STUDY PROGRAM: UNIT 1 AND UNIT

Transcription

1 IUIT ANAYSIS I MODUE ODE: EIAM4 STUDY POGAM: UNIT AND UNIT UT aal Unery of Technology

2 EIAM4 haper : Fr Order rcu Page -. FIST ODE IUITS. Summary of Bac rcu oncep and onenon eor, capacor and nducor olage and curren relaonhp and drecon = = d = d = d = d = = = = ± + () = ± + () Krchoff olage law (K) o o o 6 5 o o H H3 H o o 6 5 o = H8 H = H7 H6 H4 Krchoff curren law (K) olage Don urren Don Operaonal amplfer = Z T T Z Z Z + 3 = = + Z Z Z T Z Z Z T = = = Baery (conan olage) olage ource General ndependen A Dependen on anoher olage ource arable urren ource General ndependen curren ource Dependen on oher arable Mcellaneou herefore mply for all becaue equalen o We wll alway prefer he arrow conenon

3 EIAM4 haper : Fr Order rcu Page -. Zero Inpu epone We fr conder he uaon where an nal conon preen bu an ndependen ource no (h mean ha n he begnnng, here appear a olage on a capacor or here flow a curren hrough an nducor bu here no baery n he crcu afer = ). Example - efer o he = crcu n Fgure -. The wch cloed for < and opened a = and ay open for >. Deermne c () for all. Fgure - For < : c () = (capacor behae a an open crcu for dc). I ( olage don rule: = [ /( + )] T ) ( or I = /( + ), c = I = /( + ) ) For = : c () = (capacor olage canno change nananeouly). For : (nroducng and o faclae wh eng up equaon) By K: =. d Bu = d and = = d d d d d d K ln K () e () e e K () e K e / Seng = : () = e K e () = e K. Bu () = e K =. Therefore for : () e K e / / () e () = e / (noe he general forma: () = ()e / ) K a conan of negraon whch wll be deermned when we pecfy ha mu pa hrough he pon [, /(+ )]

4 EIAM4 haper : Fr Order rcu Page - 3 Summarzng he reul for all : () =, e /,. Or alernaely (perhap a lle more elegan), ( remember u() = for < and u() = for ): () = A + B + (ee wndow on he rgh) u() A = /(+ ) B = [/(+ )] u() = u() e / u() A+B = [/(+ )][ u()] / = e u().368 () e / mu reman connuou a = (capacor olage canno change nananeouly). / e = {[/(+ )]e -/ } u() Noe: The pecal alue =, called he me conan of he crcu and ha nan when he capacor ha lo 63% of nal olage. Suppoe now ha he wch wa opened laer han = a ay ome me =. Then for < o : () = and herefore ( ) =. d For : () e / e K a before. / A = : ( ) = e e K. - / Bu ( ) = e e K / ek e () = () = / e / e ( )/ e (agan noe ha () = ( )e ( )/ )

5 () EIAM4 haper : Fr Order rcu Page - 4 () now hfed a dance o he rgh wh repec o he preou eron of (), ha, wa replaced by. ( )/ e o Example - efer o he crcu n Fgure -. The wch cloed for < and opened a = and ay open for >. Deermne () for all. For < : () = Fgure - (nducor ac a a hor crcu o drec curren). For = : () = (nducor curren canno change nananeouly). For : (nroducng and o faclae wh eng up equaon) By K: + = d Bu = and = d d d d d ln K K a conan of negraon whch K wll be deermned () e () e K e when we pecfy ha mu pa Seng = : hrough he pon () = ek e () = e K (, / ) Bu () = e K =. Therefore for : (noe he () e K e / () e / general forma: () = ()e / )

6 EIAM4 haper : Fr Order rcu Page - 5 Alernaely:.368, (). e /, / () e u() () mu reman connuou a = (nducor curren canno change nananeouly). e / Noe: The pecal alue = /, called he me conan of he crcu and ha nan when he nducor curren ha dmnhed o 37% of he nal alue. / Suppoe we are gen ( ) a he nal conon, hen (mlar o Example -): ( - )/ () ( )e, for o Example -3 efer o he crcu n Fgure -3. Deermne an expreon for (), f () = 5A. Noe ha he curren ource no ndependen. For = : () = 5. (gen) For : By K: + + = 3 + = 3 + = 3 + = d Bu = d 3 + = d + = 3 Suden may ole h equaon from he ar, bu we can already begn o ee ha n dy general, he oluon for αy y() y()e α for and wh gen y(). () ()e (/3) = (/3) 5e for Alernaely we can ue Theenn heorem by remong he nducor and replacng he reor and dependen curren ource by Theenn equalen crcu. (Th may be done becaue he dependen curren ource doe no depend on he nducor (he elemen ha remoed) olage or curren.) Fgure -3 Whaeer he curren hrough, he curren ource deler wce ha amoun. Alo, he olage appear acro all hree elemen, he nducor, he curren ource and he reor, of h parallel connecon.

7 Theenn olage ( T ): (The Theenn olage he open crcu olage beween ermnal A and B) + = = T = = EIAM4 haper : Fr Order rcu Page - 6 B Theenn reance ( T ): (We wll apply a e olage beween ermnal A and B and calculae he reulng npu curren o, from whch he nernal reance T = o / o may be obaned. Th normally a ery ae and relable mehod o calculae T.) o = + = 3 and = o / o = 3 o / T = o / o = o /(3 o /) = /3 Hence he Theenn equalen crcu a reor wh alue /3 ohm. The / alue for h mple crcu ju (/3)/ = /3. () ()e (/3) /3 () 5e (/3) for A B A A T B o o I make ene becaue f a olage allow o flow hrough, hen / wll deler. Example -4 For he crcu n Fgure -4, deermne () and () = = ¼ F 4 6 Fgure -4 capacor open crcu olage don rule: = [ /( + )] T For < : () = (4/5) = 8 () = 8 4 For < : + 4 = d d 4 4 () ()e and () = 8e for <. () = 8e =.943 ¼ F 4

8 EIAM4 haper : Fr Order rcu Page - 7 For : +.4 = d d ( ) () = ()e () =.943e.667( ) () = 8 8e.667( -).943e d Furhermore, nce (), we alo hae for <, d () (8) = ( = 8 (a conan) for < ) 4 d for < : (8e () ) = e 4 d.667( ) and for : ().943e 4 =.6e () = e.6e.667( ).667( ) (4//6=.4) ¼ F.4 8 () 8e Noe: mu be connuou a = and =, becaue he capacor olage can neer change nananeouly, oherwe he capacor curren = (d /), wll become nfne (he lope of a ercal lne = ) whch urely mpoble ( ).9438e () ( ).6e.63 e Noe: can be dconnuou a = and =, becaue a jump n capacor curren wll only caue he lope of he capacor olage o change nananeouly a per d / = /

9 EIAM4 haper : Fr Order rcu Page - 8 Exerce P -. For he crcu n Fgure P -, fnd () and () for all. = F 3 Fgure P - An: () = 6e - u(), () = e - u() (Hn: fr ole for he olage acro he capacor hen and follow ealy) P - For he crcu n Fgure P -, fnd () and () for all me. = 5H 9 3 Fgure P - An: () = 3 + 3[e ]u(), () = 9 + 9[e ]u() P -3 For he crcu n Fgure P -3, fnd () and () for all me. = 8 6 6H 3 Fgure P -3 An: () = + [e ]u(), () = 6e u()

10 P -4 For he crcu n Fgure P -4, fnd () and () for all me. (a uble problem) Fgure P -4 < : () = 4 (and F = ) () = 4/6 = /3 A ( and F canno change nananeouly, can alo no change nananeouly and () = /3) : 3 6 F 3 Fndng he curren, may be mplfed by combnng he 6 F and F capacor n ere a a 4 F capacor wh nal olage of. Oherwe, defne an adonal olage acro he F capacor. K: 3 = Dfferenang h equaon: d/ 3d/ d / =... () and = 6d/ d/ = /6... () alo = d / d / = /... (3) () and (3) n (): /6 3d/ / = 3d/ + /4 = d/ + / = () = (/3)e / ( ()e (/) ), F EIAM4 haper : Fr Order rcu Page - 9 and from (): () = 6 + () () = 6 + () 6F () = e / = e / () = e / = / e e () = / e = (4/3)e / (4/3) + 4 () = (4/3)e / + 8/3 for (() = 8/3) {An: () = /3 + /3(e -/ )u(), () = 4 + 4/3(e -/ )u() } 4 = 3 6 F 4 3 8/3 8/3 4 3 An analogou waer ank problem 4 m (4 ) H F = 3 m ( ) (6 F) 6 m m ( F) 3 Waer draned = Waer ganed [harge lo 6F = harge ganed F] (4 H) 6 = (H ) H = 8/3 m may be obaned from = 3

