6.003 Homework #4 Solutions


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1 6.3 Homewk #4 Soluion Problem. Laplace Tranfm Deermine he Laplace ranfm (including he region of convergence) of each of he following ignal: a. x () = e 2(3) u( 3) X = e 3 2 ROC: Re() > 2 X () = x ()e d = e 2(3) u( 3)e d = e 6 e (2) d 3 = e 6 e(2) = e3 ( 2) 2 ; Re() > 2 3
2 6.3 Homewk #4 Soluion / Fall 2 2 b. x 2 () = ( ( )e 3 )u() X 2 = 4 9 ( 3) 2 ROC: Re() > Trea hi a he um of 3 ignal: x 2 () = x 2a () x 2b () x 2c (), where x 2a () = u(), x 2b () = e 3 u(), and x 2c () = e 3 u(). X 2a () = e d = e = ; Re() > X 2b () = e 3 e d = e(3) = ( 3) 3 ; Re() > 3 X 2c () = e 3 e d = e(3) ( 3) The Laplace ranfm of a um i he um of he Laplace ranfm, X 2 () = 3 ( 3) 2 = 4 9 ( 3) 2 e (3) ( 3) d = ( 3) 2 ; Re()>3 and he region of convergence i he inerecion of he 3 region: Re() >. An alernaive way o find he ranfm of x 2c = e 3 u() i o realize ha muliplying a ime funcion by i equivalen o differeniaing i ranfm by : dx() = d ( ) x()e d = x()e d d d I follow ha X 2c = d d X 2b = d d ( ) = 3 ( 3) 2
3 6.3 Homewk #4 Soluion / Fall 2 3 c. x 3 () = e X 3 = 2( 2 ) ( ) 2 ( ) 2 ROC: < Re() < The ignal x 3 () can be wrien a e u() e u(). The ranfm of he fir i ( ) 2 ; Re() > and he ranfm of he econd i ( ) 2 ; Re() <. Therefe X 3 () = ( ) 2 ( ) 2 = 2( 2 ) ( ) 2 ( ) 2 ; < Re() <.
4 6.3 Homewk #4 Soluion / Fall 2 4 d. x 4 () X 4 = e 2 2e ROC: all X 4 () = = e x 4 ()e d = e2 e d 2 e d = e e = e2 2e e Thee inegral all converge f all value of. Therefe he region of convergence i he enire plane. 2
5 6.3 Homewk #4 Soluion / Fall 2 5 e. x 5 () X 5 = e e 2 e 3 2 ROC: all Thi i no o eay o do direcly from he definiion, ince hi lead o inegral of ime e. The reul can be inegraed by par, bu i i mey. An eaier way i o realize ha he derivaive of x 5 () i imple: d d x 5() Thi funcion can be wrien a u() u( ) u( 2) u( 3) and herefe ha a Laplace ranfm equal o ( e e 2 e 3 ). Since x 5 () i he inegral of i derivaive, he Laplace ranfm of x 5 () i / ime he Laplace ranfm of i derivaive: X 5 () = 2 ( e e 2 e 3 ). The region of convergence include he whole plane ince x 5 () ha finie duraion.
6 6.3 Homewk #4 Soluion / Fall Invere Laplace ranfm Deermine all poible ignal wih Laplace ranfm of he following fm. F each ignal, indicae a cloedfm oluion a well a he region of ime f which he cloedfm oluion i valid. Three boxe are given f each par. If fewer han hree oluion exi f a given par, indicae none in he exra boxe. a. X () = 2 ( ) 2 < > x () = ( )e ( )e none none We can expand X () uing parial fracion a X () = 2 ( ) 2 = ( ) 2 Becaue boh pole are a =, here are ju wo poible region of convergence: > and <. F he lefided region, x L () = ( )e u(). F he righided region, x R () = ( )e u(). x L () x R ()
7 6.3 Homewk #4 Soluion / Fall 2 7 b. X 2 () = 2 ( ) < > x 2 () = e e e Uing parial fracion, X 2 () = 2 ( ) = 2. The wo pole a = and he pole a = break he plane ino 3 poible region of convergence: <, < <, and >. F <, all of he erm are lefided, o x 2L () = e u() u() u(). F < <, exponenial erm i lefided and he oher are righided, o x 2M () = e u() u() u(). F >, all of he erm are righided, o x 2R () = e u() u() u(). x 2L () x 2M () x 2R ()
8 6.3 Homewk #4 Soluion / Fall 2 8 c. X 3 () = < > x 3 () = e co e co none none The fac of he denomina of X 3 () are complexvalued. fracion ill wk. X 3 () = = /2 j /2 j. Neverhele, parial Boh of hee pole have he ame real par. Therefe, here are wo poible region of convergence: > and <. Boh erm are lefided f <, o x 3L () = e co()u(). Boh erm are righided f >, o x 3R () = 2 e(j) u() 2 e(j) u() = e co()u(). x 3L () x 3R () The value of x 3L () are o large ha region in i plo are clipped.
