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by: Howard Glover


Howard Glover
GPA 3.96


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This 41 page Class Notes was uploaded by Howard Glover on Monday October 19, 2015. The Class Notes belongs to ENGR 203 at Oregon State University taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/224565/engr-203-oregon-state-university in Engineering & Applied Science at Oregon State University.

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Date Created: 10/19/15
Time Domain Models of Passive Components C VRr R 1W V f if 11 I VIH VRtRiRt VL LiL ic cquotc iRtvR t iLtJvLrdriLtn vcticrdrvctn Consider the following sen39es RC circuit For 20 iv tRitVat7V tRit39 irdrva00 V40 MRItw jQZad v g vutthen 0 Let 1 be constant ie V t 1 r 2 ve 1IEjozrdr R OI il1il0 with i0vquota0 In terms of the voltage across the capacitor for t Z 0 we have vl t Rit v0t 0 or vltRCVUtVGtO 70 or vgom vx t Now consider the series RL circuit For t Z 0 v1 t Rit v0 1 v1 1 Rit Lit 0 10 d R 1 EEO 210va 1 10 1 Z 0 Let v10 VI Vt then in terms ofthe voltage V0 vltRJvardr 10 vat0 0139 0139 gun vat0 10 v1 Ri0 Consider the following sourcefree circuit tZO 80 AVAVAVAV ti ltgt 3 502 01FvC 1203 v By nodal analysis I t t vcg 415mm va 0 vim w W 0 5 12 From eq 2 v r gm 0 Substituting eq3 into eql yields 3 V I V I W 5 1 I 0 5 J 8 or 40i5t10v5t 0 Mt Also itc Hence e 4 becomes dt q 40610v5 0 or d v r dt 40 Clearly Req 4Q vt0vc0 Alternatively the Thevenin equivalent resistance seen by the capacitor is shown below R l v 2 01 F Thevenin equivalent circuit 39o l Watt0 0 I 2 3t 4 Transient response of circuit 5 t I 1 2 3 4 5 The previous circuits were described by firstorder linear ODEs of the following type d xt axt ft xt0 x0 and a 6D I where t is an external excitation function either an independent current or voltage source To solve this differential equation let us proceed as follows Taking the time derivative of emx yields d a d a d Eemxt e Ext aea xt e Ext But the quantity inside the brackets is equal to e f t ie d a 1 ER xte Hence if we integrate both sides ofthis equation we get litie x1 e quot 161 or lildiemxv e xt ewoxt0 J t o ear 1611 2 xt eeatetoxt0 61 It em 1611 eiaiiixt0 I 61104 f d x t x f t where x t 67110406 to is the unforced response due to initial conditions only and xf t J avid air is the forced response Note that if a gt 0 x t gt 0 as t gt 00 U the circuit is stable ie the effect of the initial condition xt0 dies out On the other hand if a lt 0 x t gt too as t gt 00 the circuit is unstable So for the series RC circuit with to 0 we have 7L1 1 7Ltet 1 v t eRCv 0 6 v 2 2e tgt0 Volts A A L RC Let Z39 2 RC be the time constant of the RC circuit then 7 z 1 t 7 i Z39 v0t e voj0e 0 W vi 1dl tgt0 Volts Define the unitstep function ut as follows Graphically 