MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms� � � � � � � � � � 6.013 - Electromagnetics and Applications Fall 2005 Lecture 12+13 - Transient Waves on Transmission Lines Prof. Markus Zahn October 25 and 27, 2005 I. Wave equation (Lossless) ∂v ∂i ∂2v 2 ∂2v = −L = c ∂z ∂t ∂t2 ∂z2 ∂i ∂v ⇒ 1 1 ∂z = −C∂t c = √LC = √�µ Solution: v(z, t) = V+ t − zc + V− t + zc z ∂α ∂α 1Proof: Let α = t − c ⇒ ∂t = 1, = −∂z c zSuperposition: v+(z, t) = V+ t − c ∂v+ dv+ ∂α dv+ = = ∂t dα ∂t dα ∂2v+ d2v+ ∂α d2v+ = = ∂t2 dα2 ∂t dα2 ∂v+ dv+ ∂α 1 dv+ = = −∂z dα ∂z c dα ∂2v+ 1 d2v+ ∂α 1 d2v+ ∂z2 = − c dα2 ∂z = c2 dα2 ∂2v+ d2v+2 ∂2v 2 1 d2v+ d2v+ = = c = c = ∂t2 dα2 ∂z2 c2 dα2 dα2 ∂β ∂β 1Negative z directed waves: Let β = t + zc ∂t = 1, ∂z = c⇒ ∂v dv ∂β dv− = − = − ∂t dβ ∂t dβ ∂2v d2v ∂β d2v− = − = − ∂t2 dβ2 ∂t dβ2 ∂v dv ∂β 1 dv− = − = − ∂z dβ ∂z c dβ ∂2v 1 d2v ∂β 1 d2v− = − = − ∂z2 c dβ2 ∂z c2 dβ2 ∂2v d2v2 ∂2v 1 d2v d2v− = − = c − = c 2 − = − ∂t2 dβ2 ∂z2 c2 dβ2 dβ2 II. Solution for current i(z, t) ∂v ∂i = −L ∂z ∂t ∂2i 2 ∂2i = c⇒ ∂t2 ∂z2 ∂i ∂v = −C ∂z ∂t 1� � � � � � � � � � � � � � � � � � � � � � � � � � Solution: i(z, t) = I+ t − zc + I t + zc� � � �− v(z, t) = V+ t − zc + V− t + zc +z solution: α = t − z , ∂α = 1, ∂α 1 c ∂t ∂z = −c ∂v+ ∂i+ dv+ ∂α 1 dv+ ∂z = −L ∂t ⇒ dα ∂z = − c dα di+ ∂α = −L dα ∂t di+ = −L dα dv+ di+ L di+ L di+ di+ dα = Lc dα = √LC dα = C dα = Z0 dα z z v+ = i+Z0 ⇒ I+ t − = Y0V+ t −� c c C 1 Y0 = = L Z0 −z solution: β = t + z , ∂β = 1, ∂β = 1 c ∂t ∂z c ∂v ∂i dv ∂β 1 dv− = −L − − = − ∂z ∂t ⇒ dβ ∂z c dβ di ∂β = −L dβ − ∂t = −Ldi− dβ dv di L di L di didβ − = −Lc dβ − = −√LC dβ − = − C dβ − = −Z0 dβ − z z v− = −i−Z0 ⇒ I+ t + c = −Y0V− t + c z z v(z, t) = V+ t − + V− t + � �c � �c �� z z i(z, t) = Y0 V+ t − c − V− t + c III. Transmission Line Transient Waves A. Transients on Infinitely Long Transmission Lines 1. Initial Conditions z z v(z, t = 0) = V+ − c + V− c = 0 � � � � �� z z i(z, t = 0) = Y0 V+ − c − V− c = 0 z z V+ = 0, V = 0 − c − c � � z z z > 0, t > 0 ⇒ t + c> 0 ⇒ V− � t + c � = 0 z z z t − c> 0 if t > c to allow V+ t − c = 0 �2� � � � � z � � z � v(z, t)With V− t + c = 0 ⇒ v(z, t) = V+ t − c ⇒ i(z, t)= Z0 z i(z, t) = Y0V+ t − c 2. Traveling Wave Solution v(z = 0, t) = V (t) = V+(t) v(z = 0, t) = Z0 V (t) = V+(t)Z0 + RS V (t)i(z = 0, t) = Y0V+(t) = RS + Z0 Z0 z v(z, t) = Z0 + RS V t − c 1 � z � i(z, t) = RS + Z0 V t − c 3 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � � � � � � � � � B. Reflections from Resistive Terminations 1. Reflection Coefficient l lAt z = l : v(l, t) = V+ t − c + V− t + c = i(l, t)RL � � � � �� l l = Y0RL V+ t − c − V− t + c ΓL = V− � t + cll � = RL − Z0 V+ t − c RL + Z0 Special cases: a. RL = Z0 ⇒ ΓL = 0 (matched line) b. RL = 0 ⇒ ΓL = −1 (short circuited line) If RL < Z0, ΓL < 0 c. RL = ∞ ⇒ ΓL = +1 (open circuited line) If RL > Z0, ΓL > 0 2. Step Voltage At z = 0: v(z = 0, t) + i(0, t)RS = V0 V+(z = 0, t) + V−(z = 0, t) + Y0RS [V+(z = 0, t) − V−(z = 0, t)] = V0 V+(z = 0, t) = ΓS V (z = 0, t) + Z0V0 , ΓS = RS − Z0 −Z0 + RS RS + Z0 a. Matched Line: � RL = � Z0, ΓL = 0; RS = Z0, ΓS = 0 ΓL = 0 ⇒ V− t + zc = 0, V+ (z = 0, t) = V2 0 , in steady state after time T = cl b. Short circuited line: RL = 0, ΓL = −1, RS = Z0, ΓS = 0 ΓL = −1 V+ = −V−. When V+ t − zc and V t + zc ove rlap in space, ⇒ 2l − v(z, t) = 0. For t ≥ 2T = , v(z, t) = 0, i(z, t ) = V0 . c Z0 c. Open circuited line: RL = ∞, ΓL = +1, RS = Z0, ΓS = 0 ΓL = +1 ⇒ V+ = +V−. For t ≥ 2T = 2cl , v(z, t) = V0, i(z, t) = 0 4 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.5 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.From Electromagnetic Field Theory: A Problem Solving …
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