hapter 9 omplete Response of rcuts wth Two Storage Elements In hapter 8, we had rreducble storage element and a frst order crcut. In hapter 9, we wll hae rreducble storage elements and therefore, a second order crcut. That s, these crcuts wll ether be R, R, or R crcuts. Smlar to the complete response of frst order crcuts, the complete response of nd order crcuts also hae natural and forcng responses: x(t) = x natural + x forcng Interpretaton of the Forced Response The force response of second order crcuts behae dentcal to the force response of frst order crcuts. Howeer, as we wll see, derng the complete response for second order crcuts s more challengng. To reduce ths complexty, we wll only consder constant sources n ths chapter. Therefore, the force response wll always be a constant ( B) snce the effecte source term s always constant. Interpretaton of the Natural Response An amazng aspect of second order crcuts s they behae exactly as harmonc oscllators (mass-sprng system) wth dampng. Why? The mathematcal equatons that goern both harmonc oscllators and second order crcuts are exactly the same and so, hae the same complete response form and are broken down nto three categores:. Oerdamped lots of dampng. rtcally damped crtcal pont between oerdamped and underdamped 3. Underdamped lttle dampng The problem solng strateges for the complete response of second order crcuts are ery noled, and follow the same oerall strateges as for frst order crcuts. Problem Solng Strateges Step : Determne the second order equaton Step : Determne the general complete response (x n + x f ) Step 3: Determne the ntal condtons at t + (there are three of them) Step 4: Determne the constants A and A and put nto soluton form. Step 5: Sole the crcut problem DERIVING SEOND ORDER IRUIT EQUATIONS There are two methods n derng the second order crcut equaton; howeer, they both hae common key steps.. One uses KV or K or both to generate two frst order equatons that contan the four arables,, &. For frst order crcuts we wrote () KV or K to generate ONE frst order equaton that contaned two arables, ether (,,) or (, ). Then we () generated the frst order dfferental equaton by usng the capactor or nductor equatons. For second order equatons, we wll perform ths process two tmes snce there wll be two storage elements (R, R, R) n the crcut smultaneously. In general, we wll use () KV and K to wrte TWO frst order equatons that contan four possble arables (,, & ). From these two equatons, we () generated the second order dfferental equaton by usng the capactor or nductor equatons.. How these two equatons generate a second order equaton depends on the approach taken: () Drect substtuton method (Method ) reples on usng substtutng one derate term nto another derate term, & uses algebra manpulatons (not recommended). () The Operator method (Method ) uses an 9-
operator approach and matrx calculatons (hghly recommended). The net result s that a second order dfferental equaton wll hae the form of d x dx a bx f(t) The fact that there are now two frst order equatons really adds to the complexty n derng the second order equaton, and makes t challengng to learn ths process. So untl one understands the process, be prepared to struggle untl you get t. Problem Solng Strateges for Step Method : Drect Substtuton Method Step.: Defne the arables x and x Defne arables x and x where x s what we are solng for. x, x OR x, x solng for solng for Step.: Dere two frst order equatons usng Krchhoff's aws Apply K and KV to dere two equatons that contans,, &. Apply K to relate the currents and Apply KV to relate the oltages and Rewrte all resstor currents and oltages n terms of,, &. Step.3: Rewrte ( & ) n terms of ( & ) and defne equatons () and (). Use the capactor and nductor equatons to rewrte n terms of and nto : d d and Identfy equatons () and () The equaton that contans dx / s defned as equaton (). The dentfcaton of equaton () s mportant later on when solng for the ntal condtons n Step 3. Step.4: Dere the second order equaton a substtuton Substtute equaton () nto equaton () to get the second order equaton Step.5: heck your equaton and determne the force response B Only constant sources