11 EIAM4 haper : Fr Order rcu Page -.3 Zero Sae epone Now ha we hae dcued he repone of fr order crcu ha had no npu and nonzero nal conon, le u nex conder he cae of repone o a nonzero npu for a crcu n whch all he nal conon are equal o zero. Such a repone called a zero-ae repone. Example -5 onder he crcu n Fgure -5. Fnd (), () and () for all. u() u() Fgure -5 u() for for = u() d d u() for Snce u(), for le u conder he aboe dfferenal equaon for wo me neral < and. Fr, for < : d u() = Alhough a ral oluon, () = clearly a oluon of h equaon. Snce here no nal olage on he capacor and nce he npu zero before me = (zero ae crcu), h ral oluon wha we eek, ha : () =, for <. Nex for, he dfferenal equaon become d u() = Th equaon may be wren: d d

12 EIAM4 haper : Fr Order rcu Page - d ln( ) = + K. () ( K a conan of negraon ) Bu () = for <, and herefore () =, a he capacor olage canno change nananeouly. og rule: ln( ) = + K K = ln() ln(a/b)=ln(a) ln(b) Subung K = ln() n Equaon () aboe: ln( ) = ln() ln() ln( ) = ln e/ e x e / e/ e x e / ( e /) () ( e / ) () whch can alo be expreed a he ngle equaon: () ( e / )u(),. We can now fnd he capacor curren, (), from (): d() () d d For < : () [()] = () = ( = for < ) d d For : () [()] = ( e / d ) = e / d / () e = e/ = e / () / e Alernaely, e / () u(),. The olage acro he reor may now be found wh: () = () = () () e/u(), / () ()

13 EIAM4 haper : Fr Order rcu Page -.4 Forced and Naural epone For he preou example, he dfferenal equaon for he capacor olage wa: d u(), and he oluon of h equaon : () ( e / )u() = u() e /u(). Th expreon he um of wo erm, one hang he form of he npu and he oher hang he form of he zero npu repone. To underand ha h wll alway be he cae, conder he general fr order lnear dfferenal equaon: dx() ax() f(), where f() he npu o he yem and x() he repone. Mulplyng boh de by e a, ge: dx() ea e a ax() e a f() d [e a x()] e a f() A a conan of negraon. ea x() e a f() A e af() herefore an ndefne negral (no lm). x() e - a e a f() Ae - a Equaon - Th a ery mporan reul, ald for any f(). The complee oluon con of he um of wo par. The fr par, e - a e a f(), a funcon of he npu f() and called he forced repone (alo known a he eady ae repone) and we refer o f() a he forcng funcon. The oher par of he oluon, Ae - a, whch we recognze ha he form of he zero-npu repone or naural repone ( alo called he ranen repone). The naural repone n general (f a > ) anhe wh me. Now le u conder he pecfc cae where he forcng funcon a conan b (uch a a baery). dx() ax() b b The forced repone x f () gen by, x () e - a e a b e a b e a f and a a he naural repone x n (), gen by, x Ae - a n (). Thu he complee repone, x() = x f () + x n (). x() = b Ae a Equaon - a Th a ery mporan reul ald for he pecal cae f() = b = conan. We are now n a poon o analye zero ae, zero npu and crcu wh boh nonzero nal conon and nonzero npu, n a yemac way.

14 EIAM4 haper : Fr Order rcu Page - 3 Example -6 For he zero ae crcu n Fgure -6, fnd () and (). u() Fgure -6 By K: u() = d u() d d u() u() For < : d (u() = ) (he ral oluon, () = zero ae crcu) d For : dx() (of he form ax() b, b / f () =, refer o Equaon -) a / and n () = Ae a = Ae (/). Wh () = f () + n (), hen () = Ae /. Snce () =, = A A =. () () ( e / ) for. / () ( e / )u(),. And d (). For < : d d () [ ()] = (). ( () = for < ) For : d () [ ()]. () d () = ( e / ) d () = e / d () / e = e / = e /. () e /u(),. u()

15 EIAM4 haper : Fr Order rcu Page - 4 Example -7 For he zero ae op-amp crcu n Fgure -7, fnd () and o () f () = u(). o () = u() N F 8 c Fgure -7 By K a node N: + = Bu = (/8)d /, = / and = / = (op-amp rule = ) (/8)d / + / = d / + 4 = 8 < : d / + 4 = ( () = for < ) () = : d / + 4 = 8 ( () = for ) () = 8/4 + Ae -4 = + Ae -4 (from Equaon -) Bu () = = + A A = () = e -4 = ( e -4 ) for. () = ( e -4 )u() ol Furhermore, () = () + () = () + () () = ( e -4 )u() + u() () = (3 e -4 )u() ol (ral oluon, zero ae crcu) Example -8 For he zero ae crcu n Fgure -8, deermne () for () = u() u( ). 3 () = = + ( - ) = ( - ) + ( - ) = + = ( + ) = ( + ) + ( + ) = + = jump o bu reman Non nerng amplfer dc gan + (/) = 3 u() () Fgure -8 = = (d /) = ()d / + = d / + (/) = / u( )

16 EIAM4 haper : Fr Order rcu Page - 5 For <, = and he dfferenal equaon become: d / + (/) =, and nce () nally zero, () = for <. For <, = and () =, o he dfferenal equaon become: d / + (/) = / () = + Ae /, ( Equaon - ) and wh () =, A = () = ( e / ). For, = and ( ) = ( e / ), o he dfferenal equaon become: d / + (/) = () = + Ae / ( Equaon - wh b = ) () = Ae / Therefore ( ) = Ae /. Bu a he end of he he preou me neral we alo had ( ) = ( e / ). ( e / ) = Ae / A = ( e / )e / () = ( e o/ )e o/ e / = ( e o/ ( o)/ )e So ha () = ( o )e ( )/ ( a expeced ) () ( e / ) ( o / o )/ [ ( e )] e for for o for o Example -9 For he crcu n Fgure -9, deermne () f he npu curren gen by: () for for 3 F Fgure -9 () - c By K: + = 3d/ + / = d/ + /6 = /3 ( = ) For < : d/ + /6 = /3. Becaue () wa equal o (a conan) for uch a long me, from = o =, hen when we are near o =, () mu hae eled o a conan alue, ay K. Bu h conan alue, () = K, mu ll afy he dfferenal equaon. dk/ + K/6 = /3 + K/6 = /3 K = or () = for <. For : d/ + /6 = /3 ( = ) () = [( /3)/(/6)] + Ae /6 (Eq. -) () = 4 + Ae /6. Or alernaely for <, when all he curren and olage are no changng anymore (a dc cenaro), hen when we are near =, ay when =, he capacor mu behae a an open crcu. A learly () = ol.

17 EIAM4 haper : Fr Order rcu Page - 6 Bu () = = 4 + A A = 6. () = 4 + 6e /6 for. Or for () 4 6e /6. for () = u() +( 4 + 6e /6 )u() = + 6(e /6 )u(). (Noe: A mlar argumen a for < wll confrm for >, ha () end o 4, when end o nfny: dk/ + K/6 = /3, or K = 4 (+) 4 ol.) Example - For he crcu n Fgure -, fnd he zero ae repone () and (). () -4 = e -4 u() 6 H Fgure - By K: For < : For : = /6 + = (d /)/6 + d / + 3 = 3 d / + 3 = 6e -4 u() () = (zero ae) d / + 3 = 6e -4 u() d / + 3 = 6e -4 (u() = for ) () e 3 e 3 6e 4 Ae 3 (from Equaon -) () 6e 3 e Ae 3 () = 6e 3 ( e ) + Ae 3 () = 6e 4 + Ae 3 Bu () = = 6 + A A = 6 () = 6e 3 6e 4 for () = 6(e 3 e 4 )u(). And () = 6( ) = 6(e 4 u() 6(e 3 e 4 )u() () = e 4 u() 36e 3 u() + 36e 4 u() () = 48e 4 u() 36e 3 u() () = (4e 4 3e 3 )u(). d [f()g()] df() g() f() Or alernaely: () = d / = d/[6(e 3 e 4 )u()] = d/[(e 3 e 4 )u()] () = [( 3e 3 + 4e 4 )u() + (e 3 e 4 )()] (du()/ = () ) () = (4e 4 3e 3 )u() Noe: Accordng o he amplng propery of he mpule funcon (), f()() per defnon equal o f()(). Therefore (e -3 e -4 )() = (e e ) () = () =. dg()