1 I Unitstep function If v1 t Au t then the step response ofthe circuit is given by A t 4r tar v0tet v00 TJ0e u1dl tZO e trvo 0 84 e lr 1d 139 e trvo 0 Ae tr err 1 e trvo 0 A1 e4 t 2 0 Volts For the series RL circuit with to 0 the current in is given by fit t 75071 1 ite L i0J0e L Zvi 11 tZO Amps Let T be the time constant of the circuit then it e tri0 KaiWWI ld t gt 0 Amps If v1 t Au t then the step response ofthe circuit is given by it 67470 jelttgtru 1611 t 2 0 Ar e r e I et 10T1 et 470 1 8quot t Z 0 Amps Example Find 71 2 0 4 I Hi 4A I UT 20 2 v r vv 4 O At t 0 before the switch closes the circuit behaves as follows 4A 292 vLO 5Vc07248V lt 6 For t 2 0 T 2F H H w nf AAA wv 44 2133 4953 110 or U 40 v i ByKVL vcz4izvcz42vcz0 VC08 0 v d l or EDAZH VCZ CO 2vcljvc0e 8 8e7 8120 Nowvclval0 2 val7vcl 8 Graphmally van 78242 2 20 Volts 15 desmbed as follows n 1U 2U an 4n n Examgle Fmdz2220 m 2112 N At 2 0 before the much closes 1111 MM 04 A20zl0zl0 1A szzo 21quot 2A0 H r I w mm AA vv y 1 2an my gt l 103139Lz2x10 3iz0 or il5x105iLl0 iL01A T2x10 ssec2usec and 139LliL0e tM 6 94 076 Z20 Amps Graphically iL t is given by as v m u 2 M 1 HHHHIH WWW D x y x 1 1 U DEIDEIUDZ DDUUEIEM EIEIUEIEIEIE UDDDDDE El ElElEH IUDEIEHZ EI EIUEIM I Example Find vO t for the circuit below For t g 0 no energy is applied to the circuit Therefore iL 0 0 For tgt0 lZH By KCL V 1 52VA1Lzo vAt57vo t v0t75257v0tiLt0 or 7votiL t50 1 1 d Also by KVL eve z3E1Lz1Lz 0 1 d or ve t1L IEE1L 2 Substituting 2 into 1 yields 1 d d 73E1Lt5 0 or E1Lz10 z gt 0 Now Irdib 11517 flOdr n q n iLt7WU10t iLt10t zgt0 Amps Finally from 1 vot5iLt10t5 tgt0 Volts RLC Circuits Consider the parallel RLC circuit Iquot oir 49 R L C VG r ByKCL itutiRtiLtiCt 0or z tut JCT iL t CVC 0 0 This last equation implies that at t 0 gl i 0 But iL t 5ng 1011 iL 0 gm So VCIEI IVC 7017 1IL 0 CVC 2 ii l d l d2 d gt FEDC Zvc I 0 CWDC 2 ED 3 vcl vCIvCI iilut iL0 1100 or To obtain a unique solution we need to know vC 0 and t i0 Let s now analyze the series RLC circuit R L vur ByKVL 7v lulvRZvLZVCZ 0 or d 7v lulRiLlLE z39LzvCz0 But vclJicrdrvc0JiLTdTvC0 2RiLzL 1415rdrvcov zuz d dZ 1 d gtRE1LlL1LZEZLZ0EV ZHZ 371142 ultzgti nltzgtiigrvmuw 140 mol OI Again to uniquely describe the circuit we needto know 120 and t 0 39 Both parallel and series RLC circuits are described by a 2 1 order linear constant coef cient ODE of the form 37192w xltzgtlazrxltzgtkfltzgtgt xltogt xltogtnzo ltgt Again the solutionhas the form xl xul xfl Let fl 0gt the xfl 0 and xl xul Let xulke then 5790 aging a2 xt ksze a1kse azke 0 or ke s2 alsaz 0gt for anyt20 either k 0 or s2 alsaz 0 Ifk 0 then x t 0 the trivial solution If on the other hand s2 als a2 0 then x1t kle l and x2 I kzexz k1 k2 5 0 are also solutions of when ft 0 2 where s1 s2 are roots of s als a2 0 le al lalz 4az and s al lalz 4az 