CHAPTER 8 COMPLETE RESPONSE OF RC & RL CIRCUITS

Transcription

1 HAPTER 8 OMPETE RESPONSE OF R & R IRUITS The order of the crcut s defned by the number of rreducble storage elements n the crcut. In ths chapter, we wll fnd the complete response of a frst-order crcut; a frstorder crcut means there s only one storage element n the crcut, ether an R or R crcut. A second-order crcut means there are two storage elements n the crcut; for example, R,, or. Ths s the subject of hapter 9. There are two types of sources that wll be appled to capactors or nductors n ths chapter: onstant sources frst-order response to a constant nput source (easer) Tme-dependent sources frst-order response to a tme ared nput source (harder) Response of a Frst-order rcut to a onstant Input Regardless of how complcated the geometry of an R or R crcut s, we wll always conert them nto Théenn and Norton equalents: R crcut conert nto a Théenn Equalent crcut R crcut conert nto a Norton Equalent crcut Why ths approach? It turns out that ths approach s more physcally ntute and easer to see the answer than other approaches that are more mathematcal. R IRUIT A typcal R crcut wll not be n Théenn form. So we wll conert any R crcut nto a Théenn crcut by remong the load capactor and usng the standard trcks of the trade. Pcturewse, we hae Our prmary focus s analyzng the behaor of the capactor s oltage after a swtch has been thrown. That s, we are most nterested n determnng ts behaor (0 < t < 5) for t = 0. Physcally, ths means the capactor wll ether charge up or dscharge from ts ntal to ts fnal steady state alue and (t) wll descrbe ths behaor completely. The sudden change n the crcut due to a swtch beng thrown, breaks the R behaor nto three dstnct tme perods. t < 0 (rcut #): Before the swtch s closed, the oltage source has charged the capactor to ts ntal steady state alue (0 ). 0 < t < 5 (rcut # & #3): The swtch s thrown at t = 0 and the crcut s conerted nto a Théenn equalent. (Ths requres two crcuts, () V O and () R Th). The capactor s oltage ether chargng or dschargng from ts ntal alue (0 ) and ths transent behaor s tme dependent and exponental: (0 < t < ) (t). t = : When suffcent tme has elapsed, the capactor has fully charged and has reached ts fnal steady state alue defned by ( ) = V Th = V O. Solng the crcut problem (rcut #4). If the capactor s oltage s not what s beng soled for, one has to unfold the crcut to sole for the approprate parameter of the problem. 8.

2 In summary, there are two steady state alues but t s what s n-between these two were the capactor s transent behaor s obsered and nterests us the most. When the capactor's oltage s completely known t s called the complete response: (t < 0) (0 ) Intal steady-state alue (t) omplete Response = (0 < t < 5 ) transent-state (t = ) V oc Fnal steady-state alue ookng ahead, we wll fnd that ths behaor s summarzed n the followng equaton: (t) = V ( (0 ) V )e oc t/rth oc omplete Response Forced Response Natural Response et's start analyzng an R crcut. Eery tme a swtch s thrown, a new crcut s created and the capactor naturally respondng to ts enronment (how the resstors are organzed) by ether chargng or dschargng to the sudden change. Oerall ew of solng R crcuts wth a constant source. BEFORE the swtch s thrown, determne the ntal steady state oltage (0 ).. AFTER the swtch s thrown, conert the R crcut nto a Théenn crcut. 3. Sole the st order equaton for the capactor s oltage (0 < t < ) (t). 4. Undo the Théenn crcut and sole the crcut problem and nterpret. Step : BEFORE the swtch s thrown, we sole the crcut for (0 ). VDR leads the way and ges R3 (0 ) = R R R 3 V S Step : AFTER the swtch s thrown, the R crcut s conerted nto a Théenn equalent usng standard technques. We remoe the load capactor and determne V O and R Th: The resultng Théenn equalent s Step 3: To fnd the oltage (t), apply KV to the Théenn equalent and sole for (t): Voc = R (t) Th = d / Rewrtng t, we fnd d d Voc Voc = RTh (t) = R R Th Th 8.

3 Ths frst-order response equaton s a form we wll encounter agan wth the R crcut, so t s useful to wrte ths n a general form so that we can use t agan. Defne the parameters x = d Voc dx x = RTh = = k RTh RTh k = Voc RTh where s the tme constant of the crcut and k s the effecte source term snce t s drectly proportonal to V O. To ntegrate, I must solate the x and t terms: dx (x k ) dx t = = exp[ n(x k )] = exp ntegraton constant (x k ) (x k ) = e e Ae t/ x(t) = k Ae t/ constant t/ Applyng the Intal onons (Is) to determne A and k To determne the complete response x(t), constants A and k must be ealuated wth the Intal onons (Is) establshed by the crcut. These Is are the ntal x(0 ) and fnal x( ) steady state alues of the crcut. Besdes derng the frst order equaton, the Is become one of the most mport steps n determnng the complete response. Why? It s the Is that "ft" the soluton to the partcular crcut you are solng and forces the soluton to be realstc and fnte. In other words, the general response has to match the ntal and fnal steady states alues determned by the crcut. In ths process, we must setup two equatons n order to sole all constants relate to x(0 ) and x( ). Of course, a Math 7 course cannot do ths because these course cannot be expected to teach ether physcs or engneerng whle teachng the fundamentals of problem solng. Before the swtch s thrown, the capactor s oltage has reached ts ntal steady-state alue of x(0 ). learly, the general response has to satsfy ths alue at t = 0 and we force ths I onto the general response: x(0 ) = k A A = x(0 ) k After the swtch s thrown, the capactor s oltage has eoled and reaches ts fnal steady-state alue x( ). Once agan, we force the general response to satsfy ths I at t = : x( ) k t/ x( ) = k Ae = = k A = x(0 ) x( ) So the applcaton of the Is on the general response places strong conons on ts shape: A = x(0 ) k A = x(0 ) x( ) k= x( ) k= x( ) ombnng all the terms together, the frst-order response takes on the form of x(t) = k Ae x(t) = x( ) [x(0 ) x( )]e t/ t/ A= x(0) x( ) k= x( ) complete response forced response natural response Now translatng ths to the complete response of the capactor s oltage ges d V R R oc = = (t) V [ (0 ) V ]e t/rth oc oc Th Th x(t) = (t) complete response forced response natural response 8.3