wll be consdered n ths chapter. To sole for the forced response, substtute n x f B nto the second order equaton and sole for B: d x f f a dx a db xf k a db ab k B f k/a f B Example Determne the second order dfferental equaton usng the drect method for the current of the nductor. Step.: Defne the arables x and x Solng for, so defne x & x Step.: Dere two frst order equatons usng Krchhoff's aws Apply K to relate the currents and Snce the nductor and capactor are n seres, they hae the same current: = Apply KV to relate the oltages and The oltage across the Ω s parallel to both the nductor and capactor. Usng KV on the rght loop ges Ω = + Rewrte all resstor currents and oltages n terms of and. The Ω resstor oltage Ω has to be rewrtten n terms of,, &. Howeer, the preference s always to rewrte t n terms of and. Applyng K to the top node ges ( S ) 9-
ollectng all terms together, and ( ) S Hnt: If the numercal alues of R, and are standard numbers, wat untl the end to substtute them n snce carryng these numbers throughout a calculaton s ery cumbersome. Step.3: Rewrte ( & ) n terms of ( & ) and dentfy equatons () and (). Use the capactor and nductor equatons to rewrte n terms of and nto : d d d and (S ) d Identfy equatons () and () where the equaton that contans d / s (): d d () S () Step.4: Use substtutons to determne the second order equaton & put n proper form Sole for n equaton () d sole for d S S and substtute t nto () (we elmnate so that only -terms reman): d d d S d S d d d S Puttng the second order equaton nto ts proper, we get d S d d Step.5: heck your equaton and determne the force response B Snce we are only dealng wth constant sources, ( ) = constant. For t =, the capactor s fully charged and klls off the nductor s current ( ( ) = ). Does the second order equaton agree wth ths? d ( ) d ( ) ds ( ) ( ) B So the second order equaton agrees wth the fnal steady alue, as expected. Example 9. Determne the dfferental equaton for the capactor s oltage (t) usng the drect method. Soluton Step.: Defne x and x Solng for, so defne x as : x & x Step.: Dere two frst order equatons usng Krchhoff's aws Apply K to the top node and KV around the outsde loop: K: and KV: S The Ω s n seres wth nductor ( Ω = ) whle the capactor s parallel to the Ω ( Ω = ): and S Step.3: Rewrte ( & ) n terms of ( & ) and dentfy equatons () & (). 9-3
Rewrte and by applyng the capactor and nductor equatons (snce the alues of and are not nce ones, wat tll the end to substtute them n): d d and S The equaton that contans d / s equaton (): d d () () S Step.4: Use substtutons to fnd the second order equaton Sole for n () and substtute t nto (): d S d d d d d ( ) Reorganze () nto proper form and substtutng n the numercal alues for and : d ( ) d S =mh =μf d 3. d 8 8 Step.5: heck your equaton and determne the force response B The capactor s oltage fnal steady state alue s ( ) = constant. At t =, use VDR to determne ( ) = Ω ( ): ( ) ( ) S S. Does the second order equaton agree wth ths? d ( ) d ( ) 8 8 3. ( ) S ( ) S B. So the second order equaton agrees wth the fnal steady alue, as expected. Method : ramer s Operator Method The operator method takes the two equatons (KV and K) and conerts them nto two smultaneous lnear algebrac equatons. By solng ths system of lnear equatons usng matrx technques, one generates a second order dfferental equaton. Although the drect and operator methods are ery smlar, the operator method s both more ersatle and powerful. Ths should be your method of choce when derng a second order equaton. Furthermore, there are certan crcuts that usng the drect method s consderably much harder to use; on the other hand, the dffcultly of the operator method stays the same. The operator method conerts dfferentals and ntegrals nto algebrac expresson (ery smlar to aplace operators): d s s s Wth these defntons, we can rewrte the capactor and nductor currents and oltages as d d s s s s Step.: Defne arables x and x (Same as Drect Substtuton Method) S 9-4