2 s 1 2 2 Also x t x1 I x2 I kleslf kzeszf t 2 0 s11 are solutions of s2 als a2 0 To get a unique solution 1 At t0 xu0k1 k2 2 At t 0 1h s1k1 szk2 dt H Let us now introduce a new tool to overcome the difficulties presented by the timedomain approach The Laplace Transform Let f t be a function such that J Owlftequot dt k lt oo where Oquot is areal number De ne the onesided Laplace transform Fs Of by Fs E Ifte dt where s Oquot ja is a complex variable 039 and a are real variables and j V l Example Let f t u t the unit step function then Fs U s Lu t 2 Leon te dt fl ei dt 00 J wei rjwtdt 1 67mm 0 039 10 H 00 Ieim eijat 039 10 0 But le ja l ICOS wt jsin ml cos2 wt Sil l2 wt 1 Vt that is e j remains bounded in magnitude for IE 0 This means that e m 67 limequot 67 lime vim 0ja H 0ja Hm H0 2 lime m 67 l 0a Hm Now 7 00 as t gtoo when 039lt0 e 0 as t gtoo when 039gt0 Therefore 1 00 1 1 equot e ja O 1 039gt0 0a H 0a 0a or usl if sagto The implication here is that the Laplace transform does not may not exist everywhere in the complex splane For the unit step function example the region of convergence is shown in the picture below Region of convergence ROC of Us Define the Dirac delta function 5 t as the function with the following properties 1 gtoo as t gt 0 l 39f b 2 Ib5t todl 106p a 0 1f to e ab 3Ifft39 39 t h Ibft5t dz fc ifcea b t t t 1s con muousa C en a c 0 if caab This last property is known as the sifting property of the Dirac delta function Remark In reality the Dirac delta function is a mathematical artifice trick used to approximate very narrowband pulses in order to facilitate analysis Example Let ft5t to to 20then FS L5t 10 00050 toeiszdt 6405 using the sifting property ofthe Dirac delta function Let us now consider other common but important functions Example A sin wot ut then 00 mm 0 11 4911111 Help 17 lt o 10 0 lt 17 o J 006 ooex Aauln Ueq1lt17 gt 9 10 co gt 17 0 H l Aamn Jepysuoo n3 putaq Jeqlo up HQ 3900 1 03 10949 xngga JUW9Aa Ina 0 17 0f o 0 1277f349 f me E 1P1DXAQI 11 1 1049 1 r047 2 SM 120 17 S 1z7s49 I uelu Olt17 1n 010 s 010 s z mm 1 2 310 zf Omfs Omf s 0lto OWNS OWNS V I I V 010 10 9 010 m o z 0lt9 1 0 1 0 ueq1 0lt 7329 191 Omzf 0 mum49 1349 mm3 w 00 10mf4vf349 I m mf 0 00f mf o z 49 010 0f 0 010 0f 0 10mfmf349 1pqnrmrg49 1fvf3 a 1 Zf 1plmf3ga OIV 119E 1 10 uys 177E 1172149 100219 zpuganzmsrI n10mmsv7 w 1 ia a 77 a Fs 11me me 11me Hm y Ujcoa H Hf 0 1 Ugt a Ujcoa 1 Resgt a sa So the region of convergence is given by fa m s ROC for Le 39ut a gt 0 Transform theorems 1 Lkft J Owlg te dtkfte dt kFs k is a constant 2 Let Cl and k2 be two constants Then L k1f1 t i szz t jo k1f1 t i szz ew k1 I f1 newt i k2 f2 newt lel s i kze s This is the linearity property of the Laplace transform operator d 60 d is 60 7x 3 ngtflttgt1logtflttgtie W Integrating by parts Let u e and