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 11 - Generalized Reflection Coefficient Prof. Markus Zahn October 18, 2005 � � VˆLi(z = 0, t) = iL(t) = Re IˆLejωt , IˆL = ZL vˆ(z) = Vˆ+e−jkz + Vˆ−e +jkz V+e−jkz +jkz ˆi(z) = Y0 ˆ− Vˆ−e vˆ(z = 0) = VˆL = Vˆ+� + Vˆ− � Vˆ+ + Vˆ− = VˆL ˆi(z = 0) = IˆL = VL = Y0 ˆV⇒ ˆV = VL ZˆL V+ − ˆ− V+ − ˆ− Y0ˆZL Add: 2 Vˆ+ = VˆL � 1 + 1 � Vˆ+ = VˆL � Y0ZL + 1 � Y0ZL ⇒ 2 Y0ZL Subtract: 2 Vˆ= VˆL � 1 − 1 � Vˆ= VˆL � Y0ZL − 1� − Y0ZL ⇒ − 2 Y0ZL Load reflection coefficient: ΓL = Vˆ−= Y0ZL − 1 Vˆ+ Y0ZL + 1 ZL Z0 − 1 ZL − Z0 = = ZL + 1 ZL + Z0Z0 I. Arbitrary Impedance Terminations iL (t)ZLZυL (t)+_0v(z = 0, t) = vL(t) = Re � ˆVLejωt � 1 Image by MIT OpenCourseWare.� � � � � � � � � � � � Generalized reflection coefficient: Γ(z) = Vˆ−e+jkz = Vˆ−e 2jkz = ΓLe 2jkz Vˆ+e−jkz Vˆ+ vˆ(z) = Vˆ+e−jkz [1 + Γ(z)] ˆi(z) = Y0Vˆ+e−jkz [1 − Γ(z)] Z(z) vˆ(z) 1 + Γ(z)Zn(z) = Z0 =ˆi(z)Z0 =1 − Γ(z) Normalized impedance Γ(z) = Zn(z) − 1 Zn(z) + 1 Properties A. |Γ(z)| = |ΓL| ≤ 1 B. λΓ z ± 2 = Γ(z) λ Zn z ± 2= Zn(z) C. λΓ z ± 4= −Γ(z) λ 1 Y (z)ˆi(z) Zn z ± 4= Zn(z)= Yn(z) = Y0 = Y0vˆ(z) D. If line is matched, ZL = Z0, ΓL = 0, Zn(z) = 1 II. Load Impedance Reflected Back to the Source Zn(z = 0) = ZL = RL + jXL Z0 Z0 4Zn z = − λ =1= Z0 = Z z = −λ 4 Zn(z = 0) RL + jXL Z0 2 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � � � = � � V0cosωtYRsZ � z = − λ 4 � = Z0Zn Y � z = − λ 4 � = Z YT = Y + Y To tune the line, choose Y = −jXL Z2 0⇒ YT = RL Z2 0 . Y is usually created by a variable length short circuited transmission line called a stub. There is maximum power into the line if RS = Z2 0 RL ⇒ �P �max = 1 2 (1 2V0)2 V2 0RL III. Quarter Wavelength Matching λ Z02 z = − = 4 RL + jXL 1 RL + jXL z = −λ 4 Z02 λ RL + jXL z = − = Y +4 Z02 20V1 1= = . 20RS 8 RS 8 Z= √Z1RL. Z22Z2To match Z1 to RL ⇒ Z1 = RL/Z2 = , Z2RL 3 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.Image by MIT OpenCourseWare.IV. Smith Chart Zn(z) = r + jx Γ(z) = Γr + jΓi (z) = 1+Γ(z) 1+Γr +jΓiZn1−Γ(z) ⇒ r + jx = 1−Γr −jΓi 1−Γ2 r −Γ2 i 2Γir = x = (1−Γr)2+Γ2 (1−Γr)2+Γ2 i i � �2 � 1. Γr − 1+rr + Γi 2 = (1+1 r)2 Orthogonal 2. (Γr − 1)2 + � Γi − x 1 �2 = x1 2 Circles 1. Circle of radius 1+1 r Center at Γi = 0, Γr = 1+rr 2. Circle of radius 1 Center at Γi = 1 , Γr = 1 |x| x 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.i(t) = |Iˆ| sin(ωt − φ) |Iˆ| = |50 + Z� V(z 0 = −l)| � φ = tan−1 Im [Z(z = −l)] 50 + Re [Z(z = −l)] 6 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � ��� ��� ������ ��� ������ ��� V. Standing Wave Parameters vˆ(z) = Vˆ+e−jkz [1 + Γ(z)] vˆ(z) = Vˆ+||1 + Γ(z)|ˆi(z) = Y0Vˆ+e−jkz [1 − Γ(z)] ⇒ |ˆi(z)| | ⇒ | | = Y0|Vˆ+||1 − Γ(z)| max |vˆvˆ((zz))|min = 1 + |ΓL| = VSWR (Voltage Standing Wave Ratio) | |1 − |ΓL|VSWR − 1 |ΓL| = VSWR + 1 ΓL = |ΓL|ejφ 4dmin 2kdmin − 1 = φ + π φ = π⇒ λ dmin is the shortest distance from load to first voltage minimum (at B in Figure 8-21 above) Special Cases: A. Matched Line: ΓL = 0, VSWR = 1 B. Short or open circuited line: vˆ(z) vˆ(z)ΓL|; vˆ(z=0) |ΓL| = 1, VSWR = ∞ C. = 1 + = 1 −|ΓL|| Vˆ+Vˆ+ min D. Vˆ+ peak = |1 + ΓL|; ˆi(z=0) Y0Vˆ+ = |1 − ΓL| E. If ZL = RL (real), then ΓL real. If ZL > Z0, VSWR = ZL . If ZL < Z0, VSWR = Z0 Z0 ZL 7 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.Z0 �2 �1 − jVSWR tan φ [1 jVSWR tan(kdmin)]� � �� 1 + ΓLejφ Load Impedance: ZL = Z0 |ΓL|ejφ 1 �− | |jφ �VSWR + 1 + (VSWR − 1)e= Z0 VSWR + 1 − (VSWR − 1)ejφ VSWR − j tan φ 2 = −= Z0 [VSWR − j tan(k dmin)] Example: Z0 = 50 Ω, VSWR = 2 d = distance between succes sive voltage minima 8 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � �� λ= 40 cm = 2 dmin = distance from load to first minimum = 10 cm 2π 2π λ = 2d = 80 cm k = = = 2.5π⇒ λ 0.8 π dmin = 10 cm = 0.1 m kdmin = 2.5π(0.1) = ⇒ 4 VSWR − 1 1 =|ΓL| = VSWR + 1 3 φ 4dmin 4(0.1) 1 π π = λ − 1 = 0.8 − 1 = −2 ⇒ φ = − 2 ΓL = |ΓL|ejφ = 31 e−jπ/2 = −3 j [1 − jVSWR tan(kdmin)]ZL = Z0 VSWM − j tan(k dmin) 50 1 − j(2) tan +π = � � 4 2 − j tan +π 4 50(1 − 2j)=2 − j = 40 − 30j ohms
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