Lecture 15 Metallic circular walled waveguides

# Lecture 15 Metallic circular walled waveguides - Lecture...

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cture 5 Lecture 15 Metallic circular walled waveguides ayt H 14 Hayt CH 14 1 McGill ECSE 352, Fall 2011, D. Davis

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ircular Waveguide Circular Waveguide e circular waveguide is a hollow metal The circular waveguide is a hollow metal cylinder (constant inner radius “a”) that guides fields along the length (defined as z direction). The basic relationship used to model the circular waveguide is the Helmholtz vector wave equation. The solution of the wave equation will use cylindrical coordinates (since the system is cylindrical). McGill ECSE 352, Fall 2011, D. Davis 2
ircular Waveguide Circular Waveguide e can write the radial (rho) and rotational We can write the radial (rho) and rotational (phi) components of the field as a function of e z irected component the z directed component. + = ωμ β ρ z z H E j E 2 = ωε z z H E j H 2 φ c k c k = z z c H E k j E 2 + = z z c H E k j H 2 McGill ECSE 352, Fall 2011, D. Davis 3

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ircular Waveguide Circular Waveguide order to model the fields inside the circular In order to model the fields inside the circular waveguide, the engineer only needs to find e z irected electric or magnetic fields the z directed electric or magnetic fields. The solution will be obtained in the same anner as with the rectangular waveguide manner as with the rectangular waveguide. The field is assumed to have the form: In this case electric field was used. z McGill ECSE 352, Fall 2011, D. Davis 4
ircular Waveguide Circular Waveguide e solution can be broken into TE modes (z The solution can be broken into TE modes (z directed electric field is zero) and TM modes irected magnetic field is zero) (z directed magnetic field is zero). The TE case uses: 2 ݖ 2 ݖ The TM case uses: To produce a solution for the electric and 2 ݖ 2 ݖ magnetic fields within the guide. McGill ECSE 352, Fall 2011, D. Davis 5

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ircular Waveguide Circular Waveguide sing the separable differential equation Using the separable differential equation (function of radius and rotation) the two quations used to construct the solution are: equations used to construct the solution are: d 2 Φ ݀߶ 2 ൅݇ ߶ 2 Φൌ0 For the phi dependent portion and: ߩ 2 ݀ 2 ܲ ൅ߩ ݀ܲ ൫ߩ 2 ݇ ܿ 2 െ݇ ߶ 2 ൯ൌ0 For the radial dependent portion.
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Lecture 15 Metallic circular walled waveguides - Lecture...

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