Lecture notes in Laplace Transform

what is laplace transform of impulse function, what is laplace transform in control system, how laplace transform works,how to laplace transform step function pdf free download
Prof.SteveBarros Profile Pic
Prof.SteveBarros,United Kingdom,Teacher
Published Date:28-07-2017
Your Website URL(Optional)
Comment
Chapter 7 Laplace Transform The Laplace transform can be used to solve di erential equations. Be- sides being a di erent and ecient alternative to variation of parame- ters and undetermined coecients, theLaplace method is particularly advantageous for input terms that are piecewise-de ned, periodic or im- pulsive. Thedirect Laplace transform or the Laplace integral of a function f(t) de ned for 0t1 is the ordinary calculus integration problem Z 1 st f(t)e dt; 0 succinctly denoted L(f(t)) in science and engineering literature. The Lnotation recognizes that integration always proceeds over t = 0 to st t =1 and that the integral involves an integrator e dt instead of the usual dt. These minor di erences distinguish Laplace integrals from the ordinary integrals found on the inside covers of calculus texts. 7.1 Introduction to the Laplace Method The foundation of Laplace theory is Lerch's cancellation law R R 1 1 st st y(t)e dt = f(t)e dt implies y(t) =f(t); 0 0 (1) or L(y(t) =L(f(t)) implies y(t) =f(t): In di erential equation applications, y(t) is the sought-after unknown while f(t) is an explicit expression taken from integral tables. Below, we illustrate Laplace's method by solving the initial value prob- lem 0 y =1; y(0) = 0: The method obtains a relation L(y(t)) =L(t), whence Lerch's cancel- lation law implies the solution is y(t) =t. The Laplace method is advertised as a table lookup method, in which the solution y(t) to a di erential equation is found by looking up the answer in a special integral table.7.1 Introduction to the Laplace Method 247 R 1 st Laplace Integral. The integral g(t)e dt is called the Laplace 0 R N st integral of the functiong(t). It is de ned by lim g(t)e dt and N1 0 depends on variable s. The ideas will be illustrated forg(t) = 1, g(t) =t 2 and g(t) =t , producing the integral formulas in Table 1. R 1 t=1 st st (1)e dt = (1=s)e Laplace integral of g(t) = 1. 0 t=0 = 1=s Assumed s 0. R R 1 1 d st st (t)e dt = (e )dt Laplace integral of g(t) =t. 0 0 ds R R R 1 d st d d = (1)e dt Use F(t;s)dt = F(t;s)dt. 0 ds ds ds d = (1=s) UseL(1) = 1=s. ds 2 =1=s Di erentiate. R R 1 1 2 st d st 2 (t )e dt = (te )dt Laplace integral of g(t) =t . 0 0 ds R 1 d st = (t)e dt ds 0 d 2 2 = (1=s ) UseL(t) = 1=s . ds 3 = 2=s R 1 st 2 Table 1. The Laplace integral g(t)e dt for g(t) =1, t and t . 0 R R R 1 1 2 1 1 1 st st 2 st (1)e dt = (t)e dt = (t )e dt = 0 0 0 2 3 s s s n n In summary, L(t ) = 1+n s An Illustration. The ideas of the Laplace method will be illus- 0 trated for the solution y(t) =t of the problem y =1, y(0) = 0. The method, entirely di erent from variation of parameters or undetermined coecients, uses basic calculus and college algebra; see Table 2. 0 Table 2. Laplace method details for the illustration y =1, y(0) = 0. 0 st st 0 st y (t)e =e Multiply y =1 by e . R R 1 1 0 st st y (t)e dt = e dt Integrate t = 0 to t =1. 0 0 R 1 0 st y (t)e dt =1=s Use Table 1. 0 R 1 st s y(t)e dty(0) =1=s Integrate by parts on the left. 0 R 1 st 2 y(t)e dt =1=s Use y(0) = 0 and divide. 0 R R 1 1 st st y(t)e dt = (t)e dt Use Table 1. 