Chaper 9 Sinusoidal Seady Sae Analysis 9.-9. The Sinusoidal Source and Response 9.3 The Phasor 9.4 pedances of Passive Eleens 9.5-9.9 Circui Analysis Techniques in he Frequency Doain 9.0-9. The Transforer 9. Phasor Diagras
Overview We will generalize circui analysis fro consan o ie-varying sources (Ch7-4). Sinusoidal sources are paricularly iporan because: () Generaion, ransission, consupion of elecric energy occur under sinusoidal condiions. () can be used o predic he behaviors of circuis wih nonsinusoidal sources. Need o work in he real of coplex nubers.
Key poins Wha is he phase of a sinusoidal funcion? Wha is he phasor of a sinusoidal funcion? Wha is he phase of an ipedance? Wha are in-phase and quadraure? How o solve he sinusoidal seady-sae response by using phasor and ipedance? Wha is he refleced ipedance of a circui wih ransforer? 3
Secion 9., 9. The Sinusoidal Source and Response. Definiions. Characerisics of sinusoidal response 4
Definiion A source producing a volage varying sinusoidally wih ie: v()= cos( +). : Phase angle, deerines he value a =0. : Apliude. : Angular frequency, relaed o period T via =/T. The arguen changes radians (360) in one period. 5
More on phase angle Change of phase angle shifs he curve along he ie axis wihou changing he shape (apliude, angular frequency). Posiive phase (>0), he curve is shifed o he lef by in ie, and vice versa. cos(+) cos() 6
Exaple: RL circui () Consider an RL circui wih zero iniial curren i( 0 ) 0 and driven by a sinusoidal volage source v ( ) cos( ) : s d L d By KL: i Ri cos( ). 7
Exaple: RL circui () The coplee soluion o he ODE and iniial condiion is (verified by subsiuion): i( ) i ( ) i ( ), r ss i i r ss ( ) ( ) R cos( ) R L L e ( R L) cos( ) Transien response, vanishes as. Seady-sae response, lass even. an L R. 8
Characerisics of seady-sae response i ss () of his exaple exhibis he following characerisics of seady-sae response: i ss ( ) R L cos( ). reains sinusoidal of he sae frequency as he driving source if he circui is linear (wih consan R, L, C values).. The apliude differs fro ha of he source. 3. The phase angle differs fro ha of he source. 9
Purpose of Chaper 9 Direcly finding he seady-sae response wihou solving he differenial equaion. According o he characerisics of seady-sae response, he ask is reduced o finding wo real nubers, i.e. apliude and phase angle, of he response. The wavefor and frequency of he response are already known. Transien response aers in swiching. will be deal wih in Chapers 7, 8,, 3. 0
Secion 9.3 The Phasor. Definiions. Solve seady-sae response by phasor
Definiion The phasor is a consan coplex nuber ha carries he apliude and phase angle inforaion of a sinusoidal funcion. The concep of phasor is rooed in Euler s ideniy, which relaes he (coplex) exponenial funcion o he rigonoeric funcions: e cos sin. cos Re, sin e e.
