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

Problem 7.3.3

Designing a guide for 9 GHz

An air-filled rectangular waveguide with $a = 2 b$ is to carry $9 GHz$ in the TE $_{10}$ mode only. The operating frequency must lie at least $25 %$ above the TE $_{10}$ cutoff and at least $15 %$ below the cutoff of the next mode.

Find the allowed range of $a$ .
Take $a = 2.5 cm$ . Find $f_{co, 10}$ , $λ_{g}$ , $v_{g}$ , $Z_{T E}$ and $α_{c}$ for copper walls.
Where is the cutoff of TM $_{11}$ ? Does it matter?
A dielectric filling makes a guide smaller. Fill the guide of (b) with PTFE ( $ε_{r} = 2.1$ ): which modes propagate at $9 GHz$ ? What range of $a$ meets the specification with PTFE?

Try it yourself first.

Strategy

With $a = 2 b$ the next modes, TE $_{20}$ and TE $_{01}$ , share the cutoff $v / a = 2 f_{co, 10}$ . The two margins become $f \geq 1.25 f_{co, 10}$ and $f \leq 0.85 \times 2 f_{co, 10}$ , two inequalities for $a$ .

Bookkeeping

$f = 9 GHz$ , $λ = 3.33 cm$ , $k = 188.5 rad/m$ . Copper at $9 GHz$ : $R_{s} = π f μ_{0} / σ_{c} = 24.8 mΩ$ .

\frac{c}{2 a} \leq \frac{f}{1.25} f \leq 0.85 \frac{c}{a} \Rightarrow a \geq \frac{1.25 c}{2 f} = 2.08 cm, \Rightarrow a \leq \frac{0.85 c}{f} = 2.83 cm .

$a = 2.5 cm$ , $b = 1.25 cm$ : $f_{co, 10} = 6 GHz$ , $f = 1.5 f_{co, 10}$ , $s = 1 - (2/3)^{2} = 5 /3 = 0.745$ :

λ_{g} v_{g} Z_{T E} α_{c} = \frac{3.33}{0.745} = 4.47 cm, = 0.745 c = 2.24 \times 1 0^{8} m/s, = \frac{377}{0.745} = 505.8 Ω, = \frac{0.0248 [ 1 + ( 4/9 )]}{377 \times 0.0125 \times 0.745} = 0.0102 Np/m = 0.088 dB/m .

$f_{co, 11} = 1.5 \times 1 0^{8} 4 0^{2} + 8 0^{2} = 13.42 GHz$ for TE $_{11}$ and TM $_{11}$ . It lies above $f_{co, 20} = f_{co, 01} = 12 GHz$ , so it never limits the band.

With PTFE $v = c / 2.1 = 2.07 \times 1 0^{8} m/s$ : $f_{co, 10} = 4.14 GHz$ and $f_{co, 20} = f_{co, 01} = 8.28 GHz < 9 GHz$ . Three modes propagate (TE $_{10}$ , TE $_{20}$ , TE $_{01}$ , while TE $_{11}$ is at $9.26 GHz$ ), so the specification fails. All cutoffs scale with $v$ , hence the allowed range of $a$ shrinks by $2.1 = 1.449$ :

1.44 cm \leq a \leq 1.96 cm .

Answer

2.08 cm \leq a \leq 2.83 cm (air), 1.44 cm \leq a \leq 1.96 cm (PTFE) a = 2.5 cm : f_{c} = 6 GHz, λ_{g} = 4.47 cm, v_{g} = 2.24 \times 1 0^{8} m/s, Z_{T E} = 506 Ω, α_{c} = 0.088 dB/m

Check

At the two limits: $a = 2.08 cm$ gives $f_{co, 10} = 7.2 GHz = 9/1.25$ , and $a = 2.83 cm$ gives $f_{co, 20} = 10.59 GHz = 9/0.85$ .

(a) Cutoffs against a (a=2b). — (a) Cutoffs against $a$ ( $a = 2 b$ ).

(b) Mode chart for a=2.5\,\mathrm{cm}, line at 9\,\mathrm{GHz}. — (a) Cutoffs against $a$ ( $a = 2 b$ ).

Remark

WR-90, for example, has its band $8.2$ – $12.4 GHz$ between $1.25$ and $1.9 f_{co, 10}$ . A dielectric filling makes the guide $ε_{r}$ times smaller, at the price of a dielectric loss and of a lower power rating (Exercise 7.3.5).

Problem 7.3.3: Designing a guide for 9 GHz