20.2. The Transform and its Inverse. Introduction. Prerequisites. Learning Outcomes

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1 The Trnform nd it Invere 2.2 Introduction In thi Section we formlly introduce the Lplce trnform. The trnform i only pplied to cul function which were introduced in Section 2.1. We find the Lplce trnform of mny commonly occurring ignl nd produce tble of tndrd Lplce trnform. We lo conider the invere Lplce trnform. To begin with, the invere Lplce trnform i obtined by inpection uing tble of trnform. Thi pproch i developed by employing technique uch prtil frction nd completing the qure introduced in.6. Prerequiite Before trting thi Section you hould... Lerning Outcome On completion you hould be ble to... undertnd wht cul function i be ble to find nd ue prtil frction be ble to perform integrtion by prt be ble to ue the technique of completing the qure find the Lplce trnform of mny commonly occurrring cul function obtin the invere Lplce trnform uing technique involving (i) tble of trnform (ii) prtil frction (iii) completing the qure (iv) the firt hift theorem HELM (28): Section 2.2: The Trnform nd it Invere 11

2 1. The Lplce trnform If f(t) i cul function then the Lplce trnform of f(t) i written L{f(t) nd defined by: L{f(t) e t f(t) dt. Clerly, once the integrl i performed nd the limit ubtituted the reulting expreion will involve the prmeter lone ince the dependence upon t i removed in the integrtion proce. Thi reulting expreion in i denoted by F (); it precie form i dependent upon the form tken by f(t). We now refine Key Point 1 (pge 4). Key Point The Lplce Trnform of Cul Function L{f(t)u(t) e t f(t)u(t) dt F () To begin, we determine the Lplce trnform of ome imple cul function. For exmple, if we conider the rmp function f(t) t.u(t) with grph f(t) t u(t) 45 t we find: L{t u(t) e t t u(t) dt [ ] t e t ( ) [ te t ( ) Figure 11 e t t dt ince in the rnge of the integrl u(t) 1 ] e t ( ) dt [ ] e t ( ) 2 uing integrtion by prt Now we hve the difficulty of ubtituting in the limit of integrtion. The only problem rie with the upper limit (t ). We hll lwy ume tht the prmeter i o choen tht no 12 HELM (28): Workbook 2: Lplce Trnform

3 contribution ever rie from the upper limit (t ). In thi prticulr ce we need only demnd tht i rel nd poitive. Uing thi rule of thumb : [ ( )] 1 L{t u(t) [ ] ( ) Thu, if f(t) t u(t) then F () 1/ 2. A imilr, but more tediou, clcultion yield the reult tht if f(t) t n u(t) in which n i poitive integer then: L{t n u(t) n! n+1 [We remember n! n(n 1)(n 2)... ()(2)(1).] Find the Lplce trnform of the tep function u(t). Begin by obtining the Lplce integrl: You hould obtin L{u(t) e t dt ince in the rnge of integrtion, t > nd o u(t) 1 leding to e t u(t) dt e t dt Now complete the integrtion: You hould hve obtined: L{u(t) e t dt [ ] e t ( ) [ ] 1 1 ( ) where, gin, we hve umed the contribution from the upper limit i zero. HELM (28): Section 2.2: The Trnform nd it Invere 1

4 A econd exmple, we conider the decying exponentil f(t) e t u(t) where i poitive contnt. Thi function h grph: f(t) e t u(t) t In thi ce, L{e t u(t) e t e t dt e (+)t dt [ e (+)t ( + ) ] 1 + Figure 12 (zero contribution from the upper limit) Therefore, if f(t) e t u(t) then F () 1 +. Following thi pproch we cn develop tble of Lplce trnform which record, for ech cul function f(t) lited, it correponding trnform function F (). Tble 1 give limited tble of trnform. The linerity property of the Lplce trnformtion If f(t) nd g(t) re cul function nd c 1, c 2 re contnt then L{c 1 f(t) + c 2 g(t) e t [c 1 f(t) + c 2 g(t)] dt c 1 e t f(t) dt + c 2 e t g(t) dt c 1 L{f(t) + c 2 L{g(t) Key Point 4 Linerity Property of the Lplce Trnform L{c 1 f(t) + c 2 g(t) c 1 L{f(t) + c 2 L{g(t) 14 HELM (28): Workbook 2: Lplce Trnform

