Presentation on theme: "Rectangular Waveguides"— Presentation transcript:
1 Rectangular Waveguides ECE 3317Prof. David R. JacksonSpring 2013Notes 20Rectangular Waveguides
2 Rectangular Waveguide bxy, Cross sectionWe assume that the boundary is a perfect electric conductor (PEC).We analyze the problem to solve for Ez or Hz (all other fields come from these).TMz: Ez onlyTEz: Hz only
4 TMz Modes (cont.) Define: We then have: Note that kc is an unknown at this point.We then have:Dividing by the exp(-j kz z) term, we have:We solve the above equation by using the method of separation of variables.We assume:
5 TMz Modes (cont.) where Hence, Divide by XY : Hence This has the form Both sides of the equation must be a constant!This has the form
22 Wavenumber PlotGeneral behavior of the wavenumber“Light line”
23 Dominant ModeThe "dominant" mode is the one with the lowest cutoff frequency.Assume b < aabxy, Cross sectionLowest TMz mode: TM11Lowest TEz mode: TE10TEzTMzThe dominant mode is the TE10 mode.
24 Dominant Mode (cont.) Formulas for the dominant TE10 mode At the cutoff frequency:
25 Dominant Mode (cont.)What is the mode with the next highest cutoff frequency?abxy, Cross sectionAssume b < a / 2The next highest is the TE20 mode.useful operating regionTE01fcTE10TE20
26 Dominant Mode (cont.) Fields of the dominant TE10 mode Find the other fields from these equations:
27 Dominant Mode (cont.) E H b a From these, we find the other fields to be:wherexyabEH
28 Dominant Mode (cont.) Wave impedance: Note: This is the same formula as for a TEz plane wave!Define the wave impedance:
29 Example Standard X-band* waveguide (air-filled): a = inches (2.286 cm)b = inches (1.016 cm)Note: b < a / 2.Find the single-mode operating frequency region.UseHence, we have* X-band: from 8.0 to 12 GHz.
30 Example (cont.) Standard X-band* waveguide (air-filled): a = inches (2.286 cm)b = inches (1.016 cm)Find the phase constant of the TE10 mode at 9.00 GHz.Find the attenuation in dB/m at 5.00 GHzRecall:At 9.00 [GHz]: k = [rad/m]At 9.00 GHz: = [rad/m]At 5.00 GHz: = [nepers/m]
31 Example (cont.) At 5.00 [GHz]: = 88.91 [nepers/m] Therefore, A very rapid attenuation!Note: We could have also used
32 Guide WavelengthThe guide wavelength g is the distance z that it takes for the wave to repeat itself.(This assumes that we are above the cutoff frequency.)From this we haveHence we have the result
33 Phase and Group Velocity Recall that the phase velocity is given byHenceWe then haveFor a hollow waveguide (cd = c):Hence:vp > c !(This does not violate relativity.)
34 Phase and Group Velocity (cont.) The group velocity is the velocity at which a pulse travels on a structure.The group velocity is given by(The derivation of this is omitted.)tA pulse consists of a "group" of frequencies (according to the Fourier transform).+-Vi (t)waveguiding system
35 Phase and Group Velocity (cont.) If the phase velocity is a function of frequency, the pulse will be distorting as it travels down the system.Vi (t)+-waveguiding systemA pulse will get distorted in a rectangular waveguide!
36 Phase and Group Velocity (cont.) To calculate the group velocity for a waveguide, we useHence we haveWe then have the following final result:For a hollow waveguide:vg < c
37 Phase and Group Velocity (cont.) For a lossless transmission line or a lossless plane wave (TEMz waves):We then have(a constant)For a lossless transmission line or a lossless plane wave line there is no distortion, since the phase velocity is constant with frequency.Hence we have
38 Plane Wave Interpretation of Dominant Mode Consider the electric field of the dominant TE10 mode:whereSeparating the terms, we have:#1#2This form is the sum of two plane waves.
39 Plane Wave Interpretation (cont.) Picture (top view):xzak1k2TE10 modeAt the cutoff frequency, the angle is 90o. At high frequencies (well above cutoff) the angle approaches zero.
40 Plane Wave Interpretation (cont.) Picture of two plane wavesak2k1zxCrests of wavesgcdvpWaves cancel outThe two plane waves add to give an electric field that is zero on the side walls of the waveguide (x = 0 and x = a).
41 Waveguide Modes in Transmission Lines A transmission line normally operates in the TEMz mode, where the two conductors have equal and opposite currents.At high frequencies, waveguide modes can also propagate on transmission lines.This is undesirable, and limits the high-frequency range of operation for the transmission line.
42 Waveguide Modes in Transmission Lines (cont.) Consider an ideal parallel-plate transmission line (neglect fringing)(This is an approximate model of a microstrip line.)xywhEHPMCPECAssume w > hRectangular “slice of plane wave” (the region inside the waveguide)A “perfect magnetic conductor” (PMC) is assumed on the side walls.
43 Waveguide Modes in Transmission Lines (cont.) Some comments about PMCA PMC is dual to a PEC:PECPMCPECEPMCH
44 Waveguide Modes in Transmission Lines (cont.) From a separation of variables solution, we have the following results for the TMz and TEz waveguide modes:xywhPMCPECAssume w > hTMzTEz
45 Waveguide Modes in Transmission Lines (cont.) The dominant mode is TE10The cutoff frequency of the TE10 mode is