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

Problem 7.3.5

A TM $_{11}$ mode

In an air-filled rectangular waveguide with $a = 4 cm$ and $b = 3 cm$ a TM $_{11}$ mode is excited at $f = 11 GHz$ ,

E_{z} = E_{0} sin \frac{π x}{a} sin \frac{π y}{b} e^{- j τ z}, E_{0} = 1 kV/m .

Find the cutoff frequency of TM $_{11}$ , $τ$ , $λ_{g}$ and $Z_{T M}$ .
Write the transverse field components with numbers.
Find the power carried by the mode.
How many modes can propagate at $11 GHz$ ? Why is TM $_{11}$ hardly ever used to carry power?

Try it yourself first.

Strategy

Take one term of the TM $_{mn}$ series (7.3.17), $m = n = 1$ and $C_{11} = E_{0}$ . Every transverse component is $\pm j$ times a constant times a product of one sine and one cosine. The power follows from the general TM formula (7.3.38).

Bookkeeping

$a = 0.04 m$ , $b = 0.03 m$ , $f = 11 GHz$ : $λ = 2.73 cm$ , $k = 230.4 rad/m$ , $ω ε_{0} = 0.6111 S/m$ . $E_{0} = 1000 V/m$ .

$f_{co, 11} = 1.5 \times 1 0^{8} 2 5^{2} + 33. 3^{2} = 6.25 GHz$ , $f_{c} / f = 0.568$ , $s = 0.823$ :

τ Z_{T M} = k s = 189.6 rad/m, λ_{g} = \frac{2 π}{τ} = 3.31 cm, = 377 \times 0.823 = 310.2 Ω.

$h^{2} = (π / a)^{2} + (π / b)^{2} = 78.5 4^{2} + 104.7 2^{2} = 17135 m^{- 2}$ , $τ / h^{2} = 0.01106 m$ , $ω ε_{0} / h^{2} = 3.566 \times 1 0^{- 5} S$ . Then $E_{x}$ has the amplitude $(π / a) (τ / h^{2}) E_{0} = 869 V/m$ , $E_{y}$ the amplitude $(π / b) (τ / h^{2}) E_{0} = 1159 V/m$ , and so on:

E_{x} E_{y} H_{x} H_{y} = - j 869 cos \frac{π x}{a} sin \frac{π y}{b} e^{- j τ z} V/m, = - j 1159 sin \frac{π x}{a} cos \frac{π y}{b} e^{- j τ z} V/m, = j 3.735 sin \frac{π x}{a} cos \frac{π y}{b} e^{- j τ z} A/m, = - j 2.801 cos \frac{π x}{a} sin \frac{π y}{b} e^{- j τ z} A/m .

P = \frac{ab}{8} \frac{ω ε _{0} τ}{h ^{2}} E_{0}^{2} = \frac{( 0.04 ) ( 0.03 )}{8} \times 0.6111 \times 0.01106 \times 1 0^{6} = 1.01 W .

All modes with $f_{co} < 11 GHz$ : TE $_{10}$ ( $3.75$ ), TE $_{01}$ ( $5$ ), TE $_{11}$ and TM $_{11}$ ( $6.25$ ), TE $_{20}$ ( $7.5$ ), TE $_{21}$ and TM $_{21}$ ( $9.01$ ), TE $_{02}$ ( $10$ ), TE $_{12}$ and TM $_{12}$ ( $10.68 GHz$ ): ten modes. TM $_{11}$ shares the third place with TE $_{11}$ . Any bend, step or flange error converts part of its power into the other modes, which travel with other phase and group velocities.

Answer

f_{c} = 6.25 GHz, τ = 189.6 rad/m, λ_{g} = 3.31 cm, Z_{T M} = 310 Ω P = 1.01 W, 10 modes at 11 GHz

Check

$869/2.801 = 1159/3.735 = 310.2 Ω = Z_{T M}$ . The power from the transverse fields directly, each $cos^{2} sin^{2}$ product averaging to $\frac{1}{4}$ : $P = \frac{ab}{8} \frac{86 9 ^{2} + 115 9 ^{2}}{Z _{T M}} = \frac{0.0012}{8} \times \frac{2.098 \times 1 0 ^{6}}{310.2} = 1.01 W$ .

(b) All cutoffs below 13.2\,\mathrm{GHz}. — (a) The transverse field.

Remark

A TM $_{11}$ wave is useful in special devices (mode filters, test fixtures, and accelerating structures, which need an $E_{z}$ ), but power is always sent as TE $_{10}$ , in a guide used below its second cutoff, here $3.75$ – $5 GHz$ .

Problem 7.3.5: A TM_{11} mode