Lect24MultiLayer

# Lect24MultiLayer - ECE 3030 Electromagnetic Fields and...

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1 ECE 3030 Electromagnetic Fields and Waves Fall 2009 Lecture 24 2009/10/23 Multi-Layer Structures Dielectric Anti-Reflection (AR) Coatings Dielectric High Reflection (HR) Coatings 1 Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 24: Instructor: Dr. Wesley E. Swartz Dielectric High-Reflection (HR) Coatings Photonic Band-Gap Structures E +o E +3 3 η E -o 2 1 2 1 2 1 2 1 o Transmission Line Junctions and Discontinuities (1) Transmission line discontinuities generate reflections. 1 o Z 2 o Z 1 + V 1 V 2 + V 0 = z ( ) z k j z k j z e V e V z V 1 1 1 1 0 + + < + = ( ) z k j z e V z V 2 2 0 + > = 2 Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 24: Boundary conditions: Continuity of voltage at z =0 : Continuity of current at z =0 : 1 2 1 2 1 1 o o o o Z Z Z Z V V + = = + Γ 2 1 1 + + = + V V V 2 2 1 1 1 1 o o o Z V Z V Z V + + = 1 2 2 1 2 2 o o o Z Z Z V V T + = = + + Transmission Line Junctions and Discontinuities (2) The infinite transmission line on the right 1 o Z 2 o Z 1 + V 1 V 2 + V 0 = z () z k j z k j z e V e V z V 1 1 1 1 0 + + < + = z k j z e V z V 2 2 0 + > = 3 Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 24: can be modeled with a lumped impedance: which gives: 1 2 1 2 1 1 o o o o Z Z Z Z V V + = = + 1 o Z 1 + V 1 V 2 o Z 0 = z + - 2 + V 1 2 2 1 2 1 1 1 2 2 1 1 0 o o o Z Z Z V V T V V V z V V + = + = = + = + = = = + + + + + Unmatched Transmission Lines (1) Consider the following 3 transmission line segments: Question : How does one solve a problem like this? 1 o Z 3 o Z 2 o Z 0 = z = z 1 + V 1 V 3 + V 2 + V 2 V 4 Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 24: + + + + < + = z k j z k j z e V e V z V 1 1 1 1 z k j z e V z V 3 3 0 + > = z k j z k j z e V e V z V 2 2 2 2 0 + + < < + = • In each segment (except the right most one), the wave is written such that the phase is zero at the right end of the segment. • In each segment, the phase has the wavevector corresponding to that segment. Unmatched Transmission Lines (2) Question : How does one solve a problem like this? STEP 1: Replace the last transmission line with a lumped equivalent impedance (corresponding to an infinite line) 1 o Z 3 o Z 2 o Z 0 = z = z 1 + V 1 V 3 + V 2 + V 2 V 1 + V 2 + V 2 2 o o Z Z V 5 Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 24: STEP 2: Calculate the impedance Z ( z =- ) : STEP 3: Replace the middle line with the impedance Z ( z =- ) . 1 o Z 2 o Z 0 = z = z 1 V 2 V 3 o Z 2 3 3 2 o o Z Z V + = = + 2 2 2 2 2 1 1 k j k j o e e Z z Z Γ Γ + = = 1 1 1 1 o o Z z Z Z z Z V V + = = = = + 1 o Z 1 + V 1 V = z Z Matching Transmission Lines (1) A quarter-wave length of transmission line can impedance match two dissimilar transmission lines so that there is no reflection. 0

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Lect24MultiLayer - ECE 3030 Electromagnetic Fields and...

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