Chapter 9 - The Laplace Transform

Transcription

1 Chaper 9 - The Laplace Tranform Selece Soluion. Skech he pole-zero plo an region of convergence (if i exi) for hee ignal. ω [] () 8 (a) x e u = 8 ROC σ ( ) 3 (b) x e co π u ω [] ( ) () (c) x e u e u ROC 5 σ = 5 =. Saring wih he efiniion of he Laplace ranform, Soluion 9-

2 ( ())= = () L g G g e, fin he Laplace ranform of hee ignal. () (a) x e u (b) x e co π u () (c) x ramp Uing () x e u () = () = X x e ramp e e ax ax e xe x = ( ax ) a e X ( ) =, Re > ( ) σ () 3. Uing he ime-hifing propery, fin he Laplace ranform of hee ignal. () ( ) (a) x u u ( ) 3 (b) x 3e u ( ) 3 (c) x 3e u 3 3u() L Uing he ime hifing propery, L 3u( ) 3e 3 L 3e 3e u( ) 3 ( 3) Alernae oluion: 6 3( ) x 3e e u ( ) Uing he ime hifing propery, Soluion 9-

3 ( ) X () x 5in π u 3e e 3 3e = Uing he complex-frequency-hifing propery, fin an kech he invere Laplace ranform of X ( j ) 3 ( j) Uing he ime-caling propery, fin he Laplace ranform of hee ignal. (a) x δ( ) (b) x u u (), Re> L L u ( ) =, Re> 6. Uing he ime-iffereniaion propery, fin he Laplace ranform of hee ignal. (a) x u () (b) x u e () L ( g() ) G g ( ) u (), Re> L ( u() ) u, ( )= All (c) x in π u () x co 5π u L = Soluion 9-3

4 7. Uing muliplicaion-convoluion ualiy, fin he Laplace ranform of hee ignal an kech he ignal veru ime. () () (a) x e u u () (b) x e in π u u e in π u u L π π () π π π π A B π = Muliply hrough by, le approach infiniy an olve for A. Afer fining A, le = an olve for B, X π π π ( ) ( π) π ( ) ( π ) π x e co π π π (c) x 8co u u u [ ] () () ( ) [ ] () ( ) () x 8co π u u u Afer ime, =, he oluion i zero. in π π u () 8. Uing he iniial an final value heorem, fin he iniial an final value (if poible) of he ignal whoe Laplace ranform are hee funcion. Iniial Value Theorem Final Value Theorem g= limg() limg limg (), if limg() exi (a) X 8, One pole in open LHP x( )= lim 8 = Soluion 9-

5 lim x lim = 8 an he limi exi becaue he only pole of X 8 i in he open LHP (b) X 3, Pole a 3± j 3 (c) X, Pole a ± j x( )= lim = Final-value heorem oe no apply becaue here are wo pole on he ω axi () (e) (f) X X X, Pole a 5 ± j , Pole a an - ( ) 8, Double pole a zero. 9. Fin he invere Laplace ranform of hee funcion. (a) X 8 X ( )= () 8 x 3 e u (b) X 3 (c) X 5 6 () X ( 6 ) X = ( 3) 6 ( 3) ( 3) 6 Soluion 9-5

6 (e) X 6 X A B C D 6 6 = Uing he cover up meho, A =. 58. Uing K qk = ( m k)! m k m ( p ) H, k =,,, m m k q p q [ ( )] = B = = 6 6! ( ) 6 C D X 6 6 = Muliply hrough by an le approach infiniy, C C. 5 = =. 5 X D ( ) = 6 ( 3) 6 Then le =. X 8 8 = D X 6 8 D. 78 = = = = ( 3) 6 8 ( 3) 6 Soluion 9-6

