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電路學 - [第二章] 電路分析方法

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電路分析方法

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電路學 - [第二章] 電路分析方法

  1. 1. Department of Electronic Engineering National Taipei University of Technology
  2. 2. • • • (Node Voltage Analysis) • (Mesh Current Analysis) Department of Electronic Engineering, NTUT2/26
  3. 3. • N KVL N = Department of Electronic Engineering, NTUT 1 2 Nv v v v= + + +⋯ 1 1 2 2, , , N Nv R i v R i v R i= = =⋯ 1 2 Nv R i R i R i= + + +⋯ 1 2 N v i R R R = + + +⋯ 1 2 1 N s N n n R R R R R = = + + + = ∑⋯ + + − + − + − v i v1 − vN v2 R1 R2 RN 3/26
  4. 4. • N • Department of Electronic Engineering, NTUT 1 2 1 2, , , N N s s s RR R v v v v v v R R R = = =⋯ 1 2 1 2: : : : : :N Nv v v R R R=⋯ ⋯ + + − + − + − v i v1 − vN v2 R1 R2 RN 4/26
  5. 5. 1 • 1: i v1 v2 v3 ? Department of Electronic Engineering, NTUT (a)(a)(a)(a) + − 16 Ω + − 6 Ω + − 8 Ω + − 2 Ω 12 Ω 4 Ω v1 i ′a b c ′b ′c a v2 v3 ( )2 30 Vt e− (b)(b)(b)(b) + − + − + − i a b ′a ′b 2 Ω 12 Ω 16 Ω 4 Ωv1 v2( )2 30 Vt e− (a) (b) 5/26
  6. 6. 1 Department of Electronic Engineering, NTUT (b)(b)(b)(b) + − + − + − i a b ′a ′b 2 Ω 12 Ω 16 Ω 4 Ωv1 v2( )2 30 Vt e− (c)(c)(c)(c) + − + − i a ′a ( )2 30 Vt e− 2 Ω 8 Ω v1 (b) (c) • (c) KLC 2 30 2 8 0t e i i− − + + = 2 3 t i e− = ( )2 2 1 8 30 24 V 2 8 t t v e e− −  = ⋅ =  +  ( )2 2 1 4 6 V 12 4 t v v e−  = ⋅ =  +  ( )2 3 2 8 4 V 8 4 t v v e−  = ⋅ =  +  6/26
  7. 7. • : N i = i1 + i2 + ... + iN • R1 R2 Department of Electronic Engineering, NTUT 1 1 2 2, , , N Ni G v i G v i G v= = =⋯ 1 2 Ni G v G v G v= + + +⋯ 1 2 N i v G G G = + + +⋯ 11 2 1 1 1 1 1N np N nR R R R R= = + + + = ∑⋯ 1 2 1 2 1 2 ||eq R R R R R R R = = + + − vi G1 G2 GN i1 i2 iN 7/26
  8. 8. • Department of Electronic Engineering, NTUT 1 2 1 2 1 2 , , ,p p pN N p p p N R R RGG G i i i i i i i i i G R G R G R = = = = = =⋯ 1 2 1 2 1 2 1 1 1 : : : : : : : : :N N N i i i G G G R R R = =⋯ ⋯ ⋯ + − vi G1 G2 GN i1 i2 iN 8/26
  9. 9. 2 • 2: i1 i2 ? Department of Electronic Engineering, NTUT 6 Ω3 Ω 6 Ω 3 Ω 4 Ω 6 Ω a b a′ b′ (a)(a)(a)(a) (A)12 i1 i2 3 Ω 3 Ω3 Ω6 ΩA12 a b a′ b′ (b)(b)(b)(b) i1 (a) (b) 9/26
  10. 10. 2 Department of Electronic Engineering, NTUT 3 Ω 3 Ω3 Ω6 ΩA12 a b a′ b′ (b)(b)(b)(b) i1 a A12 a′ (c)(c)(c)(c) 6 Ω2 Ω i1 ( )1 2 12 3 A 2 6 i   = ⋅ =  +  • (c) ( )2 1 4 3 A 4 6 6 4 i i   = ⋅ =  + +  • (a) 10/26