18 EIAM4 haper : Fr Order rcu Page - 7 Exerce P -5 For he crcu n Fgure P -5, fnd he zero ae repone (). () = u() Fgure P -5 An: ½( e -/ )u() P -6 For he crcu n Fgure P -6, fnd he zero ae repone (). 6 () = u() 5 F 3 An:.486( e -7.5 ) Fgure P -6 P -7 Fnd he zero ae ep repone, (), for he crcu n Fgure P -7 o Fgure P -7 An: () = - / ( e -/ )u()

19 EIAM4 haper : Fr Order rcu Page - 8 P -8 Fnd he zero ae ep repone () and (), for he crcu n Fgure P F 6 Fgure P -8 An: () = /3(-e -4 )u() () = /3e -4 u() P -9 For he crcu n Fgure P-9, deermne () for all, f () gen by: () - for for () 5 F - Fgure P -9 An:. = /5 + d/, for <: = /5 + d/ () = r < And for >= : d/ + /5 = - = [-/(/5)] + Ae -/5 () = = -5 + A A = 5 () = e -/5 for >= []

20 EIAM4 haper : Fr Order rcu Page - 9 P - Deermne he zero ae ep repone () for he crcu n Fgure P -. F 8 6 () = u() Fgure P - An:. For <, () = ; zero ae repone For > : - = / + (/8)d/( -) 8-8 = 4 + d / - d/ 8 = + d /-d/ and / = -/6 = -/3 8 = -4 - (/3)d/-d/ 8 = -4 - (4/3)d/ 6 = -3 - d/ d/ + 3 = -6 () = - + Ae -3 bu () = = - + A A = () = (e -3 ), for > P - For he crcu n Fgure P -, deermne () f he npu olage gen by: for (). - 4 for 3 5 () F -4 Fgure P - An: () = e -/6 P- efer o he crcu n Fgure P - and deermne he zero ae ep repone () and (). = u() H Fgure P -

21 EIAM4 haper : Fr Order rcu Page - An: For < : () = and () = For > : = / and = d/( ) = d/ = [ d/]/ d/ + = = + Ae bu () = = + A A = () = e for > and = d/ = e for > P -3 efer o he crcu n Fgure P -3 and calculae () and () for all. = 4 6 A /7 F < : One could ealy ue he uperpoon heorem o calculae = () or replace he 6A / 4 Noron crcu wh a 4 / 4 Theenn equalen crcu. On he oher hand, defnng an auxlary curren S, could perhap alo be calculaed whou oo much rouble. By K (loop ): 4(6 S) S = + 6S = 4 () By K (loop ): 4 + 4S = 4S = () () + 4 (): 5 = 4 = 8 ol. () = 8 and () = (cap open crcu) Fgure - : Agan, one could mplfy he problem by //4 3 replacng he 6A/4 Noron crcu wh 6 a 4/4 Theenn crcu. Bu we can 3 alo ju ole he crcu a. 6 4(6 ) By K: 4(6 ) 3 = 7 + = 4 Bu = (/7)d/ 7 (/7)d/ + = 4 4 /7 F d/ + = 4 () = 4 + Ae. Bu () = 8 A = -6 () = 4 6e () = 8 8u() + (4 6e )u() = 8 + 6( e )u() () = (/7)d/ = (/7) 6e = (6/7)e and () = for < : () = (6/7)e u() S 4(6 S) S 4 S 4 4S

22 EIAM4 haper: Second Order rcu Page -. SEOND ODE IUITS. Zero Inpu rcu Suppoe ha for he zero npu crcu n Fgure -, he nal curren n he nducor () and he nal olage acro he capacor (). Fgure - d From he crcu, by K : d Ung he fac ha, we ge he econd order dfferenal equaon: d d d d Equaon - d Agan from he crcu: Dfferenang once yeld he econd order dfferenal equaon: d d d d Equaon - Boh Equaon - and - hae he ame form and we wll now conder he oluon o he general econd order dfferenal equaon: d y() dy() n y() The oluon may ake one of hree form (pleae noe ha h ONY he naural zero npu oluon): ) If > n (oer damped cae): e y() A e A, where ) If = n (crcally damped cae): n and y() = (A + A ) e 3) If < n (under damped cae): n, y() = e (A co d + A n d ), where d = n The conan A and A, may be calculaed from he nal alue y() and dy().

23 EIAM4 haper: Second Order rcu Page - Example - For he crcu n Fgure -, uppoe ha = 5, = ½ H and = ⅛ F. Alo uppoe ha () = A and () =. Deermne () and (). = (/) = 5 and n = /() = 4 So > n and he crcu oer damped. = n = 8 and = n = 8 () = A e + A e Wh () = : A + A =...() 8 And d()/ = 8A e A e The econd nal conon gen wa () = and no d()/. Howeer, referrng o he crcu: d()/ = () () d()/ = (/)() (/)() = () () d()/ = = d()/ = () () = () () = 4 Bu d()/ = 8A A 8A A = 4...() From () and (): A = and A = 8 () = e e for A Equaon - ha he ame form a Equaon -, 8 () = B e + B e Seng = yeld: B + B =...(3) 8 And d()/ = 8B e B e From he crcu, () = d()/ d()/ = (/)() = 8 8B B = 8...(4) From (3) and (4): B = and B = 4 8 () = e + 4e for Of coure () may alo hae been oled from: () = d()/ () 8 = ( /)( 6e + e 8 ) 5(e e ) 8 = e + 4e () = 4e - e -8 () = e - + e -8

24 EIAM4 haper: Second Order rcu Page -3 Example - For he crcu n Fgure -, fnd () and () f () = A and () =.4. (/) (/5) H F Fgure d/ = and = (/5)d/ (/5)d/ + +(/5)d / = d / + 6d/ + 5 = = 3 and n = 5 o n > under damped cae wh d = 4 r/. () = e 3 (B co4 + B n4) () = B = and d/ = e 3 ( 4B n4 + 4B co4) 3e 3 (B co4 + B n4) d()/ = d/ = = 3B + 4B = 4B 3 From he crcu: = d/ d/ = (/) = 5 d()/ = 5() d()/ = 5(.4) = 7 4B 3 = 7 B = () = e 3 (co4 n4) for = e 3 co(4 + /4) Noe: B co d + B n d = Bco( d ) where B = B B and = an - B B And () = d/ = (/5)d/[e 3 (co4 n4)] = (/5)[e 3 ( 4n4 4co4) 3e 3 (co4 n4)] = (/5)e 3 ( 7co4 n4) = (/5)e 3 [ (7) () co(4 )] an = 7 auon when calculang - refer o dagram. 7 = an = /7 = an - (/7) =.49 = an - ( / 7) = + = or = = =.9997 = () = ( 5 /5) e 3 3 co[4 (.9997)] = (/)e 3 co(4 + 3) for d()/ = 7 ()= e -3 co(4+/4) ()= (/)e -3 (co(4 + 3) -.4

25 EIAM4 haper: Second Order rcu Page -4 Example -3 For he crcu n Fgure -3, deermne () and () f () = A and () =. 3 H F Fgure d/ + = and = d/ + 3(d/ + ) + d/(d/ + ) + = d / + 4d/ + 4 = = and n = crcally damped cae. () = (A + A )e - () = A = d/ = A e - (A + A )e - d()/ = A A = A 4 Bu d/ = d()/ = () () = = = A 4 = A = 3 () = (3 + )e - for and () = d()/ + = [3e - (3+)e - ] + (3 + )e - = 3e - + e - = ( 3 + )e - for () = (3+)e - () = ( 3+)e -

26 EIAM4 haper: Second Order rcu Page -5 Exerce P - For he crcu n Fgure P -, fnd () and () for. 6 = /3 F A 3 /4 H Fgure P - An: () = -e e -, () = e -6 e - A P - For he crcu n Fgure P -, fnd () and () for all. u() F H Fgure P - An: -[-e - co(+/4)]u(), -[-e - co(-/4)]u() P -3 For he crcu n Fgure P -3, deermne () and () for. u() H F Fgure P -3 An: () = e -/ co(3/) + (/3)e -/ n(3/) A () = -(4/3)e -/ n(3/)

