Chapter 6: AC Circuits

Transcription

1 Chaper 6: AC Crcus

2 Chaper 6: Oulne

3 Phasors and he AC Seady Sae

4 AC Crcus A sable, lnear crcu operang n he seady sae wh snusodal excaon (.e., snusodal seady sae. Complee response forced response naural response. n he seady sae, naural response.

5 A cos( θ A cos( θ Same frequency, dfferen ampludes and phases

6 Snusods Three parameers are needed o deermne a snusod. x(x m cos(φre[x m e (φ ]. X m : amplude, πfπ/τ: angular frequency, φ: phase angle (radan.

7 Phasors The hree parameers can be represened by a roang phasor n a wo-dmensonal plane. A a gven me (e.g.,, he nonroang phasor s represened by X X m φ. The frequency nformaon s no ncluded.

8 AC Forced Response The forced response of any branch varable (curren or volage s a he same frequency as he excaon frequency for a lnear crcu. n oher words, any branch varable has he general form y(y m cos(φ y. Crcu analyss becomes manpulaon of complex numbers.

9 Complex Numbers n he Complex Plane a a r A a A a φ φ sn cos Roang phasor ( an 8 an / < ± > r r a r r a r a a a a a a a a A φ φ

10 Complex Addon and Subracon Re m [ A ± B] Re[ A] ± Re[ B] [ A ± B] m[ A] ± m[ B]

11 Complex Conugae A A * A A φ A A φ * a a r a a a r a r A a a

12 Complex Mulplcaon A B f k ( a r b r a b s real, Re[ kb] k ( a r b a b r Re Re[ B], m[ kb] k m[ B] [ AB] m[ AB]

13 Complex Dvson (Raonalzaon B A BA AA * * b r a a r r b a a b r a a r b a a r

14 Complex Number n Exponenal Form Euler s formula: Complex number n exponenal form: α α α cos sn e ± ± a Ae A φ 9 B A B A B A B A b a b a φ φ φ φ

15 Phasor Represenaon A snusod can be represened by a phasor: Xe e Xe Xe X φ φ φ Re Re Re cos( The sum of wo snusods a he same frequency can be represened by anoher phasor. The new phasor s smply he sum of he wo orgnal phasors. e X X e X e X e X e X ( Re Re Re Re

16 Phasor Represenaon The seady-sae response of any branch varable n a sable crcu wh a snusodal excaon wll be anoher snusod a he same frequency (forced response n Chaper 5 Krchhoff s laws hold n phasor form (only addons are nvolved.

17 Phasor wh Dfferenaon [ ] X X Xe d dxe d aon dfferen Xe d φ φ φ Re Re Re

18 Example 6.3: Parallel Nework wh an AC olage Source C d dv C v v C c C R R ( ( cos(4 ( ( cos(4 ( 46.6( 6.7 A A C R

19 Example 6.4: Parallel Nework wh an AC Curren Source C C R, 5 3 3cos 4 ( ( cos(4 (.. 3. (. 3 v C R

20 mpedance and Admance

21 Phasor Represenaon Under ac seady-sae, boh he volage and he curren of a branch are snusods a he same frequency. e v m v φ Re cos( ( e m φ Re cos( (

22 Ressors Curren and volage are collnear (n phase. Re e R e v Re Re Re m R v m R φ φ v m R m φ φ,

23 nducors Curren lags volage by 9 degrees. e Le d de L e d d L d d L v Re Re Re Re L, 9 o v m L m φ φ

24 Capacors Curren leads volage by 9 degrees. e Ce d de C e d d C d dv C Re Re Re Re C C, 9 o v C m m φ φ

25 Phasor Relaons (Ressor

26 Phasor Relaons (nducor

27 Phasor Relaon (Capacor

28 mpedance n general, we can defne a quany Z and Ohm s law for ac crcus as Z Z R Z L Z c R / L L 9 o C / C 9 o

29 Tme Doman vs. Frequency Doman

30 Admance Smlarly, anoher quany admance Y can be defned. Y / Z Y

31 Equvalen mpedance and Admance Seres equvalen mpedance: Z Z Z L ser Z N ( L N Parallel equvalen mpedance: Y par Y Y L Y N ( L N

32 Load Nework Z eq /

33 mpedance and Admance mpedance and admance are complex funcons of frequency. [ Z ] m[ Z ] R( X ( Z Z( Re Ressance (Ω Reacance (Ω nducors and capacors are reacve elemens, nducve reacance s posve and capacve reacance s negave.