Step.: Dere two frst order equatons usng Krchhoff's aws (Same as Drect Substtuton Method) Step.3: Rewrte ( & ) n terms of ( & ) and defne equatons () and (). Use the capactor and nductor equatons to rewrte n terms of and nto : s and s Identfy equatons () and (). Equaton () s the one wth the ŝx term. Step.4: Dere the second order equaton a ramer s Rule Wrte the KV & K equatons as system of equatons nto matrx form ax ax c a a x c ax ax c a a x c coeffcent matrx where the a-matrx s called the coeffcent matrx. Use ramer s rule to sole for x. That s, replace the x -terms n the coeffcent matrx wth the source terms (c & c ): a c a a Dx a c a c where D a a a a a c a a determnant determnant of coeffcents where the coeffcent matrx D s a rank two operator equaton. That s, there are no operator hgher than a ŝ. One can also easly sole for x as shown on the rght. If one desres, to conert back nto dfferental forms replace ŝ d/. Important note: the determnant of coeffcents s dentcal for both x and x matrx equatons. What ths tells us s that the nductor and capactor wll both hae the same natural response. Step.5: heck your equaton and determne the force response B Set x f ( ) B, substtute nto second order equaton and sole for the constant B. Example Use the operator method to determne the second order dfferental equaton for the nductor current. Soluton Step.: Defne x and x ; solng for. x & x Step.: Dere two frst order equatons usng Krchhoff's aws Apply K to the top node and we see = : K: R S and KV: Snce all elements are parallel so defne the oltage as and rewrte Ω = : and S Step.3: Rewrte ( & ) n terms of ( & ) and dentfy equatons () & (). s s and s S s eq () Step.4: Dere the second order equaton a ramer s Rule Set up the two equatons nto matrx form ( s) S s S s s coeffcent matrx c Dx c a a 9-5
Use ramer s rule to sole for s D ( s) S S S where the operator matrx D s s D s( s) ( s s ) s ombnng both expresses together, we get D s s S S Puttng ths nto proper form and conertng back to derate terms ( s d/ ) d d s s S S Step.5: heck your equaton and determne the force response B Set ( ) B, substtute nto second order equaton and sole for the constant B. d ( ) d ( ) ( ) S ( ) S B For t =, the nductor shorts eerythng out so ( ) = S and ( ) =. Ths s what as expected. I also could hae easly dered the equaton for the capactor. From the matrx equaton, I get a dfferent rght-hand sde: s S S D ss s(constant) s s Snce the rght-hand sde s zero, the second order equaton for the capactor oltage s d s s For the capactor equaton, set ( ) B, substtute nto second order equaton and sole for the constant B. d ( ) d ( ) ( ) ( ) B Ths s what as expected. Example 9. Determne the second order equaton for the capactor s oltage usng the operator method. Soluton et s streamlnng ths process by combnng the frst 3 steps together. Steps.,., &.3: We are solng for and hae x, x The Ω resstor and the capactor are n seres ( = Ω ); the oltage across the Ω resstor s S. Applyng Krchhoff s laws ges K: ( ) s s KV: s s rewrte S s S resstor terms s d Step.4: Dere the second order equaton a ramer s Rule Set up the two equatons nto matrx form 9-6
( s) s ( s) s S S s ( s) s ( s) Use ramer s rule to sole for coeffcent matrx ( s) S D ( s) s S S constant ŝ where the operator matrx D s ( s) s 3 D ( s) ( s) s( s) s s s (s) ombnng both expressons together, we get D s s Puttng ths nto proper form and conertng back to derate terms ( s d/ ) 4 d 4 d s 3 s 3 3 3 Step.5: heck your equaton and determne the force response B Set ( ) B, substtute nto second order equaton and sole for the constant B. d ( ) 4 d ( ) 3 3 ( ) ( ) B At t =, the nductor shorts the Ω resstor and capactor: Ω ( ) = ( ) = B. Ths s what as expected What about the equaton for the nductor current? From the matrx equaton, I only get a dfferent rght-hand sde: ( s) s S S s D S ss S s ( s) ( s) Snce the rght-hand sde s S, the second order equaton for the nductor current s ŝ 4 3 ŝ 3 3 S S 3 ( ) ( ) B The second order equaton for the nductor also checks out. STEP : Determnng the omplete Response In the oerall problem solng strateges, we are currently on Step. Step : Determne the second order equaton Step : Determne the complete response (x n + x f ) Step 3: Determne the ntal condtons at t + (there are three of them) Step 4: Determne the constants A and A and put nto soluton form. Step 5: Sole the crcut problem Now that you folks are experts at derng second order equatons, the complete response has the form x(t) = x natural + x forcng. As we saw n hapter 8, the natural response s how the capactor s or nductor s responded when there are no sources n the crcut. On the other hand, the forcng response s when there s a source present n the crcut. d xn dxn no sources: a axn d x dx a ax f(t) d xf dxf sources: a axf f(t) 9-7
