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/terms6.013 - Electromagnetics and Applications Fall 2005 Lecture 10 - Transmission Lines Prof. Markus Zahn October 13, 2005 I. Transmission Line Equations A. Parallel Plate Transmission Line E ¯ must b e perpendicular to the electrodes and H ¯ must b e tangential, so E ¯= Ex(z, t)¯ix H ¯= Hy(z, t)¯iy ∂H ¯ ∂Ex ∂HyE = −µ = −µ� × ¯ ∂t ⇒ ∂z ∂t ¯ ∂E ¯ ∂Hy ∂Ex � × H = � ∂t ⇒ ∂z = −� ∂t 1 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.������� 2 v(z, t) = E ¯ d ¯ l = Ex(z, t)d· 1 z = constant i(z, t) = Kz(z, t)w = Hy(z, t)w ∂v ∂i µd= −L L = henries / meter Inductance per unit length∂z ∂t w ∂i ∂v �w = −C C = farads / meter Capacitance per unit length∂z ∂t d µd�w 1 LC = = µ� = 2wd cB. Transmission Line Structures 2 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.C. Distributed Circuit Representation with Losse s i(z, t) − i(z + Δz, t) = CΔz ∂v(z, t)+ GΔz v(z, t)∂t v(z, t) − v(z + Δz, t) = LΔz ∂i(z + Δz, t)+ i(z + Δz, t )RΔz ∂t i(z + Δz, t) − i(z, t) ∂i ∂v lim = = −C ∂t − Gv Δz 0 Δz ∂z →v(z + Δz, t) − v(z, t) ∂v ∂i lim = = −L∂t − iR Δz 0 Δz ∂z →R is the series res istance per unit length, measured in ohms/meter, and G is the shunt conductance per unit length, measured in siemens/meter. 3 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.����� ����� � ����� ����� � ��� � � � � � � � � � If the line is lossless (R = G = 0), we have the Telegrapher’s equations: ∂i ∂v = −C ∂z ∂t ∂v ∂i = −L ∂z ∂t Including loss, Poynting’s theorem for the circuit equivalent form is: ∂i ∂v ∂z = −C ∂t − Gv ∂v ∂i v · i = −L∂t − iR · ∂z ∂ �1 = −∂t 2 ∂(vi) 1 ∂i ∂v Cv2 + Li2 − Gv2 − i2RAdd: v + i = 2 ∂z ∂z ∂z D. Wave Equation (Lossless, R = 0, G = 0) ∂2i ∂2∂ ∂t ∂i ∂v v = −C ∂t ⇒ ∂z∂t = −C ∂t2∂z ∂2i 1 ∂2v ∂ ∂z ∂v ∂i ∂z = −L∂t ∂z∂t = −L ∂z2 2 2∂tc� 1 ∂2v ∂2v ∂2v ∂2v 1 ∂2v = −C = LC = − ⇒ ∂z2 Wave equation II. Sinusoidal Steady State A. Complex Amplitude Notation ∂z2 ∂t2 ∂t2L jωtv(z, t) = Re vˆ(z)ei(z, t) = Re ˆi(z)ejωt Substitute into the wave equation: ∂2v 1 ∂2v d2vˆ ω2 ω ∂z2 = c2 ∂t2 ⇒ dz2 = − c2 vˆ(z), let k = c d2vˆdz2 + k2 vˆ = 0 ⇒ vˆ(z) = Vˆ+e−jkz + Vˆ−e +jkz dvˆ 1 −��jk Vˆ+e−jkz + ��jk Vˆ−e= −Ljωˆi ⇒ ˆi(z) = − +jkz dz L��jω √LCk ω k = �= √LC C = Y0 is the Line Admittance= = ω ωc�⇒ ωL L L 1 L Z0 = = is