5 Tble 1. Tble of Lplce Trnform Rule Cul function Lplce trnform 1 f(t) F () 2 u(t) t n u(t) 4 e t u(t) 5 in t. u(t) 6 co t. u(t) 7 e t in bt. u(t) 8 e t co bt u(t) 1 n! n b ( + ) 2 + b 2 + ( + ) 2 + b 2 Note: For convenience, thi tble i repeted t the end of the Workbook. Tht i, the Lplce trnform of liner um of cul function i liner um of Lplce trnform. For exmple, L{2 co t. u(t) t 2 u(t) 2L{co t. u(t) L{t 2 u(t) ( ) ( ) Obtin the Lplce trnform of the hyperbolic function inh t. Begin by expreing inh t in term of exponentil function: inh t 1 2 (et e t ) Now ue the linerity property (Key Point 4) to obtin the Lplce trnform of the cul function inh t.u(t): HELM (28): Section 2.2: The Trnform nd it Invere 15

6 You hould obtin /( 2 2 ) ince { e t e t L{inh t.u(t) L.u(t) L{et.u(t) 1 2 L{e t.u(t) 1 [ ] 1 1 [ ] 1 (Tble 1, Rule 4) [ ] 2 2 ( )( + ) 2 2 Obtin the Lplce trnform of the hyperbolic function coh t. You hould obtin 2 2 ince { e t + e t L{coh t.u(t) L.u(t) L{et.u(t) L{e t.u(t) 1 [ ] [ ] 1 (Tble 1, Rule 4) [ ] 2 2 ( )( + ) HELM (28): Workbook 2: Lplce Trnform

7 Find the Lplce trnform of the delyed tep-function u(t ), >. Write the delyed tep-function here in term of n integrl: You hould obtin L{u(t ) L{u(t ) e t u(t ) dt e t dt (note the lower limit i ) ince: e t u(t ) dt + In the firt integrl < t < nd o (t ) <, therefore u(t ). e t u(t ) dt In the econd integrl < t < nd o (t ) >, therefore u(t ) 1. Hence L{u(t ) + Now complete the integrtion: e t dt. L{u(t ) [ e e t t dt ( ) ] e Exercie Determine the Lplce trnform of the following function. () e t u(t) (b) u(t ) (c) e t in t.u(t) (d) (5 co t 6t ).u(t) () 1 + (b) e (c) ( + 1) (d) HELM (28): Section 2.2: The Trnform nd it Invere 17

8 2. The invere Lplce trnform The Lplce trnform tke cul function f(t) nd trnform it into function of, F (): L{f(t) F () The invere Lplce trnform opertor i denoted by L 1 nd involve recovering the originl cul function f(t). Tht i, Key Point 5 Invere Lplce Trnform L 1 {F () f(t) where L{f(t) F () For exmple, { uing tndrd trnform from Tble 1: L 1 co 2t. u(t) ince L{co 2t. u(t). (Tble 1, Rule 6) Alo { L 1 t u(t) ince L{t u(t). (Tble 1, Rule ) 2 2 Becue the Lplce trnform i liner opertor it follow tht the invere Lplce trnform i lo liner, o if c 1, c 2 re contnt: Key Point 6 Linerity Property of Invere Lplce Trnform L 1 {c 1 F () + c 2 G() c 1 L 1 {F () + c 2 L 1 {G() For exmple, to find the invere Lplce trnform of { 2 L L 1 { { 2 L we hve 1 t u(t) in 2t. u(t) (from Tble 1) Note tht the frction hve hd to be mnipulted lightly in order tht the expreion mtch preciely with the expreion in Tble HELM (28): Workbook 2: Lplce Trnform

