Section 6 5 TRƯỜNG ĐIỆN TỪ

No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]

Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, Urbana, Illinois, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India 6.5 Lines with Initial Conditions 6.5-3 Line with Initial Conditions I(z, 0) ++++++++ Z 0, v p V(z, 0) V  (z, 0)  V – (z, 0) V(z, 0) I  (z, 0)  I – (z, 0)  I(z, 0)  – V V I  , I – – Z0 Z0 V  (z, 0) – V – (z, 0) Z0 I(z, 0) 6.5-4 V  z ,    V  z ,   Z0 I  z ,   V  z ,    V  z ,   Z0 I  z ,   6.5-5 Example: V(z, 0), V I(z, 0) ++++++++ Z0, vp V(z, 0) z = Z 50  zz =ll 50 l z I(z, 0), A l z 6.5-6 V +(z, 0), V I +(z, 0), A B 50 A C l z l z I –(z, 0), A V –(z, 0), V 50 C D l z –1 l z 6.5-7 l t 2vp V +, V 50 V –, V 50 B C D l z B A I +, A l I –, A z z V, V –1 l 100 z I, A 50 l l z l z 6.5-8 t  vl p V +, V I +, A ’ ’ 50 D C l z V –, V I –, A ’ 50 z l ’ B C A l z l z –1 V, V I, A 50 l z –1 l z 6.5-9 I(z, 0) t=0 +++++++ Z0, vp - RL = Z0 = 50  V(z, 0) z=0 z=l V(z, 0), V I(z, 0), A 50 l z l z 6.5-10 V +(z, 0) V V –(z, 0) V B 50 50 C D A C l z l z [V]RL, V 50 A B l/2vp C l/vp D 3l/2vp t 6.5-15 Bounce Diagram Technique for Uniform Distribution + I+ RL + V0 + V + – z=0 V0  V   RL 0  I    V I  Z0  z=l B.C 6.5-16 RL   V0  V – V Z0  RL   V 1  – V0  Z0  V  – V0 Z0 R L  Z0 For V0 100 V, Z0 50 , and RL = 150 , 50  V – 100 – 25 V 150  50 6.5-17 = 75 100 V –25 –12.5 t, mS 37.5 50 –12.5 –6.25 18.75 z=0 100 –25 75 =1 25 z=l z [V]RL 37.5 18.75 9.375 t, mS 6.5-18 Energy Storage in Transmission Lines we, Electric stored energy density = CV l We, Electric stored energy = z0 CV dz 2  CV l (for uniform distribution) 2  CV v p T  CV T 2 LC 1V0  T Z0 6.5-19 wm, Magnetic stored energy density = LI l Wm, Magnetic stored energy = z0 LI dz 2  LI l (for uniform distribution) 2  LI v p T  LI T 2 LC = I Z0 T 6.5-20 Check of Energy Balance Initial stored energy We  Wm 1V0  T  I Z0 T Z0 (100)2  10 –3  50 0.1 J