34 mpedance and Admance [ Y ] m[ Y ] G( B( Y Y ( Re Conducance (Semens Suscepance (Semens nducors and capacors are reacve elemens, nducve reacance s posve and capacve reacance s negave.

35 mpedance Trangle

36 Example 6.6: mpedance Analyss of a Parallel RC Crcu. Z Y 5 5 Z Y 4.47Ω A S 6.6

37 Example 6.7: Frequency Dependence of a Parallel RC Nework. ( ] m[ ( ( ] Re[ ( ( ( CR CR Z X CR R Z R CR CR R Z Z Z Z Z C R C R

38 Example 6.8: AC Ladder Calculaons AC ladder neworks can be analyzed by seres-parallel reducon (by replacng ressance wh mpedance. L C ( cos5( ma

39 Example 6.8: (Con ( ( Z C L

40 AC Crcu Analyss

41 AC Crcu Analyss Sources a he same frequency: Phasor ransform mehod: he me doman snusods are ransformed o he frequency doman and represened by phasors. Tme doman Frequency doman

42 AC Crcu Analyss Sources a he same frequency: Wh he ransformaon, all ressve crcu analyss echnques are applcable. Ressance s replaced by mpedance and conducance s replaced by admance. Proporonaly Thévenn-Noron Node Analyss Mesh Analyss

43 AC Crcu Analyss Sources a he same frequency: Afer analyss, he resulan phasors are convered back o he me-doman snusods. Frequency doman Tme doman

44 AC Crcu Analyss Sources a dfferen frequences: Due o he lneary, he proporonaly mehod s sll applcable. The phasor analyss s performed a each ndvdual frequency

45 Example 6.9: AC Nework wh a Conrolled Source ( cos v / /( 8 ( Le Z x A A A Z cos( / ( 9 / Use proporonaly

46 Example 6.: Phase-Sf Oscllaor v x v n /K Oscllaor: Generaor a snusodal oupu whou an ndependen npu source wh nal sored energy. Desgn goal: A one parcular frequency, v ou v n.

47 Example 6.: (Con. Use proporonaly and le n L L / CR C x, R Oscllao n requres OSC when / LC OSC, n x : L K / C L CR

48 Superposon An ac source nework s any wo-ermnal nework ha consss enrely of lnear elemens and sources. f here are more han one ndependen source, all of hem mus be a he same frequency so ha he phasor mehod can be appled.

49 oc Frequency Doman Thévenn Parameers Frequency-doman Thevenn parameers: he open-crcu volage phasor: he shor-crcu curren phasor: Thévenn mpedance: oc sc Z oc / sc

50 Example 6.: Applcaon of an AC Noron Nework. Thévenn parameers: Maxmze Z 4 OC SC Z OC Ω A 8.8

51 Example 6.: (Con ( (.4 mnmum s maxmum f s / and S Y B G Y Y Y Z Z Y eq eq eq SC eq Y eq

52 AC Mesh Analyss By usng phasors, mpedance and admance, node analyss and mesh analyss are sll applcable assumng all ndependen sources are a he same frequency. AC mesh analyss: [ ][ ] [ ] Z [ or Z Z ][ [ ] ] s ~ ~ s wh conrolled sources

53 Example 6.: Sysemac AC Mesh Analyss Fnd Two sources a he same frequency

54 Example 6.: (Con cos( ( ( mesh equaon : Sngle Ω A Z Z S S

55 AC Node Analyss [ Y ][ ] [ ] ~ or s ~ s [ Y Y ][ ] [ ] wh conrolled sources

56 Example 6.3: Sysemac AC Node Analyss Fnd and Z Consran equaon: (- /4

57 Example 6.3: (Con. [ ] [ ] Ω Ω / Z A Y s

58 Chaper 6: Problem Se 7, 7, 4, 3, 36, 4, 44, 47, 5, 53, 57, 59