Waveguide Loaded Cavity: TEAM Problem 18

Figure 5.1: Cutaway diagram of waveguide loaded cavity.

5.2 Rectangular Cavities 99

X Y Z MODEL: TEST2

FEMGEN/FEMVIEW 2.3-07 Cambridge University (HP) 23 MAY 95

team18_2 mesh

Figure 5.2: Mesh used for TEAM problem 18, with t=a/32 (see Figure 5.1).

gave rise to a total of 18,135 edges of which 13,523 were free to vary. The problem was solved in the frequency domain since a good frequency accuracy is required in order to accurately calculate resonant frequency. A TE₁₀ mode was prescribed on the open end of the waveguide and the reflection coefficient calculated as from the field solution (§4.2). The SIC-QMR method required approximately 260 iterations to reduce the residual norm to 5×10⁻⁷ · kbk, the exact number varying slightly with the excitation frequency. The small number of iterations meant that a single solution at one frequency took only a few minutes to complete. The calculated phase of the reflection coefficient at the input plane is shown in Figure 5.3(a). The calculations were performed for 50 different frequencies and took a total of 109 minutes on a Sparc-10 workstation. This gives an average of 2.18 minutes per frequency. Figure 5.3(a) also shows the results due to Bardiet al. [1994b] which show the resonance at a slightly lower frequency. The field distribution is shown in Figure 5.4. The peak magnitude of the field inside the cavity is nearly 30 times the value of the field applied at the input plane.

The magnitude of the reflection coefficient is unity at all frequencies since the system is lossless. The phase, however, varies with frequency undergoing a 360^◦ shift as the resonance is passed. The reflection coefficient is measured at a plane in the waveguide but because of the intrinsic impedance of the waveguide the phase will be dependent upon the position of this plane. Since the position of the input plane is somewhat arbitrary Bardiet al. [1994b] suggest the use of a reference plane, see Figure 5.5, which is positioned such that the phase angle of the reflection coefficient, measured at this plane,

new inp

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9 9.05 9.1 9.15 9.2 9.25 9.3

Frequency, GHz Angle of ρ

Frequency Domain

Time Domain:

20 steps/cycle

Time Domain:

40 steps/cycle

Time Domain:

60 steps/cycle

(a) Phase angle at the input plane._Sref

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Frequency, GHz

Angle of ρ

This Study (l=3.98 mm)

Bardi et al. [1994b]

This Study (l=27.44 mm)

(b) Phase angle at the reference plane.

Figure 5.3: Phase angle of the reflection coefficient for t = a/32 and d = 2a/8 (see Figure 5.1).

5.2 Rectangular Cavities 101

Figure 5.4: Magnitude of the electric field at the resonance, t=a/32 andd= 2a/8.

ρ ρ

Figure 5.5: One port model of the waveguide and cavity (after Bardi et al. [1994b]).

is symmetric about the zero axis. This makes identification of the resonant frequency of the cavity straightforward. Finding the position of this plane is computationally cheap since it does not require any further field calculations. Having found the reflection coefficient at the input plane using the techniques described in Chapter 4 the phase at any plane,z, can be found from

ρ_ref =ρ_inpe^2βz. (5.1)

It can be seen from equation (5.1) that there will be series of planes, λ_g/2 apart for which the reflection coefficient will have zero phase at the resonant frequency of the cavity. Figure 5.3(b) shows the phase of the reflection coefficient at two different planes, one at z = 3.98 mm and the other at z = 27.44 mm. The latter is actually inside the iris, so it is debatable whether this is a valid choice of plane since it is no longer in the waveguide. Bardi et al. do not give the position of the reference plane used for their results, which are also shown in Figure 5.3(a). One can note from the graph that when the plane at 27.44 mm is chosen the frequency response either side of the main resonance is flat. This is to be expected since here only the cavity is affecting the response. At 3.98 mm the phase angle away from the parallel resonance reduces with increasing frequency due to the impedance of the waveguide. The calculated resonant frequency for this problem is 9.180 GHz, whereas Bardiet al. [1994b] give 9.1685 GHz, which corresponds to a difference of only 0.125% indicating good agreement between the two methods.

A second problem was modelled, this time with t = 0 and d = 2a/8, the results are shown in Figure 5.6. Here the resonant frequency was calculated as 9.1734 GHz whereas Bardi et al. [1994b] give 9.1519 GHz in this case, a difference of 0.235%. In addition to the frequency domain method this problem was also solved using the time domain technique in order to compare the results and observe the effect of changing the size of the time step. Three different time steps were used, corresponding to 20, 40 and 60 steps per cycle at 9.15 GHz. Figure 5.6 compares the time domain results with the frequency domain results for the phase at both the input plane and the reference plane. The time domain solution was excited by a Gaussian pulse modulated at 9.15 GHz and then run for 500 cycles at this frequency. This allows a frequency resolution of 18.3 MHz, giving 16 different frequencies in the range 9.0 to 9.3 GHz. One of the major

5.2 Rectangular Cavities 103

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(a) Phase angle at the input plane

new ref

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9 9.05 9.1 9.15 9.2 9.25 9.3

Frequency, GHz

Angle of ρ

Frequency Domain

Time Domain:

20 steps/cycle

Time Domain:

40 steps/cycle

Time Domain:

60 steps/cycle

(b) Phase angle at the reference plane

Figure 5.6: Comparison of time domain and frequency domain results for the phase angle of the reflection coefficient for t= 0 andd= 2a/8 (see Figure 5.1).

problems with solving this type of problem in the time domain is that the fields decay very slowly, requiring a large number of time steps to get good resolution. It is clear from Figure 5.6 that as the time step is reduced the accuracy of the solution is increased, which is to be expected. Table 5.1 shows the discrepancy between the frequency domain solution and the time domain solution, even at 20 time steps per cycle the difference between them is less than one percent. It is interesting to note that as the number of steps per cycle is increased the number of iterations required per time step is reduced significantly. This is due to the reduction in the contribution from the [S] matrix to the matrix that requires solution. This has the effect reducing the solution time for the smaller time steps: doubling the steps per cycle does not double the solution time.

Steps / cycle Iterations per Resonant Discrepancy Solution Time

time step Frequency (minutes)

Frequency Domain — 9.1734 — 109

20 8 9.1035 0.762% 253

40 4 9.1589 0.158% 341

60 3 9.1706 0.031% 445

Table 5.1: Comparison of frequency domain (50 frequencies) and time domain calcula- tions (500 cycles).