4 Physcal nterpretaton. Mathematcally, the complete response satsfes the Is: 0 (0 ) = Voc [ (0) Voc ] e = (0) ( ) = V oc [ (0) V oc ] e = Voc. hargng/dschargng of a capactor. What does t mean for a capactor to charge up? It means that (0 ) s less than ( ) = V O: V O > (0 ). Mathematcally, the complete response reads t/ t/ (t) = V oc [ (0 ) V oc ]e Voc Ve negate where V = (0 ) V oc That s, the mnus sgn ndcates that the capactor s chargng to a hgher oltage alue. Physcally (and n the words of Deane), when negate work s done on the capactor t ncreases the electrcal potental of the capactor snce the capactor moes up hgher on the electrcal nclne. What does t mean for a capactor to dscharge? It means that V O s less than (0 ): (0 ) > V O. Mathematcally, the complete response tells us (t) = V [ (0 ) V ]e V Ve t/ t/ oc oc oc poste That s, the plus sgn ndcates that the capactor s dschargng to a lower oltage alue. Physcally, when poste work s done t decreases the electrcal potental of the capactor and moes lower down the electrcal nclne. 3. Tme onstant Another way to nterpret the complete response and the role that the tme constant plays, s to rewrte the complete response as t/ t/ t/ t/ t/ (t) = V oc [ (0 ) V oc ]e = (0 )e Voc e Vntale Vfnal e These two terms show that t/ Vntale dscharges from ntal steady-state alue t/ Vfnal ( e ) charges to the fnal steady-state alue If we only focus on the exponental behaor accordng to the table below, ( ) ( ) the capactor has dscharged from ts ntal oltage (0 ) to essentally zero (0.7% of ts orgnal alue) after 5 tme constants (5). the capactor has charged to ts fnal steady state alue of V O after (5 ) V e = V e = V V V V ( ) ntal fnal ntal fnal fnal oc 8.4

5 The chargng/dschargng depends ONY on the crcut elements: R Th and. Problem Solng Strateges for Frst Order rcuts wth constant sources The four step process for solng for the capactor oltage s. Determne the capactor s ntal steady state oltage (0 ) before the swtch s thrown.. After the swtch s thrown, determne the Théenn parameters (V O & R Th) and the nerse tme constant /. 3. Put the oltage nto soluton form and nterpret the soluton: t/rth (t) = V oc [ (0) V oc]e 4. To sole the crcut problem for the approprate parameter, go the crcut before the R crcut was conerted nto a Théenn crcut and relate unknown parameter to (t) usng K or KV. et s apply ths procedure to the followng two examples. Example 8. The crcut s n steady state before the swtch closes at t = 0 s. a. Descrbe the behaor of the capactor for all tme? Is t chargng or dschargng after the swtch s closed? b. () Determne the capactor oltage (t) for t > 0 s. () Plot (t) s. t for - < t <. Soluton I wll follow the problem solng strateges from aboe. Step : Determne (0 ) = (0 ) To determne the ntal oltage (0 ) before the swtch s thrown, we replace the capactor wth ts steady state equalent and use whateer method s most conenent to get (0 ). Snce s crcut s completely open, the oltage across the capactor s the same as the 3V-source: (0 ) = 3V snce 3Ω = 6Ω = 0. Step : Determne the Théenn parameters (V O & R Th) and / The swtch s thrown and the R crcut must be conerted nto a Théenn equalent. There are two crcuts needed to fnd V O & I S: To determne V O, we see that 6Ω = V O use VDR to sole for t: 6 = 6Ω 3V V Voc 6 3 = = To determne R TH, snce there are only ndependent sources, we can short out the oltage source and sole for R TH: R = = 8Ω Determne the tme constant / Th 8.5