Howeer, I already stated that only constant sources wll be consdered n ths chapter. So, we already know the force response from Step : x f ( ) = B. The complete response now looks lke x(t) x x A x A x B n f n n All that s left s to get the form of the natural response and the constants A and A. Physcal Interpretaton of the Natural Response of Two-storage element crcuts The key to understandng the natural response (or behaor) les n understandng the characterstc equaton and ts roots. Ths nformaton s contaned n the coeffcent matrx D from Step. Analogy: A undamped/damped harmonc oscllator has the same characterstc natural response behaor as an /R crcut. If an mpulse (but not a contnuous force) s appled to a harmonc oscllator, ts natural response s d xn d xn F kxn ma m xn x Acos( t ) n Wth no energy loss due to frcton, the oscllator wll oscllate foreer wth natural frequency of ω = k/m. Howeer, as soon as frcton (or dampng) s ntroduced, the ampltude x(t) decreases to zero but the manner n how t des off s determne the dampng factor b (frcton alue) and the mass m of the oscllator. There are two obous cases: () t damps out wth no oscllates (t turns out there are two types of ths) or () contnues to oscllate wth ts ampltude decreasng to zero. Snce frcton s usually elocty dependent, the natural response (there are forcng terms stll) of the oscllator now behaes as dxn d xn d xn b dxn F kxn b ma m xn m n bt/m x (t) Ae cos( t ) The type of responses of the oscllator depends on the rato b/m. large b/m-alue natural response s oerdamped small b/m -alue natural response s underdamped moderate b/m -alue natural response s called crtcally damped Applet http://www.lon-capa.org/~mmp/applst/damped/d.htm rtcal frequency dampng constant s 6.6 kg/s R rcut Oscllatons. For an crcut (R = ) the capactor and nductor wll oscllate ts electrc and magnetc energy between themseles, smlar to an oscllator wth no dampng. Applet http://www.walter-fen.de/ph4e/osccrc.htm. For an R crcut, the resste element dsspates energy away from the system, and consequently, the energy s damped out ery smlar to the oscllator wth dampng. That s, dependng on the geometrcal confguraton of the R, the energy (or oltage/current) wll ether be oerdamped s when the energy s mmedately dsspated away (smlar to chapter 8). underdamped s when the energy s dsspated slowly enough so that some oscllatons between U E and U B occur before decayng to zero. 9-8
crtcally damped s the pont that separates the oerdamped and underdamped natural responses, howeer, goes to zero more quckly than oerdamped. To dstngush between the types of dampng n the R crcut, we must sole the characterstc equaton (contaned n the coeffcent matrx) to get the characterstc roots. The characterstc roots contan the nformaton necessary to determne the type of natural response the crcut wll hae. That s, f there are no sources n the crcut, all of the c-terms are zero: ax ax a a Dx x ax ax a a x Dx coeffcent matrx Startng wth the proper form of the second order equaton, we defne two new constants: Naper frequency s s xn Natural frequency We assume a natural response of the form x n = Ae st (same form as n chapter 8) where s s a constant to be determned. Substtutng ths nto the second order equaton, we get a quadratc equaton that s called the characterstc equaton: st st s s Ae s s Ae s s operator equaton s s the alue to be determned haracterstc equaton The solutons (there must be two of them) to the characterstc equaton are called the characterstc roots (s, s ) tells us the form of the natural response: s s s x A e A e, haracterstc Roots n s t s t General Natural Response Therefore, the form of the alues s and s wll determne the type of natural response, and t s the square root term that clearly ddes the nto ts three categores: > o, o, or < o. ase : If o, the characterstc roots are dentcal (s s ), and the crcut s natural response behaes crtcally damped: s t s x crtcal(t) e (At A ) o ase : If > o, the characterstc roots are real and dstnct (s s ), and the crcut s natural response behaes oerdamped: s ; s o st st oerdamped x (t) A e A e ase 3: If < o, the characterstc roots are complex and dstnct (s s ), and the crcut s natural response behaes underdamped: 9-9