9 Although the invere Lplce trnform cn be exmined t deeper mthemticl level we hll be content with thi imple-minded pproch to finding invere Lplce trnform by uing the tble of Lplce trnform. However, even thi pproch i not lwy trightforwrd nd coniderble lgebric mnipultion i often required before n invere Lplce trnform cn be found. Next we conider two tndrd rerrngement which often occur. Inverting through the ue of prtil frction The function 1 F () ( 1)( + 2) doe not pper in our tble of trnform nd o we cnnot, by inpection, write down the invere Lplce trnform. However, by uing prtil frction we ee tht F () ( 1)( + 2) nd o, uing the linerity property: { L 1 1 ( 1)( + 2) { 1 { 1 L 1 L et 1 e 2t (Tble 1, Rule 4) Find the invere Lplce trnform of ( 1)( 2 + 1). Begin by uing prtil frction to write the given expreion in more uitble form: ( 1)( 2 + 1) Now continue to obtin the invere: HELM (28): Section 2.2: The Trnform nd it Invere 19

10 { L 1 ( 1)( 2 + 1) { 1 2 L 1 { 1 2 L { 1 2 L [ e t co t in t ] u(t) (Tble 1, Rule 4, 6, 5). The firt hift theorem The firt nd econd hift theorem enble n even wider rnge of Lplce trnform to be eily obtined thn the trnform we hve lredy found. They lo enble ignificntly wider rnge of invere trnform to be found. Here we introduce the firt hift theorem. If f(t) i cul function with Lplce trnform F (), i.e. L{f(t) F (), then we hll ee, the Lplce trnform of e t f(t), where i given contnt, cn eily be found in term of F (). Uing the definition of the Lplce trnform: But if L{e t f(t) F () L{f(t) ] e [e t t f(t) dt e (+)t f(t) dt e t f(t) dt then imply replcing by + on both ide give: F ( + ) e (+)t f(t) dt Tht i, the prmeter i hifted to the vlue +. We hve then the ttement of the firt hift theorem: Key Point 7 Firt Shift Theorem If L{f(t) F () then L{e t f(t) F ( + ). 2 HELM (28): Workbook 2: Lplce Trnform

11 For exmple, we lredy know (from Tble 1) tht L{t u(t) 6 4 nd o, by the firt hift theorem: L{e 2t t u(t) 6 ( + 2) 4 Ue the firt hift theorem to determine L{e 2t co t.u(t). You hould obtin 2 ( 2) nd o by the firt hift theorem (with 2) L{e 2t co t.u(t) 2 ( 2) obtined by imply replcing by 2. ince L{co t.u(t) (Tble 1, Rule 6) We cn lo employ the firt hift theorem to determine ome invere Lplce trnform. Find the invere Lplce trnform of F () Begin by completing the qure in the denomintor: ( 1) 2 9 HELM (28): Section 2.2: The Trnform nd it Invere 21

12 Reclling tht L{inh t u(t) 2 9 the firt hift theorem: (from the on pge 15) complete the inverion uing You hould obtin { L 1 e t inh t u(t) ( 1) 2 9 Here, in the nottion of the hift theorem: f(t) inh t u(t) F () 2 9 nd 1 Inverting uing completion of the qure The function: 4 F () doe not pper in the tble of trnform nd, gin, need mending before we cn find it invere trnform. In thi ce, becue doe not hve nice fctor, we complete the qure in the denomintor: nd o ( + 1) F () ( + 1) Now the numertor need mending lightly to enble u to ue the pproprite rule in the tble of trnform (Tble 1, Rule 8): { F () ( + 1) ( + 1) { ( + 1) ( + 1) [ ] 4( + 1) ( + 1) ( + 1) Hence { { L 1 {F () 4L L 1 2 ( + 1) ( + 1) e t co 2t. u(t) 2e t in 2t. u(t) e t [4 co 2t 2 in 2t]u(t) 22 HELM (28): Workbook 2: Lplce Trnform

13 Find the invere Lplce trnform of Begin by completing the qure in the denomintor of thi expreion: ( 2) Now obtin the invere: You hould obtin: { { [ ] L 1 L 1 2 ( 2) ( 2) e 2t in 2t.u(t) (Tble 1, Rule 7) Exercie Determine the invere Lplce trnform of the following function. () 1 4 (f) (b) 2 ( + 1)( 2 + 1) (c) (d) + ( 1)( + 2) (e) () 1 6 t u(t) (b) (e 4t co t 5e 4t in t)u(t) (c) ( co t 7 in t)u(t) (d) (2e t + e 2t )u(t) (e) ( e 4t )u(t) (f) (e t co t + in t)u(t) HELM (28): Section 2.2: The Trnform nd it Invere 2