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Rectangular Waveguides

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1 Rectangular Waveguides
ECE 3317 Prof. David R. Jackson Spring 2013 Notes 20 Rectangular Waveguides

2 Rectangular Waveguide
b x y ,  Cross section We 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 only TEz: Hz only

3 TMz Modes TMz (Helmholtz equation) (PEC walls) Guided-wave assumption:

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

6 TMz Modes (cont.) Denote General solution: Boundary conditions: (1)

7 TMz Modes (cont.) The second boundary condition results in
This gives us the following result: Hence Now we turn our attention to the Y (y) function.

8 TMz Modes (cont.) We have Hence Denote Then we have General solution:

9 TMz Modes (cont.) Boundary conditions: (3) (4)
Equation (4) gives us the following result:

10 TMz Modes (cont.) Hence Therefore, we have New notation:
The TMz field inside the waveguide thus has the following form:

11 TMz Modes (cont.) Recall that Hence
Therefore the solution for kc is given by Next, recall that Hence

12 TMz Modes (cont.) Summary of TMz Solution for (m,n) Mode (Hz = 0)
Note: If either m or n is zero, the entire field is zero.

13 TMz Modes (cont.) Cutoff frequency
Note: The number kc is the value of k for which the wavenumber kz is zero. We start with where Set

14 TMz Modes (cont.) Hence which gives us
The cutoff frequency fc of the TMm,n mode is then This may be written as

15 TEz Modes TEz We now start with
Using the separation of variables method again, we have where and

16 TEz Modes (cont.) y b ,  x a Boundary conditions: cross section
The result is This can be shown by using the following equations:

17 TEz Modes (cont.) Summary of TEz Solution for (m,n) Mode (Ez = 0)
Same formula for cutoff frequency as the TEz case! Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law:

18 Summary TMz TEz TMz TEz Same formula for both cases

19 Wavenumber General formula for the wavenumber TMz or TEz mode: with
Note: The (m,n) notation is suppressed here. Above cutoff: Recall: Hence

20 Wavenumber (cont.) Below cutoff: Hence Hence

21 Wavenumber (cont.) Recal that Hence we have

22 Wavenumber Plot General behavior of the wavenumber “Light line”

23 Dominant Mode The "dominant" mode is the one with the lowest cutoff frequency. Assume b < a a b x y ,  Cross section Lowest TMz mode: TM11 Lowest TEz mode: TE10 TEz TMz The 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? a b x y ,  Cross section Assume b < a / 2 The next highest is the TE20 mode. useful operating region TE01 fc TE10 TE20

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: where x y a b E H

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. Use Hence, 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 GHz Recall: 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 Wavelength The 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 have Hence we have the result

33 Phase and Group Velocity
Recall that the phase velocity is given by Hence We then have For 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.) t A 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 system A pulse will get distorted in a rectangular waveguide!

36 Phase and Group Velocity (cont.)
To calculate the group velocity for a waveguide, we use Hence we have We 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: where Separating the terms, we have: #1 #2 This form is the sum of two plane waves.

39 Plane Wave Interpretation (cont.)
Picture (top view): x z a k1 k2 TE10 mode At 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 waves a k2 k1 z x Crests of waves g cd vp Waves cancel out The 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.) x y w h E H PMC PEC Assume w > h Rectangular “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 PMC A PMC is dual to a PEC: PEC PMC PEC E PMC H

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: x y w h PMC PEC Assume w > h TMz TEz

45 Waveguide Modes in Transmission Lines (cont.)
The dominant mode is TE10 The cutoff frequency of the TE10 mode is

46 Waveguide Modes in Transmission Lines (cont.)
x y w h Example Microstrip line Assume: fc  32 [GHz]

47 Waveguide Modes in Transmission Lines (cont.)
Dominant waveguide mode in coax (derivation omitted) TE11 mode: Example: RG 142 coax a b

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