lecture23

lecture23 - Lecture 23 Multilayer Structures In this...

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1 ECE 303 – Fall 2005 – Farhan Rana – Cornell University Lecture 23 Multilayer Structures In this lecture you will learn: • Multilayer structures • Dielectric anti-reflection (AR) coatings • Dielectric high-reflection (HR) coatings • Photonic Band-Gap Structures ECE 303 – Fall 2005 – Farhan Rana – Cornell University 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 + > = 1 1 1 2 1 2 1 1 + = = Γ + o o o o Z Z Z Z V V Transmission Line Junctions and Discontinuities - I Boundary conditions: 2 1 1 + + = + V V V 2 2 1 1 1 1 o o o Z V Z V Z V + + = 1) Continuity of voltage at z=0: 1) Continuity of current at z=0: 1 2 1 2 1 2 1 2 + = = + + o o o o Z Z Z Z V V T Transmission line discontinuities generate reflections
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2 ECE 303 – Fall 2005 – Farhan Rana – Cornell University 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 + + < + = Transmission Line Junctions and Discontinuities - II Can also replace the infinite transmission line on the right by a lumped impedance 1 o Z 1 + V 1 V 2 o Z 1 1 1 2 1 2 1 1 + = = Γ + o o o o Z Z Z Z V V Which gives: 0 = z ( ) z k j z e V z V 2 2 0 + > = + - 2 + V ( )( ) 1 2 1 1 0 1 2 1 2 1 2 1 1 1 2 + = Γ + = = Γ + = + = = = + + + + + o o o o Z Z Z Z V V T V V V z V V ECE 303 – Fall 2005 – Farhan Rana – Cornell University Unmatched Transmission Lines - I 1 o Z 3 o Z 2 o Z 0 = z l = z 1 + V 1 V 3 + V 2 + V 2 V Question: How does one solve a problem like this? ( ) l l l + + + + < + = 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 + + < < + = l • 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
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3 ECE 303 – Fall 2005 – Farhan Rana – Cornell University Unmatched Transmission Lines - II 1 o Z 3 o Z 2 o Z 0 = z l = z 1 + V 1 V 3 + V 2 + V 2 V Question: How does one solve a problem like this? 1 o Z 2 o Z 0 = z l = z 1 + V 1 V 2 + V 2 V 3 o Z Now calculate the impedance Z ( z =- ): 1 1 2 3 2 3 2 2 + = = Γ + o o o o Z Z Z Z V V () l l l 2 2 2 2 2 1 1 k j k j o e e Z z Z Γ Γ + = = STEP 2: Replace the middle line with the impedance Z ( z =- ) 1 o Z 1 + V 1 V ( ) l = z Z ( ) 1 1 1 1 1 1 + = = = = Γ + o o Z z Z Z z Z V V l l STEP 1: Replace the last line with a lumped equivalent impedance (corresponding to an infinite line) ECE 303 – Fall 2005 – Farhan Rana – Cornell University Question: Is it possible to use a transmission line to perfectly match two dissimilar transmission lines so that there is no reflection?
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This note was uploaded on 02/02/2008 for the course ECE 3030 taught by Professor Rana during the Fall '06 term at Cornell.

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lecture23 - Lecture 23 Multilayer Structures In this...

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