v + By Norton s theorem I N Given v(0)=v 0, R what is v(t) for t >0? i R (t)?

Transcription

1 11/27/217 haper 7 Frs order rus A ru wh one apaor or nduor 17 By Theenn s heorem By Noron s heorem V h Need o know how o sole he aboe wo smple rus ef : When V h =, soure free ru ; Oherwse, sep response of ru gh: When I N =, soure free ru Oherwse, sep response of ru I N 7.2 The soure free rus a Gen ()=V, wha s () for >? ()? b By K: = (1) Need o use ru laws o dere a dfferenal equaon for (). Then use mah mehod o sole for. By propery of ressor and apaor: d = ; = d d Plug no (1) o oban: d d 1 (2) d A frs order dfferenal equaon 17 1

2 11/27/217 The soure free rus an be desrbed wh a 1 s order dfferenal equaon 17 a Gen ()=V, wha s () for >? ()? d 1, (2) d b ombned wh nal ondon ()=V, unque soluon () an be found. d 1 (2) an be rearranged as d (3) Inegrang boh sdes d 1 = d = (4) Sne d ( ) ln ( ) = ln ( ) ln () ln () () ln = () () = e () () =() e () =V e In summary: For soure free ru, a Gen ()=V, he soluon s b The naural response of he ru. I depends on: V : nal ondon : a onsan () =V e 3.5 V V s alled he me onsan, denoed as () V () = = e When, () =V e 2

3 11/27/217 s alled he me onsan, denoed as A =, ()=V e 1 =.368V A = 2, () =V e 2 =.13534V ; A =5, () =.674V apaor almos fully dsharged () 4 V = 2 = 3 Smaller =, faser deay () = e V () =V e =.5 = 1 The power dsspaed n : V p = = e Energy absorbed by ressor from o V V e pd = e (1 ) 2 As, all energy sored n wll be dsspaed a. 17 3

4 11/27/217 Example: Gen ()=15V, fnd (), x () for >. a 8 For ru, always hoose apaor olage as key arable Fnd () frs, hen express oher 5.1F 12 arables n erms of. x For hs ru, 12 b x() = () =.6 () 128 To fnd (), onsder he res as a sngle ressor eq, We all eq wh respe o (w.r.) he apaor eq =5//(812) = 4.1F a b Wh () = 15V, 1/ = 1/(4.1)=2.5 () = 15e () =.6 15e x 2.5 V e V Queson: How nal ondon s esablshed? () = () e Generally, here s a soure o sore energy n he apaor for <. A =, he soure s urned off by a swh and he ru beomes a soure free ru, for >. eall wo mporan fas abou a apaor: 1. A D ondon, a apaor behaes lke an open ru 2. () s a onnuous funon of me. These fas wll be used o deermne he nal ondon (). Operaons of swh: Or, = = For <, For >, Or, = = For <, For >, 4

5 11/27/217 Example: The swh has been losed for a long me before s open a =. Fnd () for >. = 3 a 1 Inerpre he ondon: Swh losed for a long me 9 2V means ha a D ondon 2mF has been reahed a =. b The alue of rgh before swh, ( ), an be found by onsderng he ru before swh under D ondon. Sne () s onnuous, () = ( ) = () Sep 1: Fnd () from he D ru before swh ( < ). Sep 2: Fnd eq wh respe o for ru afer swh (>) Sep 3: plug n he general formula () = ()e eq 17 Sep 1: For <, or before swh. 3 2V 9 1 2mF onsder he ru under D ondon apaor behaes lke an open ru No urren hrough 1Ω, hus no olage drop olage aross 9 = By olage dson: 9 215V 9 3 () 15V Sep 2: For >, (or, afer swh) Volage soure dsonneed from he rgh sde. I has no effe o he apaor 1 eq =19=1 Sep 3 : form () 9 2mF eq 3 5 () = ()e 5 () = 15e V 17 eq 5

6 11/27/217 Soure free ru gudelne For >, () = ()e Two key parameers: Swh urns off ndependen soures for >. For >, by Theenn s heorem, he wo ermnal ru behaes lke a sngle ressor eq eq eq For >, 1. (), nal ondon for apaor olage. Esablshed by ndependen soures before swh (<). Deermned from he ru before swh. Assume a D ondon has been reahed. apaor behaes lke open ru 2. eq, equalen ressane wh respe o he apaor for ru afer swh ( > ). 17 Afer () of apaor s found, oher arables an be obaned usng bas laws Example: 6

7 11/27/217 Prae 1: Fnd () and 1 () for > V = mF 6 6 Prae 2: Fnd () and 1 () for >. 17 = A mF 4 7

