LaPlace Transforms in Design and Analysis of Circuits Part 2: Basic Series Circuit Analysis

Transcription

1 LaPlace Tranfrm in Deign and Analyi f Circuit Part : Baic Serie Circuit Analyi Cure N: E- Credit: PDH Thma G. Bertenhaw, Ed.D., P.E. Cntinuing Educatin and Develpment, Inc. 9 Greyridge Farm Curt Stny Pint, NY 98 P: (877-8 F: ( inf@cedengineering.cm

2 LaPlace Tranfrm in Deign and Analyi f Circuit Part by Tm Bertenhaw Baic Circuit Analyi - Serie Circuit Serie C Circuit A erie C circuit i a baic electrical building blck. Frequently thee circuit are cnfigured t be either a lw pa r a high pa filter. In later mdule we will invetigate the deign f active filter, but an undertanding f the underlying principle i fundamental. Analyi begin with undertanding the rle f the tranfer functin, hw t develp the tranfer functin and it utility t predicting the time repne f the circuit. Cnider the fllwing circuit: V( I( /C and uppe we wih t knw the vltage drp acr any f the cmpnent, and further uppe we wih t d by uing tranfer functin. Uing Ohm' Law we can write: V in ( I( I ( C C further, V ( I( and V C ( I ( C T identify the tranfer functin fr either cmpnent we frm the rati f vltage acr the cmpnent divided by the vltage applied. Fr example: V V in ( ( C (Tranfer functin evied: Aug,

3 r V ( V ( in ( C that ay the vltage acr the reitr equal the derivative f the applied vltage divided by the denminatr. It i clear that if the applied vltage i a DC, the vltage acr the reitr i a the derivative f a cntant i ; in ther wrd all the vltage appear acr the capacitr. That make ene becaue a capacitr preent an pen t DC. Al, pleae nte that the factr C i the "time cntant". Further, thi value i cmpletely determined by the chice f value fr bth & C. Aume that ( Ain( ωt, and ince the LaPlace Tranfrm f that ine functin i then, V in Aω ( ω Aω V ( Eq. ( ω C We have nw pitined urelve t predict the utput a a functin f the input. In the "" dmain, the tranfer functin time the driver equal the utput (nt in the time dmain, be careful with thi, the time dmain require the ue f the cnvlutin integral. Frm Table at the end f thi dicuin, it i apparent that the driver in Eq. will invert t Kω c( ωt where K i yet t be determined, and ω c( ωt i clearly the derivative f in( ω t. T cmpletely invert V ( t V (t, it i firt neceary t determine the apprpriate numeratr f a decmped left ide: Aω B C V ( ( ω ( ω Eq. (re-written C C evied: Aug,

4 where B & C are determined by partial fractin expanin. The technique f partial fractin expanin can be fund in almt any Algebra text, r fr cnvenience, there i a hrt tutrial included here a Appendix A. A an example, frm the abve let then, V ( ( A ω C A B ( The ple ("ple" are the value( that caue the denminatr t be zer f the abve equatin are & ± j, which require we re-write a: V ( A B C j j Multiplying bth ide by and repeating at j and again at j, and then evaluating at, we get: 7 A j ( j ( j B j C ( j( j V.. j.8. j.8 ( eq. j j The tw cmplex denminatr n the right hand ide d nt match any f the pair in Table, they mut be ratinalized. V (.. (.68 and that yield (nte that ω wa factred frm the numeratr f the third in anticipatin f inverin a a ine functin in the time dmain: evied: Aug,

5 r re-writing: f ( t t. c(t.68in(t.e eq. f ( t.9in(t.e t Eq Amplitude Time in.ec Nte that in(t c(t 9, atifying the aertin that f (t will cntain a functin f the derivative f the driver. A hrt dicuin n btaining phae angle i included in Appendix B. The abve prcedure i lng and tediu and pen t errr, let u ee if we can find a "hrt cut" methd fr dealing with lutin f plynmial that cntain cmplex rt (inuid. Firt, merely factr the denminatr; ( ( B V ( C Secnd, multiply bth ide by the cmplex factr, in thi cae ( Third, et ; jω, and imprtantly, factr and ilate ω frm the numeratr reult: j ( and C. j B..9 evied: Aug,

