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# lecture25 - Lecture 25 Guided Waves in Parallel Plate Metal...

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1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Lecture 25 Guided Waves in Parallel Plate Metal Waveguides In this lecture you will learn: • Parallel plate metal waveguides •TE and TM guided modes in waveguides ECE 303 – Fall 2007 – Farhan Rana – Cornell University Parallel Plate Metal Waveguides d W z • Consider a parallel plate waveguide (shown above) • We have studied such structures in the context of transmission lines • We know that they can guide TEM waves ( T ransverse E lectric and M agnetic) in which both the electric and magnetic fields point in direction perpendicular to the propagation direction • But these structures can guide more than just the TEM waves that we have considered so far ………….

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2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Basic Wave Equations d z x o µ ε Consider a parallel plate waveguide: The electric field of any guided wave will satisfy the complex wave equations: () () r H j r E o r r r r ω = × r E j r H r r r r = × ( ) ( ) r E r E o r r r r 2 2 = ( ) ( ) r H r H o r r r r 2 2 = We look for solutions of the equation, ( ) ( ) r E r E o r r r r 2 2 = where the z-dependence is that of a wave going in the z-direction, and where the E-field is pointing in the y-direction: () ( ) z k j z e x F y r E = ˆ r r Some unknown function of “x” ECE 303 – Fall 2007 – Farhan Rana – Cornell University TE Guided Modes - I d z x o The assumed solution form: ( ) ( ) z k j z e x F y r E = ˆ r r represents a TE guided wave ( T ransverse E lectric) since the direction of E-field is transverse to the direction of wave propagation Plugging the assumed solution into the equation gives: () x F k x x F r E r E x z r E r E z o o o 2 2 2 2 2 2 2 2 2 2 2 = = + = r r r r r r r r Perfect metal boundary conditions ( ) ( ) 0 0 = = = = d x F x F
3 ECE 303 – Fall 2007 – Farhan Rana – Cornell University TE Guided Modes - II x Need to solve: () ( )() x F k x x F z o 2 2 2 2 = ε µ ω With boundary conditions ( ) ( ) 0 0 = = = = d x F x F Solution is: () ( ) x k E x F x o sin = But the value of k x cannot be arbitrary – boundary condition at x = d dictates that: K K , 3 , 2 1 : where , m d m k x = = π Automatically satisfies the boundary condition: ( ) 0 0 = = x F Solution becomes: = x d m E x F o sin And: { K K r r , 3 , 2 , 1 sin ˆ = = m e x d m E y r E z k j o z d z o m = 1 m = 2 E y E y ECE 303 – Fall 2007 – Farhan Rana – Cornell University d z x E TE Guided Modes - III { K K r r , 3 , 2 , 1 sin ˆ = = m e x d m E y r E z k j o z d z x E E-field: m=1 mode E-field: m=2 mode z k z k

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4 ECE 303 – Fall 2007 – Farhan Rana – Cornell University
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lecture25 - Lecture 25 Guided Waves in Parallel Plate Metal...

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