  11. 11. (Node Voltage Analysis) • KCL • • va , vb , vc va , vb , vc Department of Electronic Engineering, NTUT a : 1 2 1 0si i i+ − = ( )1 2 1 0a b a sG v v G v i− + − = 1 3 4 0i i i− + + = ( ) ( )1 3 4 0a b b b cG v v G v G v v− − + + − =b : 4 5 2 0si i i− + + = ( )4 5 2 0b c c sG v v G v i− − + + =c : ( )1 2 1 10a b sG G v G v i+ − + = ( )1 1 3 4 4 0a b cG v G G G v G v− + + − = ( )4 4 5 20 b c sG v G G v i− + + = − is1 a b d G2 i2 G3 i3 i5 i4i1 G1 G4 c is2 G5 11/26
  12. 12. • • G Department of Electronic Engineering, NTUT v1 . . . vN N ( ) Gjj j j Gjk = Gkj , j ≠ k , j k j k ij j G G G G G G G G G v v v i i i N N N N NN N N 11 12 1 21 22 2 1 2 1 2 1 2 −−−− −−−− −−−− −−−− −−−− −−−−                         ====             ...... ...... ...... ...... ... ... ... ... ... 12/26
  13. 13. 3 • 3: v1 , v2 , v3 , i1 , i2 , i3 , i4 Department of Electronic Engineering, NTUT Cramer’s rule A2 3 A v1 i22 Ω 1 Ω 4 Ω v2 v3 i3i4 i1 11 3G = ℧ 22 5G = ℧ 33 5G = ℧ 12 21 1G G= = ℧ 13 31 2G G= = ℧ 23 32 2G G= = ℧ 1 2 3 3 1 2 2 1 5 0 3 2 0 5 3 v v v − − −         − =         − −     3 1 2 1 5 0 50 2 0 5 − − ∆ = − = − 2 2 1 2 3 5 0 65 3 0 5 − − − ∆ = = − − 3 3 2 2 1 3 0 17 2 3 5 − − ∆ = − = − − 4 3 1 2 1 5 3 56 2 0 3 − − ∆ = − = − − − 1 1 1.3 Vv ∆ = = − ∆ 2 2 0.34 Vv ∆ = = ∆ 3 3 1.12 Vv ∆ = = − ∆ ( )1 1 21 1.64 Ai v v= − = − ( )2 1 21 0.36 Ai v v= − = 3 3.36 Ai = − 4 1.36 Ai = 13/26 Ω 3
  14. 14. 3 • Department of Electronic Engineering, NTUT = 3 × 5 × 5 + (-1) × 0 × (-2) + (-1) × 0 × (-2) −(-2) × 5 × (-2) − 0 × 0 × 3 − (-1) × (-1) × 5 = 75 + 0 + 0 − 20 − 0 − 5 = 50 502 051 213 − − −− =∆ (4) 3 1 2 1 5 0 2 0 5 2 3 3 1 2 3 −−−− −−−− −−−− −−−−                     ==== −−−− −−−−           v v v (1) For node 1: 2+1× (v1 v2)+2×(v1 v3) = 0 ⇒ (1+2) × v1-1×v2 2 × v3 = 2 ⇒ G11v1+G12v2+G13 = vs1 (2) For node 2: 1× (v1 v2) + 4 × (v2 0) 3= 0 ⇒ −1× v1+(1+4)v2+ 0 × v3 = 3 ⇒ G21v1+G22v2+G23v3 = vs2 (3) For node 3: 2× (v3 v1) + 3 × (v3 0) + 3= 0 ⇒ −2× v1+ 0 × v2 +(2+3)v3 = 3 ⇒ G31v1+G32v2+G33v3 = vs3 A2 3 A v1 i22 Ω 1 Ω 4 Ω v2 v3 i3i4 i1 14/26 Ω 3
  15. 15. • M (M-1) (M-1) (M-1) (Supernode) KCL Department of Electronic Engineering, NTUT15/26
  16. 16. • v1 v2 v3 v4 v5 5 v1 = vs1 , v5 – v4 = vs2 v2 v3 v4 KCL Department of Electronic Engineering, NTUT 2 3 ( )1 2 4 2 1 1 2 3 4 5 0G G G v G v G v G v+ + − − − = ( )2 3 5 3 2 2 5 4 0G G G v G v G v+ + − − = ( ) ( )4 5 2 5 4 3 6 5 0G v v G v v G v− + − + = + − 1 2 3 4 5vs1 G1 G4 G6 G5 G3 G2 vs2 16/26
  17. 17. • KVL KCL KCL 1 2 3 KCL ( ) Department of Electronic Engineering, NTUT + − 1 2 3 44 sv v= ( )1 2 3 1 2 2 3 3 1 0sG G G v G v G v G v+ + − − − = ( ) ( )2 1 2 5 2 1 3 0G v G G v v vβ− + + + − = ( ) ( )3 1 3 4 3 1 3 0G v G G v v vβ− + + − − = vs G1 G2 G3 G4 ( )1 3v vβ − G5 17/26