6 r then inverting: V V (.9 ( ( t.9in(t..e t The anwer are cnitent, and the "hrt cut" i eaier. A purit might reject the hrt cut a being intellectually lazy, but the authr i a charter member f that grup we will adpt that methd a ur mdu perendii. Cntinuing with uncvering the vltage drp acr each f the cmpnent, we will nw purue the drp at the capacitr, f cure uing the ame driver f in(t. The tranfer functin at the capacitr i develped a fllw; V V C in i( ( C ( i( SC i( C Frming the tranfer functin, then manipulating thee exprein t btain functin f the driver, we get: Letting, a befre: C Uing the uual methd, then, and V C V C ( ( ( C C 7 ( ( ( 7 A. B ((.7 9 ω factred ( ω j. ((.7 9 VC ( A B V C ( a a evied: Aug,

7 V C ( t.e.7 in(t 9 t eq. Anther quick check i that V ( t & V C( t are 9 ut f phae, a they huld be. A prficiency i develped with thi frm f circuit analyi, peed and accuracy increae rapidly, and btaining quantitative anwer fr the utput in the time dmain becme far le tediu. Eq Amplitude Time in.ec VrVc 6 Amplitude - Serie Time in.ec evied: Aug, 6

8 The abve figure i the um f the teady tate um f Equatin & ; the vltage acr the reitr and the capacitr. It gratifying that it mirrr the driver, verifying that ur cmputatin are crrect. The Impule epne What i the native repne f a circuit? That quetin interet u becaue the native repne i the tranient prtin f the utput (the driver frm the teady tate prtin f the utput - δ (t r a pule being an exceptin a there i n teady tate utput. Since the tranfrm f δ ( t we will ue the impule a a driver, the utput i then the native, natural r impule repne i.e., V ( C C ( tranfer functin with impule driver C f ( t e C t C again fr the ake f illutratin nly, letting C Impule epne.. Amplitude Time in Hundredth f a Secnd evied: Aug, 7

9 The Step epne The tep repne ilate the repne f the tranfer functin t DC. Like the Impule epne, the utput i expreed a the tranfer functin time the input, in thi cae a unit tep, i.e. C V C ( C C V C ( t e t C Step epne. Amplitude Hundredth f a Secnd It i intereting t nte the relatinhip between the impule and unit tep repne, bth mathematically and graphically. But the real ignificance i that the impule repne i the native time repne f the circuit, and the tep repne i the repne t a DC driver (r DC cmpnent f a driver. evied: Aug, 8

10 The Serie LC Circuit The erie LC circuit i a fundamental building blck in circuitry, even thugh the deired circuit repne can ften be btained uing active circuit. T undertand LC like behavir, a well a t analyze and/r deign a circuit t btain a pecific repne, it i very deirable that a thrugh grunding in the fundamental i well undertd. One f baic parameter f electrical deign, the circuit renant frequency, i uncvered by undertanding LC behavir. Serie LC circuit are metime referred t a "erie tank circuit", becaue they d pe an inherent renant frequency. At renance ( jx L jx the impedance f the netwrk i at the minimum. C A uual ur apprach will be via the tranfer functin. Thi apprach nt nly prvide fr a predictin f utput in the time dmain, but al pitin u fr analyi and deign wrk in the frequency dmain, a we hall eventually ee in later mdule. L V(in I( /C Circuit vltage i (Kirchhff' vltage lp: V ( I ( ( L C We need t re-arrange the vltage lp exprein a it will becme the denminatr f any tranfer functin and it MUST be in a frm cmpatible with Table fr inverin. L V ( I ( L LC A nice quadratic! That i imprtant a a quadratic ha either a pair f real rt r a pair f cmplex rt; in either cae it i invertible after PFE. And a we will ee a time ge n, that thi quadratic play a majr rle, ften the dminant rle, in any circuit that the lutin t the quadratic i cmplex. The vltage acr the capacitr i (Ohm' law: Vc( I ( * C and acr the inductr: V L ( I( * L and acr the reitr: evied: Aug, 9

