Laplace transfom: t-translation rule , Haynes Miller and Jeremy Orloff
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1 Laplace ransfom: -ranslaion rule 8.03, Haynes Miller and Jeremy Orloff Inroducory example Consider he sysem ẋ + 3x = f(, where f is he inpu and x he response. We know is uni impulse response is 0 for < 0 w( = e 3 = u(e 3. for > 0 This is he response from res IC o he inpu f( = δ(. Wha if we shifed he impulse o anoher ime, say, f( = δ( 5? Linear ime invariance ells us he response will also be shifed. Tha is, he soluion o is x( = w( 2 = ẋ + 3x = δ( 2, wih res IC ( 0 for < 2 e 3 for > 2 = u( 2e 3( 2. In words, his is a sysem of exponenial decay. The decay sars as soon as here is an inpu ino he sysem. Graphs are shown in Figure below. Figure. Graphs of w( and x( = w( 2. We know ha L(δ( a = e as. So, we can find X = L(x by aking he Laplace ransform of Equaion. (s + 3X(s = e 2s X(s = e 2s s + 3 = e 2s W (s, where W = Lw. So delaying he impulse unil = 2 has he effec in he frequency domain of muliplying he response by e 2s. This is an example of he -ranslaion rule ranslaion rule The -ranslaion rule, also called he -shif rule gives he Laplace ransform of a funcion shifed in ime in erms of he given funcion. We give he rule in wo forms. L(u( af( a; s = e as F (s (2 L(u( af(; s = e as L(f( + a; s. (3 For compleeness we include he ranslaion formulas for u( a and δ( a: L(u( a = e as /s (4 L(δ( a = e as. (5
2 8.03 Laplace ransfom: -ranslaion rule 2 Remarks:. Formula 3 is ungainly. The noaion will become clearer in he examples below. 2. Formula 2 is mos ofen used for compuing he inverse Laplace ransform, i.e., as u( af( a = L ( e as F (s. 3. These formulas parallel he s-shif rule. In ha rule, muliplying by an exponenial on he ime ( side led o a shif on he frequency (s side. Here, a shif on he ime side leads o muliplicaion by an exponenial on he frequency side. Proof: The proof of Formula 2 is a very simple change of variables on he Laplace inegral. L(u( af( a; s = = = 0 a 0 u( af( ae s d f( ae s d (u( a = 0 for < a f(τe s(τ+a dτ (change of variables: τ = a = e as f(τe sτ dτ 0 = e as F (s. Formula 3 follows easily from Formula 2. The easies way o proceed is by inroducing a new funcion. Le g( = f( + a, so We ge f( = g( a and G(s = L(g = L(f( + a. L(u( af(; s = L(u( ag( a = e as G(s = e as L(f( + a; s. The second equaliy follows by applying Formula 2 o g(. ( ωe Example. Find L as s 2 + ω 2. answer: Firs ignore he exponenial and le ( ω f( = L s 2 + ω 2 = sin(ω. Using he shif Formula 2 his becomes ( ωe L as s 2 + ω 2 = u( af( a = u( a sin ω( a. Example 2. L (u( 3; s = e 3s L( + 3; s = e 3s ( s s. Example 3. L(u( 3 ; s = e 3s L(; s = e 3s /s.
3 8.03 Laplace ransfom: -ranslaion rule 3 Example 4. Find L(f for f( = 0 for < 2 2 for > 2. answer: In order o use he -shif rule we have o wrie f( in u-forma: f( = u( 2 2. So, Formula 3 says L(f = e 2s L(( ; s = e 2s ( 2 s s s. Example 5. Find L(f for f( = cos( for 0 < < 2π 0 for > 2π. answer: Again we firs pu f( in u-forma. Noice he he funcion for 0 < < 2 u( u( 2 = 0 elsewhere. Therefore f( = (u( u( 2π cos( = u( cos( u( 2π cos(. Using he -ranslaion formula 3 we ge L(f = s s 2 + e 2πs L(cos( + 2π = The las equaliy holds because cos( + 2π = cos(. s s 2 + s e 2πs s A longer example The fish populaion in a lake is no reproducing fas enough and he populaion is decaying exponenially wih decay rae k. A program is sared o sock he lake wih fish. Three differen scenarios are discussed below. Example 6. A program is sared o sock he lake wih fish a a consan rae of r unis of fish/year. Unforunaely, afer /2 year he funding is cu and he program ends. Model his siuaion and solve he resuling DE for he fish populaion as a funcion of ime. answer: Le x( be he fish populaion and le A = x(0 be he iniial populaion. Exponenial decay means he populaion is modeled by ẋ + kx = f(, x(0 = A (6 where f( is he rae fish are being added o he lake. In his case r for 0 < < /2 f( = 0 for /2 <.