0 0 y(t) =t Apply Lerch's cancellation law.248 Laplace Transform In Lerch's law, the formal rule of erasing the integral signs is valid pro- vided the integrals are equal for large s and certain conditions hold on y and f see Theorem 2. The illustration in Table 2 shows that Laplace theory requires an in-depth study of a special integral table, a table which is a true extension of the usual table found on the inside covers of calculus books. Some entries for the special integral table appear in Table 1 and also in section 7.2, Table 4. TheL-notation for the direct Laplace transformproducesbrieferdetails, aswitnessed bythetranslation ofTable2into Table3below. Thereader is advised to move from Laplace integral notation to the Lnotation as soon as possible, in order to clarify the ideas of the transform method. 0 Table 3. Laplace method L-notation details for y = 1, y(0) = 0 translated from Table 2. 0 0 0 L(y (t)) =L(1) ApplyL across y =1, or multiply y = st 1 by e , integrate t = 0 to t =1. 0 L(y (t)) =1=s Use Table 1. sL(y(t))y(0) =1=s Integrate by parts on the left. 2 L(y(t)) =1=s Use y(0) = 0 and divide. L(y(t)) =L(t) Apply Table 1. y(t) =t Invoke Lerch's cancellation law. SomeTransformRules. Theformalpropertiesofcalculusintegrals plustheintegrationbypartsformulausedinTables2and3leadstothese rules for the Laplace transform: L(f(t)+g(t)) =L(f(t))+L(g(t)) The integral of a sum is the sum of the integrals. L(cf(t)) =cL(f(t)) Constants c pass through the integral sign. 0 L(y (t)) =sL(y(t))y(0) The t-derivative rule, or inte- gration by parts. See Theo- rem 3. L(y(t)) =L(f(t)) implies y(t) =f(t) Lerch's cancellation law. See Theorem 2. 1 Example (Laplace method) Solve by Laplace's method the initial value 0 problem y = 52t, y(0) = 1. Solution: Laplace's method is outlined in Tables 2 and 3. The L-notation of 2 Table 3 will be used to nd the solution y(t) = 1+5tt .7.1 Introduction to the Laplace Method 249 0 0 L(y (t)) =L(52t) ApplyL across y = 52t. 5 2 0 L(y (t)) = Use Table 1. 2 s s 5 2 sL(y(t))y(0) = Apply the t-derivative rule, page 248. 2 s s 1 5 2 L(y(t)) = + Use y(0)= 1 and divide. 2 3 s s s 2 L(y(t)) =L(1)+5L(t)L(t ) Apply Table 1, backwards. 2 =L(1+5tt ) Linearity, page 248. 2 y(t) =1+5tt Invoke Lerch's cancellation law. 2 Example (Laplace method) Solve by Laplace's method the initial value 00 0 problem y = 10, y(0) =y (0) = 0. 2 Solution: TheL-notationofTable3willbeusedto ndthesolutiony(t) = 5t . 00 00 L(y (t)) =L(10) ApplyL across y = 10. 0 0 0 sL(y (t))y (0) =L(10) Apply thet-derivative rule to y , that is, 0 replace y by y on page 248. 0 ssL(y(t))y(0)y (0) =L(10) Repeat the t-derivative rule, on y. 2 0 s L(y(t)) =L(10) Use y(0) =y (0) = 0. 10 L(y(t)) = Use Table 1. Then divide. 3 s 2 L(y(t)) =L(5t ) Apply Table 1, backwards. 2 y(t) =5t Invoke Lerch's cancellation law. R 1 st Existence of the Transform. The Laplace integral e f(t)dt 0 1 is known to exist in the sense of the improper integral de nition Z Z 1 N g(t)dt = lim g(t)dt N1 0 0 provided f(t) belongs to a class of functions known in the literature as functions of exponential order. For this class of functions the relation f(t) (2) lim = 0 at t1 e is required to hold for some real number a, or equivalently, for some constants M and , t (3) jf(t)jMe : In addition, f(t) is required to bepiecewise continuous on each nite subinterval of 0t1, a term de ned as follows. 1 An advanced calculus background is assumed for the Laplace transform existence proof. Applications of Laplace theory require only a calculus background.250 Laplace Transform De nition 1 (piecewise continuous) A function f(t) is piecewise continuous on a nite interval a;b pro- vided there exists a partition a = t  t = b of the interval a;b 0 n and functions f , f , ..., f continuous on (1;1) such that for t not 1 2 n a partition point 8 f (t) t t t ; 1 0 1 . . . . (4) f(t) = . . : f (t) t t t : n n1 n The values of f at partition points are undecided by equation (4). In particular, equation (4) implies that f(t) has one-sided limits at each point of a t b and appropriate one-sided limits at the