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Unformatted text preview: 30 Guide impedance and TL analogies TE mode fields: H x = jk z H z x k 2 k 2 z , H y = jk z H z y k 2 k 2 z , E y = j o H z x k 2 k 2 z , E x = j o H z y k 2 k 2 z . TM mode fields: H x = j o E z y k 2 k 2 z , H y = j o E z x k 2 k 2 z , E y = jk z E z y k 2 k 2 z , E x = jk z E z x k 2 k 2 z . The above relations between the transverse components of TE and TM mode fields imply that TE case: E x H y = E y H x = o k z = o /k 1 f 2 c f 2 = o 1 f 2 c f 2 TE . TM case: E x H y = E y H x = k z o = k 1 f 2 c f 2 o = o 1 f 2 c f 2 TM . 1 The guide impedances defined above can be used to set up transmission line models for waveguide circuits in which the parameters TE and TM for each mode play the same role as the characteristic impedance Z o in TL theory. For example, two waveguides in cascade with different values of TE can be quarterwave matched by inserting a quarterwave section having a guide impedance equal to the geometric means of the two guides. For dielectricfield guides replace o by the appropriate , and also in calculating the length of the quarterwave section use g = 2 k z appro priate for that section (see HW). Note that, using the cutoff wavelength, we have TE case: TE = o 1 f 2 c f 2 = o 1 2 2 c TM case: TM = o 1 f 2 c f 2 = o 1 2 2 c . 2 Example 2: Consider an airfilled rectangular waveguide with a = 3 cm and b = 1 cm. Determine the TE 10 mode fields for the guide from the results of Example 1 of Lect 29 assuming that at the operation frequency the freespace wavelength is = 3 cm....
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This note was uploaded on 09/27/2011 for the course ECE 450 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 Electromagnet, Impedance

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