6 = RTh = 8 ( 0.05) = 0.4 s = =.5 Hz = 0.4 Step 3: Put the oltage (t) nto soluton form and nterpret the soluton t/ t/0.4.5t (t) = V [ (0 ) V ]e = 3 e = e V Puttng ths together, the capactor oltage behaes as oc oc ( ) 3V t < 0 (t) =.5t e V t 0 Interpretaton: the capactor s oltage starts off at 3V and dscharges to V. After 5, the capactor has almost arred at ts fnal oltage of 3V. The open crcut oltage acts lke the termnal elocty of an object fallng n ar: t does not change t alue agan. Does the capactor behae as t s suppose? That s, s (0 ) = (0 ) and (0 ) (0 )? 0 (0 ) = (0 ) 3V = e = 3V d (0 ) 0 (0 ) (0 ) 0 = 0.05(0.5e ) = 5 ma So when the swtch was thrown, the capactor s oltage doesn t change nstantly but the capactor s current does snce t jumped from 0A to 5 ma; there s a dscontnuty! Example 8. a. Descrbe the behaor of the capactor s oltage: () ntal and fnal steady states, () chargng or dschargng? Assume steady state at t = 0. b. () Determne the current x(t) for t > 0 s. () Plot x(t) s. t for - < t <. Soluton Does the capactor charge up or dscharge when the swtch s closed? I would clam that the capactor dscharges because before the swtch s thrown, both sources wll be contrbutng to the capactor s oltage. Howeer, when the swtch s closed, t cuts out the 4V-source and therefore, the capactor sees less source addng to t. Step : Determne (0 ) Wth the crcut at ntal steady state, the resultng crcut s shown on the rght. To sole for (0 ), I dentfy the nodes and note that there are two loops but one of the loops contans an solated 4mA-source. Ths leads me to choose Smple rcuts n solng for (0 ). Settng the ground at the bottom, I am gong to follow the current startng wth the 4mAsource. I see that the 4mA-source s n seres wth the rght most kω-resstor. Snce the 4V-source s connected to the ground, and I know the oltage across the kω-resstor, the node oltage for the poste termnal of (0 ) can be calculated: ( ) = 4V kω = 4 k 4m = V To fnd the negate termnal of (), I contnue to follow the current through the ground and see that at the bottom node, the current s splt eenly between the two branches snce both branches hae 4kΩ resstors. So ma goes up the kω, gng ( ) = k m = 4V 8.6

7 Therefore, (0 ) s (0 ) = ( ) ( ) = ( 4) = 6V = (0 ) Step : Fnd V O, R Th and / The swtch has now been thrown and the crcut now looks lke onertng ths R crcut nto a Théenn equalent, the V O- and R Th-crcuts are below. Determne V O Usng smlar thnkng as when we determne (0 ), I am gong to label the nodes and follow the current. The poste node V O() s grounded and the 4mA-source splts between the two 4kΩ resstors such that half of the current (ma) splt at the bottom node agan. The negate V O() termnal of the open crcut s equal to kω. From Ohm's law, the oltage kω s -4V and V O s V = V ( ) V ( ) = 0 ( 4V) = 4V = V oc oc oc oc Determne R Th Openng the current source, the Théenn resstance s Determne / The tme constant s R = 6 = kω= R Th Th 3 0 = = = Hz = R 3k 00µ 3 Step 3: Wrtng out the soluton (t), we hae Th ( ) t/ 0t/3 0t/3 (t) = V oc [ (0 ) V oc ]e = e = 4 e V Puttng ths together, the capactor oltage behaes as 6 V t < 0 (t) = 0t/3 4 e V t 0 Does ths agree wth our ntal assumptons whether the capactor was gong to ether dscharge or charge up? Because of the plus-sgn on the transent soluton, the capactor dscharges from 6V to 4V, as expected. Step 4: Solng the crcut problem for x(t), we need to determne () x(0 ) and () x(t). If we go back to the calculaton for (0 ), we fnd that x(0 ) was opposte n drecton to the current: x(0 ) = ma. To determne x(t), we to go the crcut before we conerted the R crcut nto a Théenn crcut and relate x(t) to (t). In dong so, we see that (t) s antparallel (opposte polarty) to x(t); x =. Usng Ohm s law to determne x(t), the current s ma t 0 x(t) = (t) 0t/3 = 6e ma t > 0 k 8.7

8 Does ths make physcal sense? If we redraw the crcut and focus on the capactor and the kω n queston, t becomes clear that they are opposte n polarty and the reason why the current x(t) s negate. So why s there a sudden dscontnuty jump n the kω resstor? Snce the capactor s beng forced to dscharge, the current x(t) s also beng force to go to a lower (more negate) alue as well. heckng on the capactor s current at t = 0, we fnd that d (0 ) 4 0 (0 ) = = 0 = 8mA 3 As expected, we expected a dscontnuty jump on the capactor s current from 0mA to 8mA. Furthermore, as the 4mA-source takes, the current reaches ts fnal steady state alue of x(5 = 0.75 s) = ma. R rcut Hang now studed the R crcut n detal, smlar type of thnkng occurs wth the R crcut and so of nstead of repeatng these steps, I wll refer to the R crcut for detals. An R crcut s conerted nto a Norton equalent to fnd the current through an nductor: Our prmary focus s analyzng the behaor of the nductor s current after a swtch has been thrown: (0 < t < 5). That s, we want to answer the queston of whether the nductor ether stores up current or decays t from some ntal steady state alue n ths parallel R crcut. Smlar to the R behaor, three dstnct tme perods. t < 0 (rcut #): Before the swtch s closed, the source stores up current n the nductor and establshes ts ntal steady state alue (0 ). 0 < t < 5 (rcut # & #3): The swtch s thrown at t = 0 and the crcut s conerted nto a Norton equalent. (Ths requres two crcuts, () I S and () R Th). Here, the nductor s current ether stores current or decays t from ts ntal alue (0 ). Ths transent behaor of the nductor s current s tme dependent and exponental: (0 < t < ) (t). t = : When suffcent tme has elapsed, the nductor s saturated and has reached ts fnal steady state alue defned by ( ) = I N (= I S). Solng the crcut problem (rcut #4). If the nductor s current s not what s beng soled for, one has to unfold the crcut to sole for the approprate parameter of the problem. In summary, there are two steady state alues and t s n-between these two steadystates that the nductor s transent current behaor s what we wll be solng for: (t < 0) (0 ) Intal steady-state alue (t) omplete Response = (0 < t < ) transent-state (t = ) IS Fnal steady-state alue ookng ahead, we wll fnd that ths behaor s summarzes as t/(/r Th ) (t) = IS ( (0) IS )e omplete Response Forced Response Natural Response 8.8