dv 2 df t then ch se dt and v This implies that Lft fem IOWftsemdt zliirgfte x f0sJ fte dt m j fte dt g I0wft L frl diva If the Laplace transform of exists then J Ooolf tequot 7 st 6 dt I0w f te j quot dt Mldt I0w f tequot dt dt 7m e dt lt 00 But this implies that for a at range ofvalues of Oquot I is notunbounded as t gt 00 Therefore limfte 0 00 whenever the Laplace transform of eXists and sFs Generalizing this result to the nth derivative of t yields dquot n r171 r172 dH LdtnfrsFs s fo s EWHHWftl0 or ftsquotFssn1iftt0 do keep1ng 1n mind that 4 LJ fxdx J OOOJ fxdxe dt I Again integrating by parts let u 2 I0 fxdx and dv ei dt then du dt d l J or du ftdtand V e Therefore Iflf fltxgtdxldt e Ignxwxlzw HJW m t 7m 39 1 is f If J Ofxdx grows slower than 6 as t gt 00 for some 039 then lgge J Ofxdx 0 This is directly related to the existence of the Laplace transform of f t 5 Let a gt 0 then Lft aut a J wat aut ae dt J wft ae dt Let Tt athen dTdt and J wft aequotdt jwfre dr e J wfre dr a 0 0 F Lft aut ae Fs for some constant a gt 0 6 Le flJ OooeiarfteendtJ wateesatdt Leimfl Fsa 739 Letagt0athen Lfat Ifatemdt Let T at then d r zadt and J wfate dt lJ wfTeSardr 0 a F 511 Therefore LfatF a gt0 Example Let gtsinw0t tdut td Then if ftsina0tut gtft td andGse FsorGse 2 039gt0 SQ Example Let gt e mdr a gt 0 Let e wu I then GsFslsa ms 0quotgt a S Applications of the Laplace Transform Consider the following circuit 29 AAA I 2H 1 00 12A 49 f 0 30 6Q J Find 1390 I l2 0 At t 0 prior to closing the switch Since the switch has been open for a long time the inductor acts like a short circuit in the presence of DC Therefore at l 0 MO 110 17A 9 3Q 6Q 28 ByKCL 7120 gt vx 24Volts and iL0394A orbycmrentdivision iL0 12 3 4A 3 6 For I 2 0 39 3 7 29 I I 2H quot10 IHI ILU 1140 39 60 gt 39 9 Again by KCL VXTWVXTWHLZ0 or 5vxt6iLt0andvxz7 iLz Also by KVL of WC ZEiLt 6amp0 Substituting vxt from above diLz6iLz 2 101iLz36iLz0 6 7 g1Ll 2dl dz d orE1Ll361Ll0 1L01L0 4A Inthe sdomain ie LiLl 36139Ll 0 or SILSZIL0 361LS 0 s36ILsiL04 gt 1Lsm sgt736 4 4e uz and 151 ivXTm 77e 3 6911 e 3 6341 Amps 29 Consider now a circuit with a dependent source Forl07 49 g 2 o i 3A VA V00 J 2 at l 07 no current ows through the capacitor steady state DC condition since it acts like an open circuit Thus VA 2 43 2 12 Volts Using KVL around the second loop yields vA 24 2VA v000 2 v00 2431260 Volts For t Z 0 24V 2v r ByKCLili2iC0 t t t V L V Lzct20 3 Igor W120 30 iclttgtcgwtgt1 lm 2vot 330 Again using KVL around the second loop yields vAt 24 2VAIVOI 0 gt 3vAtv0t 24 But W 4vot 12v0tvot 24 or ditvotvot 2 with v00 60 Volts Using the Laplace Transform we get Lghd h vi d 2 sns vooivs orsijvs3voo26o S S 0 12 3 1475 21 601Resgt0 37 3 12 12 Z Vos k21 601 RIZsgtO s Si si 12 12 24 31 SDomain Models of Passive Comgonents vt Vs R R it I s 00 Vs IsCsVs7Cv0 