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電路學 - [第七章] 正弦激勵, 相量與穩態分析

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正弦激勵, 相量與穩態分析

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電路學 - [第七章] 正弦激勵, 相量與穩態分析

  1. 1. Department of Electronic Engineering National Taipei University of Technology
  2. 2. • • ( ) • ( ) • ( ) • • Department of Electronic Engineering, NTUT2/41
  3. 3. (I) • Department of Electronic Engineering, NTUT ( ) sinmv t V tω= Vm t (sec) T v(t) −Vm 2 π ω π ω 3 2 π ω 2π ω 5 2 π ω 6 2 π ω 0 T ω : Vm (volt) : T (sec) : f (Hz, cycle/sec) : (rad/sec) 2 fω π= 3/41
  4. 4. (II) • ( ) ( )sinmv t V tω φ= + ω : Vm (volt) : T (sec) : f (Hz, cycle/sec) : (rad/sec) 2 fω π= φ: (rad)檔案中找不到關聯識別碼 rId12 的圖像部分。 Vm t (sec) v(t) −Vm 2 π ω π ω 3 2 π ω 2π ω 5 2 π ω 6 2 π ω 0 T φ 1t Department of Electronic Engineering, NTUT4/41
  5. 5. ( ) • vR = iR ( ) ( ) • , • R, L, C = + ( ) ( / ) L di v L dt = 1 t Cv idt C −∞ = ∫ Department of Electronic Engineering, NTUT5/41
  6. 6. 1 • RL i(t) if (t) , KVL + − L Ri(t) ( ) ( )cos Vs mv t V tω= ( ) ( ) cosm di t L Ri t V t dt ω+ = ( ) cos + sinfi t A t B tω ω= ( ) ( ) cosm di t L Ri t V t dt ω+ = mRA LB Vω+ = 0LA RBω− + = ( ) ( )sin cos cos sin cosmL A t B t R A t B t V tω ω ω ω ω ω ω− + + + = , 2 2 2 mLV B R L ω ω = +2 2 2 mRV A R Lω = + ( )fi t ( ) ( ) ( )1 2 2 2 2 2 2 2 2 2 cos sin cos tan cos A R m m m f m RV LV V L i t t t t I t R L R L R L ω ω ω ω ω ω φ ω ω ω −  = + = − = +  + +  + 2 2 2 m m V I R Lω = + 1 tan L R ω φ − = − ( ) ( ) ( )f ni t i t i t= +( ) 1 RtL ni t Ae− = ( )sv t Department of Electronic Engineering, NTUT6/41
  7. 7. 2 • RLC v(t) i(t) KVL + − + − 5 3 R = Ω 5 HL = ( )i t ( )v t 1 F 25 C =( )sv t ( ) 17cos3sv t t= ( ) ( ) ( ) ( )2 2 1 sd v t dv t v tR v t dt L dt LC LC + + = ( ) ( ) ( ) 2 2 5 85cos3 d v t dv tR v t t dt L dt + + = ( ) ( ) ( ) ( )20cos3 +5sin3 =5 17 cos 3 2.9 5 17 cos 3 166 Vv t t t t t= − − = − ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 3 17 3 17 60sin3 15cos3 cos 3 1.33 cos 3 76 A 25 25 5 5 dv t dv t i t C t t t t dt dt = = = + = − = − , ( ) 1 2cos3 sin3v t A t A t= + ( ) ( )1 2 1 24 cos3 4 sin3 85cos3A A t A A t t− + + − − = 1 24 85A A− + = 1 24 0A A− − = 1 20A = − 2 5A = Department of Electronic Engineering, NTUT7/41