7 3 55 x e co( ) in( ) u () (f) X 3 (g) X 3 (h) (i) X X (j) X j5 j5 j j j X 5 x e e u j j (). Uing a able of Laplace ranform, fin he CTFT of hee ignal. () (a) x e u 5 (b) x 3e co π u. Uing he Laplace ranform, olve hee ifferenial equaion for. (a) x ( ) x u(), x = () X x X X = 9 e X 9 x u () Soluion 9-7

8 Checking iniial coniion, x = which agree wih he iniial coniion, x( )=. For hi yem an hi exciaion he repone canno change inananeouly., x ( )=, x() = (b) x ( ) x ( ) x u() π, x = (c) x ( ) x in( ) u( ). Uing he Laplace ranform, fin an kech he ime-omain repone, y(), of he yem wih hee ranfer funcion o he inuoial exciaion, x Aco π u. = (a) H Y A π π A = π ( π ) π π π ( π ) (b) H ( ) 6 Y A = A ( ) 6 ( π) ( ) 6 ( π) 6 π Y ( ) π π π 3 Y Uing MATLAB,»X Tranfer funcion: ^ 987»H Soluion 9-8

9 Tranfer funcion: ^ -»Y Tranfer funcion: ^ ^ - ^3 7 ^ e»[z,p,k] = zpkaa(y,'v') ;»z z =»p p = i i..i. -.i r = i i i i Y Y Y. A. 6 j. 6. j. 6 6 jπ jπ A. 6 j j. 3 j j A ( π ) ( ) 6 Y. A ( π ) π π π [ ] ( π ) { } () y. A e co. 95in co π 5. in π u Soluion 9-9

10 y() Wrie he ifferenial equaion ecribing hee yem an fin an kech he inicae repone. (a) x u (), y x() () i he repone, y = y() The oluion i coninuou a = becaue, if i were no he iconinuiy woul caue an impule on he lef-han ie of he equaion which coul no be equae o he ep exciaion on he righ-han ie. (b) v =, v() i he repone R = kω C = µf v() - CV v Cv ( ) () = R [ () ] () = V v R V C = C R RC Soluion 9-

11 () > RC v e u e u, ( )= = ( ). Check. v v The oluion i coninuou a = becaue he capacior volage canno change inananeouly. x(). () () (). Fin he hree par, x ac, x an xc, of he following ignal. () ( ) (a) x e u e u (b) x K x e u, x, x e u ac = () c () (c) x u () x u () 5. Fin he bilaeral Laplace ranform of hee ignal. () ( ) 7 (a) x 3e u e u 7 3 xc 3e u X c, Re 7 () ()> x X xac e u( ) xac( )= e u() X ac( )=, Re()> X ac, Re()< 3 X = 3 5, < Re()< Soluion 9-

12 (b) x e 5 6. Fin he repone, y(), of hee yem o hee exciaion. () 5 (a) h e u () ( ) 7, x 3e u e u (), x e u() (b) h ri Uing L e e ri (), All e e H, Therefore e Y All an X, Re()> e e e = ( ), Re()> Y ( e e ), Re()> y y () ( ) ramp ramp ramp u( ) u() u( ) ( ) ( ) e u( ) e u() e u( ) [ ramp( ) e ] u( ) [ ramp() e ] u() ( ) [ ramp( ) e ] u( ) () (c) h e u, x e 5 7. Skech he pole-zero plo an region of convergence (if i exi) for hee ignal. ( ) () (a) x e u e u ( ) () (b) x e u e u Soluion 9-

