EECE 31 Signal & Syem Prof. Mark Fowler Noe Se #27 C-T Syem: Laplace Tranform Power Tool for yem analyi Reading Aignmen: Secion 6.1 6.3 of Kamen and Heck 1/18
Coure Flow Diagram The arrow here how concepual flow beween idea. Noe he parallel rucure beween he pink block (C-T Freq. Analyi) and he blue block (D-T Freq. Analyi). New Signal Model Ch. 1 Inro C-T Signal Model Funcion on Real Line Syem Properie LTI Caual Ec Ch. 3: CT Fourier Signal Model Fourier Serie Periodic Signal Fourier Tranform (CTFT) Non-Periodic Signal Ch. 2 Diff Eq C-T Syem Model Differenial Equaion D-T Signal Model Difference Equaion Zero-Sae Repone Ch. 5: CT Fourier Syem Model Frequency Repone Baed on Fourier Tranform New Syem Model Ch. 2 Convoluion C-T Syem Model Convoluion Inegral Ch. 6 & 8: Laplace Model for CT Signal & Syem Tranfer Funcion New Syem Model New Syem Model D-T Signal Model Funcion on Ineger New Signal Model Powerful Analyi Tool Zero-Inpu Repone Characeriic Eq. Ch. 4: DT Fourier Signal Model DTFT (for Hand Analyi) DFT & FFT (for Compuer Analyi) D-T Syem Model Convoluion Sum Ch. 5: DT Fourier Syem Model Freq. Repone for DT Baed on DTFT New Syem Model Ch. 7: Z Tran. Model for DT Signal & Syem Tranfer Funcion New Syem Model 2/18
Wha we have een o far. Diff. Equaion decribe yem Differenial Eq. for CT Difference Eq. for DT Convoluion wih he Impule Repone can be ued o analyze he yem An inegral for CT A ummaion for DT Fourier Tranform (and Serie) decribe wha frequencie are in a ignal CTFT for CT ha an inegral form DTFT for DT ha a ummaion form There i a connecion beween hem from he ampling heorem The Frequency Repone of a yem give a muliplicaive mehod of analyi Freq. Repone = CTFT of impule repone for CT yem Freq. Repone = DTFT of impule repone for DT yem We now look a wo power ool for yem analyi: Laplace Tranform for CT Syem Exenion of CTFT Z Tranform for DT Syem Exenion of DTFT 3/18
Ch. 6 Laplace Tranform & Tranfer Funcion Back o C-T ignal and yem We ve een ha he FT i a ueful ool for Bu only if: -ignal analyi (underanding ignal rucure) -yem analyi/deign 1. Syem i in zero ae 2. Impule repone aifie 3. Inpu aifie x ( ) d < h ( ) d Called Aboluely Inegrable < Well here are a few ignal ha we can handle wih FT ha do no aify hi: Sinuoid and uni ep are wo of hem So frequency repone i a ool ha can only be ued under hee hree condiion! The Laplace Tranform i a generalizaion of he CTFT i can handle cae when hee hree condiion are no me. 4/18
There are wo analyi mehod ha he Laplace Tranform enable: Zero ae (Sec. 6.5) Non zero-ae (Sec. 6.4) LT & Tranfer Funcion x() and h() may or may no be aboluely inegrable So hi ju allow u o do he ame hing ha he FT doe bu for a larger cla of ignal/yem LT-baed oluion of differenial equaion x() and h() may or may no be aboluely inegrable Thi no only admi a larger cla of ignal/yem i alo give a powerful ool for olving for boh he zero-ae AND he zero-inpu oluion ALL AT ONCE Fir we ll define he LT Nex See ome of i properie Then See how o ue i in yem analyi in hee wo way 5/18
Secion 6.1: Define he LT There are 2 ype of LT: Two-ided (bilaeral) One-ided (unilaeral) We ll only ue he one-ided LT Two-Sided LT X 2( ) = x( ) e d wih = σ + jω complex variable The book doen do hi One-Sided LT complex variable X 1 ( ) = x( ) e d wih = σ + jω One-ided LT defined hi way even if x(), < Bu we will moly focu on caual yem and caual inpu 6/18
One place he LT i mo ueful i when Caual ignal 1. The yem ha Iniial condiion a = 2. Inpu x() i applied a = x() = < (Thi will be our focu in hi coure) For hi cae: X 1 () = X 2 () Ju ue X() noaion (drop he 1 ubcrip) a complex valued funcion Noe ha X() i: of a complex variable = σ + jω Mu plo on a plane he -plane X () { } j Im jω plane Re{ } σ Similarly for X() 7/18