  8. 8. (Complex Number) • θ a b A a jb= + Re b Im | |A +A a jb= j A A e Aθ θ= = ∠ [ ]Re cosa A A θ= = [ ]Im sinb A A θ= = 2 2 A a b= + 1 tan b a θ − = 1j = − 1 1 1A a jb= + 2 2 2A a jb= + ( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2+A A a jb a jb a a j b b= + + + = + + + ( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2A A a jb a jb a a j b b− = + − + = − + − • +A a jb= A a jb∗ = − • ( )( ) ( ) ( ) 2 1 2 1 1 2 2 1 2 1 2 2 1 1 2 1 2 1 2 1 2 2 1 A A a jb a jb a a ja b ja b j b b a a b b j a b a b = + + = + + + = − + +    ( )( ) ( )1 2 1 1 2 2 1 2 1 2A A A A A Aθ θ θ θ= ∠ ∠ = ∠ + • ( )( ) ( )( ) 1 1 2 21 1 1 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a jb a jbA a jb a a b b b a a b j A a jb a jb a jb a b a b + −+ + − = = = + + + − + + ( )11 1 2 2 2 AA A A θ θ= ∠ − Department of Electronic Engineering, NTUT8/41
  9. 9. (a+jb) (I) • (a) A = 4 + j3 , A (b) A = –4 + j3 A θ∠ 2 2 4 3 5A = + = 1 3 = tan 36.9 4 θ − = 5 36.9A = ∠ 1 13 3 = tan 180 tan 180 36.9 143.1 4 4 θ − −  = − = − =  −      ( ) 2 2 4 3 5A = − + = 5 143.1A = ∠ (a) 4 θ = − tan13 4 | |A =5 Re 3 j o o A A jA 9.365 9.364 3 tan 534 34 1 22 ∠=⇒ ==⇒ =+=⇒ += − θ Re j θ = − 180 3 4 1 tan −4 | |A =5 3 ( ) oo oo A A jA 1.14351.143 9.361804 3 tan 534 34 1 22 ∠=⇒= −=−=⇒ =+−=⇒ +−= − θ (b) Department of Electronic Engineering, NTUT9/41
  10. 10. (a+jb) (II) (c) A = –4 – j3 ( ) o oo oo A A jA 9.365 9.361.323 9.36360 4 3 tan 534 34 1 22 −∠=⇒ −== −=−=⇒ =−+=⇒ −= − θ j | |A =5 Re −3 4 θ = − 360 3 4 1 tan −3 Re j −4 | |A = 5 ( ) ( ) o oo oo A A jA 1.1435 1.1439.216 9.36180 4 3 tan 534 34 1 22 −∠=⇒ −== += − − =⇒ =−+−=⇒ −−= − θ θ = − 180 + 3 4 1 tan A θ∠ ( ) ( ) 2 2 4 3 5A = − + − = 1 13 3 = tan 180 tan 4 4 180 36.9 216.9 143.1 θ − −−  = −  −  = + = = −     5 143.1A = ∠ − (d) A = 4 – j3 ( ) 22 4 3 5A = + − = 1 13 3 = tan =360 tan 4 4 323.1 36.9 θ − −−  −    = = −  5 36.9A = ∠ − (c) (d) Department of Electronic Engineering, NTUT10/41
  11. 11. (I) • • • RL (Complex excitation) (Real excitation) i(t) KVL: cos sinj t m m mV e V t jV tω ω ω= + cos Re sin Im j t m m j t m m V t V e V t V e ω ω ω ω   =     =   1 j t mv V e ω = ( ) [ ]1cos Reg mv t V t vω= = + − L Ri(t)cosmV tω 1 1+ j t m di L Ri V e dt ω = 1 j t i Ae ω = ( ) j t j t mj L R Ae V eω ω ω + = -1 tan 2 2 2 L j m m R V V A e R j L R L ω ω ω − = = + + 1 tan 1 2 2 2 L j t RmV i e R L ω ω ω −  −    = + i1 i(t) ⇒⇒⇒⇒ Department of Electronic Engineering, NTUT11/41