4 8.03 Laplace ransfom: -ranslaion rule 4 Firs we wrie f in u-forma: he equaion. f( = r ( u( /2 and find he Laplace ransform of F (s = L(f(s = r s r s e s/2. Nex we find he Laplace ransform of he equaion and solve for f. sx x(0 + kx = F (s (s + kx A = r s ( e s/2 X(s = A s + k + r s(s + k ( e s/2. To find x( we emporarily ignore he facor of e s/2 and ake Laplace inverse of wha s lef. (using parial fracions. ( ( A L = Ae k, L r = r s + k s(s + k k ( e k. The -ranslaion formula hen says ( L re s/2 s(s + k = r k u( /2 ( e k( /2. Puing i all ogeher we ge (in u and cases forma. x( = Ae k + r ( e k r ( k k u( /2 e k( /2 Ae k + r ( = k e k for 0 < < /2 Ae k r ( k e k + e k( /2 for /2 <. Example 7. (Periodic on/off The program is refunded and hey have enough money o sock a a consan rae of r for he firs half of each year. Find x( in his case. answer: All ha s changed from Example 6is he inpu funcion f(. We wrie i in cases-forma and ranslae ha o u-forma so we can ake he Laplace ransform. r for 0 < < /2 f( = 0 for /2 < < r for 0 < < 3/2 0 for 3/2 < < 2 = r ( u( 2 + u( u( The compuaions from here are essenially he same as in he previous example. We skech hem ou. L(f = r ( e s/2 + e s e 3s/2 +..., so X = A s s + k + r ( e s/2 + e s.... s(s + k
5 8.03 Laplace ransfom: -ranslaion rule 5 Thus, x( =Ae k + r [ ( e k u( /2( e k( /2 k ] + u( ( e k( u( 3/2( e k( 3/ and in cases forma: Ae k + r k r k e k for 0 < < 2 Ae k r ( k e k e k( /2 for 2 < < x( = Ae k + r k r ( k e k e k( / e k( n for n < < n + 2 Ae k r ( k e k e k( / e k( n /2 for n + 2 < < n + Noe ha he paern in he formula for he response alernaes beween he periods of socking and no socking. In paricular, noice ha he consan erm r/k is only presen during periods of socking. Example 8. (Impulse rain The answer o he previous example is a lile hard o read. We know from experience ha impulsive inpu usually leads o simpler oupu. In his scenario suppose ha once a year r/2 unis of fish are dumped all a once ino he lake. Find x( in his case. answer: Once again, all ha s changed from Example 6 is he inpu funcion f(. In his case we have f( = r (δ( + δ( + δ( 2 + δ( This is called an impulse rain. Is Laplace ransform is easy o find. F (s = r 2 ( + e s + e 2s + e 3s One nice hing abou dela funcions is ha hey don inroduce any new erms ino he parial fracions par of he problem. sx(s x(0 + kx(s = r ( + e s + e 2s + e 3s X(s = A s + k + r ( + e s + e 2s + e 3s (s + k Laplace inverse is easy: ( ( e L = e k L ns = u( ne k( n. s + k s + k Thus, x( = Ae k + r 2 e k + r 2 u( e k( + r 2 u( 2e k( 2 + r 2 u( 3e k( Here are graphs of he soluions o examples 6 and 8 (wih A = 0, k =, r = 2. Noice how hey sele down o periodic behavior.
6 8.03 Laplace ransfom: -ranslaion rule Fig.. Graphs from example 2 (lef and example 3 (righ.
Notes 04 largely plagiarized by %khc
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