Phasor represenaion A sinusoidal funcion can be represened by he real par of a phasor ies he coplex carrier. A phasor can be represened in wo fors:. Polar for (good for, ): e,. Recangular for (good for +, -): cos sin. Re ( ) ) Re e e e e Re cos( phasor carrier ag. real 3
Phasor ransforaion A phasor can be regarded as he phasor ransfor of a sinusoidal funcion fro he ie doain o he frequency doain: P cos( ) e. ie doain freq. doain The inverse phasor ransfor of a phasor is a sinusoidal funcion in he ie doain: P - Re e cos( ). 4
5 Tie derivaive Muliplicaion of consan ). cos( ) cos( ), 90 cos( ) sin( ) cos( d d d d Tie doain:. ) ( ) cos( ) cos( 90 ) 90 ( d d, e e e d d P P Frequency doain:
How o calculae seady-sae soluion by phasor? Sep : Assue ha he soluion is of he for: Re Ae e Sep : Subsiue he proposed soluion ino he differenial equaion. The coon ie-varying facor e of all ers will cancel ou, resuling in wo algebraic equaions o solve for he wo unknown consans {A, }. 6
Exaple: RL circui () Q: Given v ( ) cos( ), calculae i ss (). d L d L cos( ) R cos( ) s sin( ) R Assue ( ) cos( ) i ss cos( ). d L iss( ) Ri d cos( ), ss ( ) cos( ), cos( ), 7
8 Exaple: RL circui () By cosine convenion:., R L e e R L A necessary condiion is:. Re Re, Re Re Re, Re Re Re ), cos( ) cos( ) 90 cos( ) 90 ( e e R L e e R e L e e e e R e e L R L
9 Exaple: RL circui (3) A ore convenien way is direcly ransforing he ODE fro ie o frequency doain:. i.e., R L e e R L The soluion can be obained by one coplex (i.e. wo real) algebraic equaion:., ), cos( ) ( ) ( R L R L Ri i d d L ss ss
Secion 9.4 pedances of The Passive Circui Eleens. Generalize resisance o ipedance. pedances of R, L, C 3. n phase & quadraure 0
Wha is he ipedance? For a resisor, he raio of volage v() o he curren i() is a real consan R (Oh s law): v( ) R. resisance i( ) For wo erinals of a linear circui driven by sinusoidal sources, he raio of volage phasor o he curren phasor is a coplex consan :. ipedance
The i-v relaion and ipedance of a resisor i() and v() reach he peaks siulaneously (in phase), ipedance =R is real.
The i-v relaion and ipedance of an inducor () Assue i( ) cos( i ) d v( ) L i( ) d L sin( ) i L cos( 90 ). By phasor ransforaion: i L L. 3
The i-v relaion and ipedance of an inducor () v() leads i() by T/4 (+90 phase, i.e. quadraure) ipedance = L is purely posiive iaginary. d v( ) L i( ) d 4
The i-v relaion and ipedance of an capacior () d i( ) C v( ). d C, C. 5
The i-v relaion and ipedance of a capacior () v() lags i() by T/4 (-90 phase, i.e. quadraure) ipedance C is purely negaive iaginary. 6
More on ipedance pedance is a coplex nuber in unis of Ohs. pedance of a uual inducance M is M. X Re R, are called resisance and reacance, respecively. Alhough ipedance is coplex, i s no a phasor. n oher words, i canno be ransfored ino a sinusoidal funcion in he ie doain. 7
Secion 9.5-9.9 Circui Analysis Techniques in he Frequency Doain 8
Suary All he DC circui analysis echniques:. KL, KCL;. Series, parallel, -Y siplificaions; 3. Source ransforaions; 4. Thévenin, Noron equivalen circuis; 5. NM, MCM; are sill applicable o sinusoidal seady-sae analysis if he volages, currens, and passive eleens are replaced by he corresponding phasors and ipedances. 9
30 KL, KCL KL: KCL: i () + i () + + i n () = 0, 0.... n 0.... n, 0 Re Re ) cos( 0, ) ( ) ( ) ( ) ( ) ( n q q n q q n q q q n q q n e e e v v v v q q
Equivalen ipedance forulas pedances in series ab pedances in parallel ab 3
Exaple 9.6: Series RLC circui () Q: Given v s ()=750 cos(5000+30), i()=? L C s L C 75030 (5000)(30, (5000)(5 3 0 ) 6 ) 60, 40, 3
