{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Helvetica" 0 14 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R 3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 0 12 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Ohlfs" 0 14 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 3" -1 258 1 {CSTYLE "" -1 -1 "Ohlfs" 0 12 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 81 " Un it Step Function and Laplace Transform (MapleV Release 4)" }}{PARA 0 " " 0 "" {TEXT -1 216 "This worksheet uses Maple to explore Laplace tran sforms.\nSince the book uses u(t) for the unit step function we will\n set Maple's unit step function to be u. First some\npreliminary stuff \+ -read Maple's laplace library." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "r estart;\nwith(inttrans):\nalias(u=Heaviside);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 204 "Now let's make sure that we understand how the u func tion works.\n \nDefine function that is zero everywhere except that\n \+ it is equal to 1 between 1 and 2: \n\nThe result is u(t-1)-u(t-2) wh ich we can plot:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(u(t-1)-u(t -2),t=0..3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "Use Maple to comp ute some Laplace transforms\nand invert other Laplace transforms..." } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "L1 :=laplace(exp(3*t)*sin(4*t),t,s );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Now invert the Laplace tran sform that was just computed." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "in vlaplace(L1,s,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Try another \+ pair:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "L2 := laplace((u(t-3)-u(t- 8))*exp(t-3),t,s);\ninvlaplace(L2,s,t);\nsimplify(\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 169 "Let's look at Laplace transforms of func tions and their derivatives.\nNotice that derivatives are turned into \+ multiplication by s:\nWe use D(y)(t) in place of diff(y(t),t)" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "laplace(D(y)(t),t,s);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "To make the result more readable we defin e the alias Y(s)=laplace(y(t),t,s)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "alias(Y(s)=laplace(y(t),t,s));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Now find the Laplace Transform of: y'' + 3 y' - 4 y = 0" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "laplace((D@@2)(y)(t)+3*D(y)(t)-4*y( t)=0,t,s);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Now solve for the L aplace tranform Y(s):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "solve(\",Y (s));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "If we want to see the r esult in a form that makes it easy to see the\ninverse transform we mi ght use the\nMaple commands such as expand or convert" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "convert(\",parfrac,s);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 44 "We now use Maple to invert Y(s) to get y(t):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "invlaplace(\",s,t);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 122 "Try a more complicated problem where the nonhomog eneous\nterm has a step function and the initial conditions are specif ied:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "ode1 := (D@@2)(y)(t)+3*D(y) (t)-4*y(t)=u(t-2)*(t-2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "laplace(ode1,t,s);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "subs (y(0)=2,D(y)(0)=3,\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "s olve(\",Y(s));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "expand(\" );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "invlaplace(\",s,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 81 } {VIEWOPTS 1 1 0 1 1 1803 }