endpoints. Therefore,f hasatworstajumpdiscontinuityateach partition point. t 3 Example (Exponential order) Show that f(t) = e cost +t is of expo- nential order, that is, show thatf(t) is piecewise continuous and nd 0 t such that lim f(t)=e = 0. t1 Solution: Already, f(t) is continuous, hence piecewise continuous. From t L'Hospital's rule in calculus, lim p(t)=e = 0 for any polynomial p and t1 any 0. Choose =2, then f(t) cost t lim = lim + lim =0: 2t t 2t t1 t1 t1 e e e Theorem 1 (Existence of L(f)) Let f(t) be piecewise continuous on every nite interval int 0 and satisfy t jf(t)jMe for some constants M and . ThenL(f(t)) exists for s and lim L(f(t)) = 0. s1 Proof: It has to be shown that the Laplace integral of f is nite for s . Advanced calculus implies that it is sucient to show that the integrand is ab- (s )t solutely bounded above by an integrable function g(t). Take g(t) =Me . Then g(t) 0. Furthermore, g is integrable, because Z 1 M g(t)dt = : s 0 t Inequality jf(t)j  Me implies the absolute value of the Laplace transform st integrand f(t)e is estimated by st t st f(t)e Me e =g(t): R M 1 The limit statement follows from jL(f(t))j  g(t)dt = , because the 0 s right side of this inequality has limit zero at s =1. The proof is complete.7.1 Introduction to the Laplace Method 251 Theorem 2 (Lerch) R 1 st Iff (t) andf (t) are continuous, of exponential order and f (t)e dt = 1 2 1 0 R 1 st f (t)e dt for all ss , then f (t) =f (t) for t 0. 2 0 1 2 0 Proof: See Widder ?. Theorem 3 (t-Derivative Rule) st 0 If f(t) is continuous, limf(t)e = 0 for all large values of s and f (t) t1 0 0 is piecewise continuous, then L(f (t)) exists for all large s and L(f (t)) = sL(f(t))f(0). Proof: See page 276. Exercises 7.1 P N Laplace method. Solve the given 18. f(t) = c sin(nt), for any n n=1 initial value problem using Laplace's choiceoftheconstantsc ,...,c . 1 N method. Existence of transforms. Let f(t) = 0 1. y =2, y(0) = 0. 2 2 t t te sin(e ). Establish these results. 0 2. y =1, y(0) =0. 19. The function f(t) is not of expo- 0 nential order. 3. y =t, y(0) =0. 0 20. The Laplace integral of f(t), 4. y =t, y(0) = 0. R 1 st f(t)e dt, converges for all 0 0 5. y =1t, y(0) = 0. s 0. 0 6. y =1+t, y(0) = 0. Jump Magnitude. For f piecewise 0 7. y =32t, y(0) =0. continuous, de ne the jump at t by 0 J(t) = lim f(t+h) lim f(th): 8. y =3+2t, y(0) =0. h0+ h0+ 00 0 9. y =2, y(0) =y (0) = 0. Compute J(t) for the following f. 00 0 10. y = 1, y(0) =y (0) = 0. 21. f(t) =1 for t 0, else f(t) =0 00 0 11. y = 1t, y(0) =y (0) = 0. 22. f(t) =1 for t1=2, else f(t) =0 00 0 12. y = 1+t, y(0) =y (0) = 0. 23. f(t) =t=jtj for t6=0, f(0) = 0 00 0 13. y = 32t, y(0) =y (0) = 0. 24. f(t) = sint=jsintj for t 6= n, n f(n) =(1) 00 0 14. y = 3+2t, y(0) =y (0) = 0. Taylor series. The series relation P P Exponential order. Show that f(t) 1 1 n n L( c t ) = c L(t ) often n n n=0 n=0 is of exponential order, by nding a n holds, in which case the resultL(t ) = constant  0 in each case such that 1n ns can be employed to nd a f(t) lim = 0. series representation of the Laplace t t1 e transform. Use this idea on the fol- 15. f(t) = 1+t lowing to nd a series formula for t L(f(t)). 16. f(t) =e sin(t) P 1 2t n P N 25. f(t) =e = (2t) =n n n=0 17. f(t) = c x , for any choice n n=0 P 1 t n of the constants c , ..., c . 