes and useful tips ² mistake t should NOT appear in the Laplace transform U(s) of u(t). U(s) only depends on s. The confusion sometimes occur because the dt in the R 1 ¡st integral(16.1)ismissing. Forexample, whenwriting e it becomes easy 0 to forget that the integrand is a function of t and integrate as a function of s instead. Another opportunity for an error occurs when evaluating an expression such as (16.2b). Students sometimes substitute 0 and 1 for s instead of t. ² mistake The Laplace transform of a product f(t)g(t) of two functions is NOT the product F(s)G(s) of the Laplace transforms. For example 1 f(t)=g(t)=1 ) F(s)=G(s)= ; s but 1 f(t)g(t)=1 ) Lff(t)g(t)g(s)= = 6 F(s)G(s): s A mistake is often made the other way around. Consider for example 2 U(s)= : 2 (s+1)(s +1) It is tempting, but INCORRECT, to do the following: 2 1 U(s)= 2 s+1 s +1 ¡1 ¡1 L L ¡t ¡t 2e sint )u(t)=2e sint: (3) (4) 2 2 If this were the case we would have U(s) = = according to 2 2 (s+1) +1 s +2s+2 formula11, whichis notthecase. Thecorrect function u(t)canbeobtained74 Chapter 16. Basic Laplace Transform Techniques here using a PFD of U(s). From Example ... we have 1 ¡s+1 1 ¡s 1 U(s)= + = + + 2 2 2 s+1 s +1 s+1 s +1 s +1 ¡1 ¡1 ¡1 L L L ¡t e ¡cos(t) sin(t) (3) (5) (4) ¡t )u(t)=e ¡cost+sint: 0 0 ² mistake The Laplace transform of f (t) is NOT F (s) (but is sF(s)¡f(0)). ² tip The solution of an IVP obtained using the Laplace transform can be checkedusingtheIC(s). RememberthattheICsareusedearlyinthesolution process. Manyerrorscanoccurinthecalculationsuntilthe¯nalanswer. Thus check whether your solution satis¯es the ICs (at least the ¯rst one, u(0). For example the solution of the IVP (16.14) given by (16.17) satis¯es u(0)=¡3¡0+8¡4=1; X 0 u(0)=0¡1¡4+4=¡1: X ² tip The Laplace inversion process of a rational function F(s) can be carried out without explicit knowledge of the coe±cients of the PFD, provided it is set-up correctly. In Example 16.18 the set-up (16.17) yields as+b c d F(s)= + + 2 s +1 2s+1 s+1 s 1 1 1 c =a +b + + d 2 1 2 2 s +1 s +1 s+1 s+ 2 ¡1 ¡1 ¡1 ¡1 L L L L 1 ¡ t ¡t 2 cost sint e e (5) (4) (3) (3) c 1 ¡1 ¡ t ¡t 2 )f(t)=L fF(s)g=acost+bsint+ e +de : 2 So you can still proceed if you get stuck in the system solution for a, b, c, d. ² tip Large values of s correspond to small values of t. This remark is some- times useful to check whether a Laplace transform or inverse transform makes sense. The diagram 1 L at e ¡¡¡¡ s¡a ? ? ? ? ¼ for s large ¼ for t'0y y 1 L 1 ¡¡¡¡ s illustrates this. If s = i, large s means large , i.e., high frequency. A function which varies with a high frequency must be observed on a small time scale.Exercises 75 Exercises 16.1. Use bt ¡bt a) either the expression of sinh(bt) and cosh(bt) in terms of e and e b) or Problem 1.54 as well as Table 16.1 to show the following formulas: 2bs b at (i) Lftsinh(bt)g= (iii) Lfe sinh(bt)g= 2 2 2 2 2 (s ¡b ) (s¡a) ¡b 2 2 b s +b at (iv) Lfe cosh(bt)g= (ii) Lftcosh(bt)g= 2 2 2 2 2 (s¡a) ¡b (s ¡b ) InProblems16.2-16.3determinetheinverseLaplacetransformofthegivenfunction. 4s 2 16.2. 16.3. 2 2 2 (s ¡4) (s¡2) ¡4 Z t 16.4. Consider the function g(t)= f(¿)d¿. 0 0 a) Verify that g (t)=f(t). What is g(0)? b) Apply the derivative formula (16.7) to g(t) and show that 1 Lfg(t)g=F(s) ; s i.e., Z t ¡1 1 L F(s) ¡¡¡¡ f(¿)d¿: (16.21) s 076 Chapter 16. Basic Laplace Transform Techniques Table 16.1. Short table of Laplace transforms u(t) U(s) Reference 1 1 1 s 1 2 t 2 s 1 at 3 e s¡a b 4 sin(bt) 2 2 s +b s 5 cos(bt) 2 2 s +b a 6 sinh(at) 2 2 s ¡a s 7 cosh(at) 2 2 s ¡a 1 at 8 te 2 (s¡a) 2bs 9 tsin(bt) 2 2 2 (s +b ) 2 2 s ¡b 10 tcos(bt) 2 2 2 (s +b ) b at 11 e sin(bt) 2 2 (s¡a) +b s¡a at 12 e cos(bt) 2 2 (s¡a) +b 3 2b 13 sin(bt)¡btcos(bt) 2 2 2 (s +b ) 14 ®f(t)+¯g(t) ®F(s)+¯G(s) 0 15 f (t) sF(s)¡f(0) 00 2 0 16 f (t) s F(s)¡sf(0)¡f (0)Chapter 17 Advanced Laplace Transform Techniques 17.1 Laplace transform of special functions 17.1.1 Piecewise functions Let c¸0 ¯xed and consider the function u (t): c 6 ½ 0 if t·c u (t) c 