s j ; s j x (t) e (A cos t A sn t) t underdamped d d where j. where d Problem Solng Strateges for Step Step.: Determne the characterstc roots Usng the coeffcent matrx, wrte down the characterstc equaton, defne &, and sole for the characterstc roots: s s s,s characterstc equaton characterstc roots Step.: Defne the type of natural response: t : s s x (t) e (A t A ) crtcal st st : s,s x oerdamped(t) Ae Ae : s,s j ; x (t) e (A t A t) t underdamped cos d sn d where Step.3: Wrte out the general complete response st st x(t) x x A e A e B n f Example 9.3 Assume (-) =, (-) = and that all sources turn on at t =. d a. Determne the general response of the nductor current and plot ts behaor, gen b. Determne the general response of the nductor current and plot ts behaor, gen c. Determne the general response of the capactor oltage and plot ts behaor, gen Soluton a. From the second order equaton, determne the force response B. s B sb B s B Ths makes physcal sense snce the open capactor klls of the nductor current. Step.: Determne the characterstc roots S 9-
Startng from the second order equaton, we pck-off the characterstc equaton and determne α and ω : s s s n sn n st n Ae s s 4, ; underdamped In the underdamped case, we need to calculate the addtonal undamped frequency ω d :.968 d 4 Step.: The natural response accordng to the and s t/4 n(t) e (Acos.968t Asn.968t) Step.3: Wrte out the complete response to the nductor current and plot ts behaor. (t) n f t/4 e (Acos.968t Asn.968t) The soluton agrees wth our ntal predcton: the nductor des off after 5τ =.5 s snce the capactor opens up. b. From the second order equaton, determne the force response B. s B 7sB B B In ths crcut, the nductor current ( ) = 6Ω ( ), usng DR, we fnd that 6 ( ) 6( ) 5A A 4 6 Ths agrees wth the second order equaton. Step.: Determne the characterstc roots Startng from the second order equaton, we pck-off the characterstc equaton and determne α and ω : s 7s 7 s n 7sn n st n Ae s s 7, ; oerdamped Step.3: The natural response accordng to the and s st st n(t) Ae Ae In the oerdamped case, the actual roots need to be calculated. There are two ways to do ths: Method : s,s 7 49 4 ; 5 Method : s 7s (s )(s 5) s,s ; 5 "eyeng the solutons" So, the general natural response s t 5t n(t) Ae Ae Step.3: Wrte out the complete response to the nductor current and plot ts behaor. t n f 5t (t) A e A e A c. From the second order equaton, determne the force response B. 9-
s B 4sB 4B B Wth no supplyng source, both the capactor and nductor decay to zero. Ths agrees wth the second order equaton. Step.: Determne the characterstc roots Startng from the second order equaton, we pck-off the characterstc equaton and determne α and ω : s 4s 4 4 s n 4sn 4n st n Ae s s 4, Step.3: The natural response accordng to the and s (t) n f t e (At A ) ; crtcal STEP 3: Determne the Intal ondtons x( ) and dx( )/ We are now on the thrd step of the problem solng strateges: Step : Determne the nd order equaton Step : Determne the complete response Step 3: Determne the ntal condtons at t + (there are three of them) Step 4: Determne the constants A and A and put nto soluton form. Step 5: Sole the crcut problem The complete general response has two unknown constants (A, A ) that need to be soled for st st x(t) xn x f Ae Ae B As we saw wth frst order crcuts, the Is are appled to the complete general response to force the constants A and A to match the condtons at t =. The mnmum number of Is s three: ( ), ( ) and ŝ ( ) or ŝ ( ). In specal stuatons, all four Is wll be requred. Solng Strateges for the Intal ondtons (Is) Step 3.: Determne the numercal alues of ( ) and ( ) for the crcut Step 3.: Determne the numercal alue of ŝ x ( ) usng the two frst order equatons n Step. In general, equaton () s where we get the expresson for ŝ x ( ). Example 9.4 a. Determne the two frst order equatons for (t) for t >. b. Determne the Is ( ), ( ) and ŝ ( ) Soluton et s only do Steps and 3: dere the two frst order equatons and determne the Is. Step : Dere the two frst order equatons Defne x, x and dere the two frst order equatons: 8 s 8 5 5 K: s () KV: s s () ; 6 6 6 s 8 Step 3: Determne the Is ( ), ( ) and ŝ ( ) At t =, the nductor shorts out the capactor s oltage and the nductor s current s equalent to the current source alue: 9-