0 N 26. f(t) =e = (t) =n n=0252 Laplace Transform 7.2 Laplace Integral Table The objective in developing a table of Laplace integrals, e.g., Tables 4 and 5, is to keep the table size small. Table manipulation rules appear- ing in Table 6, page 257, e ectively increase the table size manyfold, making it possible to solve typical di erential equations from electrical and mechanical problems. The combination of Laplace tables plus the table manipulation rules is called the Laplace transform calculus. Table 4 is considered to be a table of minimum size to be memorized. Table 5 adds a numberof special-use entries. For instance, the Heaviside entry in Table 5 is memorized, but usually not the others. Derivationsarepostponedtopage270. Thetheoryofthegammafunc- tion (x) appears below on page 255. Table 4. A minimal Laplace integral table with L-notation R n n 1 n st n (t )e dt = L(t ) = 0 1+n 1+n s s R 1 1 1 at st at (e )e dt = L(e ) = 0 sa sa R s s 1 st (cosbt)e dt = L(cosbt) = 0 2 2 2 2 s +b s +b R b b 1 st (sinbt)e dt = L(sinbt) = 0 2 2 2 2 s +b s +b Table 5. Laplace integral table extension as e L(H(ta)) = (a 0) Heaviside unit step, de ned by  s 1 for t 0; H(t) = 0 otherwise. as L((ta)) =e Dirac delta, (t) =dH(t). Special usage rules apply. as e L( oor(t=a)) = Staircase function, as s(1e ) oor(x) = greatest integerx. 1 L(sqw(t=a)) = tanh(as=2) Square wave, s oor(x) sqw(x) = (1) . 1 L(atrw(t=a)) = tanh(as=2) Triangular wave, 2 R s x trw(x) = sqw(r)dr. 0 (1+ ) L(t ) = Generalized power function, 1+ R s 1 x (1+ ) = e x dx. 0 r  p 1=2 L(t ) = Because (1=2)= . s7.2 Laplace Integral Table 253 3t 4 Example (Laplace transform) Letf(t) =t(t1)sin2t+e . Compute L(f(t)) using the basic Laplace table and transform linearity properties. Solution: 2 3t L(f(t)) =L(t 5tsin2t+e ) Expand t(t5). 2 3t =L(t )5L(t)L(sin2t)+L(e ) Linearity applied. 2 5 2 1 = + Table lookup. 3 2 2 s s s +4 s3 5 Example (Inverse Laplace transform) Use the basic Laplace table back- wards plus transform linearity properties to solve for f(t) in the equation s 2 s+1 L(f(t)) = + + : 2 3 s +16 s3 s Solution: s 1 1 1 2 L(f(t)) = +2 + + Convert to table entries. 2 2 3 s +16 s3 s 2s 3t 1 2 =L(cos4t)+2L(e )+L(t)+ L(t ) Laplace table (backwards). 2 3t 1 2 =L(cos4t+2e +t+ t ) Linearity applied. 2 1 3t 2 f(t) = cos4t+2e +t+ t Lerch's cancellation law. 2 6 Example (Heaviside) Find the Laplace transform of f(t) in Figure 1. 5 Figure 1. A piecewise de ned 1 function f(t) on 0t1: f(t) = 0 except for 1t 2 and 3t4. 1 3 5 Solution: The details require the use of the Heaviside function formula  1 atb; H(ta)H(tb) = 0 otherwise: The formula for f(t): 8   1 1t 2; 1 1t2; 1 3t 4; 5 3t 4; f(t) = = +5 0 otherwise 0 otherwise : 0 otherwise Then f(t) = f (t) + 5f (t) where f (t) = H(t 1) H(t 2) and f (t) = 1 2 1 2 H(t3)H(t4). The extended table gives L(f(t)) =L(f (t))+5L(f (t)) Linearity. 1 2 =L(H(t1))L(H(t2))+5L(f (t)) Substitute for f . 2 1254 Laplace Transform s 2s e e = +5L(f (t)) Extended table used. 2 s s 2s 3s 4s e e +5e 5e = Similarly for f . 2 s 7 Example (Dirac delta) A machine shop tool that repeatedly hammers a P N die is modeled by the Dirac impulse model f(t) = (tn). Show n=1 P N ns thatL(f(t)) = e . n=1 Solution:   P N L(f(t)) =L (tn) n=1 P N = L((tn)) Linearity. n=1 P N ns = e Extended Laplace table. n=1 8 Example (Square wave) A periodic camshaft force f(t) applied to a me- chanical system has the idealized graph shown in Figure 2. Show that 1 f(t) = 1+sqw(t) andL(f(t)) = (1+tanh(s=2)). s 2 Figure 2. A periodic force f(t) applied 0 to a mechanical system. 1 3 Solution:  1+1 2nt 2n+1, n =0;1;:::, 1+sqw(t) = 11 2n+1t 2n+2, n =0;1;:::,  2 2nt 2n+1, n =0;1;:::, = 0 otherwise, = f(t): 1 tanh(s=2) By the extended Laplace table,L(f(t)) =L(1)+L(sqw(t)) = + . s s 9 Example (Sawtooth wave) Express the P-periodic sawtooth wave repre- sented in Figure 3 as f(t) =ct=P c oor(t=P) and obtain the formula Ps c ce L(f(t)) = : 2 Ps Ps sse c Figure 3. A P-periodic sawtooth 0 wave f(t) of height c0. P 4P7.2 Laplace Integral Table 255 Solution: The representation originates from geometry, because the periodic function f can be viewed as derived from ct=P by subtracting the correct con- stant from each of intervals P;2P, 2P;3P, etc. The technique used to verify the identity is to de ne g(t) =ct=Pc oor(t=P) and then show that g is P-periodic and f(t) = g(t) on 0  t P. Two P- periodic functions equal on the base interval 0  t P have to be identical, hence the representation follows. The ne details: for 0 t P, oor(t=P) = 0 and oor(t=P +k) = k. Hence g(t + kP) = ct=P + ck c oor(k) = ct=P = g(t), which implies that g is P-periodic and g(t) =f(t) for 0tP. c L(f(t)) = L(t)cL( oor(t=P)) Linearity. P Ps c ce = Basic and extended table applied. 