ECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 10. Waveguides Part 7: Transverse Equivalent Network (TEN)

Similar documents
ECE Microwave Engineering

1.0 Electrical Systems

T-Match: Matching Techniques For Driving Yagi-Uda Antennas: T-Match. 2a s. Z in. (Sections 9.5 & 9.7 of Balanis)

e t dt e t dt = lim e t dt T (1 e T ) = 1

P441 Analytical Mechanics - I. Coupled Oscillators. c Alex R. Dzierba

ENGR 1990 Engineering Mathematics The Integral of a Function as a Function

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

4.8 Improper Integrals

MTH 146 Class 11 Notes

September 20 Homework Solutions

MATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)

FM Applications of Integration 1.Centroid of Area

Solutions to Problems from Chapter 2

More on Magnetically C Coupled Coils and Ideal Transformers

ECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 8. Waveguides Part 5: Coaxial Cable

Magnetostatics Bar Magnet. Magnetostatics Oersted s Experiment

Contraction Mapping Principle Approach to Differential Equations

5.1-The Initial-Value Problems For Ordinary Differential Equations

SOME USEFUL MATHEMATICS

Properties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)

Minimum Squared Error

ECE Spring Prof. David R. Jackson ECE Dept. Notes 20

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 9: The High Beta Tokamak

Chapter Direct Method of Interpolation

Chapter 5 Waveguides and Resonators

Physics 2A HW #3 Solutions

Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges.

Minimum Squared Error

The solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.

REAL ANALYSIS I HOMEWORK 3. Chapter 1

INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).

( ) ( ) ( ) ( ) ( ) ( y )

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

A Kalman filtering simulation

Mathematics 805 Final Examination Answers

S Radio transmission and network access Exercise 1-2

CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES

CHAPTER 12 DIRECT CURRENT CIRCUITS

Physics 1402: Lecture 22 Today s Agenda

6. Gas dynamics. Ideal gases Speed of infinitesimal disturbances in still gas

Probability, Estimators, and Stationarity

Linear Circuit Elements

0 for t < 0 1 for t > 0

potentials A z, F z TE z Modes We use the e j z z =0 we can simply say that the x dependence of E y (1)

ECE Microwave Engineering

Let us start with a two dimensional case. We consider a vector ( x,

on the interval (x + 1) 0! x < ", where x represents feet from the first fence post. How many square feet of fence had to be painted?

Average & instantaneous velocity and acceleration Motion with constant acceleration

f t f a f x dx By Lin McMullin f x dx= f b f a. 2

Introduction to AC Power, RMS RMS. ECE 2210 AC Power p1. Use RMS in power calculations. AC Power P =? DC Power P =. V I = R =. I 2 R. V p.

Chapter 2. First Order Scalar Equations

EE243 Advanced Electromagnetic Theory Lec # 13: Waveguides and sources

ES 250 Practice Final Exam

INDEX. Transient analysis 1 Initial Conditions 1

Week 8 Lecture 3: Problems 49, 50 Fourier analysis Courseware pp (don t look at French very confusing look in the Courseware instead)

An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.

Honours Introductory Maths Course 2011 Integration, Differential and Difference Equations

THE SINE INTEGRAL. x dt t

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218

f(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2

6.01: Introduction to EECS I Lecture 8 March 29, 2011

Problem Set #1. i z. the complex propagation constant. For the characteristic impedance:

A Simple Method to Solve Quartic Equations. Key words: Polynomials, Quartics, Equations of the Fourth Degree INTRODUCTION

- If one knows that a magnetic field has a symmetry, one may calculate the magnitude of B by use of Ampere s law: The integral of scalar product

Name: Total Points: Multiple choice questions [120 points]

A LOG IS AN EXPONENT.

On Hadamard and Fejér-Hadamard inequalities for Caputo k-fractional derivatives

Collision Detection and Bouncing

Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x,

Chapter 10 INDUCTANCE Recommended Problems:

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Exponential and Logarithmic Functions -- ANSWERS -- Logarithms Practice Diploma ANSWERS 1

LAPLACE TRANSFORMS. 1. Basic transforms

Fuji Power MOSFET Power calculation method

3. Renewal Limit Theorems

V L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode.

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.

NMR Spectroscopy: Principles and Applications. Nagarajan Murali Advanced Tools Lecture 4

Version 001 test-1 swinney (57010) 1. is constant at m/s.

RESPONSE UNDER A GENERAL PERIODIC FORCE. When the external force F(t) is periodic with periodτ = 2π

Circuit Variables. AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters per second to miles per second: 1 ft 12 in

Y 0.4Y 0.45Y Y to a proper ARMA specification.

THE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

graph of unit step function t

EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A SECOND-ORDER ITERATIVE BOUNDARY-VALUE PROBLEM

Hamilton- J acobi Equation: Weak S olution We continue the study of the Hamilton-Jacobi equation:

f t te e = possesses a Laplace transform. Exercises for Module-III (Transform Calculus)

Logarithms Practice Exam - ANSWERS

..,..,.,

Laplace transfom: t-translation rule , Haynes Miller and Jeremy Orloff

PARABOLA. moves such that PM. = e (constant > 0) (eccentricity) then locus of P is called a conic. or conic section.