Example Obtain 13910 and 13920 in the circuit below 8H 10 H 336139 112525 42 At 1 0 a Igor 336T 1129 a 429 31 482 Ji 1 i2 7A 112 y 42 48 But i20 z39z 7A and i10 iyz39z 8715A For lgt0 201 m 70139 By KVL 11211s 8s11s712042Ils712s 0 or 8s15411s74212s 120 1 and 42Izs71s10s12s7704812s0 or742I1s10s902s70 2 To solve for 15 and 25 we must solve equations 1 and 2 simultaneously In matrix fmm gs 4125 4 101290128 2 1720 er AsIs Vs where As 8s 54 42 1s 42 10s90 120 and Vs 70 This means that if the inverse of As exists then 1 I 41 V d tA V s s s dams a m 6 6 Let As1 detAs Then As 80 s2 2260 s 12096 Now 11s 1 10s9 42 120 12s 80s22260s12096 42 8s154 70 18 120s 1374 12s s22825s1512 56s1582 15s17175 or Ms s2 28 25s1512 and Ms 7s19775 s2 2825s151239 Now 15s17175 k1 k2 2 7174 1 s7174s21076 s7174S21076 R435 15s17175 s 21076 i 97174 k2 M 210386 s 7174 5321076 k1 4614 So i1tL1IlsLl 4614 Lli 10386 s7174 s21076 4614e 739174 ut10386e 2139 76 u t Amps Likewise 7s19775 11 12 Resgt7174 2 s7174s21076 s7174 s21076 39 3612 39 10612 zes s7174 s21076 34 7 1 7 1 4614 1 10386 121 11S 7173 s21076 izl10612e397 1 uz 3612e z m u 1 Amps Example Obtain VLZ when VC0 5V and Ii 0 0A IVAl 1amp2 1amp2 IS AAA AAA u vvv vvv vvv solr m 111 4quot In the sdomain 50 ByKVL7 112K11171211s110 S ors3llilz2Vx 1 S 112711 2VX112 0 S S or 411 s11Z istx 5 2 Also 7211VX 0 gt VX 2711 S S Substituting this value of VX into 1 and 2 yields s311712 7211 S S or ss111 isIZ 750 3 7s11sl1272s271175 S or s11 s11Z 95 4 Using Kramer s rule on equations 3 and 4 to solve for I1 yields 750 is 1S7 95 sl 7 45s750 1 ssl is sisz3slii s sl But ms S11S amp s 3sl s0382s2618 ms k1 1 k2 300517505 s0382 s2618 S0382 s2618 7 71 7 71 Vzmii V1S 0382 s2618 3005 1 7505 vLz73005e 0m uz7505e 26 8 uz Volts I Example If Ii 0 0A and V0 0 0V obtain i1 I for the following circuit In the sdomain ByKVL 7 211s117120 gt s2llis12 2 S s or ss2llis21Z 12 1 512712212 0 3 S1S22 s s S S or 7s211s22lz 6 2 Using Kramer s rule 12 isz NSL 6 s22 7 18s224 7k kZ k 1 I Z i Z 7 l i 7 2 2 ss2 s SS S S Slij Slj Sz 2 2 2 2 2 6 321a 321e 16ZZ 11s7 J7 l 2 i1l 6ul 6497 cos 71 622 Jul Amps Example Compute 13900 for the following circuit 4m 100m or g I 2m am lt 3va IIquot lHU Atz0 4m 3 o ro vAv g r em 0139 1mm 334 lt III39 M 4 By KVL around the inner loop V6 07 Va 07 Vb 07 Vb0 7 8 7E 3000 7300073 12k I6V2V 4k12k 2 A150 1507 mAmps By voltage division V0 07 3k A1b 1t diquot v 0 12V4V SO yVO age VISIOH b 39va039457405V For I gt 0 in the sdomain we have 0 v 0Vquot 4 3 M 31 mi 3k By KVL 79 400011 s 1200011 s710 s 0 S or 1600011s71200010s 9 S 4 12000Ios711s E10s 300010s 0 S S 4 or 71200011sboomimo 7 K S J S Using the substitution method 2 3000 5 1404 5 Rem 101 1105 1 23000 2 97331 Amps s 53 7 3000 Example Compute the initial and nal values of Va 1 for the following circuit Without explicitly computing