27 EIAM4 haper: Second Order rcu Page -6. rcu wh nonzero npu We hae condered ome econd order crcu wh zero npu and nonzero nal conon and we wll now look a econd order crcu wh nonzero npu alhough we hall lm our dcuon o forcng funcon ha are ep funcon. onder he zero ae ere crcu n Fgure -4, whoe npu a olage ep funcon. u() Fgure -4 d By K: u() d d d Snce : u() d d d u() d For < h equaon become: and he oluon () =. Howeer for, d d, Equaon -3 and a n he cae of fr-order equaon, he oluon () con of wo par he forced repone f () and he naural repone n (). Tha he oluon of he form: () = n () + f (), d d where n () he oluon o,, where = / and n = /. The forced oluon mu be of he ame form a he forcng funcon. Thu he forced repone a conan, ay K, and mu afy Equaon -3, ha, d K dk K or K =. Hence, he oluon o Equaon -3 for he oer damped cae () A e A e. For he crcally damped cae, () = + A e - + A e -. For he under damped cae, () = + A e - co d + A e - n d. n

28 EIAM4 haper: Second Order rcu Page -7 Example -4 For he crcu n Fgure -4, he under damped cae reul when =, = H and = (/5)F, nce =3 and n = 5. Alo d = 4. If = (/5), hen he oluon o Equaon -3 : () = /5 + A e -3 co4 + A e -3 n4. And () = /5 + A = A = -/5 Alo d()/ = -3A e -3 co4 4A n4 3A e -3 n4 + 4A e -3 co4 d()/ = -3A + 4A = 6/5 + 4A From he crcu () = d()/ () = (/5)d()/ Bu () = d()/ = 6/5 + 4A = A = -6/ = -3/ Therefore for : () = /5 (/5)e -3 co4 (3/)e -3 n4 = /5 (/)e -3 (4co4 + 3n4) = /5 (/)e -3 ((4 +3 ) 5co[4 + an - (3/4)] = /5 (/)e -3 co(4+.643) ombnng h wh () = for < () = [(/5) (/)e -3 co( )]u() The zero ae curren repone () = d()/ = (/5)d/{[/5 (/5)e -3 co4 (3/)e -3 n4]u()} = (/5){[/5 (/5)e -3 co4 (3/)e -3 n4]() + [(6/5)e -3 co4 + (8/5)e -3 n4 +(9/)e -3 n4 (/)e -3 co4]u()} = (/5){[(/5) (/5)]() + [(5/)e -3 n4]u()} = [(/)e -3 n4]u() ()=[.4-.5e -3 co(4+.643)]u() ()=[.5e -3 n4]u().4... Example -5 The crcu hown n Fgure -5 n eence an old-fahoned (non-elecronc) auomoble gnon yem. The npu o he crcu a auomoble baery. The ere reance con of a balla reor, he reance of he gnon wch and he ere reance of he nducor (known a he gnon col ). The capacor (called a condencer ) n parallel wh a wch (called he pon ) ha cloed for < and open for. (In acualy, he pon open and cloe perodcally he rae dependng on he engne rpm.) I he olage produced acro he col ha appled o he park plug. Th n urn produce a park ha gne he fuel mxure.

29 EIAM4 haper: Second Order rcu Page -8 3 H F = Fgure -5 For he gen crcu, for <, clearly () = and () = /3 = 4 A. Thu () = and () = 4 A. For, by K, d 6 3 d d 6 3 Hence = 3/ < n = 3 The crcu (ery) under damped wh d = n n = 3 Snce he forced repone zero, he approxmae curren ha he form, () A e -3/ co 3 + A e -3/ n 3 and () = 4 A = 4. Nex d()/=-6e -3/ co e -3/ n 3 -(3/)A e -3/ n A e -3/ co 3 d()/ = A Howeer, from he crcu (for ) by K: = 3 + d/ + d/ = 3 and d()/ = 3() () = = A = A = 6-3 The curren hen () 4e -3/ co e -3/ n 3 The econd erm conderably maller han he fr, a good approxmaon () 4e -3/ co 3 and he olage acro he nducor approxmaely () = d/ d/[4e -3/ co 3 ] -6e -3/ co n 3 In h expreon he fr erm conderably maller han he econd, o a good approxmaon () -4 3 e -3/ n 3. e u ealuae h formula he fr me he ne erm equal uny ha for = (/) -3 ec. Then we ge, () -4 3 e -3(/)-3/ -4 Alhough he alue of and gen n h example are praccal n naure, hey were ll choen o yeld compuaonal conenence. In a ypcal auomoble, he olage requred o produce a park acro he gap of a park plug beween 6 and.

30 EIAM4 haper: Second Order rcu Page -9 Exerce P -4 For he crcu n Fgure P -4, fnd () for all ()=-4+u() 6 7 H /4 F Fgure P -4 An: () = -4 + (5 + e -6 6e - )u() A P -5 For he crcu n Fgure P -5, fnd he zero ae repone () and () ()=u() H / F Fgure P -5 An: () = ( e - e - )u() A, () = e - u() P -6 For he crcu n Fgure P -6, fnd he zero ae ep repone (). F o o ()=u() /5 F Fgure P -6 An: () = [ - (5)/e - co(.464)]u()

31 EIAM4 haper: Second Order rcu Page - P -7.The wch n he crcu of Fgure P -7, cloed a me =. Deermne () and () for all. H 4 = F 4 Fgure P -7 An: For <: () = A and () = For >: (/)d/ + = and = (/4)d/ + (/)d/[(/4)d/ + ] + = (/8)d / + (/)d/ + = d / + 4d/ + 8 = = en n = 8 under damped wh d = r/ () = Ae - co + Be - n () = A = d/ = -Ae - co Ae - n Be - n +Be - co d()/ = -A+B bu ()=(/4)d()/ + () = (/4)d()/ + d()/ = -() + B= B= () = e - co + e n for > and d/ = 4e - co 4e - n 4e - n +4e - co = 8e - n () = (/4)d/ + = (/4)[ 8e - n] + e - co+e - n () = e - co for > P -8 The wch n he crcu of Fgure P -8, opened a me =. Deermne () and () for all. H F = Fgure P -8 An: For < : () = /A and () = For > : + d/ + = and = d/ d/ + d / + = d / + d/ + = = and n = crcally damped: () = + Ae - + Be - () = A = - d/ = -Ae - + Be - Be - d()/ = -A+B bu () = d()/ d()/ = ½ -A + B = ½ B = -½ [] () = e - ½e r > and = d/ = e - ½e - + ½e -

32 EIAM4 haper 3: Snuodal Analy Page 3-3. SINUSOIDA ANAYSIS 3. Tme Doman Analy The nuod an exremely mporan and ubquou funcon. To begn wh, he hape of ordnary houehold olage nuodal. onumer rado ranmon are eher amplude modulaon, n whch he amplude of a nuod changed, or frequency modulaon, n whch he frequency of a nuod changed. Snuod een occur n uble way, for a nonnuodal waeform (lke a awooh or pule ran) n eence ju he um of nuod. We are herefore now nereed o deermne he repone of a crcu o a regular nuodal npu. For a fr order crcu a lea, we could ue Equaon - o oban he oal oluon f a nuodal npu f(), appled o he npu a =. We wll howeer rerc ourele o a nuodal npu ha wa preen n he crcu for all me. The oal oluon for he repone, wll herefore only con of he forced (eady ae) componen and no ranen wll be preen. In h way we can hen relae our reul n he me doman, o ha n he frequency doman where we perform analy n a more mplfed manner ha aod he drec oluon of dfferenal equaon. Example 3- onder he nuodal crcu hown n Fgure 3-. Deermne he eady ae repone, () and () for h crcu. Snce = + and = ½d/, hen ½d/ + =, o ha d/ + =. d/ + = n.. () For any lnear crcu he forced repone mu be a nuod f he crcu dren by a nuodal gnal. For our example we herefore n ha he repone mu be of he form Aco( + ) (or An( + ) f we wh), reng he fac ha he repone may hae anoher amplude and phae angle a he npu gnal bu wll defnely be nuodal wh he ame frequency a he npu ( r/). In general, any funcon ha afe he dfferenal equaon (), wll be he requred eady ae repone. Fndng he eady ae oluon hould herefore be farly eay a we already know ha mu be of he form Aco( + ). When fndng he eady ae oluon for () from he dfferenal equaon (), we wll howeer prefer (h wll requre a lle b le manpulaon) he alernae bu equalen form o Aco( + ) namely: () = A co + A n. Subung () = A co + A n (and d/ = A n + A co) n he dfferenal equaon (), reul n, ( A + A )n + (A + A )co = n. () Equang coeffcen (ung he fac ha f Aco + Bn = co + Dn, hen A = and B = D) n equaon (), yeld he mulaneou equaon: A + A = and A + A = wh he oluon: A = 3 and A = 3. = 6n ol ½ F From Equaon - he ame reul for he forced repone - a a - e e f() = e e n = -3co+3n The eady ae olage hen: () = A co + A n = 3co + 3n. Fgure 3-

