APPENDIX 2 LAPLACE TRANSFORMS

Transcription

1 APPENDIX LAPLACE TRANSFORMS Thi ppendix preent hort introduction to Lplce trnform, the bic tool ued in nlyzing continuou ytem in the frequency domin. The Lplce trnform convert liner ordinry differentil eqution (LODE ) into lgebric eqution, mking them ey to olve for their frequency nd time-domin behvior. There re mny excellent preenttion of the Lplce trnform, in Oppenheim [997], for thoe who would like more informtion. A. Definition The Lplce trnform i generlized Fourier trnform, where given ny function f(t), the Fourier trnform F( ω ) i defined : where jωt F( ω ) = { f ( ) }( ω ) = f (t) e dt ω= π f nd f i frequency, in hz. F (A.) In the me pirit, we cn define the Lplce trnform : where i complex: + t F() = { f ( )}() = f (t) e dt L (A.) =σ+ jω, (A.3) σ nd ω re rel number which define the loction of in the complex plne, ee Figure A. below. Alo, ω= π f bove. by Chpmn & Hll/CRC

2 θ Im() Re() ω n ω σ Figure A.: σ nd ω definition in complex plne. Remrk: ) if f(t) for t <, then { } { } F f() ( ω ) = L f() (j ω) (A.4) ) The limit in the Lplce trnform definition tke cre of f(t)' which contin the δ function. 3) The integrl in the definition of the Lplce trnform need not be L f () my not exit for ll. However, if f(t) finite, i.e. { } i bounded by ome exponentil: f(t) Me σ t (A.5) then L { f }() will mke ene for uch tht Re{ } >σ. by Chpmn & Hll/CRC

3 4) The Lplce trnform i liner: A. Exmple, Lplce Trnform Tble ) Exponentil { f + f} = { f} + { f} L L L (A.6) t f(t) = e (t) = = = > + t t (+ )t F() e (t)e dt e dt [ ] (A.7,b) ) Impule t F() = δ (t)e dt = e = for ny f(t) =δ(t) [ ] (A.8,b) 3) Step ( ) () e e t F() = e dt = = > f(t) = (t) [ ] (A.9,b) Tble A. below contin Lplce trnform for few elected function in the time domin. The Region of Convergence or ROC i defined the rnge of vlue of for which the integrl in the definition of the Lplce trnform (A.) converge (Oppenheim 997). by Chpmn & Hll/CRC

4 f(t) Lplce Trnform Region of Convergence ) δ (t) ll ) δ(t T) 3) (t) T e ll Re{ } > 4) t m (t) m! Re > { } m + 5) e (t) t + { } > Re{ } Re 6) (m )! m t t e (t) ( + ) m { } > Re{ } Re 7) t ( e )(t) ( + ) { } > mx{, Re{ } } Re 8) t bt (e )(t) b ( + )( + b) { } > mx{ Re{ },Re{ b} } Re 9) in(t) (t) + ) co(t)(t) + Re{ } > Re{ } > ) e t in(bt)(t) b ( + ) + b Re{ } > ) e t co(bt)(t) + ( + ) + b Re{ } > Tble A.: Lplce trnform tble. by Chpmn & Hll/CRC

5 A.3 Dulity The following dulity condition exit: tf(t) d d F() (A.,b) d f(t) dt F() A.4 Differentition nd Integrtion Differentition nd the Lplce trnform: Suppoe then { } L x () = X() (A.) {&} L x () = X() x( ), (A.) o we cn interpret differentition opertor: Integrtion nd the Lplce trnform: Suppoe then d dt (A.3) { } L x () = X(), (A.4) t L x( τ)d τ () = X(), (A.5) nd we cn interpret / n integrtion opertor: t dt (A.6) by Chpmn & Hll/CRC