Usng equaton () from Step, ( ) ( ) and ( ) ( ) 5 A ( ) s ( ) ( ) s ( ) () 6 8 we see that t contans ŝ but does NOT depend only on ( ) and ( ); t also depends on ŝ. When ths happens, t s best to thnk of these as a system of equatons and sole for ŝ ( ). If we multplyng () by 6 and add ths to (), we solate ŝ ( ): 6 6 s 8 () 6 s s () 8 8 s s ( ) ( ) 4 ( ) solng for ŝ ( ) A/s s ( ) ( ) ( ) 5/A STEP 4: DETERMINE THE ONSTANTS A & A AND PUT INTO SOUTION FORM Now that we hae the Is from Step 3, all that s left to do s apply them to the complete general response n order to sole for A and A. Howeer, these two constants requre two equatons that are setup wth x ( ) and ŝ x ( ). Snce there are three types of dampng responses, there wll be three dfferent setups for x ( ) and ŝ x ( ): t. rtcal dampng: x (t) e (A t A ) B crtcal t x crtcal( ) e (At A ) B A t= B x crtcal( ) A d t ŝx ŝx ( ) A A crtcal( ) e (At A ) B A t A s t oer. Oerdampng: x (t) A e A e B s t crtcal st st x oer( ) Ae Ae B A A B t= x oer( ) A A B d st st ŝx ŝx ( ) s A s A oer( ) Ae Ae sa sa t 3. Underdampng: x (t) e (A cos t A sn t) B t under d d oer t x under ( ) e (Acos Asn ) B A t= B x under ( ) A B d t ŝx ŝx ( ) A A under ( ) e (Acos Asn ) B A t da I wll now do three crcuts wth each of a dfferent knd of dampng response. ontnuaton of Example 9.4 a. Determne the complete response of (t) for t >. b. Plot and nterpret (t) and (t) s. t. Soluton Step : Determne the second order equaton Defne x, x and determne the two frst order equatons: 8 s 8 5 5 K: s () KV: s s () ; 6 3 6 s 4 under d 9-3
Wrtng these nto matrx form s 5 8 3 s 4 s and now solng the two determnants: ŝ 8 6 D s 8 6 s and 3 5 5 5 4 8 s 8 D s s s s s s s The second order equaton s D s 4s 4 Solng for the B constant, we fnd that s 4s 4 B B A 5 Ths agrees wth ( ) from the crcut. 3 4 4 4 Step : Determne the complete general response: From the characterstc equaton, we determne the characterstc roots and the type of dampng: s 4s 4 (s )(s ) s s (crtcal dampng) The complete general response s crtcally damped: t (t) e (At A ) Step 3: Determne the Is ( ), ( ) and ŝ ( ) Snce we already dd ths preously, here are the alues agan: ( ), ( ) A, and s ( ) A/s 5 Step 4: Determne the constants A and A Applyng the Is to determne A and A, we get 5 5 ( ) A A d t ŝ ( ) 5 e At A t A The complete crtcally damped response s (t) te t 5 To determne the mnmum alue, we fnd the crtcal pont by settng the derate to zero and solng for tme t: d t t t 5 ŝ (t) te e te t.5s So the mnmum occurs at.5s, and has a alue of.5 5 (.5s).5 e.66 A The plot of s. t s on the rght. Step 5: Sole the crcut problem To calculate the nductor s oltage, we smply use the nductor equaton: d t 5 t s te e (t ) (t) 5 9-4
Interpret the nductor s oltage response What do we know about the behaor of the nductor s oltage?. The nductor s oltage s always dscontnuous snce t always starts off as a short, ( ) =, and jumps to ts nonzero alue at ( ): ( ) e ( ) V. The nductor s oltage has a maxmum/mnmum when ts slope s zero. To calculate ths max/mn, we determne the crtcal pont by settng ts derate to zero and solng for t: d t t ŝ e (t ) 4 e (t ) e t t.s Substtutng n ths crtcal pont of t =. s back nto the nductor s oltage response, ts maxmum alue s (t.s) e ().7V 3. Another addtonal alue s that we can easly calculate when the nductor s oltage s zero. By lookng when the (t ) =, we fnd the tme t =.5 s: (t s) e (.5 ) V Plottng the nductor s oltage (wth the help of Maple), we see what the nductor s response looks lke. There s a clear dscontnuty jump from V at t = to V at t =, and a maxmum nductor s oltage at t =. s. Example 9.5 a. Determne the complete response of 8 (t) for t >, gen ( ) = 5V. b. Plot and nterpret 8 (t). Soluton What s the easest what to sole for 8 (t), solng for or? I clam snce K mples the 8Ω = S ; otherwse, usng requres a complcated KV wth seeral terms. Step : Determne the second order equaton Defne x, x and fnd two equatons: K: s () KV: 8( ) 4 s () 8 4 Wrtng ths nto matrx form s s D s 8 8 Determnng the determnant of the operator coeffcent matrx: 9-5