the Line Impedance Y0 C Vˆ+e−jkzˆi(z) = Y0 V− ˆ−e +jkz 4� � � � vˆ(z) = Vˆ+e−jkz + Vˆ−e +jkz v(z, t) = Re Vˆ+ej(ωt−kz) + Vˆ−ej(ωt+kz) i(z, t) = Re Y0 Vˆ+ej(ωt−kz) − Vˆ−ej(ωt+kz) k = ω = ω√LC = ω√�µ c B. Short Circuited Line (v(z = 0, t) = 0, v(z = −l, t ) = V0 cos(ωt)) vˆ(z) = Vˆ+e−jkz + Vˆ−e +jkz ⇒ vˆ(z = 0) = 0 = � Vˆ+ + Vˆ− ⇒� Vˆ+ = −Vˆ− vˆ(z = −l) = V0 = Vˆ+e +jkl + Vˆ−e−jkl = Vˆ+ ejkl − e−jkl 5 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � � � � � ��� = 2j sin(kl)Vˆ+ ˆV0V+ = −Vˆ− =2j sin(kl) vˆ(z) = V0 � e−jkz +jkz � = V0(−2j) sin(kz) 2j sin(kl) − e 2j sin(kl) V0 sin(kz)= − sin(kl ) ˆi(z) = Y0 Vˆ+e−jkz − Vˆ−ejkz = Y0V0 e−jkz + e +jkz 2j sin(kl) 2�Y0V0 cos(kz)= 2�j sin(kl) jY0V0 cos(kz)= −sin(kl ) � � V0 sin(kz) V0 sin(kz) cos(ωt) v(z, t) = Re vˆ(z)ejωt= Re − sin(kl ) ejωt = − sin(kl ) � �� jY0V0 cos(kz) � Y0V0 cos(kz) sin(ωt)i(z, t) = Re ˆi(z)ejωt = Re −sin(kl ) ejωt = sin(kl ) We have resonance when sin(kl) = 0 kl = nπ = ωl ω = ωn ≡ nπc , n = 1, 2, 3, . . .⇒ c ⇒ l Complex impedance: Z(z) = vˆ(z) = −jZ0 tan(kz)ˆi(z) Z(z = −l) = +jZ0 tan(kl) In the following, take n = 1, 2, 3, . . .: kl = nπ Z(z = −l) = 0 short circuit π kl = (2n − 1)2 Z(z = −l) = ∞ open circuit π(n − 1)π < kl < (2n − 1)2 Z(z = −l) = +jX, X > 0 (positive reactance, inductive) 1(n − 2)π < kl < nπ Z(z = −l) = −jX, X > 0 (negative reactance, capacitive) kl � 1 Z(z) = −jZ0k⇒ � L = −j ω L��C C = −jLZ Z(z = −l) = j(Ll) inductive V0z |kz| � 1 v(z, t) = − l cos(ωt) ⇒ v(z = −l, t) = V0 cos(ωt) di = (Ll) (z = −l, t)dtV0Y0 V0 sin(ωt)i(z, t) = kl sin(ωt) i(z = −l, t) = (Ll)ω 6� � � � � � �C. Open Circuited Line (i(z = 0, t) = 0) v(z = −l, t) = V0 sin(ωt) ˆi(z) = Y0 Vˆ+e−jkz − Vˆ−e +jkz ⇒ ˆi(z = 0) = 0 = Y0 Vˆ+ − Vˆ− ⇒ Vˆ+ = Vˆ− vˆ(z = −l) = −jV0 = Vˆ+e +jkl + Vˆ−e−jkl = Vˆ+ ejkl + e−jkl = 2 Vˆ+ cos(kl) Vˆ+ = VˆjV0 − = −2 cos(kl) vˆ(z) = − jV0 � e−jkz + e +jkz � 2 cos(kl) jV0 · �2 cos(kz)= −2 cos(kl) jV0 cos(kz)= −cos(kl) ˆi(z) = − jY0V0 � e−jkz − e +jkz � 2 cos(kl) (−jY0V0)(−2j) sin(kz)= 2 cos(kl) Y0V0 sin(kz)= − cos(kl) v(z, t) = Re � vˆ(z)ejωt� = V0 cos(kz) sin(ωt)cos(kl) i(z, t) = Re � ˆi(z)ejωt � = −V0Y0 sin(kz) cos(ωt)cos(kl) πResonance: cos(kl) = 0 ⇒ (kl) = (2n − 1)2 , n = 1, 2, 3, . . . (2n − 1)π ωn = 2 …
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