8 11/27/217 Prae 3: The swh has been losed for a long me before open a =. Fnd I x () for >. 17 I x = 1 2I x.5f 4 2A The soure free ru Gen ()=I. Need o fnd (), for all >. Assgn and aordng o passe sgn onenon. Then d =, = d By KV, = ompared wh equaon for soure free ru d = d d = 1 d Same mah problem. Soluon: () = ()e d 1 d () = ()e 18 8

9 11/27/217 The soure free ru Gen ()=I. Soluon: () = ()e Tme onsan: () = ()e Oher arables: ) = 2 Energy sored n nduor: w () = ()= ()e Energy onsumed by ressor: w () = () ()e w () w ( ) ()= onsan 2 Two key parameers: 1. (), nal nduor urren. Obaned from ru before swh, (<), by reang nduor as a shor ru (under D ondon). Induor urren s onnuous funon of me:. 2., equalen ressane wh respe o nduor for ru afer swh ( > ) Example: Gen ()=1A, fnd (), x () for >..5H x 2 Supply = 1A urren, Then eq = /1= =? Use nodal analyss 4 K a, = 1/3V eq =1/3 1 3 Need o fnd eq wh respe o nduor. eall Theenn s equalen ressane from haper 4. Due o he dependen soure, need o supply exernal soure 2 x 2 () () e? 4 3 = 1; x = ; = 2 = 4 4 1/ e 1e A Sne 2 n parallel wh nduor: /2 / ( ).51 ( ) e / 2 e A x eq?

10 11/27/217 Example: The swh has been losed for a long me before open a =. Fnd, for >. 2 4V = H Imporan fas for fndng nal ondon : 1. D ondon for < 2. Under D ondon, nduor shor ru 3. Induor () onnuous 18 Sep 1: Fnd from ru before swh ( < ). 2 4V //= No urren, no olage drop aross 16 4 and 12 n parallel Volage dson: 4//12 2 4// = =6A Sep 2: For >, 4V dsonneed H eq wh respe o nduor eq =16//(412)=8 eq /=4 Sep 3: Sep 4: d d 2 6 ( 4) ()= = 4 4 e e A A smpler mehod: By urren dson

11 11/27/217 Example: Fnd for >. = A 4 6.5H For >, wh respe o Ω Fnd 4 From ru before swh: 4.5A H Pu ogeher: urren dson: // A Sngulary funons We saw swhes n preous rus. They hange he sruure of he ru, e.g., a =. A swhng me =, some arables, suh as (), () are onnuous, bu (), () may be dsonnuous. 18 Sngular funons are hose hang dsonnuous derae, or, beng dsonnuous Sep funon: a bas sngular funon ab, < () V, > ab = a V b ab b A sep funon an be realzed by swh, n oher words, The operaon of swh an be desrbed by a sep funon 11

12 11/27/217 The un sep funon:, < u () 1, > 1 u() 18 General sep funon an be expressed n erms of un sep:, < ab() Vu() V, > Un sep wh a me shf:, < u ( ) 1, > 1 u( ) Tme reersed un sep 1, < u( ), > 1 u( ) The mpulse funon: he derae of un sep (), < ( ) undefned, =, The un ramp funon: he negraon of un sep 18, < r (), > r() 12

13 /27/217 Prae 4: Fnd (), 1 (), 2 () for > = 12V.5H A: () 1.5e A ( ) ( ).75e A 13

14 11/27/217 Prae 5: Fnd I 1 () and 2 () for >. 18 = 2 1 I 1 2 2H 4 3A Sep response of an ru V s = a b V s u() Assume () = V. Fnd () for >. () s he response o a sep funon V s u() alled sep response. For >, V s u() = V s : To dere a dfferenal equaon for, assgn and V s d d = ; = = d d By KV, = V s Sep funon: d V s Wh nal ondon: ()=V d, < u () 1, > 14

15 11/27/217 V s For >, V s u() = V s : d V s Wh nal ondon: ()=V d d 1 ( V s ) d d 1 d -V s ln( V s ) () Vs ln V V s ()-V - s = e V-V s s s () = V (V V ) e for > Sep response Sep response: V, < () = - V s (V-V s) e, > A sep response brngs a ru from one D ondon o anoher D ondon V s V V < V s V s V V > V s Speal ases: 1. When V s =, 2. When V =, - () = V e for > edue o soure free ru s s - () = V V e for > 15