9 Skppng ahead (assumng we e already determne (0 )), we mmedately go to the Norton crcut and sole for the nductor s current (t) by applyng K: I = (t) S RTh Snce parallel elements hae equal oltages, we use of Ohm s law to rewrte the current through R Th d (t) d R = Th (t) = R (t) = = Th RTh RTh Insertng ths n the K equaton aboe, ges the frst-order response equaton: I d d = (t) = (t) = S S RTh RTh / RTh / RTh Snce ths has the same form as that of the Théenn R crcut dfferental, t clearly hae the same soluton form: dx x t/ = k x(t) = x( ) [x(0 ) x( )]e Translatng ths for the nductor, the completer response of the nductor s current s t/ (t) = I [ (0 ) I ]e where = /R S S Th omplete Response Forced Response Natural Response Because of the frst order equaton s the same, the Théenn R and Norton R crcut hae all the same physcal trats and behaors. So I wll only hghlght how the nductor stores/decays the current. Physcal nterpretaton Storng/Decayng of an nductor s current What does t mean for an nductor to store up current? It means that (0 ) s less than ( ) = I S: I S > (0 ). Mathematcally, the complete response reads t/ t/ (t) = IS [(0) IS]e IS Ie negate where I = (0 ) IS That s, the mnus sgn ndcates that the nductor s storng a hgher current alue. What does t mean for an nductor to decay current? It means that (0 ) > I S and now reads (t) = IS [(0) IS]e IS Ie t/ t/ poste That s, the plus sgn ndcates that the nductor s decayng from a hgher to a lower current alue. Example 8.3 a. Descrbe the behaor of the nductor s current: () ntal and fnal steady states, () storng or decayng current? Assume steady state at t = 0. b. () Determne the oltage x(t) for t > 0 s. () Plot x(t) s. t for - < t <. Soluton Is the nductor storng current or decayng t? At the ntal steady state, the nductor shorts out eerythng left of the nductor. When the swtch s thrown, t klls of the 500mA-source and now the nductor has to share the current wth the rest of the resstors. So I expect a I 8.9

10 decayng current! We want to determne the oltage across the 600Ω resstor, not the current of the nductor. Howeer, the nductor controls the current flow n the crcut and we frst sole for the nductor current and then relate (t) to x(t) to sole the crcut problem. Step : Determne (0 ) At the ntal steady state, the nductor shorts out the resstors on the left. So the current of the nductor and the current source are the same: (0 ) = 0.5A Step : Determne I S, R Th and / When the swtch s thrown, we now conert the R crcut nto a Norton equalent. The two resultng crcut wll ge us I S and R Th. Determne I S Replacng the nductor wth a short crcut, the current through the 400Ω = I S. Usng VDR and Ohm's law to get 400Ω, we get Ω = ( 00V) = 37.5V Ω 400Ω = = 93.8mA = I S 400Ω Determne R Th By shortng out the oltage source R = = = 640Ω= R Th Determne / The nerse tme constant s R 640 Th Th = = = 6400 Hz = 0. Step 3: Wrtng out the soluton to the current (t) ges us (t) = 500 ma t < t e ma t 0 Interpretaton: the nductor s current starts off at 500mA and decays to a lower fnal steady-state alue of 93.8mA (at 5 = 0.78 ms) snce t now shares the current wth the other resstors when the 00V-source takes oer. Ths s what we expected. Step 4: Solng the crcut problem for 600Ω We need to determne () 600Ω (0 ) and () 600Ω(t). If we go back to the calculaton for (0 ), we mmedately note that the 600Ω resstor was shorted out: 600Ω(0 ) = 0. To determne 600Ω(t), we to go the crcut before we conerted the R crcut nto a Norton crcut and relate 600Ω(t) to (t). Applyng KV to the rght-handed loop n the crcut, we get d 600Ω = 400Ω (t) = 400 (t) 0 Substtutng n the nductor s current (t) nto KV ges 8.0