33 Therefore () n he form: () = Aco + Bn = co( ), where =, = (A +B ), = an - (B/A), A = 3 and B = + 3. Thu = (A +B ) = 3/4 B = +3 = (3 + 3 ) /4 +3 A = -3 -/4 = ( 9) = 3-3 and = an - (B/A) = an - (+3/ 3) = 3/4 (ee kech). Some calculaor may erroneouly calculae: an - (+3/ 3) a an - ( 3/+3) = /4. The correc angle may ll be obaned wh: = /4 +. () = 3co( 3/4) ol. EIAM4 haper 3: Snuodal Analy Page 3- Ueful rgonomerc dene: Aco + Bn = (A +B )co[ an - (B/A)] = (A +B )n[ + an - (A/B)] co = n( + /) n = co( /) co = co( + ) add o n = n( + ) change gn n(a + B) = nacob + coanb co(a + B) = coacob nanb The eady ae curren : () = ½d/ = ½ (6n + 6co) = 3n + 3co. () = 3co( /4) amp. ( or () = ½d/(3co( 3/4)) = 3n( 3/4) ) (() and () may alo be expreed a () = 3n( /4) and () = 3n( + /4)). () = 6n () = 3n( + /4) (capace curren () leadng upply olage () by 45, and olage () acro capacor, by 9, a hould.) Fgure 3- () = 3n( /4) Example 3- onder he econd order crcu n Fgure 3-3. Deermne he repone () and (). 5/3 5 H () = 7co3 ol = (5/3) + 5d/ + and = (/5)d/ 7co3 = (5/3)[(/5)d/] + (/5)d / + d /d + (/3)d/ + 5 = 85co3 The forced repone ha he form: () =A co3 + A n3. ( 4A + A )co3 + ( A 4A )n3 = 85co3, yeldng: 4A + A = 85 and A 4A =, wh he oluon: A = and A = 5. () = co3 + 5n3 =.6co(3.89). And () = (/5)d/ =.474n(3.89) {() may alo be expreed a: /5 F Fgure 3-3 an ( 5/+) =.5 () =.474co(3.89 /) =.474co( ) =.474co( ) =.474co(3.3) (add o change gn) } orrec we are lookng for an (+5/ ) =.5 + (or ) =.89 (or 3.39) +5 alculaor ge a

34 EIAM4 haper 3: Snuodal Analy Page Frequency Doman Analy Gen a lnear crcu whoe npu he nuod Aco(+), he forced repone a nuod of frequency, ay Bco(+). If he npu delayed by 9, ha, f he npu An(+), hen by he me-narance propery, he correpondng repone Bn(+). Furhermore, f he npu caled by a conan K, hen by he propery of lneary, o wll be he repone; ha, he repone o KAn(+) KBn(+). Agan by lneary, he repone o: Aco( + ) + KAn( + ), Bco( + ) + KBn( + ). onder now he pecal cae ha he conan K = - = j. Thu he repone o: Aco( + ) + jan( + ) = Ae j(+), Bco( + ) + jbn( + ) = Be j(+). We herefore ee ha wheher we wh o fnd he repone Bco(+) or Be j(+) o he excaon Aco(+) or Ae j(+), we mu deermne he ame wo conan B and. In Example 3- we hae found he repone o he real nuod Aco(+). In he nex example we wll fnd he repone o he complex nuod Ae j(+). (Noe: I he conenon o aocae he real nuod Aco(+) wh he complex nuod Ae j(+), when conerng beween real and complex nuod. Alhough wa conenen o ue he more famlar ne wae for comparon n Fgure 3-, we wll from now on alo adop h rule and alway prefer o ue he cone funcon a reference, n faour of he ne funcon. I herefore mporan o change a real nuodal gnal no he forma Aco( + ) before conerng o he complex nuod Ae j(+) and lkewe, a complex nuod Ae j(+) mu be underood o repreen a real cone gnal Aco( + ).) Example 3-3 econder he crcu n Example 3- (redrawn agan n Fgure 3-4). We hae already een ha d/ + = and ha f he npu () = 6n = 6co( /) (nng on he forma Aco( + ) ) hen he forced repone of he form () = Bco( + ). Euler rule: If y = co + jn ( y() = ) hen dy/d = n + jco = j n + jco = j(co + jn) dy/d = jy or dy/d jy = Soluon: y = Ke j. Bu y() = y = e j e j = co + jn Ae j = Aco + jan = A Now, nead of ung 6co( /), we wll ue () = 6e j( /) a he npu and nead of Bco( + ), we wll ue () = Be j( + ) a he forced repone. eplacng () wh Be j( + ) and () wh Be j( + ) n he aboe dfferenal equaon, yeld: d/(be j( + ) ) + (Be j( + ) ) = e j( /) jbe j( + ) + Be j( + ) = e j( /). jbe j e j + Be j e j = e j/ e j ( e (a+b) = e a e b ) jbe j + Be j = e j/. ( ddng by e j ) (In he preou ep we dded boh de by e j whch mean ha we cancelled ou me and frequency, hereby ranformng he nuodal (ac) problem no a phaor problem. Th really he eence of phaor analy, dcued n he nex econ.) jbe j + Be j = 6e j/ (j + )Be j = 6e j/ Be j = 6e j/ /(j + ) j( /) = 6e Fgure 3-4 ½ F

35 EIAM4 haper 3: Snuodal Analy Page 3-4 Be j = (6 /)/(/4) = (6/) 3/4 = (3) 3/4 = 3e j3/4. Be j = 3e j3/4 whch mean ha B = 3 and = 3/4. We conclude herefore ha he repone o () = 6e j( /) () = 3e j( 3/4), and herefore, he repone o () = 6co( /) (or a wa gen, () = 6n), () = Bco( + ) = 3co( 3/4) ( compare wh Example 3- ) Phaor Im Ae j(+) Ju a he nuod Aco(+) ha frequency, amplude A and phae angle, we can ay ha he complex nuod, A Ae j(+) = Ae j e j (a roang ecor) + alo ha frequency, amplude A, and phae angle. e u denoe he complex number A = Ae j by A = A whch a Aco(+) e ac ecor (no roang) and known a he phaor repreenaon of he real (Aco(+)) or complex (Ae j(+) ) nuod. The phaor preere he eenal nformaon (A and ), of he nuod bu dcard frequency and me (e j ). Ung phaor o repreen nuodal quane, open he pobly o ole problem n he frequency doman by hdng frequency and me n he background, and ung only complex algebra o ole he crcu n nearly a ac dc-lke manner. Noe: I cuomary o repreen me aryng nuod wh mall leer (for example ()) and phaor wh bold capal leer ( for example). onder a reor wh a complex nuod olage = e j+) acro. The curren ha flow hrough wll hen alo be a complex nuod ay = Ie j(+). By Ohm law, = e j(+) = Ie j(+) e j e j = Ie j e j e j = Ie j (ddng by e j ). Thu, we hae he phaor equaon: = I, where = and I = I. For a reor, of coure, he olage and curren are n phae (Z = ) and =, a mpled by = I. (If 3 = hen = 3.) Smlarly for an nducor, f = e j(+) and = Ie j(+), hen d = e j(+) d = (Ie j(+) ) e j(+) = jie j(+) e j e j = jie j e j. e j = jie j, and ung phaor form: = ji, where = and I = I. Becaue = ji, he angle of mu be equal o he angle of ji. angle() = angle(/ I) = angle(i(/ + )) = + /. Thu we ee ha for an nducor, he curren lag he olage by 9 and from = ji, Z = /I = / r = 9. For a capacor, f = e j(+) and = Ie j(+), hen d = Ie j(+) d = (e j(+) ) Ie j(+) = je j(+) Ie j e j = je j e j. Ie j = je j, and ung phaor form: I = j, where I = I and =. Becaue I = j, he angle of I mu be equal o he angle of j. ang(i) = ang(/ ) = ang((/ + )) = + /. We deduce herefore ha for a capacor, he curren lead he olage by 9 and from I = j, Z = /I = /j = // = (/)/ r = (/) 9.