  12. 12. (II) 6 RL i(t) i1 ( vg(t) v1 ) i1 v1 if (t)= Re[i1] vg = Re[v1] 1 tan 1 2 2 2 L j t RmV i e R L ω ω ω −  −    = + ( ) [ ] 1 tan 1 1 2 2 2 2 2 2 Re Re cos tan L j t Rm mV V L i t i e t RR L R L ω ω ω ω ω ω −  −  −       = = = −     + +   [ ]1 1 1Re Re di L Ri v dt    + =      [ ]( ) [ ]( )1 1Re Re cosm d L i R i V t dt ω+ = ( ) ( ) [ ]1Refi t i t i= = Department of Electronic Engineering, NTUT12/41
  13. 13. • 11 : • ? KVL: 1 1+ j t m di L Ri V e dt ω = 1 j t i Ae ω = ( ) j t j t mj L R Ae V eω ω ω + = -1 tan 2 2 2 L j m m R V V A e R j L R L ω ω ω − = = + + 1 tan 1 2 2 2 L j t RmV i e R L ω ω ω −  −    = + A ” ” ” ” j t e ω j t e ω jω j t e ω 1 jω Department of Electronic Engineering, NTUT13/41
  14. 14. (I) • (Euler’s formula) (Phasor) • rms • ”−” ( ) ( )cos Re Rej j t j t m mv t V t V e e Veθ ω ω ω θ    = + = =    j m mV V e Vθ θ= = ∠ 2 j m rms rms V V V e Vθ θ θ= = ∠ = ∠ j m mV e Vθ θ= = ∠V j m mV V e Vθ θ= = ∠ j t e ω j t e ω 1: Department of Electronic Engineering, NTUT14/41
  15. 15. (II) I R + = − V RI i R + = − v Ri I + − i L + − di v L dt = V j LIω= j Lω + − V 1 I j CV V I j C ω ω = ⇒ = 1 j CωC + − v dv i C dt = R R L j Lω C 1 j Cω ( ) ( ) 2: Department of Electronic Engineering, NTUT15/41
  16. 16. (III) + − L Ri(t)cosmV tω + − RI0mV V= ∠ j Lω j LI R Vω + =1 2 2 2 0 tan L m mV VV L I R j L R j L RR ω ω ω ω −∠ = = = ∠ − + + + ( ) 1 1 tan tan 1 2 2 2 2 2 2 2 2 2 Re Re cos tan L L j j t j tR Rm m mV V V L i t e e e t RR L R L R L ω ω ω ω ω ω ω ω ω − −    − −    −          = = = −       + + +       2: 1: , ! 3:! ! Department of Electronic Engineering, NTUT16/41
  17. 17. R L C I R + = − V RI (a) i R + = − v Ri t v i, v i (b) i L + − I + − (a) (b) di v L dt = V j LIω= j Lω v i t v i, v i t v i, Department of Electronic Engineering, NTUT C + − v + − I 1 j Cω dv i C dt = (a) (b) 1 V I j Cω = 17/41
  18. 18. (I) Z (Impedance) ( ) ( )1cosmv t V tω φ= + ( ) ( )2cosmi t I tω φ= + 2mI I φ= ∠ 1mV V φ= ∠ ( ) ( )1 2= m m VV Z Z R jX I I θ φ φ= = ∠ ∠ − = + Ω (a) + − + − V I (b) ( ) ( )v t ( )i t [ ]ImX Z= [ ]ReR Z= (Reactance) (Resistance) Z ω ω ( )Z jω ( )R R ω= ( )X X ω= θ R X Z R jX= + Re Im 2 2 Z R X= + 1 tan X R θ − = cosR Z θ= sinX Z θ= Department of Electronic Engineering, NTUT18/41
  19. 19. (II) • • (Admittance) • Z Y Y Y = G + jB G = Re[Y] B = Im[Y] (Susceptance) Y G B R V Z R I = = LX Lω= L L V Z jX j L I ω= = = 1 CX Cω = − 1 C C V Z jX j I Cω = = = − 1 Y Z = ( )( ) 2 2 2 2 1 1 = R jX R X Y j G jB Z R jX R jX R jX R X R X − = = = − = + + + − + + Department of Electronic Engineering, NTUT19/41