1 u (t)= c 1 if tc - t 0 c The function u (t) is piecewise constant and is a classical model for physical phe- c nomena which can toggle between \o®" (0) and \on" (1) states (e.g., a switch). Thefunctionu (t)iscalleda Heaviside function andcanbeusedtoexpressgeneral c piecewise functions. Example 17.1 The function 6 2 8 0 if t·1 f(t) 1 if 1t·2 1 f(t)= 2 if 2t·3 3 4 - t ¡1 if 3t·4 0 1 2 : 0 if t4 ¡1 7778 Chapter 17. Advanced Laplace Transform Techniques is expressed as f(t)=0+u (t)(1¡0)+u (t)(2¡1)+u (t)((¡1)¡2)+u (t)(0¡(¡1)) 1 2 3 4 =u (t)+u (t)¡3u (t)+u (t): 1 2 3 4 For example f(2:5) = 1+1¡3(0)+0 = 2 since u (t) = 1 for t 1, u (t) = 1 for 1 2 t2, u (t)=0 for t·3 and u (t)=0 for t·4. 3 4 In general the expression of f(t) starts with the ¯rst expression of the function (at t = 0). Whenever the function f(t) exhibits a discontinuity at some time t = c a jump term u (t)(right function¡left function) is added. c Example 17.2 The function 6 2 8 t if t·2 g(t) 1 8¡3t if 2t·3 g(t)= t¡4 if 3t·4 3 4 - t : 0 1 2 0 if t4 ¡1 is expressed as g(t)=t+u (t)(8¡3t¡t)+u (t)(t¡4¡(8¡3t))+u (t)(0¡(t¡4)) 2 3 4 =t¡4u (t)(t¡2)+4u (t)(t¡3)¡u (t)(t¡4): 2 3 4 For example, for 3t·4, g(t)=t¡4(t¡2)+4(t¡3)=t¡4. ThereasonwhyweexpresspiecewisefunctionsintermsofHeavisidefunctions u (t) is to determine their Laplace transform. One common mistake is to believe c that this transform can be computed directly from the expression of the function on each subinterval. We start with u (t) itself using the only too available to do c17.1. Laplace transform of special functions 79 this, i.e., the de¯nition (16.1): Z 1 ¡st Lfu (t)g= u (t)e dt c c 0 Z Z c 1 ¡st ¡st (split integral) = u (t)e dt+ u (t)e dt c c 0 c Z Z c 1 ¡st ¡st (substitute expression on each interval) = 0¢e dt+ 1¢e dt 0 c Z 1 ¡st = e dt c · ¸ 1 ¡st e = ¡s c µ ¶ µ ¶ ¡st ¡cs e e = lim ¡ t1 ¡s ¡s ¡cs e (assume s0) = : s This formula is recorded as entry 17 of Table 17.1. It can be used, together with the linearity property 14 of Table 16.1, to determine the Laplace transform of linear combinations of Heaviside functions, such as the function from Example 17.1. Example 17.3 Lfu (t)+u (t)¡3u (t)+u (t)g=Lfu (t)g+Lfu (t)g¡3Lfu (t)g+Lfu (t)g 1 2 3 4 1 2 3 4 ¡s ¡2s ¡3s s 4 =e +e ¡3e +e : An additional formula, discussed in Section 17.2.1 is needed to determine the Laplace transform of a function such as the one considered in Example 17.2. 17.1.2 Point distributions Let c¸0 ¯xed, h0, and consider the function f (t): h80 Chapter 17. Advanced Laplace Transform Techniques 6 ¾ ± (t) c ¾ f (t) as h0 h ( 0 if t·c or tc+h f (t)= 1 h f (t) if ct·c+h 1 h h h - t c+h 0 c We are interested in the limit of the function f (t) as h 0. Call it ± (t) h c (\delta-c"). The \function" ± (t) must vanish everywhere except t = c, where it c must be in¯nite. It is called the Dirac distribution, and is an example of a point distribution. The Dirac distribution is often used as a mathematical model for physical phenomena which occur in a very short amount of time (h 0) but with a ¯nite transmission of energy, momentum, etc (¯nite positive \area" under the \curve"). Think for example of two cars colliding (ouch). The \function" ± (t) is, in some sense, the derivative of the piecewise function c u (t) introduced in Section 17.1.1: c 0 1. u (t)=0=± (t) for all t6=c c c 2. Since the area below f (t) is 1 independently of h 0, the area below ± (t) h c must also be equal to 1 (this is an instance where 0£1=1). 3. The cumulative area below ± (t) thus remains 0 for t · c and jumps to 1 c whenever tc. This is u (t) c Formally, we write ½ Z t 0 if t·c 0 ± (t)=u (t) in the sense that ± (¿)d¿ = =u (t): h c c c 1 if tc 0 Useful properties of ± (t) include: c ² ± (t)f(t)=± (t)f(c)foranytandfunctionf, sincebothsidesvanishfort6=0 c c and are obviously equal for t=c. Z Z Z 1 1 1 ² ± (t)f(t)dt= ± (t)f(c)dt=f(c) ± (t)dt=f(c). c c c 0 0 0 ¡st ² With f(t)=e we obtain ¡sc Lf± (t)g=e c (recorded as entry 18 in Table 17.1).

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