2 Ps Ps sse 10 Example (Triangular wave) Express the triangular wave f of Figure 4 in 5 terms of the square wave sqw and obtainL(f(t)) = tanh(s=2). 2 s 5 Figure 4. A 2-periodic triangular 0 2 wave f(t) of height 5. R t= Solution: The representation of f in terms of sqw is f(t) =5 sqw(x)dx. 0 Details: A 2-periodic triangular wave of height 1 is obtained by integrating the square wave of period 2. A wave of height c and period 2 is given by R R t 2t=P ctrw(t) = c sqw(x)dx. Then f(t) = ctrw(2t=P) = c sqw(x)dx where 0 0 c =5 and P = 2. Laplace transform details: Use the extended Laplace table as follows. 5 5 L(f(t)) = L(trw(t=)) = tanh(s=2): 2  s Gamma Function. In mathematical physics, the Gamma func- tion or the generalized factorial function is given by the identity Z 1 t x1 (1) (x) = e t dt; x 0: 0 This function is tabulated and available in computer languages like For- tran, C and C++. It is also available in computer algebra systems and numerical laboratories. Some useful properties of (x): (2) (1+x) = x(x) (3) (1+n) = n for integers n 1:256 Laplace Transform R 1 t Details for relations (2) and (3): Start with e dt = 1, which gives 0 (1) = 1. Use this identity and successively relation (2) to obtain relation (3). To prove identity (2), integration by parts is applied, as follows: R 1 t x (1+x) = e t dt De nition. 0 R t=1 1 x t t x1 x t = t e j + e xt dt Use u =t , dv =e dt. t=0 0 R 1 t x1 =x e t dt Boundary terms are zero 0 for x 0. =x(x). Exercises 7.2 Laplace transform. Compute Inverse Laplace transform. Solve L(f(t)) using the basic Laplace table the given equation for the function and the linearity properties of the f(t). Use the basic table and linearity transform. Do not use the direct properties of the Laplace transform. Laplace transform 2 21. L(f(t)) =s 1. L(2t) 2 22. L(f(t)) =4s 2. L(4t) 2 3 23. L(f(t)) =1=s+2=s +3=s 2 3. L(1+2t+t ) 3 24. L(f(t)) =1=s +1=s 2 4. L(t 3t+10) 2 25. L(f(t)) =2=(s +4) 5. L(sin2t) 2 26. L(f(t)) =s=(s +4) 6. L(cos2t) 27. L(f(t)) =1=(s3) 2t 7. L(e ) 28. L(f(t)) =1=(s+3) 2t 8. L(e ) 2 29. L(f(t)) =1=s+s=(s +4) 9. L(t+sin2t) 2 30. L(f(t)) =2=s2=(s +4) 10. L(tcos2t) 31. L(f(t)) =1=s+1=(s3) 2t 11. L(t+e ) 32. L(f(t)) =1=s3=(s2) 2t 12. L(t3e ) 2 3 33. L(f(t)) =(2+s) =s 2 13. L((t+1) ) 2 34. L(f(t)) =(s+1)=s 2 14. L((t+2) ) 2 3 35. L(f(t)) =s(1=s +2=s ) 15. L(t(t+1)) 3 36. L(f(t)) =(s+1)(s1)=s 16. L((t+1)(t+2)) P 10 1+n 37. L(f(t)) = n=s P n=0 10 n 17. L( t =n) n=0 P 10 2+n 38. L(f(t)) = n=s P n=0 10 n+1 18. L( t =n) n=0 P n 10 39. L(f(t)) = P 10 n=1 2 2 s +n 19. L( sinnt) n=1 P s P 10 10 40. L(f(t)) = 20. L( cosnt) n=0 2 2 n=0 s +n7.3 Laplace Transform Rules 257 7.3 Laplace Transform Rules In Table 6, the basic table manipulation rules are summarized. Full statements and proofs of the rules appear in section 7.7, page 275. The rules are applied here to several key examples. Partial fraction expansions do not appear here, but in section 7.4, in connection with Heaviside's coverup method. Table 6. Laplace transform rules L(f(t)+g(t)) =L(f(t))+L(g(t)) Linearity. The Laplace of a sum is the sum of the Laplaces. L(cf(t)) =cL(f(t)) Linearity. Constants move through theL-symbol. 0 L(y (t)) =sL(y(t))y(0) The t-derivative rule. 0 DerivativesL(y ) are replaced in transformed equations.   