  20. 20. • (a) RL (b) KVL: + − L Ri( )cos VmV tω (a) + − R j Lω I0mV ∠ (b) 0L mZ I RI V+ = ∠ ( ) 0mj L R I Vω + = ∠ ( ) 1 2 2 2 cos tanmV L i t t RR L ω ω ω −  = −   + 1 2 2 2 0 tanm mV V L I R j L RR L ω ω ω −∠   = = ∠ −  +  + Z j L Rω= + 0mVV I Z R j Lω ∠ = = + Department of Electronic Engineering, NTUT20/41
  21. 21. 3 • (a) v(t) i(t) 1. (a) (b) 2. (c) 3. (c) 10 0V = ∠ 2 rad secω = ( )Z 2L j L jω= = Ω ( ) 1 1CZ j j Cω = − = − Ω 1 1.5 2Z j= + ( )( ) 2 1 1 0.5 0.5 1 1 j Z j j − = = − − + ( ) ( )0.283sin 2 81.9 Vv t t= − 1 2 10 0 10 0 4 36.9 2.5 36.9 TI Z Z ∠ ∠ = = = ∠− + ∠ ( ) 1 4 36.9 2.83 8.1 A 1 1 2 45 TI I j ∠ − = = = ∠ − ∠− ( ) ( ) ( )2.83sin 2 8.1 Ai t t= + + − + −( )10sin2 Vt 1.5 Ω 1 H 0.5 F 1 Ω ( )Ti t ( )i t ( )v t (a) + − + − 10 0∠ 1 Ω 2j Ω 1j− Ω 1.5 Ω TI I V (b) + − + − TI 10 0∠ V 1Z 2Z (c) , , , ( ) ( ) 2 1 2 0.5 0.5 10 0 10 0 1.5 2 0.5 0.5 Z j V Z Z j j − = × ∠ = × ∠ + + + − 0.283 81.9 V= ∠− Department of Electronic Engineering, NTUT21/41
  22. 22. 4 • (a) i(t) 1. (a) (b) a KCL: + − a 1 2 1 8F + − v1 i t( ) 4Ω( )10cos4 At v1 (a) + − a 1 2 −j2 Ω + − V1 I 4Ω( )10 0 A∠ V1 (b) ( ) 1 1 2 1 4 8 C j Z j j Cω = − = − = − Ω       4rad secω =10 0I = ∠ 1 1 1 2 10 0 2 V V I j − + = ∠ − ( )1 1 10 0I j+ = ∠ ( ) 10 0 7.07 45 A 1 1 I j ∠ = = ∠ − + ( ) ( )( )7.07cos 4 45 Ai t t= − , , Department of Electronic Engineering, NTUT22/41
  23. 23. (Phasor Diagram) • (Vector) • Vg= VR + VL + VC (b) VR , VL , VC V|VL| |VC| + − Vg R + −VR + −VL + − VC 1 j Cω j Lω VL VC+ VR (a) RLC I (a) RLC 0I I= ∠ I I I (b) VR VL VC (a) VL VC (c) VR + VL + VC = Vg Department of Electronic Engineering, NTUT23/41
  24. 24. Department of Electronic Engineering, NTUT24/41
  25. 25. 5 − • (a) i(t) 1. (a) (b) 2. V V + 3000I KVL: 4. 3. KCL: (a) +− + − 1 k 2 Ω 2 kΩ 1 F 5 µ ( )3000 Vi t ( ) ( )3000v t i t+ 2 kΩ 1 F 5 µ ( ) ( )4cos5000 Vsv t t= ( )sv t I V + − +− 1 k 2 Ω 3000V I+ 3000I 4 0sV = ∠ ( ) 2 1 2 k 5 j− Ω ( )2 1 kj− Ω (b) ( ) ( )( ) ( )( ) 4 3000 0 1 2 2 1 103103 1 2 103 2 5 V V V I jj − + + + = −− ( ) 4 1 103 2 V I − = 3 24 10 53.1 A=24 53.1 mAI − ° = × ∠ ∠ ( ) ( )24cos 5000 53.1 mAi t t ° = + Department of Electronic Engineering, NTUT25/41 ( )i t ( )v t
  26. 26. 6 − • (a) i1(t) i2(t) 1. ω = 2 rad/s : (b) (b) + − 1I 2I 3 Ω 2j Ω 1 Ω 1j Ω 2j− Ω 1 Ω18 0sV = ∠ 2. ( ) 1 24 2 18 0j I I ° + − = ∠ ( )1 22 1 0I j I− + − = 2 2 0 AI ° = ∠ 1 4.47 26.6 AI ° = ∠ − 3. ( )2 2sin2 Ai t t= ( ) ( )1 4.47sin 2 26.6 Ai t t ° = − Department of Electronic Engineering, NTUT26/41 + − 3 Ω (a) 1 H 1 Ω 1 Ω ( ) ( )18sin 2 Vsv t t= ( )sv t ( )1i t 1 H 2 1 F 4 ( )2i t