16 11/27/217 Sep response: V, < () = - V s (V-V s) e, > You an use () o fnd oher arables: V s d 1 ()= =(V V s)( )e d Vs V = e Smpler approah by KV: s s V () V V ()= e Sep response of general ru One or more swhes hanges sruure of ru a = (or = ). Problem: need o fnd () or, oher arables for >. For <, ru behaes lke a D ru. Fnd he nal ondon () from hs D ru. For >, use Theenn s heorem: V h h Noe ha: V h V s, h () = V (V V ) e h h () = V (V V ) e = V h h h ( ) = V (V Soluon an also be wren as h h V ) e V h () = ( ) (() ( )) e h 16

17 11/27/217 Soluon an also be wren as () = ( ) (() ( )) e h ( ) : The fnal alue. apaor olage under D ondon for ru afer > A shor u o form sep response, whou derng dfferenal equaon Three key parameers: 1. Inal ondon ()=V, Esablshed from ru before swh. Obaned by solng a D ru, < (apaor = open ru) 2. Fnal ondon ( Esablshed from ru afer swh. Obaned by solng anoher D ru, > (apaor = open ru) 3. Equalen ressane h, or eq wh respe o apaor for ru afer swh Sne ( ) = V h, (V h, h ) an be ogeher onsdered as he Theenn s equalen, wh respe o he apaor, from ru afer swh. The problem of fndng sep response breaks down o seeral D ru problems n hapers 2 and 4. Example: Fnd he apaor olage () for >. 3k 24V 5k =.5mF 4k 3V Sep 1: For (), use ru before swh ( < ) 3k 4k 24V 5k.5mF 3V Sep 2: For h, V h, w.r. apaor, use ru afer swh 3k 24V 5k.5mF 4k 3V h =4k; V h =3V h Sep response: By olage dson, 5 () 24 15V 5 3 () 3 (153) e.5 = 3 15 e V.5 17

18 11/27/217 Example: Fnd he apaor olage () for >. 3u()V F Sep 1: For (), use ru for < = 1V eall:, < 3 u ( ) 3, F 1V ()=1V ( )= 1A Sep 2: For V h, h, use ru for > 3V F 1V 3V 1 2 V h =2V= h =1//2=2/3Ω / 3 h () 2 (12) e.6 = 2 1 e V.6 Sep 3: How o express () n erms of ()? 3 ( ).6 () 1e A 1 Is () onnuous a =? ()=2A ( )= 1A In ru, he only arable ha s guaraneed o be onnuous s apaor olage. Oher arables, suh as apaor urren, ressor olage/urren an be dsonnuous. Don make generalzaon lke: 18

19 11/27/ Sep response of rus One or more swhes hanges sruure of ru a = (or = ). Problem: need o fnd () or, oher arables for >. For <, ru behaes lke a D ru. Fnd he nal ondon () from hs D ru. For >, by Noron s heorem d By K, = I N = = N N d d N I d N N N d ( I N ) d I N Same mah problem as ru Sep response: () = I (() I )e Noe ha: ( = I N. As, a D ondon wll be reahed. ( ) s nduor urren for he D ru for > N N N Sep response of rus Sep response: () = I (() I )e N N N Sne ( ) = I N, an also be wren as: () = ( ) (() ( ))e N Three key parameers: 1. (), nal ondon, from he D ru for <. 2. ( )=I N, fnal alue, from D ru for >. 3. = N = h, equalen ressane w.r. nduor from ru for > (I N, N ) ogeher as Noron s equalen, w.r. nduor, for >. 19

20 11/27/217 Example: Fnd () for >. 2 1V = 3 1 H 3 Sep 1: fnd (). onsder D ru for <. Sep 2: Fnd ( ) = I N from D ru for >. 2 1V 3 ( ) = 1/5=2A N w.r.. for > : ( ) 2 1V 3 () N = 23=5Ω N Sep response: /3 () = 1/2=5A Example: Fnd () for >. Sep 2: Fnd ( = I N from D ru ( > ) ( ) 5 = 1.5H 1 3A 5 1 3A Sep 1: Fnd (). onsder D ru ( < ) 5 () 1.5H ()=3A 1 3A By urren dson: 1 ( ) 3 2A N w.r.. for > : N 1 N =15Ω, N /=

21 x /14 /14 x /27/217 Prae 6: Fnd (), x () for > V 2F = x 4 x 2A 3 A: () 48e V () 2 (8/7) e A Prae 7: Fnd x () for >. 19 1k 3mA = 1k x.25mf 2k A: () 15(1 e ) V ( ) e A 21

22 () 2 A, ( ) 3 A, N () 3e A /2 11/27/217 Prae 8: Fnd () for all. 19 1V 24V 2H A: 2 = 6 3 Prae 9: Swh onneed o a for a long me before onneed o b a =. Fnd () for >. 19 6Ω.4F 1Ω 4Ω b a = 12V A: () 2e V 3Ω 4Ω 22