36 EIAM4 haper 3: Snuodal Analy Page 3-5 Example 3-4 econder he crcu for Example 3- (redrawn agan n Fgure 3-5). Ue phaor o ole for and I. = 7 = 3 ( = 7co(3 r )) 5/3 j=j5 = 59 /j=5/j3 = 5/3-9 K: = (5/3) I + 59 I + (5/3) 9 I = ( ] I 7 = ( ) I I = 7/ = amp. and = = ol. { (() =.474co(3.36 r ) and () =.6co(3.897 r )} Example 3-5 For he crcu n Fgure 3-6, he frequency doman repreenaon hown. alculae he phaor o, and from o, conruc he me doman oluon o (). I Fgure 3-5 /9 F -j3/ = o 3 N M o 8co6 (=6r/) 8 =6 r/ (f=.9549hz) Tme doman Fgure 3-6 Frequency doman By K a node N: ( 8)/3 + ( )/ + ( o )/.5 9 = ( ) o.667 = ( ) (.66679) o = () By K a node M: ( )/ + ( o )/ = = o / () Subung from () n (): ( ) o (.66679) o =.667 (. 3.7) o =.667 o =.93.7 ol Therefore () =.9co( r ) ol π radan = degree 8 (Noe: Traonally, phaor angle are expreed n degree. For he me expreon aboe, we had o coner 3.7 o.59 rad, becaue he frequency 6 rad/ec. and he angle 6 n () aboe, herefore really an angle n radan. Neerhele, he cuom o upply he angle n he me expreon n radan and preenng an angle a = r + 3, would be aburd (ju ry o fnd co(rad + 3) wh your calculaor). Alo wh phaor analy, he frequency may be gen n Hz, and f o, wll hae o be conered o rad/ec. o form he me expreon. A a fnal noe, for mplcy we aumed ha he magnude of he phaor.93.7 wa equal o he peak alue of he nuodal expreon.9co(6+.59). Wh phaor, he normal pracce acually o ue he rm alue a he phaor magnude.)

37 EIAM4 haper 3: Snuodal Analy Page 3-6 Example 3-6 efer o he crcu n Fgure 3-7. a) Fnd he eady ae repone () and () n he me doman, olng he dfferenal equaon for and, ung he real nuod. 3 =.8co + 9.6n Fgure 3-7 H b) Fnd he eady ae repone () and () n he me doman, olng he dfferenal equaon for and, ung he complex nuod. c) Ue phaor n he frequency doman o calculae I and. a) K: = 3 + d/ d/ + (3/) = / d/ + (3/) =.4co + 4.8n The forced repone ha he form: () =A co + A n 3 =.8co + 9.6n H A n + A co + (3/)A co + (3/)A n =.4co + 4.8n [(3/)A + A ]co + [ A + (3/)A ]n =.4co + 4.8n (3/)A + A =.4 3A + 4A =.8 A =. A + (3/)A = 4.8 4A + 3A = 9.6 A =.6 () =.co +.6n ampere = ((-.) +.6 ))co[ an - ( +.6/.)] = (. +.6 )co[ ( +.43 r )] = co( ( +.43 r )) +.6 () = co(.43 r ) ampere (phaor I = 6.87 amp) And () = d/ = d/(.co +.6n) () = (+.4n + 3.co) () = 6.4co + 4.8n = ( )co[ an - ( + 4.8/ + 6.4)] = 8co( ( r )) () = 8co(.6435 r ) ol (phaor = ol) b) =.8co + 9.6n = co( an - (+9.6/+.8)) = co( ( +.87 r )) = co(.87 r ). The complex nuod ha correpond o he real nuod co(.87 r ) herefore e j(.87). (phaor = ol) an - ( +.6/.) = = +.43 r.973 (calculaor alue) an - ( + 4.8/ +6.4) = r an - ( +9.6/+.8) = +.87 r

38 We already know ha, d/ + (3/) = /. Wh of he form, = e j(.87). The forced repone () mu hae he form: () = Be j( + ). EIAM4 haper 3: Snuodal Analy Page 3-7 Subung he repone () and he upply olage no he dfferenal equaon d/ + (3/) = /, yeld: jbe j( + ) + (3/)Be j( + ) = e j(.87) / = 5e j(.87). jbe j e j + (3/)Be j e j = 5e j.87 e j jbe j + (3/)Be j = 5e j.87 [j + (3/)]Be j = 5.87 r (ddng by e j ) Be j = 5.87/[(3/) + j] = 5.87/ Be j =.43 = e j.43. So f Be j = e j.43 hen B = and =.43 () = e j(.43) amp aocaed wh () = co(.43 r ) a expeced. and () = d/ = d/[e j(.43) ] = [j4e j(.43) ] = j8e j(.43). Bu j = (/) () = (/) 8e j(.43) = e j(/) 8e j(.43). () = 8e j(.43 + /) = 8e j(.6435) ol. aocaed wh () = 8co(.6435 r ) a expeced. c) We hae already een n he preou econ ha, =.8co + 9.6n = co(.87 r ). The phaor repreenaon of herefore =.87 r = Z = = I = 73.74/553.3 = 6.87 ampere. And = = ol. 3 = I 49 I eay o reconruc () and () from he phaor I and f he frequency known o be rad/ec. and all he phaor lengh repreen peak alue raher han he rm alue of he correpondng nuodal gnal. I = 6.87 A () = co(.4 r ) ampere. = () = 8co(.6435 r ) ol. 3 j(.87) = e H j = / = e j(/) Summary of reul obaned wh hree echnque Phaordagram 3 I 6.87 A e j(.43) A co(.43 r ) A I e j(.87) co(.87 r ) e j(.6435) 8co(.6435 r )

39 Example 3-7 efer o he crcu n Fgure 3-8. a) Ue frequency doman analy o fnd o (). b) Fnd he mpedance een by he olage ource EIAM4 haper 3: Snuodal Analy Page 3-8 F Fgure 3-8 =3co5 F o N M. 9 = 3 = 3. 9 a) K a node N: (3 )/ = ( )/ + ( )/ (3 ) =. 9 ( ) + ( ).6 9 (. 9) = (. 9) (. 9) + (. 9 + ) = ( ).6 9 = (.77.8/..3) (.6 9/..3) = () K a node M: ( )/ = / ( ) =. 9 = ( +. 9) = (..3/. 9) = () () n (): = ( ) = = /

40 EIAM4 haper 3: Snuodal Analy Page 3-9 = () =.53co(5.747 r ) ol. b) From (): = = = ol I = ( )/ =.94.3 amp Z n = /I = 3/.94.3 =..3 = j. I = 3 Example 3-8 The upply olage o he crcu n Fgure 3-9, gen by () = co. a) Deermne he me doman oluon for he eady ae curren repone, (), by wrng down and olng a dfferenal equaon for (). Ge your anwer n he form Aco( + ). b) Ue frequency doman analy o deermne he phaor curren I. econruc () agan from he phaor curren I. = co F H = co ( + r ) = Fgure 3-9 a) c = d c / + en c = d/ d/ = d / + d / + d/ + / = (/)co = Aco + Bn, d/ = An + Bco en d / = Aco Bn -Aco Bn An +Bco + (A/)co + (B/)n = (/)co [B A/]co + [ A B/]n = (/)co A = -/5 en B = /5 () = (/5)co + (/5)n =.447co(.344) b) Z = + ( j//j) = j = I = / = A I = [ j/( j + j)] = ( 9/9) A = A () =.447co(.344) A

41 EIAM4 haper 3: Snuodal Analy Page 3- Exerce 3- Ung me doman analy, fnd he eady ae repone () and () for he crcu n Fgure P 3-. =5co4 A 3 (/) F (/8) H Fgure P 3- An: () = 3n4, () = 5co4 3- efer o he crcu n Fgure P 3-. a) Ue frequency doman analy (phaor analy) o fnd o (). b) Deermne he mpedance een by he olage ource. = 8co 3 H F o Fgure P 3- An: a) co( /) b) 3. + j For he crcu n Fgure P 3-3, fnd he phaor curren I and I. {an I = A, I = } I 5 3 Z Z = 78 j45 Z 5 9 Z I Fgure P 3-3

