22-Elec-A7 Electromagnetics · May 2018

Question 7 of 8: Characteristic Impedance and Velocity of a Two-Layer Coaxial Line

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.

Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).

Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.

Question 7: Characteristic Impedance and Velocity 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 that is only partly filled: a thin dielectric sleeve on the inner conductor, with air filling the rest of the annulus.

Given data
Quantity	Symbol	Value
Inner radius of the outer pipe	$b$	5.0 mm (from a 10 mm diameter)
Outer radius of the inner pipe	$a$	2.5 mm (from a 5 mm diameter)
Dielectric sleeve thickness	$t$	1.0 mm
Radius of the layer boundary	$m = a+t$	3.5 mm
Sleeve permittivity	$\varepsilon_r$	2.25
Remainder of the annulus	—	air, $\varepsilon_r = 1$
Permeability everywhere	$\mu_r$	1

Find. The characteristic impedance $Z_0$ and the propagation velocity $v_p$ of the line.

Cross-section of the partly filled coaxial line. The same radial D flux crosses both annuli, so their capacitances combine in series.

Approach. Compute $C^{\prime}$ and $L^{\prime}$ per metre separately — the two layers behave as series capacitors, while the inductance ignores the sleeve entirely because $\mu_r = 1$ — then form $Z_0 = \sqrt{L^{\prime}/C^{\prime}}$ and $v_p = 1/\sqrt{L^{\prime}C^{\prime}}$, and bracket the answers between the all-air and all-dielectric limits.

Convert diameters to radii and locate the layer boundary. $a = 2.5$ mm, $b = 5.0$ mm, and the sleeve runs from $a$ out to $m = a + t = 3.5$ mm. Every logarithm below takes a ratio of radii; the stated 1 mm thickness never appears on its own.
Capacitance of each annulus. Gauss’s law gives the same radial flux $D = \rho_l/2\pi r$ in both layers, and the voltages across them add, so they are capacitors in series: $$C^{\prime}_{\text{in}} = \frac{2\pi\varepsilon_0\varepsilon_r}{\ln(m/a)} = \frac{2\pi(8.85\times10^{-12})(2.25)}{\ln 1.4} = 371.8\ \text{pF/m},$$ $$C^{\prime}_{\text{out}} = \frac{2\pi\varepsilon_0}{\ln(b/m)} = \frac{2\pi(8.85\times10^{-12})}{\ln 1.4286} = 155.9\ \text{pF/m}.$$
Combine them in series. $$\frac{1}{C^{\prime}} = \frac{1}{371.8} + \frac{1}{155.9} \;\Rightarrow\; C^{\prime} = 109.85\ \text{pF/m}.$$ For comparison, an all-air line of the same radii would give 80.2 pF/m and an all-filled one 180.5 pF/m, so the sleeve has done rather less than a third of what completely filling the line would do.
Inductance ignores the sleeve. The magnetic field between the conductors is $H_\phi = I/2\pi r$ regardless of the dielectric, and $\mu_r = 1$ everywhere, so $$L^{\prime} = \frac{\mu_0}{2\pi}\ln\frac{b}{a} = (2\times10^{-7})\ln 2 = 138.63\ \text{nH/m}.$$
Characteristic impedance. $$Z_0 = \sqrt{\frac{L^{\prime}}{C^{\prime}}} = \sqrt{\frac{138.63\times10^{-9}}{109.85\times10^{-12}}} \;\Rightarrow\; \boxed{Z_0 = 35.5\ \Omega} .$$
Propagation velocity. $$v_p = \frac{1}{\sqrt{L^{\prime}C^{\prime}}} = \frac{1}{\sqrt{(138.63\times10^{-9})(109.85\times10^{-12})}} \;\Rightarrow\; \boxed{v_p = 2.56\times10^{8}\ \text{m/s}} ,$$ which is $0.855c$, corresponding to an effective permittivity $\varepsilon_{r,\text{eff}} = (c/v_p)^{2} = 1.37$.
Bracket both answers. A fully air-filled line of these radii would have $Z_0 = (\eta_0/2\pi)\ln(b/a) = 41.6\ \Omega$ at $v_p = c$; a fully $\varepsilon_r = 2.25$ line would have $41.6/1.5 = 27.7\ \Omega$ at $c/1.5 = 2.00\times10^{8}$ m/s. Both computed values sit between those limits and closer to the air end, which is exactly what a 1 mm sleeve in a 2.5 mm annulus should give.

Final results
Quantity asked	Result
Capacitance of the dielectric annulus	371.8 pF/m
Capacitance of the air annulus	155.9 pF/m
Series capacitance per metre	109.85 pF/m
Inductance per metre	138.63 nH/m
Characteristic impedance	35.5 Ω
Propagation velocity	$2.56\times10^{8}$ m/s (0.855 c)
Effective relative permittivity	1.37
Air-filled / fully filled limits on $Z_0$	41.6 Ω / 27.7 Ω

Topic. Distributed parameters of a coaxial line with a layered dielectric.

Key relations. $C^{\prime}_{\text{layer}} = 2\pi\varepsilon_0\varepsilon_r/\ln(r_2/r_1)$, layers in series; $L^{\prime} = (\mu_0/2\pi)\ln(b/a)$; $Z_0 = \sqrt{L^{\prime}/C^{\prime}}$; $v_p = 1/\sqrt{L^{\prime}C^{\prime}}$; $\varepsilon_{r,\text{eff}} = (c/v_p)^{2}$.

Why this works. The TEM field of a coax is radial in $\mathbf{E}$ and azimuthal in $\mathbf{H}$. The same radial displacement flux therefore threads every cylindrical surface, so the layers see a common charge and their voltages add — the definition of series capacitance. The magnetic field, by contrast, is set by the enclosed current alone and is unaffected by a non-magnetic insulator, so $L^{\prime}$ is the plain air-line value. That asymmetry is why a partly filled line has $v_p$ between the two bulk values but is not described by any single permittivity except through the derived $\varepsilon_{r,\text{eff}}$.

Common pitfall. Combining the layers in parallel (adding capacitances), or area-weighting the two permittivities, both of which overstate $C^{\prime}$ and understate $Z_0$. Also common: putting the 1 mm thickness into the logarithm as if it were a radius ratio, and forgetting that the paper gives diameters, which halves every dimension.

Source. Pozar, Microwave Engineering, 4th ed., §2.1 and §3.5 (coaxial line parameters); Sadiku, Elements of Electromagnetics, 7th ed., §6.5 and §11.2.

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