13 8. Uing he inegral efiniion fin he he unilaeral Laplace ranform of hee ime funcion. () a (a) g e u ( ) > (b) g e a τ u τ, τ ( ) > (c) g e a τ u τ, τ () () g in u ω () (e) g rec (f) g rec aτ aτ G e, a e ( a) e = τ > a 9. Uing MATLAB (or any oher appropriae compuer mahemaic ool) o he inverion inegral of G numerically. Tha i, approximae he inverion inegral wih a ummaion of he form N n e g() L ( G() )= jπ n=n n n = jπ N σ jn ω e = Nσ jn ω n j ω, σ >. Chooe he combinaion of large N an mall ω o ha he ummaion will range over a conour from well below o well above he real axi. Plo g() veru by compuing he value of g() a every value of from he above ummaion approximaion o he inverion inegral. Compare o he analyical reul. Try a lea hree ifferen value of σ o ee he effec on he reul. (Ieally here i no effec of changing σ a long a i i greaer han -, bu acually, in hi numerical approximaion, here will be ome mall effec.) % Program o emonrae he invere Laplace ranform numerically. cloe all ; au =. ; w = ; p = -/au ; w = w*[-:]' ; = j*w*one(lengh(w),) ; allin = [] ; for igma = :5, = igma j*w ; in = [] ; v = [] ; for = -au*:au/:au*, Soluion 9-3

14 f = exp(.*)./( - p) ; v = [v;] ; in = [in;um(f.*)/(j**pi)] ; en in = real(in) ; allin = [allin,in] ; en ubplo(3,,) ; h = plo(v, allin(:,), 'k') ; e(h, 'LineWih',) ; xlabel('time, ()') ; ylabel('h()') ; ile('igma = ') ; gri ; axi([-.,., -., ]) ; ubplo(3,,) ; h = plo(v, allin(:,), 'k') ; e(h, 'LineWih',) ; xlabel('time, ()') ; ylabel('h()') ; ile('igma = ') ; gri ; axi([-.,., -., ]) ; ubplo(3,,3) ; h = plo(v, allin(:,3), 'k') ; e(h, 'LineWih',) ; xlabel('time, ()') ; ylabel('h()') ; ile('igma = ') ; gri ; axi([-.,., -., ]) ; ubplo(3,,) ; h = plo(v, allin(:,), 'k') ; e(h, 'LineWih',) ; xlabel('time, ()') ; ylabel('h()') ; ile('igma = 3') ; gri ; axi([-.,., -., ]) ; ubplo(3,,5) ; h = plo(v, allin(:,5), 'k') ; e(h, 'LineWih',) ; xlabel('time, ()') ; ylabel('h()') ; ile('igma = ') ; gri ; axi([-.,., -., ]) ; ubplo(3,,6) ; h = plo(v, allin(:,6), 'k') ; e(h, 'LineWih',) ; xlabel('time, ()') ; ylabel('h()') ; ile('igma = 5') ; gri ; axi([-.,., -., ]) ;. Uing a able of unilaeral Laplace ranform an he properie fin he unilaeral Laplace ranform of he following funcion. ( ) (a) g 5in π u ( ) (b) g 5in π u Therefore (c) g co π co π u Ue he rigonomeric ieniy, ( ) in( π)= in π L πe 5in( π) u( ) π co( π) co( π)= [ co( ππ) co( π π) ], hen complee he oluion a uual. Soluion 9-

15 () g u ( ) (e) g u τ τ τ (f) g 5e u ( τ), τ > Ue hee properie: Frequency Shifing Time Shifing Lineariy Time Differeniaion Once 5 (g) g e co π u π (h) x 5in π u 8. Given () fin he Laplace ranform of (a) g X π π π co in π L g() (b) ( g( ) ) L Time Differeniaion Once ( g() ) g ( ) ( ) ( g() ) L g ( ) = Iniial Value Theorem g limg ( g( ) ) L L 3 ( g( ) ) Soluion 9-5

16 (Thi i correc if g g (c) g( ) ( )= ( ). Tha i, if g i coninuou a ime, =.) () () () g g. Fin he ime-omain funcion which are he invere Laplace ranform of hee funcion. Then, uing he iniial an final value heorem verify ha hey agree wih he ime-omain funcion. (a) G 3 8 ( ) g e e u 5 5 () limg lim e e u() 5 5 = lim G lim, 3 8 ( ) = Check. (b) G lim g lim e e u() = 5 5 limg lim, ( 3) ( 8) = Check. (c) G 3 8 ( ) () 3. Given G e e u L () G fin he invere Laplace ranform of (a) G 3 Frequency Scaling 3e 3 L u G 3 3e u 3 3e u (b) G( ) G( ) (c). The CTFT of = () G() Soluion 9-6