R 1 t L g(x)dx = L(g(t)) The t-integral rule. 0 s d L(tf(t)) = L(f(t)) The s-di erentiation rule. ds Multiplying f by t appliesd=ds to the transform of f. at L(e f(t)) = L(f(t))j First shifting rule. s(sa) at Multiplying f by e replaces s by sa. as L(f(ta)H(ta)) =e L(f(t)), Second shifting rule. as L(g(t)H(ta)) =e L(g(t+a)) First and second forms. R P st f(t)e dt 0 L(f(t)) = Rule for P-periodic functions. Ps 1e Assumed here is f(t+P) = f(t). L(f(t))L(g(t)) =L((fg)(t)) Convolution rule. R t De ne (fg)(t) = f(x)g(tx)dx. 0 11 Example (Harmonic oscillator) SolvebyLaplace'smethodtheinitialvalue 00 0 problem x +x = 0, x(0) = 0, x(0) = 1. Solution: The solution is x(t) = sint. The details: 00 L(x )+L(x) =L(0) ApplyL across the equation. 0 0 sL(x )x (0)+L(x) =0 Use the t-derivative rule. 0 ssL(x)x(0)x (0)+L(x) =0 Use again the t-derivative rule. 2 0 (s +1)L(x) = 1 Use x(0) = 0, x (0) = 1. 1 L(x) = Divide. 2 s +1 =L(sint) Basic Laplace table. x(t) = sint Invoke Lerch's cancellation law.258 Laplace Transform 2 2 5t 12 Example (s-di erentiation rule) ShowthestepsforL(t e ) = . 3 (s5) Solution:    d d 2 5t 5t L(t e ) = L(e ) Apply s-di erentiation. ds ds   d d 1 2 =(1) Basic Laplace table. dsds s5   d 1 = Calculus power rule. 2 ds (s5) 2 = Identity veri ed. 3 (s5) 2 2 3t 13 Example (First shifting rule) Show the steps for L(t e ) = . 3 (s+3) Solution: 2 2 3t L(t e ) = L(t ) First shifting rule. ss(3)   2 = Basic Laplace table. 2+1 s ss(3) 2 = Identity veri ed. 3 (s+3) 14 Example (Second shifting rule) Show the steps for s e L(sintH(t)) = : 2 s +1 Solution: The second shifting rule is applied as follows. L(sintH(t)) =L(g(t)H(ta) Choose g(t) = sint, a =. as =e L(g(t+a) Second form, second shifting theorem. s =e L(sin(t+)) Substitute a =. s =e L(sint) Sum rule sin(a + b) = sinacosb + sinbcosa plus sin = 0, cos =1. 1 s =e Basic Laplace table. Identity veri ed. 2 s +1 15 Example (Trigonometric formulas) Show the steps used to obtain these Laplace identities: 2 2 3 2 s a 2(s 3sa ) 2 (a)L(tcosat) = (c) L(t cosat) = 2 2 2 2 2 3 (s +a ) (s +a ) 2 3 2sa 6s aa 2 (b)L(tsinat) = (d)L(t sinat) = 2 2 2 2 2 3 (s +a ) (s +a )7.3 Laplace Transform Rules 259 Solution: The details for (a): L(tcosat) =(d=ds)L(cosat) Use s-di erentiation.   d s = Basic Laplace table. 2 2 ds s +a 2 2 s a = Calculus quotient rule. 2 2 2 (s +a ) The details for (c): 2 L(t cosat) =(d=ds)L((t)cosat) Use s-di erentiation.   2 2 d s a = Result of (a). 2 2 2 ds (s +a ) 3 2 2s 6sa ) = Calculus quotient rule. 2 2 3 (s +a ) The similar details for (b) and (d) are left as exercises. 16 Example (Exponentials) Show the steps used to obtain these Laplace identities: 2 2 sa (sa) b at at (a)L(e cosbt) = (c)L(te cosbt) = 2 2 2 2 2 (sa) +b ((sa) +b ) b 2b(sa) at at (b)L(e sinbt) = (d)L(te sinbt) = 2 2 2 2 2 (sa) +b ((sa) +b ) Solution: Details for (a): at L(e cosbt)= L(cosbt)j First shifting rule. ssa   s = Basic Laplace table. 2 2 s +b ssa sa = Veri ed (a). 2 2 (sa) +b Details for (c): at L(te cosbt)= L(tcosbt)j First shifting rule. ssa   d = L(cosbt) Apply s-di erentiation. ds ssa    d s = Basic Laplace table. 2 2 ds s +b ssa   2 2 s b = Calculus quotient rule. 2 2 2 (s +b ) ssa 2 2 (sa) b = Veri ed (c). 2 2 2 ((sa) +b ) Left as exercises are (b) and (d).260 Laplace Transform 17 Example (Hyperbolic functions) Establish these Laplace transform facts u u u u about coshu = (e +e )=2 and sinhu = (e e )=2. 2 2 s s +a (a)L(coshat) = (c) L(tcoshat) = 2 2 2 2 2 s a (s a ) a 2as (b)L(sinhat) = (d)L(tsinhat) = 2 2 2 2 2 s a (s a ) Solution: The details for (a): 1 at at L(coshat) = (L(e )+L(e )) De nition plus linearity ofL. 2   1 1 1 = + Basic Laplace table. 