9 6.3 Homewk #4 Soluion / Fall 2 9 ( e ) 2 d. X 4 () = < < < < < 2 > 2 x 4 () = 2 none none none none none none none none Thi expand o ( ) e 2 X 4 () = = 2e e 2 2. Laplace ranfm of he f n u() = u() = δ() δ(τ)dτ u(τ)dτ 2 crepond o derivaive of igulariy funcion. Furherme, he exponenial erm repreen delay. If x() X() hen x( τ) e τ X(). Therefe, x 4 () i a ramp wih lope aring a = plu a ramp wih lope 2 aring a = plu a ramp wih lope aring a = 2, a hown below. x 4 () 2 Since X 4 () converge everywhere in, here i a ingle region of convergence, o he plo above how he only invere ranfm of X 4 (). [Uniquene require ha if muliple ignal have Laplace ranfm of he ame fm, hen he ranfm mu have muliple, nonoverlapping region of convergence: one f each diinc ignal.]
10 6.3 Homewk #4 Soluion / Fall 2 3. Symmery Deermine which of he following polezero diagram could repreen Laplace ranfm of even funcion of ime. X () X 2 () X 3 () X 4 () Ener a ube of he number hrough 4 (eparaed by pace) o repreen X () hrough X 4 () in he anwer box below. If none of X () hrough X 4 () apply, ener none. and/ 2 and/ 3 and/ 4 none: Explain. If x() i an even funcion of ime, hen he creponding Laplace ranfm X() = x()e d = x()e d = x()e d = X() mu be ymmeric abou he poin =. Thu he polezero diagram mu alo be ymmeric abou =. Thi mean ha X 2 () canno crepond o an even funcion of ime. F X() o be ymmeric abou =, hen he region of convergence mu be ymmeric abou he jω axi. Symmery i no poible f X 4 (), which mu be eiher righided lefided bu no boh. Boh X () and X 3 () have imilar parial fracion expanion: A B. To ge a zero a zero (a in X 3 ()), A = B. The creponding ime funcion i x 3 () = A ( e u() e u() ) which i an odd funcion of ime f all value of A. To ge no finie zero (a in X ()), A = B. The creponding ime funcion i x () = A ( e u() e u() ) which i an even funcion of ime f all value of A. Thu, only x () i an even funcion of. Thi example illurae ha ymmery of he polezero diagram i neceary bu no ufficien f ymmery of he creponding ime funcion.