33 Exaple 9.6: Series RLC circui () A. ) 3.3 5cos(5000 ) ( A, 3.3 5 53.3 50 30 750, 53.3 50 90) (0 an 0 90 0 90 40 60 90 i ab s ab
Thévenin equivalen circui Terinal volage phasor and curren phasor are he sae by using eiher configuraion. 34
Exaple 9.0 () Q: Find he Thévenin circui for erinals a, b. a b Apply source ransforaion o {0,, 60} wice o ge a siplified circui. 35
Exaple 9.0 () 00 (0 x 40 0) 00 0() 0 x, (30 40) 0 x 00() 36
Exaple 9.0 (3) 900 30 40 8 6.87 A, Th 0(00 0) 0 835. 0.7. 37
38 Exaple 9.0 (4), 0 0, 0 60) ( //, 40 0 60) ( // 40 x T b a a x T T a. 38.4 9., 0 40 6 0 0 6 0 00 T T Th T T T a a T a b a T
Secion 9.0, 9. The Transforer. Linear ransforer, refleced ipedance. deal ransforer 39
Suary A device based on agneic coupling. Linear ransforer is used in counicaion circuis o () ach ipedances, and () eliinae dc signals. deal ransforer is used in power circuis o esablish ac volage levels. MCM is used in ransforer analysis, for he currens in various coils canno be wrien by inspecion as funcions of he node volages. 40
4. ) ( 0, ) ( L s s L R M M L R Analysis of linear ransforer () Consider wo coils wound around a single core (agneic coupling): + + Mesh curren equaions:
Analysis of linear ransforer (), M M s s M M. in M s in M. 4
43 npu ipedance of he priary coil., in L r r s ab L R M M L R r is he equivalen ipedance of he secondary coil and load due o he uual inducance. ab = S is needed o preven power reflecion. in ab
Refleced ipedance r M M * * M *. Linear ransforer reflecs ( )* ino he priary coil by a scalar uliplier (M/ ). 44
Exaple 9.3 () Q: Find he Thévenin circui for erinals c, d. c d Th = cd. Since = 0, cd = M, where s (500 3000 00) (00 3600) 79.67 79.9 A. Th (79.67 79.9 ) ( 00) 95.60.7. 45
Exaple 9.3 () Shor c Th d Th =(00+600) + r, where r is he refleced ipedance of due o he ransforer: 700 3700. (500 00) (00 3600) r Th M (00 00 * 700 3700 600) r (700 3700), 7.09 4.6. 46
Characerisics of ideal ransforer An ideal ransforer consiss of wo agneically coupled coils wih N and N urns, respecively. exhibis hree properies:. Magneic field is perfecly confined wihin he agneic core, agneic coupling coefficien is k=, M L L.. The self-inducance of each coil L is i N i large, i.e. L =L. 3. The coil loss is negligible: R =R. 47
Curren raio L L By solving he wo esh equaions of a general linear ransforer: L L L M if L >> L L L L N N. 48
49 Subsiue ino olage raio. N N L L M L M M L L M., L M L + + L L L + L
50 By he curren and volage raios, npu ipedance + + L L., L ab in L ab N N N N. L ab R N N R For lossy ransforer, in-phase in
Polariy of he volage and curren raios 5
Exaple 9.4 () Q: Find v, i, v, i. s L 500 cos(400) 5
Exaple 9.4 () 5000 (0.5 ) (0.375 0, 0.05) 0 () (3.75 5000 ( ) () : 00 6.6, i 4 7,, 5) () 00 cos(400 6.6 ). By () : (3.75 5)(00 6.6 ) 47 4.37, v 53
Secion 9. Phasor Diagras 54
Definiion Graphical represenaion of -7-3 = 7.6-56.8 on he coplex-nuber plane. Wihou calculaion, we can anicipae a agniude >7, and a phase in he 3rd quadran. 55
Exaple 9.5 () Q: Use a phasor diagra o find he value of R ha will cause i R o lag he source curren i s by 45 when = 5 krad/s. -0.5 L 90, C 0.5 4 90, R R 0. 56
Exaple 9.5 () By KCL, s = L + C + R. Addiion of he 3 curren phasors can be visualized by vecor suaion on a phase diagra: To ake = 45, s 3 R = 3, R = /3. 57
Key poins Wha is he phase of a sinusoidal funcion? Wha is he phasor of a sinusoidal funcion? Wha is he phase of an ipedance? Wha are in-phase and quadraure? How o solve he sinusoidal seady-sae response by using phasor and ipedance? Wha is he refleced ipedance of a circui wih ransforer? 58