42 EIAM4 haper 4: eonance Page 4-4. ESONANE 4. Sere eonance onder he ere crcu n Fgure 4-. I The mpedance een by he ource, : Z = + j + j Fgure 4- Z = + j ohm We ay ha a crcu wh a lea one capacor and one nducor n reonance or reonan when he magnary par of mpedance (or admance) equal o zero. The crcu n fgure 4- herefore n reonance when = = Th parcular frequency called he reonan frequency r. A he reonan frequency he mpedance wll reach mnmum alue Z r =, and he curren maxmum alue of /. A plo of I eru hown n Fgure 4-. / I / r Fgure 4-

43 EIAM4 haper 4: eonance Page 4- A he frequence and, he magnude of he curren / of maxmum alue and he power dpaed n herefore half he power dpaed a reonance. The frequence and are called he half power frequence and we defne he bandwh BW of he reonan crcu by BW = Noe ha he maller he bandwh BW, he harper or narrower he amplude repone. Anoher quany ha decrbe he harpne of he frequency repone cure, he qualy facor Q, whch defned a he reonance frequency o be maxmum energy ored Q = oal energy lo n a perod For he ere reonan crcu n Fgure 4-, uppoe ha he olage ource gen by ()=co r. A reonance he mpedance a pure reor o ha ()=(/)co r. Alhough he poron n he crcu ac a a hor crcu, he nddual olage acro and are no zero, hey only add up o zero a each nan. The olage acro he capacor n parcular, c () = (()/)n r (remember ha c = IX c = (/)(/ r ) = ()/ and ha he capacor curren lead he capacor olage by 9). The energy ored n he nducor ½ = (½ / )co and he energy ored n he capacor ½ c = ½(()/) n r = (½ / )n r. The oal energy ored n he nducor and he capacor a any nan hen (½ / )co + (½ / )n r = (½ / ), whch a conan. The r.m.. curren hrough he reor (/)/ and he power aborbed by he reor P = ((/)/) = /, and he energy lo n one perod ( /)T = ( /)(/ r ) = / r. Thu he qualy facor of he crcu ½ / Q = = r / r Snce r = / r, alernaely we can wre Q = r = r = To relae he qualy facor o he bandwh, we noe ha a he half power frequence and, he curren dropped o / he maxmum a reonance whch mple ha he magnude of he mpedance mu hae ncreaed o me mnmum alue of a reonance. Thu a he half power frequence = =

44 EIAM4 haper 4: eonance Page 4-3 = = + Equaon 4- and = - Equaon 4- Equaon 4- a quadrac equaon wh a poe and negae oluon and o Equaon 4-. The wo negae alue are mply mrror mage of he poe alue and are of no phycal gnfcance. The wo poe alue of nere are = from equaon 4- and = from equaon 4- The bandwh hen gen by BW = = = r, from Q = r. Q Anoher ueful relaon nolng and ha may ealy be deduced from he aboe expreon for and, = = r Example 4- For he crcu n Fgure 4-, = mh, =5 F and he bandwh requred o be r/. a) alculae he alue of b) Deermne he half power frequence and. c) alculae he qualy facor Q d) Deermne he curren n he crcu a reonance and a he half power frequence f he upply olage gen by n r a) BW=/ =BW = e-3 = b) r = / = /e-35e-6 = r/ - =...() =...() From () = 4/ ubue n () 4/ - = + 4 = = 56. r/ and from () = 56. r/ c) Q = r / = e-3/ = d) Take he upply olage a reference: = A reonance: Z = +j( /) =

45 EIAM4 haper 4: eonance Page 4-4 I = /Z = / = A A = 56. r/: Z = +j( /) = +j(56.e-3 /56.5e-6) Z = j.9988 = I = /Z = / = A A = 56. r/. Z = +j( /) = +j(56.e-3 /56.5e-6) Z = +.7 = I = /Z = / = A 4. Parallel eonance e u now conder he cae of a parallel crcu a hown n Fgure 4-3. I Fgure 4-3 The admance een by he curren ource Y = Y + Y + Y = j = j - emen j For reonance: - = r = A he reonan frequency he admance wll reach mnmum alue Y r = /, and he olage = I/Y, maxmum alue of I. A plo of eru hown n Fgure 4-4. I I r Fgure 4-4

46 EIAM4 haper 4: eonance Page 4-5 A for he ere reonance crcu, he bandwh gen by: BW = Smlar alo o he ere reonance crcu, we hae for he qualy facor of he parallel reonan crcu: Q = r = r = A he half power frequence, he admance mu ncreae o me mnmum alue of / a reonance, o ha = From h he half power frequence may be obaned a = and = The bandwh hen gen by BW = = = r, from Q = r. Q Agan and are relaed by = = r Example 4- The amplude of he nuodal curren ource n he crcu n fgure ma. The crcu parameer are = k, = 4 mh, and =.5 F. a) Fnd r, Q,, and he amplude of he oupu olage a r, and. b) Wha alue of wll produce a bandwh of 5 r/? c) Wha he alue of Q n par b) a) r = = = krad/ 4e - 3.5e - 6 Q = r = (e3)()(.5e-6) = 5 BW = r /Q = e3/5 = r/ =...() - =...() From () and (): = r/ and = 49.9 r/ m ( r ) = I m = 5e-3 = m ( ) = m ( ) = m ( r )/ = 7.7 b) BW = / 5 = /(.5e-6) = 8 c) Q = r /BW = /5 =

47 EIAM4 haper 4: eonance Page Adonal eonance rcu onder he more realc ere-parallel rucure n Fgure 4-5. I g Fgure 4-5 The admance of he crcu gen by: Y = / g + j + /( + j) = / g + j + ( j)/( + ) = [/ g + /( + )] +j[ /( + )] For reonance: /( + ) = + = /...() = (/ )/ = / (/) r = A reonance hen he admance a pure conducance Y r = / g + /( + ) = / g + / ee equaon () Th expreon for Y r already hn on he dea ha a reonance, we may conder he econ a a reor wh alue /. To proceed wh analy of he crcu n Fgure 4-5, we mu mplfy and our raegy wll be o approxmae he crcu n Fgure 4-5 wh a mple parallel rucure mlar o he crcu n Fgure 4-3. onder he ere and parallel econ n Fgure 4-6. P P Fgure 4-6 Our nenon now o fnd alue for P and P n order o force he parallel

48 EIAM4 haper 4: eonance Page 4-7 rucure o be equalen (n an mpedance ene) o he ere rucure. Th wll only be poble a one frequency. If we dere o make her admance equal a a ceran frequency, hen we mu hae: j j P P j j ( ) P P ( ) ( ) Thu and P P If we now aume ha he ere econ n fac he econ of he reonan crcu n Fgure 4-5 and ha he nernal reance of a hgh qualy col, ha f Q=/ ery large whch mple ha much maller han, or ha +() (), hen we hae he approxmae relaon and ( ) P P from Equaon () f we e / The parallel rucure ha approxmae he crcu of Fgure 4-5, gen n Fgure 4-7. I g Fgure 4-7 Example 4-3 For he crcu n Fgure 4-5 aume ha g =, =., =. H and =. F. alculae r, Q and BW of he crcu.. r = = = 9.95 r/ ( r.... P = = = and P // g = // = Q = BW = r 995 = Q = 9.9. =.5 r/ = r/)

49 EIAM4 haper 4: eonance Page 4-8 Example 4-4 efer o he crcu n Fgure 4-8 Z F Y H F Fgure 4-8 a) Deermne an expreon for he mpedance Z of he crcu. b) Deermne an expreon for he admance Y of he crcu. c) alculae he reonance frequency of he crcu from he perpece ha he mpedance mu become real. d) alculae he reonance frequency of he crcu from he perpece ha he admance mu become real. (Noe: For conenence we wll ue he noaon AB o ndcae ha elemen A n ere wh elemen B, and he noaon A//B o mean elemen A n parallel wh elemen B.) a) Z = Z a Z b Z a = Z b = /j Z c = j Z d = /j Z Z = Z c //Z d Z = Z a Z b = Z a + Z b = + /j = ( + j)/j ;alo Z = j(/) Z = Z c //Z d /Z = /Z c + /Z d Z = Z c Z d /(Z c +Z d ) ;only for wo // mpedance Z = (j)(/j)/[j + (/j)] = /[j + (/j)] = j/( ) Z = Z Z = Z + Z = j(/) + j/( ) = + j[/( ) (/)] Z = + j{( )/[( )]} ohm b) Y = Y a Y b Y a = Y b = j Y c = /j Y d = j Y Y = Y c //Y d

