Characteristic Impedance of a Two-Layer Coaxial Line: 22-Elec-A7 Electromagnetics, December 2016, Q3

Question 3: Characteristic Impedance of a Two-Layer Coaxial Line (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A coaxial line whose inner conductor carries a 2 mm dielectric coating, leaving an air gap out to the outer conductor, so the space between the conductors is radially inhomogeneous.

Given data
Quantity	Symbol	Value
Inner conductor radius	a	5 mm
Outer radius of the coating	r₁ = a + 2 mm	7 mm
Inner radius of the outer conductor	b	1 cm = 10 mm
Relative permittivity of the coating	ε_r1	2.25
Relative permittivity of the air gap	ε_r2	1.00
Relative permeability everywhere	μ_r	1.00

Find. The characteristic impedance and the phase velocity of the composite line.

Figure 3.1 — Cross-section of the coated coaxial line: two dielectric annuli stacked in series between the conductors.

Approach. Treat the line as TEM and build it from its per-unit-length parameters: the two dielectric annuli carry the same radial flux and so behave as capacitors in series, while the inductance is unaffected by the coating because both media are non-magnetic and sees the full gap from a to b.

Capacitance of the coated annulus. For a cylindrical layer of dielectric between radii $a$ and $r_{1}$, $$C_{1}=\frac{2\pi\varepsilon_{0}\varepsilon_{r1}}{\ln(r_{1}/a)}=\frac{2\pi(8.85\times10^{-12})(2.25)}{\ln(7/5)}=\frac{1.2511\times10^{-10}}{0.33647}=371.8\ \text{pF/m}$$
Capacitance of the air gap. The same formula with unity permittivity, between $r_{1}$ and $b$: $$C_{2}=\frac{2\pi\varepsilon_{0}}{\ln(b/r_{1})}=\frac{5.5606\times10^{-11}}{\ln(10/7)}=\frac{5.5606\times10^{-11}}{0.35667}=155.9\ \text{pF/m}$$
Combine them in series. The same radial displacement flux crosses both annuli while the voltages add, which is exactly the series rule: $$\frac{1}{C'}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\qquad\Longrightarrow\qquad C'=\frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{(371.8)(155.9)}{371.8+155.9}=\boxed{109.8\ \text{pF/m}}$$ Notice that the series result sits below the smaller of the two, as it must.
Inductance per unit length. Both media are non-magnetic, so the magnetic field sees a single uniform region from $a$ to $b$: $$L'=\frac{\mu_{0}}{2\pi}\ln\frac{b}{a}=\left(2\times10^{-7}\right)\ln\frac{10}{5}=\left(2\times10^{-7}\right)(0.69315)=\boxed{138.6\ \text{nH/m}}$$ This is the step where the coating must be ignored; introducing it here is the single most common error in the question.
Characteristic impedance. For a lossless TEM line, $$Z_{0}=\sqrt{\frac{L'}{C'}}=\sqrt{\frac{138.6\times10^{-9}}{109.8\times10^{-12}}}=\sqrt{1262}=\boxed{35.5\ \Omega}$$
Propagation velocity. From the same pair, $$v_{p}=\frac{1}{\sqrt{L'C'}}=\frac{1}{\sqrt{(138.6\times10^{-9})(109.8\times10^{-12})}}=\boxed{2.56\times10^{8}\ \text{m/s}}$$
Close the loop with an effective permittivity. Comparing the composite capacitance with the same geometry filled with air, $C'_{\text{air}}=2\pi\varepsilon_{0}/\ln(b/a)=80.2$ pF/m, gives $$\varepsilon_{r,\text{eff}}=\frac{C'}{C'_{\text{air}}}=\frac{109.8}{80.2}=1.369\qquad\Longrightarrow\qquad v_{p}=\frac{c}{\sqrt{\varepsilon_{r,\text{eff}}}}=0.855\,c$$ which reproduces the velocity found in step 6 and confirms that no factor of $2\pi$ or $\ln$ has gone astray. The effective permittivity lands between 1 and 2.25 and much closer to unity, which is the expected result: the air gap occupies the outer, larger-radius part of the cross-section where most of the stored energy sits.

Question 3 — per-unit-length parameters and line constants
Quantity	Value
Capacitance of the coated annulus, C₁	371.8 pF/m
Capacitance of the air gap, C₂	155.9 pF/m
Series capacitance, C′	109.8 pF/m
Inductance, L′	138.6 nH/m
Characteristic impedance, Z₀	35.5 Ω
Propagation velocity, v_p	2.56 × 10⁸ m/s (0.855 c)
Effective relative permittivity	1.369

Topic. TEM transmission-line parameters of a radially inhomogeneous coaxial structure.

Key relations. $C=2\pi\varepsilon/\ln(r_{\text{out}}/r_{\text{in}})$ per unit length for a cylindrical layer; series capacitance for stacked annuli; $L'=(\mu_{0}/2\pi)\ln(b/a)$; $Z_{0}=\sqrt{L'/C'}$ and $v_{p}=1/\sqrt{L'C'}$.

Why this works. In the TEM mode the electric field is purely radial, so the normal component of the displacement vector $D_{r}$ is continuous across the coating boundary and each annulus carries the same flux while contributing its own share of the voltage — the defining signature of capacitors in series. The magnetic field is azimuthal and depends only on the enclosed current, and since $\mu_{r}=1$ throughout, the coating is magnetically invisible. Building $Z_{0}$ and $v_{p}$ from $L'$ and $C'$ rather than from a memorised coaxial formula is what makes the inhomogeneous case tractable at all.

Common pitfall. Averaging the two permittivities by volume or by radius instead of stacking the capacitances in series, which here would give roughly 1.6 rather than 1.369 and a noticeably wrong impedance; and inserting $\varepsilon_{r}$ into the inductance formula, which has no permittivity in it. A third slip is to use the 2 mm coating thickness directly in a logarithm — the arguments are ratios of radii (7/5 and 10/7), never thicknesses.

Source. Pozar, Microwave Engineering, 4th ed., §2.1 and Table 2.1 (coaxial-line parameters); Hayt & Buck, Engineering Electromagnetics, 9th ed., ch. 6 and ch. 10.

Question 3 of 8: Characteristic Impedance of a Two-Layer Coaxial Line

Question 3: Characteristic Impedance of a Two-Layer Coaxial Line (20 marks)