1 jordan.mcd Eigenvalue-eigenvector approach to solving first order ODEs. -- Jordan normal (canonical) form. Instructor: Nam Sun Wang

CHAPTER 6: FIRST-ORDER CIRCUITS

7. Capacitors and Inductors

MEMS 0031 Electric Circuits

Computer-Aided Analysis of Electronic Circuits Course Notes 3

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Waveguide Guide: A and V. Ross L. Spencer

MAT 266 Calculus for Engineers II Notes on Chapter 6 Professor: John Quigg Semester: spring 2017

Transcription:

EE 537-635 Microwve Engineering Fll 7 Prof. Dvid R. Jcson Dep. of EE Noes Wveguides Pr 7: Trnsverse Equivlen Newor (N)

Wveguide Trnsmission Line Model Our gol is o come up wih rnsmission line model for wveguide mode. y The wveguide mode is no M mode, u i cn e modeled s wve on rnsmission line. I + - V

Wveguide Trnsmission Line Model (con.) For wveguide mode, volge nd curren re no uniquely defined. y A B Ey ( ) Mode jωµπ π Ey A e j sin c y B jωµπ π j π V ( ) VAB ( ) E dr Ey dy A sin e V sin e A c The volge depends on! j 3

Wveguide Trnsmission Line Model (con.) For wveguide mode, volge nd curren re no uniquely defined. y y Mode H ( ) j π π H A e j sin c urren on op wll: Noe: If we inegre round he enire oundry, we ge ero curren (he op nd oom wlls hve opposie curren, s do he lef nd righ wlls). jπ π s sin c op ( ) ( ) ( ) j I J d H d A e d jπ π π A cos cos e c π I π π e cos cos j j The curren depends on he lengh of he inervl! 4

Wveguide Trnsmission Line Model (con.) Emine he rnsverse (, y) fields: Modl mpliudes ( + j + j ) E( y,, ) e( y, ) A e + A e ( + j + j ) H( y,, ) h( y, ) A e A e Noe: The minus sign ove rises from: ± ± H ( ˆ ± E ) w Wve impednce w or TM ωµ TM ωε c 5

Wveguide Trnsmission Line Model (con.) Inroduce defined volge ino he field equions: V A V A + + We my use whever definiion of volge we wish here. (In oher words, is rirry.) We hen hve: E( y,, ) e( y, ) V e + V e ( + j + j ) H( y,, ) h( y, ) V e V e + j + j ( ) 6

Wveguide Trnsmission Line Model (con.) Inroduce chrcerisic impednce (hving n rirry vlue) ino he equions: H y h y V e V e + j + j (,, ) (, ) H y h V y V e + j j (,, ) (, ) e + where 7

Wveguide Trnsmission Line Model (con.) Summry V( ) E( y,, ) e( y, ) Ve + Ve ( + j + j ) I ( ) + V j V + j (,, ) (, ) H y h y e e The dependence of he rnsverse fields ehves lie volge nd curren on rnsmission line. 8

Wveguide Trnsmission Line Model (con.) The rnsmission-line model is clled he Trnsverse Equivlen Newor (N) model of he wveguide. I( ) N + V ( ) -, E H V I Wveguide 9

Wveguide Trnsmission Line Model (con.) Power flow down he wveguide (comple power): ( ) ( ) WG * P E H ds S ˆ * * V( I ) ( ) * ( e( y, ) h( y, )) ds ˆ S ( ) ( ) * ( (, ) (, )) WG N * P P e y h y ds S ˆ omple power flowing down he N rnsmission line.

Wveguide Trnsmission Line Model (con.) Assume we choose o hve: N ( ) ( ) WG P P Then we hve he following consrin: ( (, ) (, )) e y h y ds * * ˆ S I is no necessry o me his ssumpion of equl powers, u i is useful choice h cn e mde.

Wveguide Trnsmission Line Model (con.) Summry of onsns (ssuming equl powers) ( (, ) (, )) e y h y ds * * ˆ S Once we pic, he consns re deermined. The mos common choice: w

Mode of Recngulr Wveguide We me he following choices: y hoose Assume power equliy ( ) ωµ e y h y ds * * (, ) (, ) ˆ S π ω µε c 3

Mode (con.) y ( (, ) (, )) e y h y ds * * ˆ S so π sin ds * * S π sin * * dyd π e ˆ ysin π h ˆ sin ωµ 4

Mode (con.) * * y Te he conjuge of he second one nd hen muliply he wo equions ogeher. Noe: The soluion is unique o wihin phse erm (we choose he phse o e ero here). Soluion: π e ˆ ysin π h ˆ sin ωµ 5

Mode (con.) π e ˆ ysin π h ˆ sin ωµ y Recll: V( ) E( y,, ) e( y, ) Ve + Ve ( + j + j ) I ( ) + V j V + j (,, ) (, ) H y h y e e 6

ωµ Emple: Mode (con.) π y Hence we hve for our finl modeling equions: V( ) π E(,, ) ˆ y ysin Ve Ve + ( + j + j ) I( ) + V j V + j π H(,, ) ˆ y sin e e 7

