Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 7.4.1: Cutoff and parameters of the TE_{11} mode

Problem 7.4.1

Cutoff and parameters of the TE $_{11}$ mode

A circular air-filled waveguide has radius $a = 2.0$ cm.

For TE $_{11}$ , the first root of $J_{1}^{'} (x)$ is $x_{11}^{'} \approx 1.841$ . Find the cutoff frequency and decide whether the mode propagates at $f = 6$ GHz and at $f = 3$ GHz.
Which other modes propagate at $6$ GHz?
For TE $_{11}$ at $6$ GHz find $τ$ , $λ_{g}$ , $v_{p}$ , $v_{g}$ and $Z_{T E}$ .
At $3$ GHz, how fast does the TE $_{11}$ field decay? Over what length is it reduced by $40$ dB?

TE_{11}, a=2 cm. — TE $_{11}$ , $a = 2$ cm.

Try it yourself first.

Strategy

For this guide every cutoff is $f_{c} = x c / (2 π a) = x \times 2.387$ GHz, so the table of zeros gives the whole mode list at once. Once $f_{c}$ is known, the parameters of part (c) are the universal ones of Eq. (7.4.3), and below cutoff the same square root gives the attenuation constant.

Bookkeeping

Air, $a = 0.02$ m, $c / (2 π a) = 2.387$ GHz. At $6$ GHz: $λ = 5$ cm, $k = 2 π / λ = 40 π = 125.7$ rad/m. At $3$ GHz: $k = 20 π = 62.83$ rad/m.

a) Cutoff

With $x_{11}^{'} = 1.841$ ,

f_{c, TE_{11}} = \frac{x _{11}^{'} c}{2 π a} = \frac{1.841 \times 3 \times 1 0 ^{8}}{2 π \times 0.02} \approx 4.39 GHz .

So at $f = 6$ GHz the mode propagates, but at $f = 3$ GHz it is below cutoff.

b) The other modes

TM $_{01}$ : $2.405 \times 2.387 = 5.74$ GHz $< 6$ GHz, so it propagates. TE $_{21}$ : $3.054 \times 2.387 = 7.29$ GHz, TE $_{01}$ and TM $_{11}$ : $3.832 \times 2.387 = 9.15$ GHz, all cut off. At $6$ GHz the guide carries TE $_{11}$ , in its two polarisations, and TM $_{01}$ .

c) TE₁₁ at 6 GHz

$f_{c} / f = 4.396/6 = 0.7326$ , $s = 1 - 0.732 6^{2} = 0.6807$ and $h = 1.841/0.02 = 92.06$ rad/m:

τ = k s = 125.7 \times 0.6807 = 85.5 rad/m, λ_{g} = \frac{λ}{s} = \frac{5 cm}{0.6807} = 7.35 cm, v_{p} = \frac{c}{s} = 4.41 \times 1 0^{8} m/s, v_{g} = cs = 2.04 \times 1 0^{8} m/s, Z_{T E} = \frac{η _{0}}{s} = \frac{120 π}{0.6807} = 553.8 Ω.

d) Below cutoff

At $3$ GHz $k < h$ and $τ = - j α$ with

α = h^{2} - k^{2} = 92.0 6^{2} - 62.8 3^{2} = 67.3 Np/m = 67.3 \times 8.686 = 584 dB/m .

$40$ dB is reached after $40/584$ m $= 6.8$ cm. The wave impedance is then $Z_{T E} = ω μ_{0} / τ = ω μ_{0} / (- j α) = j ω μ_{0} / α = j 352 Ω$ , purely reactive: no power flows.

Answer

f_{c, TE_{11}} = 4.39 GHz at 6 GHz : TE_{11} and TM_{01} τ = 85.5 rad/m, λ_{g} = 7.35 cm Z_{T E} = 554 Ω v_{p} = 4.41 \times 1 0^{8} m/s v_{g} = 2.04 \times 1 0^{8} m/s at 3 GHz : α = 67.3 Np/m = 584 dB/m 40 dB in 6.8 cm

Check

$2 π / τ = 2 π /85.54 = 7.35$ cm $= λ_{g}$ , and $v_{p} v_{g} = 4.41 \times 2.04 \times 1 0^{16} = 9.0 \times 1 0^{16} = c^{2}$ . Below cutoff $Z_{T E} = j η_{0} (f / f_{c}) / 1 - (f / f_{c})^{2} = j 377 \times 0.6825/0.7309 = j 352 Ω$ .

(a) Cutoffs in units of f_{c,\mathrm{TE}_{11}}=4.395 GHz. — (a) Cutoffs in units of $f_{c, TE_{11}} = 4.395$ GHz.

(b) f=\frac{c}{2\pi}\sqrt{\tau^2+h^2} of three modes. — (a) Cutoffs in units of $f_{c, TE_{11}} = 4.395$ GHz.

Remark

Part (a) is correct as stated, but at $6$ GHz TE $_{11}$ is not alone: TM $_{01}$ is just above its own cutoff, with $τ = 36.5$ rad/m and $λ_{g} = 17.2$ cm (Figure 7.4.10). In units of the TE $_{11}$ cutoff the operating point is $6/4.396 = 1.365$ , just beyond the single-mode band $1 < f / f_{c, TE_{11}} < 1.306$ . A designer who wants TE $_{11}$ alone at $6$ GHz needs $a < 2.405 c / (2 π \times 6 GHz) = 1.91$ cm.