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 7.4.3

A quarter-wave circular polariser

A round air-filled pipe has been squeezed very slightly. Its two TE $_{11}$ polarisations no longer share a cutoff: the one with $E$ along $x$ starts at $f_{c, x} = 8.50$ GHz, the one with $E$ along $y$ at $f_{c, y} = 8.80$ GHz. A linearly polarised TE $_{11}$ wave at $4 5^{\circ}$ between $x$ and $y$ enters at $z = 0$ .

Find $τ_{x}$ and $τ_{y}$ at $10$ GHz.
Find the length $L$ that makes the output circularly polarised. Which hand?
Find the phase difference and the axial ratio (in dB) at $9.5$ and $10.5$ GHz.
Is the section single-mode over $9.5$ – $10.5$ GHz?

Try it yourself first.

Strategy

The input splits into equal $x$ and $y$ components, each travelling as its own TE $_{11}$ mode with its own $τ$ . After a length $L$ the $x$ component lags by $(τ_{x} - τ_{y}) L$ . A lag of $9 0^{\circ}$ with equal amplitudes is circular polarisation (Section 6.1). For a phase difference $δ$ between equal components the polarisation ellipse has the axial ratio $AR = max (∣ tan (δ /2) ∣, ∣ cot (δ /2) ∣)$ .

$k = 2 π f / c = 209.4$ rad/m at $10$ GHz:

τ_{x} τ_{y} Δ τ = 209.4 1 - 0.8 5^{2} = 110.3 rad/m, = 209.4 1 - 0.8 8^{2} = 99.5 rad/m, = 10.85 rad/m .

$Δ τ L = π /2$ gives $L = 1.571/10.85 = 14.5$ cm. At the output, dropping the common phase, $E \propto - j a_{x} + a_{y}$ , that is $E (t) \propto a_{x} sin ω t + a_{y} cos ω t$ : the field turns from $y$ to $x$ . With the thumb along $+ z$ the fingers of the right hand would turn from $x$ to $y$ , so the wave is left-handed (LHCP) in the convention of Section 6.1. An input at $- 4 5^{\circ}$ gives RHCP.

With the same $L$ :

$f$ [GHz]	$τ_{x}$ , $τ_{y}$ [rad/m]	$δ$	AR
$9.5$	$88.9$ , $75.0$	$115. 3^{\circ}$	$1.58$ ( $4.0$ dB)
$10$	$110.3$ , $99.5$	$9 0^{\circ}$	$1$ ( $0$ dB)
$10.5$	$129.1$ , $120.0$	$75. 8^{\circ}$	$1.28$ ( $2.2$ dB)

The radii that go with the two cutoffs are $1.841 c / (2 π f_{c}) = 10.34$ and $9.99$ mm, and TM $_{01}$ starts at $1.306 f_{c}$ , between $11.1$ and $11.5$ GHz: the section is single-mode up to $10.5$ GHz.

Answer

Δ τ = 10.85 rad/m, L = 14.5 cm, LHCP AR = 4.0 dB (9.5 GHz) AR = 2.2 dB (10.5 GHz)

Check

At the centre frequency $tan 4 5^{\circ} = 1$ . Near cutoff $τ$ changes fastest with frequency, so the error is larger at the low edge ( $+ 2 5^{\circ}$ ) than at the high edge ( $- 1 4^{\circ}$ ), as Figure 7.4.13 shows.

Exercise 7.4.3. Phase difference between the x and y polarisations after L=14.5 cm of the squeezed pipe. Exactly 90^\circ (circular polarisation) only at 10 GHz. Near cutoff \tau changes fast with frequency, so the error grows faster towards the low edge of the shaded band. — **Figure 7.4.13.** Exercise 7.4.3. Phase difference between the $x$ and $y$ polarisations after $L = 14.5$ cm of the squeezed pipe. Exactly $9 0^{\circ}$ (circular polarisation) only at $10$ GHz. Near cutoff $τ$ changes fast with frequency, so the error grows faster towards the low edge of the shaded band.

Remark

The $x$ polarisation has the lower cutoff because the pipe is now slightly taller along $y$ : like TE $_{10}$ of a rectangular guide, TE $_{11}$ with its central $E$ along $x$ varies across $y$ , so the $y$ dimension sets its cutoff. The squeeze is only $3.5 %$ in radius, yet $14.5$ cm suffices because both modes run close to cutoff, where a small change of $f_{c}$ moves $τ$ a lot. The price is the narrow band: a $10 %$ band already costs $4$ dB of axial ratio. Practical polarisers use dielectric vanes or rows of screws and work further above cutoff to flatten $Δ τ (f)$ . The same physics, uncontrolled, is the polarisation drift of long round pipes noted in Section 7.4.8.

cf. Worked example 7.4.1

Exercise 7.4.3: A quarter-wave circular polariser