11 V ( I( * A we have dicued, the rati f the vltage acr any cmpnent divided by the urce vltage, by definitin, i the tranfer functin: and finally, V C ( LC V ( L LC V L ( V ( L LC V ( L V ( L LC fr the ake f illutratin and t generate example, aume L 9 LC Let invetigate the impule repne acr each f the cmpnent. In general; V X N( ( ( 9 The rt f thi denminatr are ± j, re-writing: V V C V L ( ( ( ( ( ( ( 9 ( ( evied: Aug,

12 Yu are invited t verify that: and that the rt are truly 9 ± j ( There a cuple f imprtant thing t be aware f regarding the denminatr. Firt if, then the rt f the denminatr becme ± j 9. That i apparent a the denminatr in that cae becme: ( ( 9 ω LC Thi frequency i cmmnly dented a ω and it i the highet frequency the circuit i naturally capable f; i.e., it i the circuit' renant frequency. The circuit may be driven t a higher frequency by an excitatin urce, but recalling the phyic f a cil, the higher the frequency i that it i driven by, the mre it begin t behave a an pen. Yu can alway uncver the renant frequency by etting the factr n. Clearly then i the renant frequency. All thee remark apply nly t the cae where the LC rt f the quadratic are cmplex; if they are real there i n renant frequency. Yu are invited t verify that if the rt are real, the inverted frm becme ±α t ±βt f ( t Ae ± Be. In the example abve, the frequency i and i cmmnly dented a ω d, the damped frequency ( ω d i alway < ω, an inpectin f the general quadratic will dicle why. Ntice that in the abve denminatr nce we che the value fr ne f the cmpnent, the value f the ther are et by the relatinhip f &. In ther L LC wrd, we want t deign a circuit in which ω 9, ω d and L. Ching a capacitr with a value f farad, require we che a cil f 86 mili-henry and a reitr f.86 hm. Nt realitic circuit value fr mall ignal circuit, but they erve ur purpe a an illutratin (fr eae f math rean we have chen a circuit with an ω f.8 rad/. Nte that exprein t L e L L which, in turn, identifie.aume the functin f the multiplier n t in the a the circuit time cntant. L A an exercie, why i the time cntant intead f L? A curry inpectin f the way in which the denminatr mut be arranged t guarantee inverin frm the dmain t, the time dmain prvide the anwer. evied: Aug,

13 Taking the invere tranfrm f the impule repne ne at a time, V C ( t.e t in(t V ( cannt be inverted a it tand, we will add and ubtract in the numeratr t yield an invertible pair; r V ( ( ( ( ( ( ( t t V ( t e (c(t (in(( t 6.9e in(t 8. See Appendix B fr a Wrd Abut Phae Angle. A quick review f Table hw that V L ( cannt be directly inverted a it tand, and additin/ubtractin a in the cae f V ( will nt relve the iue. S, a new rule i called fr, and it i thi: When the numeratr i the ame rder f a the denminatr, lng diviin MUST be applied a neceary t btain a remainder in which the denminatr i at leat ne rder higher than the numeratr. In the abve example f V L (, then: 9 V L 9 ( thi will further devlve t (uing additin/ubtractin f ± : and invert t: V L ( (.( ( ( f ( t δ ( t e δ ( t ( c( t.in( t ( in( t. t t.e 6 evied: Aug,

14 Impule epne Inductr Vltage Amplitude Hundredth f a ecnd The abve figure fairly well capture the effect that δ (t ha n the impule repne. t L T amplify the effect f e, ω d i arbitrarily increaed belw, imply t illutrate hw the expnential decay can dminate the impule repne. Inductr Impule many rad Amplitude Hundredth f ecnd T repeat, nte that if the factr n i (mean then the circuit cillate at ω 9. Wherea when the frequency i lwer (in thi cae - and that will be true with all cillating circuit. When the circuit will theretically cillate evied: Aug,