11 5 6400t ( ) 6400t ( ) ( ) ( ) 6400t 600Ω = e ma e ma 0 = e V = (t) 600Ω Does ths make physcal sense? Snce the 600Ω was ntally shorted out, t starts off wth no oltage drop across t. Howeer, at t = 0, we hae a huge dscontnuty: Ω (0 ) = ( e ) V = 60,000V Why s there a sudden large dscontnuty jump from 0V to 60,000V? Frst note that one expects a large oltage drop across the nductor snce that s when we expect the back emf to be the largest. alculatng the nductor s oltage at t = 0, we fnd that d (0 ) 3 (0 ) = = = 60V So at t = 0, the nductor domnates the crcut and ges x(0 ) a polarty opposte to the 00V-source. Howeer, the nductor cannot contnue to sustan ths oltage drop and des off exponentally to 0V. The 00V-source takes oer and charges the 600Ω resstor wth the opposte polarty untl t reaches ts fnal steady state alue of x(5 = 0.78 ms) = 37.5V. As one can see from ths example problem, solng these types of problems wll become tme consumng smply due to the fact that there are many steps n-between. SEQUENTIA SWITHING When a seres of sequental swtches are thrown at arous tmes, a seres of sequental charge-ups or dscharges wll occur wth the capactor. Of course, eery tme a new crcut s created wth the throwng of a swtch, t creates new tme constants and new complete responses. Intal steady state t = 0 rcut- wth t = 0 rcut- wth t = t rcut-3 wth 3 t = t Mathematcally what do we expect? rcut : The swtch s closed at t = 0 and steady state s establshed. The complete response of crcut- s t/ (0 < t < t ) = V ( (0 ) V )e oc rcut : When the swtched s thrown at t = t, a second R crcut s created that requres ts own complete response and new tme constant. Howeer, there are two new thngs that hae to be accounted for. The dschargng of the capactor starts at t = t and therefore, ts complete response s tme shfted: t s replaced by tt. The capactor s ntal oltage s the fnal steady state alue of the preous crcut- at tme t = t. Wth these modfcatons, the dschargng behaor of the capactor then looks lke oc oc (t < t < t ) = V ( (t ) V )e oc (t t )/ rcut 3: Followng the same deas as wth crcut- s complete response, we wrte 8.

12 (ttt )/ 3 (ttt )/ 3 3(t < t < ) = Voc3 ( (t ) Voc3 )e = (t )e et s apply these deas to the followng example crcut. Example 8.4 a. Descrbe the behaor of the nductor s current: () ntal and fnal steady states, () storng or decayng current? Assume steady state at t = 0. b. () Determne the current Ω(t) for t > 0. () Plot Ω (t) s. t for - < t <. Soluton Snce there are two swtches, there are three tme nterals: t = 0, 0 < t < 5ms, and t > 5ms. rcut #0: Determne (0 ) Note that Ω(0 ) = (0 ). Usng VDR to determne Ω(0 ) and then apply Ohm s law to calculate Ω(0 ), we get 6 4V Ω = 5V = 4V (0 ) = = A = (0 ) 6 (6 ) Ω Determnng the current Ω(0 ) through the Ω resstor, we get 3V Ω = 5V = 3V (0 ) = =.67A = Ω(0 ) 6 (6 ) Ω rcut #: Determne (0 < t < 0.05s) At t = 0, the left swtch s thrown close and shorts out the 5V-source, leang the crcut wthout a source. learly, the nductor wll decay t current. Determne the Norton parameters (I S, R Th, ) for 0 < t < 0.05s. Snce there s no source supplyng the crcut, we mmedately know that I S = 0. The Théenn resstance s RTh = 6 = 6Ω. so that the nerse tme constant s RTh 6Ω = = = 6 Hz = H Puttng ths nto soluton form: (0 < t < 0.05s) = I (0 ) I S t/ 6t ( S ) = e e A Determnng the current Ω(0 < t < 0.05s), we can use DR to determne t: 6 6t Ω(0 < t < 0.05s) = = e A = Ω(0 < t < 0.05s) 6 3 rcut #: Determne (t > 0.05s) The second swtch s thrown before the nductor has had the tme to completely decay away ts current. 8.

13 So the complete response after the second swtch s thrown has to be tme shfted to t = t = 5ms and s gen by (t 0.05)/ (t > 0.05s) = I (0.05s) I e ( ) S S Determne the ntal steady state alue (0.05s) The nductor s fnal steady state alue before the swtch was thrown at t = 0.05 s becomes the ntal steady state alue after the swtch s thrown: 6t (t = 0.05s) = e =.473A t= 0.05s Determne the Norton parameters (I S, R Th, ) for t > 0.05s Wth no supplyng source n the crcut agan, I S = 0. The Théenn resstance s smply those two seres resstors and we get I S = 0 and R Th = = 4Ω The nerse tme constant s RTh 4Ω = = = 4 Hz = H Puttng ths nto soluton form: (t > 0.05s) = I (0.05s) I S (t0.05)/ 4(t0.05) ( S ) = e.473e A Snce the nductor and the Ω resstor are n seres, they hae the same current: > = > = 4(t0.05) (t 0.05s) Ω(t 0.05s).473 e A Summarzng the two tme nterals, we get.67a 0 > t 6t Ω(t) = e A 0 < t < 0.05s 3 4(t0.05).473 e A 0 < t < 0.05s By lookng at the tme constants for the two tme nterals, note that = /6 s whle = /4 s. The nductor decayed at a much faster rate after the second swtch was thrown. UNIT STEP SOURE and PUSE SOURE In order to smplfy swtches n crcuts, we now mbed swtches n wth the sources usng unt step functons. A unt step functon s defned as a mathematcal functon that has an nstantaneous jump from zero to one at tme t = t 0: 0 t < t 0 Unt Step functon u(t t 0 ) = t > t 0 So how do we use ths unt step functon wth sources? onsder the crcut 8.3