2 sa s+a s = Identity (a) veri ed. 2 2 s a The details for (d):   d a L(tsinhat) = Apply the s-di erentiation rule. 2 2 ds s a a(2s) = Calculus power rule; (d) veri ed. 2 2 2 (s a ) Left as exercises are (b) and (c). 2s 18 Example (s-di erentiation) SolveL(f(t)) = for f(t). 2 2 (s +1) Solution: The solution is f(t) =tsint. The details: 2s L(f(t)) = 2 2 (s +1)   d 1 n 0 n1 0 = Calculus power rule (u ) =nu u . 2 ds s +1 d = (L(sint)) Basic Laplace table. ds =L(tsint) Apply the s-di erentiation rule. f(t) =tsint Lerch's cancellation law. s+2 19 Example (First shift rule) SolveL(f(t)) = for f(t). 2 2 +2s+2 t t Solution: The answer is f(t) =e cost+e sint. The details: s+2 L(f(t)) = Signal for this method: the denom- 2 s +2s+2 inator has complex roots. s+2 = Complete the square, denominator. 2 (s+1) +17.3 Laplace Transform Rules 261 S +1 = Substitute S for s+1. 2 S +1 S 1 = + Split into Laplace table entries. 2 2 S +1 S +1 = L(cost)+L(sint)j Basic Laplace table. sS=s+1 t t =L(e cost)+L(e sint) First shift rule. t t f(t) =e cost+e sint Invoke Lerch's cancellation law. 20 Example (Damped oscillator) Solve by Laplace's method the initial value 00 0 0 problem x +2x +2x = 0, x(0) = 1, x(0) =1. t Solution: The solution is x(t) =e cost. The details: 00 0 L(x )+2L(x )+2L(x) =L(0) ApplyL across the equation. 0 0 0 0 sL(x )x (0)+2L(x )+2L(x) = 0 The t-derivative rule on x . 0 ssL(x)x(0)x (0) The t-derivative rule on x. +2L(x)x(0)+2L(x) = 0 2 0 (s +2s+2)L(x) = 1+s Use x(0) = 1, x (0) =1. s+1 L(x) = Divide. 2 s +2s+2 s+1 = Complete the square in the de- 2 (s+1) +1 nominator. = L(cost)j Basic Laplace table. ss+1 t =L(e cost) First shifting rule. t x(t) =e cost Invoke Lerch's cancellation law. 21 Example (Recti ed sine wave) Compute the Laplace transform of the recti ed sine wave f(t) = jsintj and show it can be expressed in the form  s coth 2 L(jsintj) = : 2 2 s + Solution: The periodic function formula will be applied with period P = R P st 2=. The calculation reduces to the evaluation of J = f(t)e dt. Because 0 sint 0 on =t2=, integral J can be written as J =J +J , where 1 2 Z Z = 2= st st J = sinte dt; J = sinte dt: 1 2 0 = Integral tables give the result Z st st e cos(t) se sin(t) st sinte dt = : 2 2 2 2 s + s + Then s= 2s= s= (e +1) (e +e ) J = ; J = ; 1 2 2 2 2 2 s + s +262 Laplace Transform s= 2 (e +1) J = : 2 2 s + The remaining challenge is to write the answer for L(f(t)) in terms of coth. The details: J L(f(t)) = Periodic function formula. Ps 1e J 2 = Apply 1x = (1x)(1+x), Ps=2 Ps=2 (1e )(1+e ) Ps=2 x =e . Ps=2 (1+e ) Ps=2 = Cancel factor 1+e . Ps=2 2 2 (1e )(s + ) Ps=4 Ps=4 e +e Ps=4 = Factor out e , then cancel. 2 2 Ps=4 Ps=4 s + e e 2 cosh(Ps=4) = Apply cosh, sinh identities. 2 2 2 sinh(Ps=4) s + coth(Ps=4) = Use cothu = coshu=sinhu. 2 2 s +  s coth 2 = Identity veri ed. 2 2 s + 22 Example (Halfwave recti cation) ComputetheLaplacetransformofthe halfwave recti cation of sint, denoted g(t), in which the negative cycles of sint have been canceled to create g(t). Show in particular that    1 s L(g(t)) = 1+coth 2 2 2s + 2 Solution: The halfwave recti cation of sint is g(t) = (sint+jsintj)=2. Therefore, the basic Laplace table plus the result of Example 21 give L(2g(t)) =L(sint)+L(jsintj) cosh(s=(2)) = + 2 2 2 2 s + s + = (1+cosh(s=(2)) 2 2 s + Dividing by 2 produces the identity. s+1 3s 23 Example (Shifting rules) Solve L(f(t)) =e for f(t). 2 s +2s+2 3t Solution: The answer is f(t) =e cos(t3)H(t3). The details: s+1 3s L(f(t)) =e Complete the square. 2 (s+1) +1 S 3s =e Replace s+1 by S. 2 S +1 3S+3 =e (L(cost))j Basic Laplace table. sS=s+17.3 Laplace Transform Rules 263  3s 3 3S =e e L(cost) Regroup factor e . sS=s+1 3 =e (L(cos(t3)H(t3)))j Second shifting rule. sS=s+1 t 3 =e L(e cos(t3)H(t3)) First shifting rule. 3t f(t) =e cos(t3)H(t3) Lerch's cancellation law. s+7 24 Example () Solve L(f(t) = for f(t). 