15 undiminihed frever. But even if we left the reitr ut, there i reidual reitance in the lead (at any temperature abve ablute uch that there i an I l, and n matter hw light the l, cillatin will eventually decay t. In a wrd; in the real wrld n perpetual mtin i allwed. Argument cncerning ablute zer and quantum electrdynamic almt never apply in practice, i.e., Newtnian phyic i jut fine fr cmmn ue. Mving n, uppe the abve LC circuit i driven by a V in in(t, then, V L ( ( Diviin i n lnger required a the denminatr i nw rder higher than the numeratr. Neverthele, a PFE i required in rder t invert V L ( : 9 8 ( A B Uing the technique f Appendix A (PFE, and lving fr A & B: A j ( j ( ( j ( A B j ( j ( ( j ( and B.86 t f ( t.7e in(t.6 in(t.86 f (t i ur predictin f the utput given an input f in(t. It i imprtant t nte that element f bth the impule repne and the driver are preent in the utput. The impule repne element i the tranient repne and the driver element i the teady tate repne. In general, that tatement i true acr the bard, i.e., there will alway be element f bth the impule and the driver in the time dmain utput. In the cae abve, the tranient prtin i: evied: Aug,

16 .7 t e in(t. 6 crrepnding t the cntributin f the native r impule repne f the circuit t excitatin. The teady-tate prtin f the utput i in(t.86 Again, the teady tate repne i f the ame frm a the driver, e.g., A in( t ± φ and will remain until excitatin i terminated. Tranient & Steady State epn Amplitude Time Hundredth f a ecnd In the abve graph, the amplitude f the tranient prtin ha been ditrted purpefully t better identify it cntributin t the utput. It huld be clear that by the end f the ecnd cycle it ha fully diappeared. A an exercie, yu are invited t lve fr the impule repne fr each f the cmpnent f an LC erie circuit, uing the cmpnent value a hwn. evied: Aug,

17 ..88H V(in I(.f Kirchhff' vltage lp tranfrming directly rearranging plugging in value di V in ( t i( t L i( t dt dt C V in ( i( L C L V in ( i( L LC V in L ( i( ( (here i where yu are invited t d the wrk t find the impule repne; V V C ( t.e ( t e t t in(t ( c(t in(t t ( c(t.in(t VL ( t δ ( t e Let' cnvert V L (t t a ine functin: V ( t δ ( t.e L in(t 77 t If we aume a driver f in(t and then lve fr V L (t, we get: V ( t 7.8e L in(t 8.7 in(t 7.7 t evied: Aug, 6

18 Inductr Vltage 6 Amplitude Time in.ec Becaue erie tank circuit preent minimum impedance at renance, they are ueful a ntch filter t trap fr unwanted frequencie. Of cure active filter nw perfrm thi ame functin, and the will be dicued in later mdule. evied: Aug, 7

19 Tranfrm f (t F ( K t Ke σ K in( ω t K c( ω t σ Ke t in( ωt 6 Ke t K K σ Kω ω K ω Kω ( σ ω σ c( ωt K ( σ ( σ ω ( 7 δ (t 7a* Kδ (t K 8 Ku( t a Ke a 9 f '( t F ( f ( f ( t dt F ( f ( af ( t bg( t af ( bg( t at te ( Table a * K i preerved fr practical circuit rean, nt fr theretical rean a K i apprximately equal t Table i nt all incluive and ther pair will be examined and added when needed. But fr beginning analyi purpe Table i adequate. It i very imprtant t undertand that t be able t tranfrm any F ( t an f (t, F ( mut be reduced t ne f the frm far develped. If it i nt in ne f thee frm it cannt be perated n until it i. Study the right hand ide frm, they identify the left hand ide. Tranfrm and are fund a fllw: evied: Aug, 8

20 Fr tranfrm, aume f ( t t F( te t dt f ( t te ( a t F( te at dt ( a Finding tranfrm and i a traightfrward exercie in integratin by part. evied: Aug, 9