  27. 27. 7 − • (a) 1. (b) I i(t) I = I1 + I2 2. I1 I2 (c) (d) 3. (c) + − + − 5 Ω (a) 4 Ω1 H( )i t 1 F 20 ( )1sv t ( ) ( )1 20sin 2 Vsv t t= ( ) ( )2 10sin 2 Vsv t t= ( )2sv t + − + − 5 Ω 4 Ω2j Ω 20 0∠ 10j− Ω 10 0∠ I (b) + − 5 Ω 4 Ω2j Ω 20 0∠ 10j− Ω 1I (c) + − 5 Ω 4 Ω2j Ω 10j− Ω 10 0∠ 2I (d) ( ) ( ) 1 10 4 2 10 4 2 5 5 10 10 4 2 4 8 j j j j Z j j j − + − + = + = + = Ω − + + − 1 1 20 0 20 0 2 0 A 10 I Z ° ° °∠ ∠ = = = ∠ Department of Electronic Engineering, NTUT27/41
  28. 28. 7 − ( ) 3. (d) : 4. I ( ) 2 10 5 4 2 4 2 4 2 8 5 10 j Z j j j j − = + + = + + − = − + − 5 Ω 4 Ω2j Ω 10j− Ω 10 0∠ 2I (d) 3I ( )3 10 0 5 0 A 8 4 I ∠ = = ∠ ( )2 3 10 2 1 5 153.4 A 5 10 2 2 j j I I j − − + = − = = ∠ − ( )1 2 2 1 2 0 1 0.5 1.12 26.6 A 2 j I I I j − + = + = ∠ + = + = ∠ ( ) ( ) ( )1.12sin 2 26.6 Ai t t= + Department of Electronic Engineering, NTUT28/41
  29. 29. • Zth ( ) Zth : V Z Ioc th sc= a b a b Zth Eth Voc= + − a b a b IN Isc= Zth Department of Electronic Engineering, NTUT29/41
  30. 30. 8 − • (a) a-b 1. Voc a b + − 4 Ω 10j Ω 6 Ω I 4j− Ω (a) ( )6 0 V∠ a b + − 4 Ω 10j Ω 4j− Ω (b) ( )6 0 V∠ + − ocV a b 4 Ω 10j Ω 4j− Ω (c) thZ 4 6 0 3 3 4.24 45 V 4 4 oc j V j j ° °− = × ∠ = − = ∠ − − 2. Zth ( )4 4 10 2 8 8.246 75.96 4 4 th j Z j j j °− = + = + Ω = ∠ Ω − 4.24 45 4.24 45 0.375 90 A 2 8 6 11.3 45 I j ° ° ° ° ∠ − ∠ − = = = ∠ − + + ∠ 3. a b + − 2 8j+ Ω 6 Ω I 4.24 45 V° ∠− Department of Electronic Engineering, NTUT30/41
  31. 31. 9 − • (a) a-b I 1. Voc Zth 2. 6 a b + − 4 Ω 10j Ω 6 Ω I 4j− Ω (a) ( )6 0 V∠ 4.24 45 0.515 120.96 A 8.246 75.96 oc th V I Z ° ° ° ∠ − = = = ∠ − ∠ ( )2 8 0.515 120.96 0.375 90 A 2 8 6 j I j ° °+ = ∠ − = ∠ − + + b a 2 8j+ Ω 6 Ω I 0.515 120.96 A° ∠ − Department of Electronic Engineering, NTUT31/41
  32. 32. (AC Steady-State Power) (I) • ( ) ( ) ( ) ( ) ( )1 1cos cosm mp t v t i t V t I tω φ ω φ= ⋅ = + × + ( ) ( ) ( )1 20 0 cos cos T T m mt t W p t dt V t I t dtω φ ω φ = = = = + × +∫ ∫ ( ) ( )1 2 1 2 0 1 1 1 sin 2 cos cos 2 2 2 T m m m m t V I t t V I Tω φ φ φ φ θ ω =   = − + + + − =   ( )1 2 1 1 cos cos cos cos cos 2 2 2 2 m m m m m m rms rms V IW P V I V I V I S T φ φ θ θ θ θ= = − = = = = ( ) ( )1 2 1 20 1 cos 2 cos 2 T m mt V I t dtω φ φ φ φ = = + + + −  ∫ Peak RMS ( ) cosθ : Power Factor (PF) ( )1 2θ φ φ= − ( ) S: + − v(t) i(t) Department of Electronic Engineering, NTUT32/41