Va 1 In the sdomain ByKCL 4 V1 V1 s l 1 s2 s s 1 74 i4sle2 or EJF s2iV1 7 or V1ltSgt7 sisz3sll By voltage division 1 s 4s1s2 V V 0S 11 1S slsisz3sli S ms ls2 4s2 I s 3sl s0382s2618 Now the initial value of vat is obtained as follows 4 Z 2 va0l1msVasl1m ZS S 4 Volts H H S 3 S Clearly both poles lie on the le half of the splane Hence we can apply the nalvalue theorem to compute the nal value of Va 1 ie 4 2 him01 l1mSVaS l1m 0 Volts Hm Ho Hos 3S1 which is consistent since eventually no current will ow through the capacitor and therefore through the 19 resistor in the presence of a DC current source of 4 Amps Example Consider the following circuit Assume iL 0 0A 1H 120 quot1 21 In the sdomain l l m U llu V 2 HL s 0 2 m 5 mm 1 1 1 ss3 S 2 2 ltgt W 2 gt 10s Res gt0 S1 s3 1 L3917 i0 1 7 ul e 3tul Amps Let us consider the effect of one source at a time Sourcel 2 ls 2 23 23 1 7 0 01S 1 3 ss3 s s3gt 1111 S 2 2 71 2 2 3t 2 2010 101s ut e uIA Source2 am 2s 4 43 43 I 0 02s ls ss3 s s3 RBSgt gt 102t L 1Iozs ut e 3 ut A But iorzoltiozr ur eurA 2 10szolszozs The principle of superposition applies The total circuit response is equal to the sum of the resnonses due to the 39 quot 39 39 39 source inputs 42 Periodic Functions Afunction is periodic iffor some T gt0 ft ikT where k is an integer number k 20 Let f t for 0 SlSTand f t is zero elsewhere Then the new function fptftft Tut Tft 2Tut 2T t kTut kT is a periodic function of its argument time Now Lfp t HZ t kT ut kTe dt k0 If the sum converges for every value of t then Lfp t Z 0 f1 t kT u t kTequot dt Ze kT Fl s k0 k0 fptEse T k Let vlrr2rquot Then rvrr2r3 r 1 and 1rn1 v rvl rquot1gtv r l n k w k 1frlt1 ButvZr and hmvZr l r ngtoo H H does not eX1st 1f lrlzl 7T 39 7 7 7 4T Letrzenze Ellwe e wThenlrlze smcelejwlzl k Clearly for T gt0 and Oquot gt 0 6 lt1 and 2amp7 converges to 1 67 k0 Therefore 4120 Flm Reso gt0 and Fsl0T tequot dt 1 8775 20 The Inverse Laplace Transform By de nition 1 L 1ijse ds j27l39 3971 13900 where S is a complex variable Now IfFlts Lfltrfr The above integral can be solved using contour integration and the theory of residues For our purposes however we shall use the theory of residues without going into the theoretical details In the analysis of linear circuits we will deal with Laplace transforms that are rational functions of polynomials that is Ds bnsquot bnils391 1 b1sb0 mltn NS ams39quot amilsm 1alsa0 m S n is necessary for physical realizability purposes Suppose now that N s and D s have distinct roots then kszlszzsz F s 2 m R s gtmaX p lSlSn splsp2mspn where s ZI and s p ilm 1n are the roots ofthe polynomials Ns and D s respectively Let m lt n then expanding F s into its partial fractions yields Fs k1 k2 kquot zquot kj sp1 sp2 spn j1spj Now fr 1Fs i quotj 9 j1SPj j1 SPj 21 Remark The constants k j l n are nothing