6 A.5 Applying Lplce Trnform to LODE with Zero Initil Condition Aume we hve liner ordinry differentil eqution hown in (A.7): &&& y(t) + && y(t) + y(t) & + y(t) = b && u(t) + b u(t) & + b u(t) (A.7) 3 3 Aume && y(t) =, y(t) & =, y(t) = nd tke the Lplce trnform of both ide, uing the linerity property (A.6): {&&& y }() + {&& y }() + { y &}() + 3 { y }() = L{ &&} + L{ &} + L{ } L L L L b u () b u () b u () 3 (A.8) Reclling tht i the differentition opertor, replce dot with : Y() + Y() + Y() + Y() = b U() + b U() + b U() (A.9) We re now left with polynomil eqution in tht cn be fctored into term multiplying Y() nd U(): Y() = b + b + b U() (A.) Solving for Y(): b + b + b3 Y() = U() (A.) It cn be hown tht the term in the numertor nd denomintor bove re the Lplce trnform of the impule repone, H(): Y() = H()U(), (A.) [ ] H() = L h( ) (), (A.3) nd h( ) i the impule repone. For the exmple LODE (A.7) the Lplce trnform of the impule repone i: b + b + b3 H() = (A.4) by Chpmn & Hll/CRC

7 A.6 Trnfer Function Definition It cn be hown tht the trnfer function of ytem decribed by LODE i the Lplce trnform of it impule repone, H(), (A.3). Tking the Lplce trnform of the LODE h provided the Lplce trnform of the impule repone. If we could invere-trnform H() we could get the impule repone h(t) without hving to integrte the differentil eqution. Typiclly the invere trnform i found by implifying/expnding H() into term which cn be found in tble, uch Tble A., nd thn inverting by inpection. A.7 Frequency Repone Definition Hving obtined H() directly from the LODE by replcing dot by, we cn obtin the frequency repone of the ytem (the Fourier trnform of the impule repone) by ubtituting jω for in H(). H(j ) H() =ω j ω = (A.5) A.8 Applying Lplce Trnform to LODE with Initil Condition In A.5 we looked t pplying Lplce trnform to LODE with zero initil condition, which led to trnfer function nd frequency repone definition. Since trnfer function nd frequency repone del with tedy tte inuoidl excittion repone of the ytem, initil condition re of no ignificnce, it i umed tht ll meurement of the ytem undergoing inuoidl excittion re tken over long enough period of time tht trnient hve died out. On the other hnd, if we re olving for the trnient repone of ytem defined by LODE tht h initil condition, obviouly the initil condition will not be zero. We will ue the bic definition of the differentition opertion from (A.) to define the Lplce trnform of t nd nd order differentil eqution with initil condition x() nd x() & : t Order: {& } nd Order: {&& } L x(t) = X() x() (A.6) L x(t) = X() x() x() (A.7) & by Chpmn & Hll/CRC

8 A.9 Applying Lplce Trnform to Stte Spce We defined the form of tte pce eqution in Chpter 5 below: x& (t) = Ax(t) + Bu(t) (A.8) y(t) = Cx(t) + D u(t) (A.9) where the initil condition re et by x() = x o. The generl block digrm for SISO tte pce ytem i hown in Figure A.. Direct Trnmiion Mtrix D Input Mtrix Integrtor Block Output Mtrix u(t) Input B I x(t) & x(t) + + C y(t) Output Sytem Mtrix A clr vector Figure A.: Stte pce block digrm. Tking Lplce trnform of (A.8): { x& }() = { Ax} () + { Bu }() = { } + { } L L L X() x( ) AL x () BL u () (A.3,b) = AX() + BU() Solving for X(): X() AX() = x( ) + BU() ( I A) X() = x( ) + BU() X I A x I A B () = ( ) ( ) + ( ) U() (A.3,b,c) by Chpmn & Hll/CRC

9 The two term on the right-hnd ide of (A.3c) hve pecil ignificnce: ) ) ( ) ( ) I A x i the Lplce trnform of the homogeneou olution, the initil condition repone. ( ) U() I A B i the Lplce trnform of the prticulr olution, the forced repone. Tking the Lplce trnform of (A.9), the output eqution: Y() = CX() + D U() (A.3) Knowing X() from (A.3c) nd ubtituting in (A.3): Y () = C ( I A ) x ( ) + C ( I A ) B + D U() (A.33) If the initil condition re zero, x( ) =, then Y () = C ( I A ) B + D U(), (A.34) with the trnfer function for the ytem being defined by H(): H () = C ( I A ) B + D (A.35) When the term in H() bove re multiplied out, they will reult in the following polynomil form: b() H() = + D (A.36) () by Chpmn & Hll/CRC