17 exi bu he (unilaeral) Laplace ranform oe no. Why? x 5. Compare he CTFT an he Laplace ranform of a uni ep. Why can he CTFT no be foun from he Laplace ranform? e u() jω πδ ω an u () L F 6. Show ha he common Laplace ranform pair L α L α L u (), e u (), e u() α L ω in( ω ) u (), co ω u ω L () α ω α e in( ω ) u (), e co ω u α ω α ω L L () ( α) α can be erive from only he impule ranformaion, δ () L, an he properie of he Laplace ranform. ω u() L : Inegraion g τ τ G ( ) L e α e α u() : α L u() α L L δ λ λ u() L α Frequency Shifing e u() α : L Soluion 9-7

18 () in ω u () co ω u α e in ω u L ω ω : L ω : L ω α ( ) () : ω α e co ω u L () α ( α) ω : 7. Given an LTI yem ranfer funcion, H(), fin he ime-omain repone, y() o he exciaion, x (). (a) x in( π) u ( ), H= () 3 (b) x u, H (c) x u, H () 3 () () x u, H 5 = (e) x in π u, H 5 X π ( π ) Y Y Leing be zero, π 5 = π π ( π ) [ ] A B C D π π ( π ) [ ] = B D π = Muliplying hrough by an leing approach infiniy, = A C Soluion 9-8

19 Leing =, π = A B C D. π π 5 Leing =, π = A B π ( π) C D. Arranging he equaion in marix form, ( π ) A π B = 5 5 π ( π) ( π) C π D π ( π ) ( π ) Solving, where A B = C D. 8 Y Y ( π ) π. 7 π π ( π ).. ( ) ( ) { [ ]} [ ] () () y co π. 3353in π. 7535e co. 67in 8. Wrie he ifferenial equaion ecribing hee yem an fin an kech he inicae repone. (), y (a) x u () i he repone, y = 5, ( y( ) ) = = Soluion 9-9

20 x() y() Y ( ) ( ) 9 Boh he repone an i fir erivaive mu be coninuou in repone o a ep exciaion becaue of he ouble inegraion beween exciaion an repone. (), v (b) i u () i he repone, No iniial energy orage i() R = kω i () C = 3 µf R = kω C = µf v() - v i() C v ( ) () C v R = 3 { i() v v v C v R R 3 () () volage acro R volage acro curren ource Combining equaion, ( ( ) ) () RRCC v ( ) RC R R C v v R i RRCC V RC R R C V V ( ( ) ) () () R V ( ) = Soluion 9-

21 (c) i co π u, v() i he repone, No iniial energy orage i() R = kω i () C = 3 µf R = kω C = µf v() - From par (b) ( ( ) ) () RRCC v ( ) RC R R C v v R i RRCC V RC R R C V V R ( ) () () ( π ) 666 π V 396. π π π () () () 9. Fin he hree par, x ac, x an xc, of he following ignal. (a) x (b) x in ω (c) x gn () x δ () x, x, x ac () δ 3. Fin he bilaeral Laplace ranform of hee ignal. c () (a) x rec () () x c rec u e e e X c e =, = = Any x X Soluion 9-

22 x rec() u( ) x ( )= rec u ac ac e e e X ac( )= e =, = = Any e e X ac =, Any j j j j j in e e e e e e X = = = j j = inc, π Any ω Noice ha if we make he change of variable, = jω, we ge X( jω)= inc π which i he CTFT of x rec convergence i he enire plane. () (b) x rec in π () ( ) (c) x e u e u in π () an hi i allowe becaue he region of [ ] Soluion 9-