more than the residues of the contour integral 1 F se ds j27l39 Consider the case when D s has real and distinct roots only For example let 96s5sl2 F 2 0 s ss8s6 Ms thenFs k2 k3 s s8 s6 kzs k3s s8 s6 Lets gt0then limsFs k1 lim 96s5s12 120 HO HO ss8s6 Likewise s8Fs Mkz k3 s8 S s6 and lim s 8Fs 2 k2 limw 5778 H 8 ss 6 Finally 31316 s 6F s k3 31316 W 2 72 48 120 72 48 gtF s s s8 s6 1 1 120 72 48 F ft L L l s s8s6 71 120 71 72 1 48 lLl Fl s s 8 s 6 l20u t 72e 8 u t 486 t It has been assumed in this example that the ROC is Res gt 0 In general if Ds has real and and distinct roots only then C lim s piFs i ln S Px Suppose that Ds has compleX roots Let p a jb be a root of Ds then pig 2 a jb is also a root of Ds 22 100s3 s6 s26s25 Fszwziand s6s3 j4s3j4 s6 s3 j4 s3j4 Example Let Fs gt 3 then 100s3 e ggs6Fs 12 I I I 100s3 50 75313 kzl 34F 1 101 2 ng4s J s eigj4s6s3j4 3J394 e k3zsli13j4s34Fs hm 210615313 H7314 s 6s 3 1394 3 1394 or k3 k if Res gt 3 then L71 is equal to ft 12676 u t 106715113 6707110940 10615113 3th I But 106715313 3714 loej53l3 673j4t 210673 e1394z5313 ej4zes313 2 2063 cos4t 5313 Therefore ft 1266 20e 3 cos4t 5313 u t We should notice from the above result that k3 k and that L71 is equal to Zlkle w 005 t 9ut where lk and 6tanlg 23 Consider now the case when D s has repeated roots 10s2 Let Fs 2 2 then s5s10 Fs k1 k2 k3 2 and s5 s10 s10 10s2 6 k1 231315s5Fsi1315s102 5 Now s102FsWkzs10k3 ltSIOgtZFltSgt10 i1 2k12 s 0 ii m gammak1s10Wk2 d 2 s5 s 2 6 3 k2 ili log sm FSXE 010W1 and ft Lquot Fs e 5 utge 10 ut16te 10 ut Fs k1 L2 k r krtl kquot 5 sp sp SPr1 51quot 1 if r andki 11m 39 Hsp 11 739 24 Initial Value Theorem If f t contains no impulses at the origin 1 0 then grgflliggsFs FsgtLftgt Example Let ftlOcos5tutthen linilOcosSt 10 10s Now Fs s2 25 sgt0 2 limsFsli1n 10s lin1 10 10 500 2 500 s 25 1225 Zeros and Poles of F s Suppose mg zwmgn 193 sP1sp2spn As s ZI i lm Fs 0 andthe ZI s are calledthe zeros ofFs Also if m lt n then as the magnitude of 3 becomes very large k IF ml 5 W s and Fs is said to have n m zeros at in nity since llim 0 n In times Sgtoo On the other hand as s pi i l n Fs 0 and the pi s are called the poles of 8s5 ss2 Example Let Fs Then Fs has a zero at s 5 and a zero at in nity since 8 large Isl Fs 2H was 0 It also has two poles at s 0 and s 2 25 90w 0 He AM 3 4 3 4 1 4 1 63235 Final Value Theorem If the function SF s has all its poles strictly on the lefthalf plane LHP then the function 1 ft L Fs has a nite limit as t gt oo ie liggftlii13sFs Fs Lft V Examgle Let Fs Then SFS W which has poles at s 6 8 E LHP 10 Thus 1imftlimsFslimW 0 aw sgt0 sgt0 56s8 r 2 t I Let 1 t7 7 then with respect to the innermost integral cm dt and Fs f2 re W dld1 But 100 for 1 lt 0 Hence Fs ie f21e dldr 2e d2 f21e dr F1sF2s 27


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