2 s +4s+8 5 2t Solution: The answer is f(t) =e (cos2t+ sin2t). The details: 2 s+7 L(f(t)) = Complete the square. 2 (s+2) +4 S +5 = Replace s+2 by S. 2 S +4 S 5 2 = + Split into table entries. 2 2 S +4 2S +4 s 5 2 = + Prepare for shifting rule. 2 2 s +4 2s +4 sS=s+2 5 = L(cos2t)+ L(sin2t) Basic Laplace table. 2 sS=s+2 2t 5 =L(e (cos2t+ sin2t)) First shifting rule. 2 5 2t f(t) =e (cos2t+ sin2t) Lerch's cancellation law. 2264 Laplace Transform 7.4 Heaviside's Method Thispractical method was popularized by the English electrical engineer Oliver Heaviside (18501925). A typical application of the method is to solve 2s =L(f(t)) 2 (s+1)(s +1) t for the t-expressionf(t) =e +cost+sint. The details in Heaviside's method involve a sequence of easy-to-learn college algebra steps. More precisely, Heaviside's method systematically converts a polyno- mial quotient n a +a s++a s 0 1 n (1) m b +b s++b s 0 1 m into the form L(f(t)) for some expression f(t). It is assumed that a ;::;a ;b ;:::;b are constants and the polynomial quotient (1) has 0 n 0 m limit zero at s=1. Partial Fraction Theory In college algebra, it is shown that a rational function (1) can be ex- pressed as the sum of terms of the form A (2) k (ss ) 0 k where A is a real or complex constant and (ss ) divides the denomi- 0 nator in (1). In particular, s is a root of the denominator in (1). 0 Assume fraction (1) has real coecients. If s in (2) is real, then A is 0 k real. If s = +i in (2) is complex, then (ss ) also appears, where 0 0 s = i is the complex conjugate of s . The corresponding terms 0 0 in (2) turn out to be complex conjugates of one another, which can be combined in terms of real numbers B and C as A A B +Cs (3) + = : k k 2 2 k (ss ) (ss ) ((s ) + ) 0 0 Simple Roots. Assumethat (1) has real coecients and thedenomi- natorofthefraction(1)hasdistinctrealrootss ,...,s anddistinct 1 N complex roots +i , ..., +i . The partial fraction expansion 1 1 M M of (1) is a sum given in terms of real constants A , B , C by p q q N M n X X a +a s++a s A B +C (s ) 0 1 n p q q q (4) = + : m 2 2 b +b s++b s ss (s ) + 0 1 m p q q p=1 q=17.4 Heaviside's Method 265 Multiple Roots. Assume (1) has real coecients and the denomi- nator of the fraction (1) has possibly multiple roots. Let N be the p multiplicity ofreal roots andletM bethemultiplicity of complexroot p q +i , 1 p N, 1 q  M. The partial fraction expansion of (1) q q is given in terms of real constants A , B , C by p;k q;k q;k N M X X X X A B +C (s ) p;k q;k q;k q (5) + : k 2 2 k (ss ) ((s ) + ) p q q p=11kN q=1 p 1kMq Heaviside's Coverup Method The method applies only to the case of distinct roots of the denominator in (1). Extensions to multiple-root cases can be made; see page 266. To illustrate Oliver Heaviside's ideas, consider the problem details 2s+1 A B C (6) = + + s(s1)(s+1) s s1 s+1 t t = L(A)+L(Be )+L(Ce ) t t = L(A+Be +Ce ) The rst line (6) uses college algebra partial fractions. The second and third lines use the Laplace integral table and properties of L. Heaviside's mysterious method. Oliver Heaviside proposed to 1 nd in (6) the constant C = by a coverup method: 2 2s+1 C = : s(s1) s+1 =0 The instructions are to coverup the matching factors (s+1) on the left and right with box , then evaluate on the left at the root s which makes the contents of the box zero. The other terms on the right are replaced by zero. TojustifyHeaviside's coverup method, multiply(6)bythedenominator s+1 of partial fraction C=(s+1): (2s+1) (s+1) A (s+1) B (s+1) C (s+1) = + + : s s1 s(s1) (s+1) (s+1) Set (s+1) = 0 in the display. Cancellations left and right plus annihi- lation of two terms on the right gives Heaviside's prescription 2s+1 =C: s(s1) s+1=0

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.