21 Appendix A Partial Fractin Expanin Partial Fractin Expanin i a technique t decmpe a rati f plynmial int a um f factr; fr example: A B Ex. 9 where A & B are t be determined. In general, then, N ( D( A B N... p p p n Eq. N ( & D( are the numeratr and denminatr plynmial. p thrugh p are the rt f the factr ( p i. i ued becaue thee are al the ple, a will be hwn in the mdule n t-lcu - nte that there i n retrictin n ften i, cmplex. p i t be real, it can be and T lve fr any f factr numeratr, A fr example, a multiply bth ide f Eq. by ( p, b cancel ( p frm bth ide and frm the denminatr f A, then c et p. The reult lk like thi: n N( p ( p p...( p p n p Let' try it n Ex. : ( ( ( ( ( B( A nw et A ( B ( A A Fllwing the ame prcedure fr B evied: Aug,

22 ( ( ( ( ( ( B A ( B A etting, we get: ( B r B Let' check it: 9 Cmparing the rightide with Table we can ee that the lutin can nw be inverted t the time dmain (recall, a factr mut be in ne f the frm in Table t be inverted. In fact thi lutin invert t: t t e e t f ( Bth value f the time cntant are a direct functin f circuit cmpnent value. It i very imprtant t nte that multiplicatin in the '' dmain i nt multiplicatin in the time dmain. Checking the lutin by cr multiplying the denminatr t frm the LCD, we have: Q.E.D. Thi general prcedure i repeated fr each nn-repeating rt regardle f the number. When the rt are repeated a light mdificatin i called fr. Fr example: ( ( ( C B A The denminatr mut be aumed t be part f the lutin becaue it i a factr f the LCD (althugh ( B can be. Finding are traightfrward a in the example abve, : A & C evied: Aug,

23 A ( ( ( D N Eq. at we have:, A finding C ( C at, we have: 7 C S far, gd - but what abut B? The typical prcedure i t take the derivative f Eq. and evaluate the reult at the rt value. Fr example: ( ( ( d d evaluating that at 7 B therefre, ( ( ( 7 7 Again, we cmpare the denminatr t the frm required in Table, and find that: (t f t t t e e te t f 7 7 ( T check the anwer we cr multiply the numeratr a apprpriate and frm the LCD. ( ( Then there i the cae f cmplex rt, fr example: ( ( j C j B A Again, uing the prcedure we have etablihed, evaluated at : A evied: Aug,

24 Nw, we will find B by letting j r A. ( j ( j ( j j r B. j. 8 We need nt lve fr C a the ther cmpnent f a cmplex cnjugate pair with the cmplex cnjugate f B... B j.8. j.8 ( ( j j Hwever, the lutin abve de nt match a frm in Table, we mut find a denminatr fr the cmplex fractin that matche a tranfrm pair frm Table. The LCD f the cmplex pair i: ( j ( j ( 9 Cr multiplying the numeratr f the cmplex fractin and adjuting t extract the neceary numeratr fr the cine and ine factr, we end up with: F (...7 (..( ( (.967 ( ( F ( i nw eaily capable f being inverted: f ( t.e t.e t c(t.967e t in(t Again, nte that althugh the factr in dmain are multiplicative, they are nt in the time dmain (very imprtant pint, particularly fr deign wrk - the time dmain require the ue f the cnvlutin integral; multiplicatin in the dmain cnvlutin in the time dmain. T check, again after cr multiplying the numeratr f get: F ( t re-frm the LCD, we a it mut be, r A, B & C are nt crrect. ( ( evied: Aug,

25 Sme example, btain anwer t match the belw: 9 8 (.. ( ( ( 8 ( evied: Aug,

26 Appendix B A Wrd Abut Phae Angle Define a pace with crdinate axe a hwn belw: cine axi ine axi Nw cnider a tatement uch a: c( ω t.in( ωt We will interpret that a unit cincident with the cine axi and. unit cincident with the ine axi. In ther wrd, it repreent tw ide f a triangle with an adjacent f. unit and an ppite f unit. unit. unit That being the cae, the encled angle at rigin i abut abve um a:.8in( ω t.7 Cnider the um in the LC circuit in the bdy f Part :.7, and will re-write the evied: Aug,

27 V Applying the ame ratinale: ( t e t (c(t.(in(( t unit -. unit Then a meaured frm the ine axi (defined t be t 6.9e in(t 8., the reultant i: evied: Aug, 6