14 So nstead of usng a complcated set of swtches n order to turn on a source, we wll replace t wth the unt step functon (t easer to do t ths way). Physcally, ths appears lke the turnng on of a lght swtch n your ktchen. There s an nstantaneous oltage ncreases to V 0 from zero olts. To produce a pulse square oltage source of fnte duraton (or fnte duty), a superposton of these two sources wth unt step functons are used at dfferent tmes. RESPONSE of FIRST-ORDER IRUITS to NONONSTANT SOURES Up to now, we hae only had constant sources. We now ntroduce tme dependent sources so that the frst order equaton has a rght-handed sde that s not constant but tme dependent. dx x dx x = k s. = f(t) k= constant f(t) constant There wll be two sgnfcant changes how one deres the complete response for the tme dependent crcut.. For a constant source, we conerted our R or R crcut nto a Théenn or Norton crcut. As a result, the soluton always had the same form of t/ x(t) = x( ) x(0 ) x( ) e For tme dependent sources, we wll fnd that t s more tme effecte to not conert the crcut nto a Théenn or Norton crcut, but grab t by the horns and dere a new frst order equaton eery tme. We wll only encounter two types of tme dependent sources n ths chapter: exponental and snusodal. at K Ke f(t) effecte source = K Ksnωt (cosωt). Snce we wll not hae the same frst order equaton, the complete response wll hae to be dered eery sngle tme, and consequently, s more complcated. There s an adonal mnor complcaton that arses because there s a change n language (whch s most unfortunate). Here s the language change: t/ x( ) [x(0 ) x( )]e ) constant source force response natural response x(t) = t/ t/ complete x(0 )e x( )( e ) nonconstant sour ce response natural response force response In my personal opnon, I prefer the bottom complete response form snce I fnd t more physcal n explanng the behaor of the crcut. The mportant thng to realze s that ths s smply a change n language (and therefore, no bg deal ). Interpretaton of tme dependent complete response When a constant source s drng the crcut, the storage element s behaor s to exponentally decay from ts ntal state alue x(0 ) whle forcng t towards ts fnal state alue x( ) accordng to t/ t/ x(t) = x(0 )e x( )( e ) complete response natural response force response 8.4

15 A tme dependent source drng the crcut behaes exactly the same way but the bg change s the force response s now tme dependent t/ x(t) = x(0 )e x f(t) xn xf complete response natural response force response Furthermore, we know a lot about the force response because physcally, f the tme dependent source s snusodal, then the force response must also be snusodal. In other words, the functonal shape of the tme dependent source wll force the force response to hae the exact functonal form as the tme dependent source. There are two types of tme dependent sources (exponental and snusodal) that we wll encounter and therefore, two types of force responses (exponental and snusodal): at at K Ke Be B f(t) x f (t) K Ksnωt (or c osωt) Bsnω t Bcosω t) B3 effecte force source response A worded bref oerew of the process s as follows: Dere the frst order equaton usng Smple rcuts, and do a smply check to see f the fnal steady state agrees wth your equaton. From the frst order equaton, you wll be able to wrte the general complete response (natural force) down that contans an arbtrary constant A. In the language of dfferental equatons, one wll generate a homogeneous & partcular soluton to the frst order equaton. Dere the forcng response constants B and B ; Sole for the I x(0 ) and apply t to the general response to dere the arbtrary constant A. Wrte out the complete response (wth all soled for constants) and sole the crcut problem Problem Solng Strateges for Frst-order rcuts wth Tme Dependent Sources apactor equaton: = d / Inductor equaton: = d / Step : Dere the Frst Order Equaton Use Smple rcuts to wrte an equaton that contans x and dx/ (.e. & or & ). Rewrte any resstor oltage or current n terms of ( & or & ). Apply the capactor or nductor equaton to conert the frst order equaton nto a frst order dfferental equaton and wrte t nto t proper form : dx x = f(t) Do a smply check to see f the fnal steady state x( ) agrees wth your equaton. Step : Dere the general response equaton The general response equaton s of the form -t/ x(t) = x (t) x (t) = Ae B x B x B natural force f f 3 Forcng Response x f(t) The forcng response has the same form as the effecte source term f(t) n the frst order equaton: -at at K Ke Be B f(t) = x f (t) = K K sn( ωt) (or cos( ωt)) Bsn( ω t) Bcos( ω t) B 3 Substtute the general force response nto the frst order equaton and sole for constants B and B : dxf xf = f(t) x f (t) = Bx f Bxf B3 x f (t) = Bx f Bxf B3 8.5

16 Substtute the constants B, B and B 3 nto the general response equaton. Natural Response x n(t) From the frst order equaton, pck-off the tme constant / and wrte down the natural response: t/ xn = Ae Wrte the general complete response were the only unknown constant s A: t/ at Ae B t/ two possble e B x(t) = Ae Bx f Bxf B3 solutons t/ Ae B snω t B cosω t B 3 Step 3: Sole the I for x(0 ) and the arbtrary constant A Sole for the I [ (0 ) or (0 )] Apply the I to the general complete response equaton and sole for constant A. Wrte out the complete response Step 4: To sole the crcut problem for the approprate parameter, go to the crcut at t = 0 and relate the unknown parameter to x(t) usng K or KV or both. et's consder a straghtforward example and go through the aboe steps to get an dea as to what to do. Example Determne (t) for t > 0 when (0 ) = 4V. Soluton What do we expect for the capactor s response (t) to the sne oltage source? Snce the capactor s ntally charged to 4V, as soon as the snusodal oltage source turns on at t = 0, the oltage cannot change nstantaneous and therefore, cannot hae a dscontnuty. Howeer, the snusodal source wll force the capactor s oltage to oscllate at an angular frequency of ω = 0 rad/s. Step : Dere the frst-order equaton Snce ths s a seres crcut, apply KV to wrte down an equaton that contans both and. Next, rewrte the resstor oltage n terms of snce the 0Ω n seres wth the capactor (usng Ohm s law): = = R S R R= R Apply the capactor equaton to rewrte n terms of (ths turns t nto a frst order dfferental equaton): d d S = c R d = R = = 0 Wrte ths equaton nto ts proper form: d d /0 /0 S = 0 = 00 sn0t S = 0sn0t effecte source heck your equaton. At steady state, the capactor s n A steady state that means that ts response s snusodal. The detals we cannot answer at the moment snce ths requres us to sole the frst order equaton. Step : Dere the general response (t) = n f By lookng at the frst order equaton, the general response equaton must hae the form of 8.6