  33. 33. (II) V I ( ) (a) ( ) ( ) 1 1 2 2 V A rms rms V V V I I I φ φ φ φ = ∠ = ∠  = ∠ = ∠ 1 cos cos 2 m mP VI V Iθ θ= = cos R Z θ = ( ) 2 cos R P VI Z I I I R Z θ   = = =     I + − V Z + − v(t) i(t) j θ (a) Z R jX Z θ= + = ∠ Z [ ]Re Z R= [ ]Im Z X=θ Z Department of Electronic Engineering, NTUT33/41
  34. 34. 10 • (a) (a) + − ( )i t 10 Ω 0.5 H( )sv t ( ) ( )40sin 2 Vsv t t= + − 10 Ω 10j Ω( ) 40 0 V 2 sV = ∠ (b) I 40 0 V 2 V ° = ∠ ( )( )10 20 0.5 10 10Z R j L j jω= + = + = + 10 2 45 , 45θ° ° = ∠ Ω = ( ) 40 2 0 2 45 A 10 2 45 I ° ° ° ∠ = = ∠ − ∠ ( ) 40 cos 2 cos45 40W 2 rms rmsP V I θ °  = = =    Department of Electronic Engineering, NTUT34/41
  35. 35. (Complex Power) • S = P + jQ S VA P W Q VAR (a) Q S θ P j Q S θ Pj (b) cosP S θ=( ): sinQ S θ=( ): :( )cosθ :sinθ ( ) ( )rms rms rms V VV I I Z Z Z φ φ θ φ θ θ ∠ = = = ∠ − = ∠ − ∠ 0θ > 0θ < ( ) ( )rms rmsI I Iφ θ θ φ∗ = ∠ − − = ∠ − rms rmsS S V I VIθ θ ∗ = ∠ = ∠ = ( )rms rms rms rmsVI V I V I Sφ θ φ θ θ∗ = ∠ ⋅ ∠ − = ∠ = ∠ ReP VI∗  =   ImQ VI∗  =  S VI∗ = Department of Electronic Engineering, NTUT35/41
  36. 36. R L C • R θ = 0 = • ( ) θ +90 – 90 ( ) ( ) cos0 1PF = = 2 2 rms rms rms rms V P S V I I R R = = = = ( )cos 90 0PF = ± = j Lω 1 j Cω − Department of Electronic Engineering, NTUT36/41
  37. 37. R Z j θ −XC (a) 90 0θ− < <CZ R jX= − 0 90θ< <LZ R jX= + I V θ (b) RC RL j XL Z R θ (c) I V θ (d) cosθ θ RC RL ( ) I V I V Department of Electronic Engineering, NTUT37/41
  38. 38. 11 • (1). (2). (3). (4). (5). (6). (7). 10 15Z ° = ∠ Ω 50 0V ° = ∠ 50 0 5 15 A 10 15 V I Z ° ° ° ∠ = = = ∠ − ∠ 5 15 AI∗ ° = ∠ ( )( )50 0 5 15 250 15S VI∗ ° ° ° = = ∠ ∠ = ∠ ( )250 VAS S= = 15θ ° = cos15 0.966PF ° = = ( )cos 250cos15 241.5 WP S θ ° = = = ( )sin 250sin15 64.7 VARQ S θ ° = = = Department of Electronic Engineering, NTUT38/41
  39. 39. • ZL Zth θL = –θth ZL ( ) • Ia b + − ( Eth ) 2 2 2 2 max 2 4 th th L L L th th th th E E P I R I R R R R   = = = =    ( )L thZ Z= L thθ θ= − L thZ Z∗ = thE th th thZ Z θ= ∠ L L LZ Z θ= ∠ Department of Electronic Engineering, NTUT39/41
  40. 40. 12 • W ( )( )500 0 2000 90 / / 485.07 14.04 470.58 117.68 500 2000 th LZ R X j j ° ° ° ∠ ∠ = = = ∠ = + + 470.58 117.68L L CZ R jX j= − = − Ω 2 2 max 120 7.65W 4 4 470.58 th th E P R = = = × a b RL= ? X = ? XL= 2kΩ R = 05. kΩ RMS + −120 0thE = ∠ Department of Electronic Engineering, NTUT40/41
  41. 41. • • Department of Electronic Engineering, NTUT j t e ω j t e ω jω j t e ω jω 41/41

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