11 6.3 Homewk #4 Soluion / Fall 2 4. Iniial and final value (from Circui, Signal, and Syem by Sieber) a. Ue he iniial and final value heem (where applicable) o find x() and x( ) f he ignal wih he following Laplace ranfm:. et ( ) et 6. 2 ( ) ( ) 2 [( ) 2 ] 2 Aume ha he region of convergence include Re() >. b. Find he invere Laplace ranfm f each of he previou par and how ha he ime wavefm and iniial and final value agree. Ener your anwer in he boxe below. If he iniial final value heem canno be applied, ener X x() x( ) X X X() x() x() x( ) e T u() u( T ) et u( T ) ( ) 2 = ( ) 2 ( ( )e )u() 2 ( ) = 2 ( e )u() X 2 = j/2 j j/2 j in()u() X 2 ( ) 2 [( ) 2 ] 2 = /2 (( ) j) 2 /2 (( ) j) 2 e co()u() The final value heem doe no apply becaue x() doe no approach a limi a. Raher, he funcion grow wih, becaue here i a pole a. 2 The final value heem doe no apply becaue x() doe noe approach a limi a. Raher, he funcion ocillae becaue X() ha pole on he jω axi
12 6.3 Homewk #4 Soluion / Fall 2 2 Engineering Deign Problem 5. Impule repone Skech a block diagram f a CT yem wih impule repone h() = ( e ) e 2 u(). The block diagram hould conain only adder, gain, and inegra. Take he Laplace ranfm o find ha H() = ( ) Then ubiue A and wrie a a block diagram. X A 2 Y A A 3 3
13 6.3 Homewk #4 Soluion / Fall Impedance Mehod a. Deermine he oupu volage of he following circui, uing erie parallel rei combinaion and/ volage curren divider. v i 4 Ω 2 Ω 6 Ω v o The parallel combinaion of 2 Ω and 6 Ω i = 4 Ω. Thi 4 Ω equivalen reiance fm a volage divider wih he op rei. The volage divider halve he inpu volage. Thu v o = 2 v i. b. Generalize he reul from par a f arbirary rei value, and deermine an expreion f he reuling raio v o v i. R v i v R 2 R o 3 Now he equivalen reiance i R 2R 3 R 2 R 3. In combinaion wih R, he volage divider generae v o v i = R 2 R 3 R 2 R 3 R R 2R 3 = R 2 R 3 c. Conider he following circui. R 2 R 3 R R 2 R R 3 R 2 R 3 L v i R C v o Deermine a differenial equaion ha relae v i o v C a follow. Fir, deermine relaion among he elemen curren (i R, i L, and i C ) and elemen volage (v R, v L, and v C ) uing KVL and KCL. Second, relae each elemen volage o he creponding elemen curren uing he coniuive relaion f he elemen: i.e., v R = Ri R, i C = C dv C d, and v L = L di L d. Finally, olve your equaion o find a ingle equaion wih erm ha involve v i, v o, and derivaive of v i and v o. KCL: i L = i R i C. KVL: v R = v C and v i = v L v R. par: v R = Ri R, i C = C dv C d, and v L = L di L d. Eliminae he KVL equaion by ubiuing v L v i v C and v C v R ino he remaining equaion. Then eliminae he par equaion by ubiuing di L d v L L,
14 6.3 Homewk #4 Soluion / Fall 2 4 di R dv R d R d, and di C C d2 v C d d 2 ino he KCL equaion (afer differeniaing each erm by. The reuling equaion i v i v C L = dv C R d C d2 v C d 2 which can be implifed o v i = v C L dv C R d LC d2 v C d 2 = v o L dv o R d LC d2 v o d 2 d. Deermine he yem funcion H() = Vo() V i (), baed on he Laplace ranfm of your anwer o he previou par. V i = ( L R 2 LC)V o V o V i = L R 2 LC e. Subiue R L, R 2 R, and R 3 C ino your reul from par b. R 2 R 3 R R 2 R R 3 R 2 R 3 = R C LR L C R C = 2 LC L R Compare hi new expreion o your reul from par d. The ubiue give he ame expreion ha we found in par d. f. The impedance of an elecrical elemen i a funcion of ha can be analyzed uing rule ha are quie imilar o hoe f reiance. Explain he bai of hi mehod. The impedance mehod i equivalen o aking he Laplace ranfm of all of he nmal circui equaion. KVL and KCL are no changed, becaue he Laplace ranfm i linear, i.e., he Laplace ranfm of a um (of volage curren) i imply he um of he Laplace ranfm. Ohm law f a rei i alo unchanged: if v R = Ri R hen V R = RI R. However he coniuive law f induc and capaci change: v L = L di L d become V L = LI L and i c = C dv C d become I C = CV C. Thee laer law have he ame fm a Ohm law if we inerpre impedance a R (f rei), L (f induc), and C (f capaci).
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