s 3 D s s s s 5 s So the second order equaton s 3 proper D s s s 6s 5 form Solng for the B constant, we fnd that s 6s B B Ths makes physcal sense because the capactor opens up and klls of the nductor s current at t =. Step : Determne the complete general response The characterstc equaton has the followng roots: 6; and d 9 s 6s s s Underdamped The general underdamped response s 3t (t) e (Acost Asnt) Step 3: Determne the ntal condtons ( ), ( ) and ŝ ( ) It s gen that at t =, ( ) = 5V and ( ) =, snce the charged capactor was nserted nto the crcut wth opposte polarty to the Asource. Howeer, after the A-source turns on, the nductor s oltage has a dscontnuty gen by equaton (): ŝ ( ) 8 ( ) 3 ( ) 4 ( 5V) A/s s ( ) Step 4: Determne constants A and A Applyng the ntal condtons to get the constants A and A, we get ( ) e (A ) A 3 ŝ ( ) A A A So now the complete underdamped response s 3 3t (t) e snt A Step 5: Sole the crcut problem Usng K to determne the 8 resstor s current, t s oltage s 3t 8 8 8 5e sn t V Interpret the 8 resstor oltage response What do we know about the behaor of the 8 resstor oltage?. At t =, the oltage across the 8 resstor 8 ( ) s 8( ) ( ) 5V At 8 ( ), there s a dscontnuty snce there s a sudden burst of current from the A-source: 3t 8( ) 8 5e sn V 8V ( ) 5V ( ) 8V ( ) 8V It s satsfed as expected. 8 8 8 (t) 8 5e sn t V (t.34s).75v 3t 8 8 3 9-6
. The capactor s oltage has a maxmum/mnmum when ts slope s zero. We determne the crtcal pont by settng ts derate to zero and solng for t: d 3t ŝ 8 5e sn t 3t 3t 66e sn t 5e cos 5 66 t tan t.34s Substtutng n ths crtcal pont of t =..34 s back nto the capactor s oltage response, ts alue s 3.34t (t.34s) 8 5e sn.34.75v The capactor s oltage plot (wth the help of Maple) s shown on the rght. Example 9.6 a. Determne the complete response of (t) for t >. b. Plot and nterpret (t). Soluton Step : Determne the second order equaton The crcut at t = has no 5V-source and the capactor s oltage does not change polarty. Defne x, x and dere two frst order equatons: K: s () 8 KV: 4 4s () 4 Wrtng ths nto matrx form s 8 D V 4( s) 4( s) Determnng the determnant of the operator coeffcent matrx: s 8 5 D 4( s) s 8 s s 3 4( s) So the second order equaton s 5 proper D s s 3 s 5s 6 form Solng for the B constant, we fnd that s 5s 6 B B V 3 Ths agrees wth ( ) accordng VDR: ( ) B V 3 V 4 Step : Determne the complete general response From the characterstc equaton, the characterstc roots are s 5s 6 (s )(s 3) s,s, 3 (Oerdamped) The complete general oerdamped response s real and dstnct t 3t (t) A e A e Step 3: Determne the Is ( ), ( ) and ŝ ( ) 3 9-7
At t =, we see that the capactor s parallel to the 5V-source and the nductor s n seres wth the 4Ω-resstor. Howeer, note that the nductor s current n opposte to the 5V-source. Applyng KV or Ohm s law to get the current through the 4Ω, we hae 5V 5 ( ) 5V; 4( ) ( ) 4 A 4 Use equaton () from Step to get ŝ ( ) : 8 s s ( ) 4 ( ) 8 ( ) 3 V/s ( ) 5 ( ) 5/4 Step 4: Determne the constants A and A Applyng the Is on the complete general oerdamped response, the constants A and A are ( ) 5 A A A 3 Settng up a 5/3 8 A matrx equaton 5; A 3 3 A 3 ŝ ( ) 3 A 3A The complete oerdamped response s then gen by t 8 3t 3 3 (t) 5e e V Interpret the capactor s oltage response What do we know about the behaor of the capactor s oltage?. The capactor s oltage should be contnuous before and after the swtch s thrown: 8 ( ) V ( ) 5e e 3 3 V 3 3 It s satsfed as expected.. The capactor s oltage has a mnmum when ts slope s zero. So to calculate, we determne the crtcal pont by settng ts derate to zero and solng for t: d t 8 3t t 3t ŝ 5e e 3 3 5e 8e t e 8 t.47s 5 Substtutng n ths crtcal pont of t =.47 s back nto the capactor s oltage response, ts alue s 8 (t.47s) 5e e.47 3.47 3 3.78V 9-8