Example of Finding a LT Conider he ignal Thi i a caual ignal. x( ) By definiion of he LT: = b e u( ) b R b ( + b) = e e d = e X ( ) d X ( ) = 1 + b 1 + b [ ( + b) ] = [ ( + b) e = lim e 1] = look a hi Back when we udied he FT we had o limi b o being b > wih he LT we don need o reric ha!!! Thi i an eay inegral o do!! The limi i here by he definiion of he inegral If hi limi doe no converge hen we ay ha he inegral doe no exi So we need o find ou under wha condiion hi inegral exi. So le look a he funcion inide hi limi 8/18
e ( + b) = e [ ( σ + b) + jω ] if σ + b > σ > b if σ + b < σ < b Ha Two Main Behavior Thu, lim e ( + b) "exi" only for σ > b So, we can find hi X() for value of uch ha Re{} -b Bu for wih Re{} > -b we have no rouble. For each X() we need o know a which value hing work Thi e of i called he Region of Convergence (ROC) Don worry oo much abou ROC a hi level i kind of ake care of ielf So for x( ) b = e u( ) We have X = 1 ( ) {} > b + b Re 9/18
Thi reul and many oher i on he Table of Laplace Tranform ha i available on my web ie Pleae ue he able from he webie he one in he book have ome error on hem!!!! 1/18
If b > hen x() ielf decay: e b u() For b >, -b i negaive And we have on he -plane: jω σ = b ROC σ Thi cae can be handled by he FT and can alo be handled by he LT If b < hen x() ielf explode : e b u() For b <, -b i poiive And we have on he -plane: σ = b jω ROC σ Thi cae can be handled by he FT bu by rericing our focu o value of in he ROC, he LT can handle i!!! 11/18
Connecion beween FT & LT (for caual ignal) FT : X ( jω ω) = x( ) e d ( σ + jω) LT : X ( σ + jω) = x( ) e d I appear ha leing σ = give LT = FT Bu hi i only rue if ROC include he jω axi!!! If he ROC include he jω axi Then he FT i embedded in he LT Ge he FT by aking he LT and evaluaing i only on he jω axi i.e., ake a lice of he LT on he jω axi 12/18
13/18 Le Revii he Example Above b b X u e x b > + = = } Re{ 1 ) ( ) ( ) ( + = + = = = b j b X j j ω ω ω 1 1 ) ( If b >, hen ROC include he jω axi : jω ROC σ = b σ Same a on FT able
Secion 6.3 Invere LT Like he FT once you know X() you can ue he invere LT o ge x() The definiion of he invere LT i: 1 c+ j x ( ) = X ( ) e d 2πj c j wih c choen uch ha = c + jω i in ROC Thi i a complex line inegral in complex -plane HARD TO DO!! Bu if X ( ) b + b M M 1 M M 1 = N N 1 an + an 1 +... + b1 + b +... + a + a 1 Raio of polynomial in Raional Funcion Then i eay o find x() uing parial fracion and a able of LT pair Thi will be covered in ome oher noe 14/18
6.2 Properie of LT Becaue of he connecion beween FT & LT we expec hee o be imilar o he FT properie we already know! Lineariy: ax ( ) + by( ) ax ( ) + by ( ) Righ hif in ime (delay): x() Saed here for caual ignal (book give general cae) x( c) caual Alo caual for c > c c x( c) e X ( ) (c >, x() caual) Compare o ime hif for FT: e jωc v. e c Recall: = σ + jω Noe: There doe no exi a reul for lef if for caual ignal and he 1-ided LT 15/18
Time Scaling: Compare o FT x( a) 1 a X a a > Noe: a < make x(a) non-caual So we limi o a > Muliply by n : n N d x( ) N d N ( 1) X ( ) Muliply by Exponenial: x( ) X ( a) wih a real or complex Muliply by inuoid: e a j x( )in( ω ) 2 [ X ( + jω ) X ( j )] ω Shif in -plane j x( )co( ω ) 2 [ X ( + jω ) + X ( j )] ω Noe: Book doe no ue ω wih ubcrip Warning! So - jω above i wrien - jω Danger! Le = σ + jω - jω = σ! No wha i inended!! 16/18
Time Differeniaion: x( ) X ( ) x() Warning: If x() i diconinuou a = hen we ue x( - ) inead Very differen from FT propery Thi LT propery allow handling of IC!!! Inegraion: x( λ) dλ 1 X ( ) Thee wo properie have a nice oppoie relaionhip: Noe: Differeniaion Inegraion Muliply by Divide by oppoie oppoie Thee wo properie are crucial for linking he LT o he oluion of Diff. Eq. They are alo crucial for hinking abou yem block diagram 17/18
Convoluion: x( ) h( ) X ( ) H ( ) Same a for FT! Recall: If h() i yem impule repone hen H(ω) i yem Frequency Repone We ll ee ha H() i yem Tranfer Funcion SKIP: Iniial/Final Value Theorem Tranfer Funcion i a generalizaion of Frequency Repone Thee properie and ome oher are on he Table of Laplace Tranform Properie ha i available on my web ie Pleae ue he able from he webie he one in he book have ome error on hem!!!! 18/18