17 (t) = (t) (t) = Ae B sn0t B cos0t -t/ n f general natural response general force response My job now s to determne the constants B, B, and /. Determne the constants B and B To determne B and B, we substtute the general force response nto the frst-order equaton: df 0f = 0[ Bcos0t Bsn0t] 0( Bsn0t Bcos0t) = 00 sn0t d f / Groupng together lke sne and cosne terms: [ ] [ ] 0B 0B sn0t 0B 0B cos0t = 00sn0t The only way ths equaton can be true s f all of the sne/cosne terms on the left sde equals those on the rght sde of the equaton. [ ] [ ] 0B 0B sn0t = 00 sn0t 0B 0B = 00 0B 0B 0B cos0t = 0 0B = 0 Solng ths matrx for B and B : 0 0 B 00 = B =, B = B 0 Determne / From the proper form of the frst order equaton, we pck-off the tme constant /: dx x d = = = f(t) 0 00 sn0t 0 Hz Substtutng n these constants nto the general complete response ges 0t B (t) = n = f = Ae sn0t 4cos0t B =4 / = 0 Step 3: Sole (0 ) and the arbtrary constant A Sole for the I (0 ) Snce ths was gen n the problem, we wrte (0 ) = 4V. Determne the constant A Apply (0 ) to the general complete response equaton and sole for constant A: 0 (0 ) = 4 = Ae sn0 4cos0 = A 4 A = 8 The complete frst-order response s f 0t ( ) (t) = 8e sn0t 4cos0t V Interpret the soluton Usng PSpce, the complete response s plotted below. 8.7

18 Is (0 ) = 4V = (0 )? 0 (0 ) = 4V = (0 ) = 8e sn0 4cos0 V = 8 4 = 4V ( ) Does the capactor s oltage oscllate at a frequency of ω = 0 rad/s? ω 0 ω = π f = 0 rads/s f = = = Hz T = = 0.34 s π π f Ths s exactly the same perod found on the plot. What s the capactor s maxmum oltage at steady state? When the capactor s at steady state, that means that the exponental decay of the natural response s undetectable and the force response domnates ts behaor: 0t lm (t) = 8e t large can be gnored sn0t 4cos0t sn0t 4cos0t et s check ths. Use PSpce to pck-off the tme alue at one of the peak oltage alues. Why? The equaton that has to be soled to obtan the crtcal alues s a transcendental equaton, whch s not tral and a bg no thank you! n tryng to sole t. Pckng off the tme for one of the peaks, t =.083 s, and pluggng ths tme nto the complete response t predcts the oltage (t =.083) = 8e sn cos0.083 V ( ) ( 3.558) natual response force response = = 4.47 V = (t =.08 s) As expected, the natural response wll make no contrbuton to the complete response for large tmes. Furthermore, the theoretcal predcton by the complete response compares ery well wth the PSpce alue. Example 8.5 a. Descrbe the behaor of the nductor s current: () ntal and fnal steady states, () storng or decayng current? Assume steady state at t = 0. b. () Determne the oltage 4Ω(t) for t > 0. () Plot 4Ω(t) s. t for - < t <. Interpret the soluton. Soluton What s the ntal behaor of the nductor at t = 0? By performng a source transformaton on the 5A-source (ths becomes a 30V-source), we wll hae two opposng oltage sources such that the role of the 5A-source s to reduce the nductor s ntal current. At t = 0, the swtch klls off the 5A-source and therefore, the nductor s current ncreases or stores up 8.8

19 more current. The role of the exponental source at t = 0 s to ge a sudden burst of oltage to the crcut after the swtch. When wrtng out the soluton, we connect the nductor s current to 4Ω(t) usng Ohm s law: 4Ω(t) = 4 (t). Step : Dere the frst-order equaton Snce ths s a seres crcut, apply KV to wrte down an equaton that contans both and. Then rewrte the 4Ω resstor oltage n terms of the nductor s current usng Ohm s law: 5t 4 = 4 (t) = = S 3 8e Ω Apply the nductor equaton to rewrte n terms of (turns t nto a frst order dfferental equaton) and wrte t nto ts proper form: d d 5t 5t 4 8 = 3 8e 3 = 4 e heck your equaton. The fnal steady state alue of the nductor s current should be gen by the frst order equaton. To determne ths, set ( ) = constant and substtute ths nto the frst order equaton at t = : d d ( ) 4 3 = 3 ( ) = 4 e ( ) = A 3 ( ) The fnal steady state crcut wll hae the exponental 8e 5t -source decay away and the nductor s replaced wth ts steady state equalent. Solng for ( ), we see that the current ( ) s gen by Ohm s law: 3 4 ( ) = = A 4 3 So the frst order equaton behaes as expected. Step : Dere the general response (t) = n f. By lookng at the frst order equaton, the general response equaton must hae the form of -t/ 5t (t) n(t) f(t) Ae = = Be B general force response My job now s to determne the constants B, B, and /. Determne the constants B and B To determne B and B, we wll substtute the general force response nto the frst-order equaton: df 5t 5t 5t 5t 3f = 5Be 3( B Be ) = 3B Be = 4 e Equatng the constant and e 5t terms on both sdes of the equaton, we sole for B and B : 4 5t 5t 3B = 4 B = and B e = e B = 3 Determne / From the proper form of the frst order equaton, we pck-off the tme constant /: df 5t 3f = 4 e = 3 Hz Substtutng n the constants nto the general complete response ges 4 B 3t 5t / (t) = n = f = Ae e B = 4/3 3 / = 3 Step 3: Sole (0 ) and the arbtrary constant A 8.9