  18. 18. 4 • 4: v2 , v3 , v4 v1 = 100 V Department of Electronic Engineering, NTUT 3 (a) 2 (b) 3 (c) 4 ( ) ( )1 2 3 2 2 2 41 100 2 4i i i v v v v= + ⇒ − = + − 3 22v v= − ( ) ( ) ( )3 4 5 2 4 3 4 40 4 4 4 0 0i i i v v v v v+ + = ⇒ − + − + − = 2 2 3 3 4 4 7 0 4 100 12 V 2 1 0 0 24 V 1 1 3 0 4 V v v v v v v − =            = ⇒ = −           − = −       + − + − 1 Ω 4 32 V100 4v1 Ω 1 Ω 4 Ω 4 Ω 2 2v2 v3 v4 i5i1 i2 v2 i3 18/26
  19. 19. (Mesh Current Analysis) • KVL IR KVL • 2 2 Ia Ib KVL Department of Electronic Engineering, NTUT 1 (a-b-e-f-a) ⇒ 2 (b-c-d-e-b) ⇒ + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 1 1 1 3 3 0gv I R I R− + + = 3 3 2 2 2 0gI R I R v− + + = 1 1 3 3 1gI R I R v+ = 2 2 3 3 2gI R I R v+ = 19/26
  20. 20. • Department of Electronic Engineering, NTUT b + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 3 1 2I I I= − 1 2,a bI I I I= = 3 a bI I I= − ( ) ( ) 1 3 3 1 3 2 3 2 a b g a b g R R I R I v R I R R I v  + − =  − + + = − 11 3 3 23 2 3 ga gb vR R R I vR R R I + −      =      −− +      20/26
  21. 21. • Department of Electronic Engineering, NTUT M v1 j i1 j Rjj j Rjk=Rk j j k j k ij , ik Rjk( Rk j) ij , ik Rjk( Rk j) 。 + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 11 12 1 1 1 21 22 2 2 2 1 2 3 M M M M M MM M M R R R i v R R R i v R R R R i v ± ±           ± ±      =             ± ± ±      ⋯ ⋯ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ 21/26
  22. 22. • Ia , Ib R3 R3 R3 Ia Ib I3 = Ia + Ib Department of Electronic Engineering, NTUT + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 11 3 3 23 2 3 ga gb vR R R I vR R R I +      =     +      22/26
  23. 23. • KVL Department of Electronic Engineering, NTUT23/26
  24. 24. 5 • 5: i1 , i2 , i3 Department of Electronic Engineering, NTUT KVL KVL + − V2 2 Ω 1 Ω 3 Ω 2 Ω + − + − A2 + − + − 3 Ω + −+ − v1 v3 v5 v3v2 v6i1 i2 i3 ( )1 1 22 3 2i i i+ − = ( ) ( )1 2 2 3 3 2 3 3 2 2 A i i i i i i  − = + +  − = ( ) ( ) ( ) 1 2 1 2 3 3 5 3 0 2 1 11 7 3 4 5 0 A , A , A 3 9 9 0 1 1 2 i i i i i i −      − −    − = ⇒ = = =         −     24/26
  25. 25. 6 • 6: i1 , i2 , i3 Department of Electronic Engineering, NTUT (a) 1: KVL i1 – i3 = 1 (b) 2: KVL i1 – i2 = 2v3 = 2(3i2) (c) 3: KVL 2 3 4 KVL 2i1 + 3i2 + 4i3 = 0 4 Ω + − A1 + − + − + − + − v4 2 Ω 3 Ω v5 v2 v1 v3 1 Ω i3 i1 i2 2v3 A ( ) ( ) ( ) 1 2 1 2 3 3 1 0 1 1 28 4 17 1 7 0 0 A , A , A 45 45 45 2 3 4 0 i i i i i i −      −    − = ⇒ = = =              25/26
  26. 26. • (KCL) (KVL) • • ( ) ( ) Department of Electronic Engineering, NTUT26/26

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