50 EIAM4 haper 4: eonance Page 4-9 Y = Y a Y b /Y = /Y a + /Y b Y = Y a Y b /(Y a +Y b ) ;only for wo admance Y = (j)/( + j) = j/(+j) ;of coure Y = /Z Y = Y c //Y d = Y c + Y d = (/j) + j = ( )/j ;agan Y = /Z Y = Y Y /Y = /Y + /Y Y = Y Y /(Y +Y ) = {[j/( + j)][( )/j]}/{[(j/( + j)] + [( )/j]} = [( )/( + j)]/{[(j) + ( + j)( )]/[j( + j)]} = [( )/( + j)]{[j( + j)]/[(j) + ( + j)( )]} = [j( )]/[( ) + j( ] = {j( )[( ) j( ]}/[( ) + ( ) ] Y = [ ( ) + j( )( )]/( ) emen (From Z = + j{( )/[( )]}: Z = [( ) + j( )]/( ) /Z = ( )/[( ) + j( )] = ( )/[( ) j( )] = {( )[( ) + j( )]}/{[( )] + ( ) } = [ ( ) + j( )( )]/[ ( + 4 ) ] = [ ( ) + j( )( )]/[ ] = Y ;a expeced) c) For Im{Z} o anh: = =.77 rad./ec. d) For Im{Y} o anh: = or = = rad/ec. =.77 rad/ec

51 EIAM4 haper 4: eonance Page 4- Exerce P 4- Fnd he reonance frequency for each of he crcu hown n Fgure P 4- (a) (b) (c) Fgure P 4- An (a) r = P 4- Gen he praccal ank crcu n Fgure P 4-, f = 5, = 5 mh, and =.5 F, approxmae h crcu by a parallel crcu. Wha he qualy facor of he parallel crcu? Fgure P 4- An: 63. P 4-3 Fnd he qualy facor for he crcu n Fgure P (/6) F 4 H Fgure P 4-3

52 EIAM4 haper 5: aplace Nework Analy Page 5-5. APAE NETWOK ANAYSIS 5. The aplace Tranform Gen a funcon f(), we defne aplace ranform, [f()], o be [f()] = F() = f()e. Example 5- u() = for [u()] = e = e If we accep ha e - = and e =, = e e hen: e = =,. [u()] = and: e = =. =. Example 5- u() = for [ e a u()] = e a e = e (a) = e ( a) a [e a u()] = e a e = [ ] a = a a Example 5-3 e - = and e = j [n u()] = e e j Euler rule: u() = [e j u()] [e j u()] jθ -jθ e e j j j co = Bu ung, [e a jθ -jθ u()] = /(+a) from Example 5-, n = e e j j [n u()] = = j j j j j = ( j j Example 5-4 [ e j j a co u()] = e e e a u() = (a j (a j e u() e u() Ung agan, [(e a )u()] ( a j) ( a j) = /( + a), hen = = a j a j [(e (a j) )u()] [( a) j][( a) j] = /[ + (a j)], a a [(e (a + j) )u()] = ( a) = = /[ + (a + j)] ( a) Example 5-5 (h example llurae he lneary propery of he aplace ranform) [f () + f ()] = [f () f ()]e = f ()e + f ()e = [f ()] + [f ()] And alo [Kf()] = Kf()e = K f()e = K[f()]

53 In he nex example we wll ue he negraon by par rule and a reul from he he heory of lm Example 5-6 [u()] = e = ( ) (e ) = m Example 5-7 df() EIAM4 haper 5: aplace Nework Analy Page 5- = [ ( ) ] + e = + () e ( ) e e = [ ] = df() e = f()e f() e = [ f()] + f()e = f() + f()e. Bu f()e, per defnon equal o [f()]. df() df() Therefore = [f()] f() or = F() f(). Alo, f g() = f(), hen df() preou reul, Bu e - = m e - = m f()e- = for well behaed f() dg() = f(), g() = f() [f()] f() = g() : : b f()g() = a m e f() e dg() = f() and g() = f() =. Applyng he dg() = [f()] f(), o g(), hen = [g()] g(). and g() = o ha [f()] = Example 5-8 (an mporan reul) [Ae a co( + )u()] = j( e j( Ae a e f(). u() = A e a e j( u() ea e j( u() e j e (a j u() e j e (a j u() = A e j e j a j a j = A = A a j A a j b a a b f() g() df() g(),. (alculae e for =, 5, and, and ee how quckly e goe o zero for large.) f() g() f() g() df()/ g() g() f() g() df()/ =

54 EIAM4 haper 5: aplace Nework Analy Page 5-3 Table - A l of ome mporan aplace ranform. f() df() F() = [f()] = f()e F() f() d f() f() F() f() F() df() () u() e a u() u() a e a u() ( a) n u() co u() e a n u() ( a) e a co u() a ( a) Ae a co( + )u() a j A a j

55 EIAM4 haper 5: aplace Nework Analy Page The Inere aplace Tranform and Paral Fracon The general procedure o fnd he nere ranform f() = - [F()], from he aplace ranform F(), o expand F() no paral fracon and hen ue Table - o deermne f(). The mehod of paral fracon may be appled o fracon of whch he order of he numeraor le han he order of he denomnaor. Alo f F() = KF (), hen f() = K - [F ()], o a common facor K may be facored ou of F() and he problem reduced o fndng only he nere of F (). e u conder ome example. Example 5-9 (dnc real facor) Deermne he oluon of he dfferenal equaon: d x() dx() 3 x() 4e 3 u(), dx() ubjec o he nal conon x() = and =. Takng he aplace ranform on boh de of he equaon, reul n he algebrac equaon dx() 4 X() x() + 3[X() x()] + X() = 3 4 X() ( ) + 3[X() ] + X() = 3 ( )X() = = 4 ( 3) 5( 3) = X() = 9 ( = 9 A = 3 )( 3) ( )( )( 3) + B + 3, Where A = 9 = 5, + = ( )( )( 3) B = and = 9 ( )( )( 3) 9 ( )( )( 3) 5 X() = = 5 = + = + 3 = x() = 5e u() 5e u() + e 3 u() = (5e 5e + e 3 )u() In he nex wo example we wll demonrae ome echnque for fndng he nere ranform of a funcon wh complex econd order facor. The ranform par eablhed n Example 5-8 ery ueful n h regard and he mo raghforward and relable mehod, o organe he paral fracon n a form, ready o apply h ranform par. Unforunaely alway nole calculaon wh complex number whch could be lenghy and cumberome.

56 EIAM4 haper 5: aplace Nework Analy Page 5-5 Example 5- (complex facor) Fnd he nere aplace ranform f(), of he funcon F() = ( + )/[( + + 3)]. We could ulze wo mehod o fnd f(). ecommended mehod: (ung he fnal enry n Table - o handle complex roo) To fnd he facor of + +3, ole from = = + j.44 = = = j.44 = = Therefore: (.73.86)(.73.86) = = F() = ( 3) = (.73.86)(.73.86) A B B* = = A = (.73.86)(.73.86) B = F() = (.73.86)(.73.86) = (.73.86) j.44 j.44 = = f() =.3333u() e co( )u() = [ e co( )]u() =.3333 (Noe: Afer fndng A and B, F() may alo be pu n he form: F() = (.73.86) (.73.86) = (.73.86)(.73.86) = = The coeffcen of mu be one. If no, he coeffcen mu be facored ou a a common facor of F(). oo n polar form, are perhap more conenen a h age. Noe: The fr roo = +j.44 wll be mporan o u and alway locaed n he econd quadran. B * = r he conjugae of B = r. Therefore only B ha o be calculaed. Wha arace abou h mehod, ha n prncple he ame a he procedure wh real roo. oo mu now be n recangular form wh he focu on he j erm, n order o ue he fnal ranform par n Table -, effcenly. If he +j erm ued, alo ok becaue co( ) = co( ) /3 ( ) ( ) (.73.86)(.73.86) = = ( ) = ; compare wh nex mehod) 3 3