20 Sole for the I (0 ) Startng wth the crcut at t(0 ), the nductor s replaced wth ts steady state equalent and the crcut looks lke the one on the rght. Usng a source transformaton to conert the 5A-source nto a seres of oltage source, Ohm s law ges us 3 30 (0 ) = = A = 5 (0 ) 6 4 Determne the constant A Apply (0 ) to the general complete response equaton and sole for constant A (0) = = Ae e = A 4 A = Substtutng ths nto the complete response ges t 3t (t) = e e Interpret the soluton Usng PSpce, the complete response s plotted below. Is (0 ) = A = (0 )? (0 ) = A = (0 ) = e e V = 4 3 = A ( ) Does the nductor s current alue at t = match PSpce alue? (= s) = e e A = ( ) =.0347A = ( ) Does the nductor s current alue at t = match PSpce alue? 4 3 ( ) = ( e e 3 30 ) A =.333A = ( ) The theoretcal predcton modeled by the complete response compares ery well wth the PSpce alues. Step 4: Sole for the oltage across the 4Ω resstor. Snce the nductor and the 4Ω resstor are always n seres, ether before or after the swtch (ncludng the unt step functon turn on), they wll always hae the same current. Therefore, the behaor of 4Ω(0 ) and 4Ω(t) should be contnuous at t = 0. Applyng Ohm s law, we get 4Ω(t): 5t 9 3t ( ) 4Ω (t) = 4 (t) = 3 e e V Snce the oltage 4Ω s only scalng the nductor current by a factor of 4, the behaor of the 4Ω s exactly the same as (t). Therefore, I wll not repeat the same analyss as I dd for the nductor current

21 Example 8.6 a. Descrbe the behaor of the capactor s oltage: () ntal and fnal steady states, () chargng or dschargng? Assume steady state at t = 0. b. Determne (t) for t > 0. Soluton What s the behaor of the capactor n ths crcut? Intally, the capactor wll be charged by the 38.5V-source. Howeer, when the swtch s thrown open at t = 0, there s a burst of energy nto the crcut due to the 8e 5t -source but ths des off completely after 5 = s. So there s no source supply x (no x-source) and the capactor wll dscharge to zero. Step : Dere the frst-order equaton To sole for the capactor's oltage (t) at t = 0, not only wll we hae to use KV but also nclude a dependent conon (D) snce there s dependent source n the crcut. Wrte down an equaton that contans both and. Applyng KV around the outsde loop of the crcut, t wll nclude the Ω resstor. Next rewrte the Ω n terms of the dependent current x: 5t 5t 8e = 8e = 0 X Note that ths equaton does not contan both and. We need to brng n the D to rewrte x n terms of. To apply the D, use K to the top node and wrte = = X X X Substtutng x back nto our equaton, we now hae an equaton that contans both and : 5t 5t 8e = 8e = 0 X X= 3 Usng the capactor equaton to rewrte n terms of (as well nto ts proper form), we get 5t d 5t 4 = 0 d = 8e 9 = 7e = 36 heck your equaton. At t =, there are no acte sources and the capactor s oltage ( ) should go to zero. Substtutng ( ) = constant nto our frst order equaton ges d 9 = 0 9 ( ) = 7e = 0 ( ) = 0 ( ) So the frst order equaton behaes as expected. Step : Dere the general response (t) = n f By lookng at the frst order equaton, the general response equaton must hae the form of -t/ 5t (t) = n(t) f(t) = Ae Be My job now s to determne the constants B and /. 3 3 general force response Determne the constants B To determne B, we wll substtute the general force response nto the frst-order equaton: df 5t 5t 5t 5t 9 f = 5Be 9Be = 4Be = 7e B = 8 5t f = Be Determne / From the proper form of the frst order equaton, we pck-off the tme constant /: 8.

22 d 5t 9 = 7e = 9 Hz Substtutng n the constants nto the general complete response ges 9t 5t (t) Ae = 8e Step 3: Sole (0 ) and the arbtrary constant A Sole for the I (0 ) Startng wth the crcut at t(0 ), I use a source transformaton to conert the 38.5V-source nto a current source and redraw the crcut as Snce all of these elements are parallel to each other, the ntal oltage (0 ) can be calculated across any of the elements; I wll focus on the Ω-resstor: (0 ) = = Ω x Snce ths clearly depends on the controllng current x, I wll apply DR to the Ω-resstor 3 x = ( x.83) x =.83A 3 and the calculaton of (0 ) s (0 ) = x =.83 V = V = (0 ) Determne the constant A Apply (0 ) to the general complete response equaton and sole for constant A 0 0 (0 ) = = Ae 8e = A 8 A = 4V Substtutng ths nto the complete response ges = 5t 9t (t) 8e 4e As expected, the capactor's oltage goes to zero farly quckly. The natural response des off faster at e -9t than the force response at e -5t. 8.