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

About the further exercises of Section 7.4

Five further exercises on the circular guide, in the spirit of Problems 7.4.1–7.4.2, the worked examples and the home exercises, graded from easy to harder. They use only the results listed at the head of this subsection. (Five more, on the coaxial line, close Section 7.5.)

Exercise 7.4.2

The frequency of least loss

Section 7.4.9 says that the conductor loss of TM $_{01}$ has a minimum at $f = 3 f_{c}$ , and that TE $_{11}$ has one too.

Prove the $3$ result for any TM $_{mn}$ mode of a circular guide.
Show numerically that the TE $_{11}$ minimum lies near $3.15 f_{c}$ .
For an air-filled copper pipe with $a = 1.2$ cm find both minima (frequency and dB/km).
How many field patterns propagate at those two frequencies? What does this mean for a single-mode TE $_{11}$ link in this pipe?

\alpha_0=R_s(f_c)/(a\eta), green: single-mode band of TE_{11}. — $α_{0} = R_{s} (f_{c}) / (a η)$ , green: single-mode band of TE $_{11}$ .

Try it yourself first.

Strategy

Write $u = f / f_{c}$ . Since $R_{s} \propto f$ , $R_{s} (f) = R_{s} (f_{c}) u$ , and each attenuation constant becomes $α_{0}$ times a function of $u$ alone, with $α_{0} = R_{s} (f_{c}) / (a η)$ . Minimise that function.

For TM modes $α_{c} = α_{0} u / 1 - 1/ u^{2}$ . Its square is $α_{0}^{2} u^{3} / (u^{2} - 1)$ , and

\frac{d}{d u} \frac{u ^{3}}{u ^{2} - 1} = \frac{3 u ^{2} ( u ^{2} - 1 ) - 2 u ^{4}}{( u ^{2} - 1 ) ^{2}} = \frac{u ^{2} ( u ^{2} - 3 )}{( u ^{2} - 1 ) ^{2}} = 0 ⟹ u = 3 .

The minimum is $α_{c, m i n} = α_{0} 3^{1/4} / 2/3 = 1.612 α_{0}$ , for every TM $_{mn}$ (only $f_{c}$ differs).

For TE $_{11}$ , $α_{c} = α_{0} g (u)$ with $g (u) = u (1/ u^{2} + 0.418) / 1 - 1/ u^{2}$ . Values: $g (2) = 1.091$ , $g (3) = 0.973$ , $g (4) = 0.993$ . The minimum lies between $3$ and $4$ , close to $3$ , and refining gives $u = 3.15$ , $g = 0.972$ .

$c / (2 π a) = 3.979$ GHz: $f_{c, TE_{11}} = 7.33$ GHz, $f_{c, TM_{01}} = 9.57$ GHz.

TM_{01} : f = 3 \times 9.568 = 16.57 GHz, R_{s} = 33.6 mΩ, α_{c} = \frac{0.0336}{0.012 \times 377} \times \frac{1}{2/3} = 9.09 \times 1 0^{- 3} Np/m = 79.0 dB/km, TE_{11} : f = 3.15 \times 7.326 = 23.08 GHz, α_{c} = 41.7 dB/km .

At $16.57$ GHz, $k a = 4.17$ : TE $_{11}$ , TM $_{01}$ , TE $_{21}$ , TE $_{01}$ and TM $_{11}$ , that is $8$ field patterns. At $23.08$ GHz, $k a = 5.80$ : ten modes up to TM $_{02}$ ( $5.520$ ), $17$ field patterns. Single-mode TE $_{11}$ operation is limited to $7.33$ – $9.57$ GHz, where $g (u)$ is still falling steeply: $84.6$ dB/km at $1.25 f_{c}$ ( $9.16$ GHz).

Answer

TM : f_{m i n} = 3 f_{c}, α_{c, m i n} = 1.612 α_{0} TE_{11} : f_{m i n} = 3.15 f_{c}, α_{c, m i n} = 0.972 α_{0} a = 1.2 cm : 79.0 dB/km at 16.57 GHz 41.7 dB/km at 23.08 GHz 8 and 17 patterns

Check

Through $α_{0}$ , taken at the cutoff of each mode: TE $_{11}$ has $R_{s} (7.33 GHz) = 22.3$ m $Ω$ , $α_{0} = 4.94 \times 1 0^{- 3}$ Np/m and $0.972 α_{0} = 4.80 \times 1 0^{- 3}$ Np/m $= 41.7$ dB/km. TM $_{01}$ has $R_{s} (9.57 GHz) = 25.5$ m $Ω$ , $α_{0} = 5.64 \times 1 0^{- 3}$ Np/m and $1.612 α_{0} = 9.09 \times 1 0^{- 3}$ Np/m. Both agree with part (c).

Remark

The minima of TE $_{11}$ and TM $_{01}$ are real but useless for a single-mode guide: they lie far above the next cutoffs. In the usable band, $1.05$ to $1.3 f_{c}$ , the loss falls with frequency, so a designer picks the largest radius that still keeps TM $_{01}$ out. Only the circularly symmetric TE $_{0 n}$ modes, TE $_{01}$ first, have no minimum at all (Problem 7.4.2).

cf. Problem 7.4.2

Exercise 7.4.2: The frequency of least loss