Emple: Wveguide Disconinuiy For [V/m] (field he cener of he guide) inciden mode in guide A, find he mode fields in oh guides, nd he refleced nd rnsmied powers. B y.856 cm.6 cm ε r.54 f GH A ε ε r µ µ µ π 58. rd / m [ ] π ε r 34. rd / m [ ] 8

, V + ΓV + N, Emple (con.) TV + π e ˆ ysin π h ˆ sin ωµ onvenion: hoose Assume power equliy j ( A + ) inc E ( y,, ) e( y, ) e ωµ 499.7 Ω [ ] ωµ 59.6 Ω [ ] ( ) V/m A + ( since e y, lredy hs [ ]) V A + + 9

Emple (con.) + j + j ( ) ( +Γ ) V V e e ( ) V V Te + + j + j ( ) ( Γ ) j V I e e VT + j I ( ) e Equivlen reflecion prolem:, V + ΓV + o o Γ.36 + o N, o TV + T +Γ.684 Noe: The ove N enforces he coninuiy of volge nd curren he juncion, nd hence he ngenil elecric nd mgneic fields re coninuous in he WG prolem.

Emple (con.) V ( ) e + (.36) e V ( ) (.684) e j + j ( ) I ( ) e (.36) e j + j ( ) (.684) I ( ) e j j Recll h for he mode:, V( ) ΓV + + j + j E( y,, ) e( y, )( Ve + Ve ) I ( ) + V j V + j H( y,, ) h( y, ) e e [ ] [ ] 58. rd / m 34. rd / m V + N, TV + π e ˆ ysin π h ˆ sin ωµ

Emple (con.) y B ε r Hence, for he wveguide prolem we hve he fields s: E( y,, ) e( y, ) + j + j ( e (.36) e ) A ε H ( y,, ) h( y, ) 6 j + j ( e (.3 ) e ) j E ( y,, ) e( y, ) (. 684) e H( y,, ) h( y, ).684 ( ) e j

Emple (con.) Susiuing in, we hve (guide A): E( y,, ) e( y, ) + j + j ( e (.36) e ) π + (,, ) ˆ E y ysin e. 3 6 j j ( + ( ) e ) π e ˆ ysin π h ˆ sin ωµ 3

Emple (con.) Susiuing in, we hve (guide A): H ( y,, ) h( y, ) 6 j + j ( e (.3 ) e ) (,, ) ˆ H y sin π + j j ( e (.36) e ) 4

Emple (con.) Susiuing in, we hve (guide B): j E ( y,, ) e( y, ) (. 684) e (,, ) ˆ π E y ysin (.684) e j 5

Emple (con.) Susiuing in, we hve (guide B): H( y,, ) h( y, ) (.684) e j (,, ) ˆ π H y sin (.684) e j 6

Emple (con.) Summry of Fields E(,, ) ˆ y ysin e +.36 e π + j j ( ( ) ) H (,, ) ˆ y sin e.36 e π + j j ( ( ) ) π j E(,, ) ˆ y ysin (.684) e [ ] [ ] 499.7 Ω 59.6 Ω [ ] [ ] 58. rd / m 34. rd / m π j H (,, ) ˆ y sin (.684 ) e 7

Emple (con.) Power lculions: Noe: In his prolem, nd Γ re rel. inc + + * + Re Re * P V I V P ref Γ Γ + + * Re V I P rns + + * Re V I Γ Γ ( ) ( ) 8

Emple (con.) For [V/m] inciden mode in guide A (field he cener of he guide), find he mode fields in oh guide, nd he refleced nd rnsmied powers. Finl Resuls: P P P inc refl rns [ ] [ ] [ ].6 mw.6 mw.45 mw.856 cm.6 cm ε r.54 f GH B A ε Noe: 9.% of he inciden power is rnsmied. ε r y 9

Mching Elemens in Wveguide A qurer-wve rnsformer is shown here. d λ gt T /4 T λ gt T π β T ( ) β lossless Top view T π T ε rt ε T rt T d T ωµ T Now % of he inciden power is rnsmied. 3

Mching Elemens in Wveguide (con.) Recngulr Wveguide (end view) Noe: Plnr disconinuiies re modeled s purely shun elemens. Inducive iris pciive iris Resonn iris The equivlen circui gives us he correc reflecion nd rnsmission of he mode. 3

Mching Elemens in Wveguide (con.) Top view Γ Inducive iris in ir-filled wveguide T ωµ η π Noe: The shun inducor models he effecs of he iris nd gives he mpliudes of he mode correcly everywhere, u he N model does no ell us how srong he higher-order modes re. Higher-order mode region N Model Becuse he elemen is shun disconinuiy, we hve T +Γ Ls 3

Mching Elemens in Wveguide (con.) Much more informion cn e found in he following reference: N. Mrcuvi, Wveguide Hndoo, Peer Perigrinus, Ld. (on ehlf of he Insiue of Elecricl Engineers), 986. Equivlen circuis for mny ypes of disconinuiies Accure AD formuls for mny of he disconinuiies Grphicl resuls for mny of he cses Someimes, mesured resuls 33