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17-Phys-A3 Electromagnetics · May 2016

Question 6 of 8: Per-Unit-Length Parameters of a Parallel-Strip Transmission Line

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

Notes on this paper

98-Phys-A3, Electromagnetics — National Exam, May 2016. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the eight questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value. Aids given on the paper: ε0 = 8.85×10-12 F/m, μ0 = 4π×10-7 H/m. All eight printed questions are solved below as a complete study resource.

Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — transmission lines, waveguides, plane waves, antennas; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients; Pozar, Microwave Engineering (4th ed.) — Smith-chart-free impedance transformation and waveguide cutoff; Balanis, Antenna Theory (4th ed.) — short-dipole far field.

Question 6: Per-Unit-Length Parameters of a Parallel-Strip Transmission Line (equal value)

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. Ribbon width $w=2$ cm; dielectric thickness (plate separation) $d=0.5$ mm; dielectric $\varepsilon_r=2.25$; non-magnetic conductors/dielectric ($\mu_r=1$); fringing neglected (parallel-plate approximation valid since $w\gg d$).

Given data
QuantitySymbolValue
Ribbon width$w$2 cm
Separation$d$0.5 mm
Relative permittivity$\varepsilon_r$2.25

Find. $C'$, $L'$ (per metre), $Z_0$, and $v_p$.

Approach. Model the ribbons as an ideal parallel-plate capacitor/inductor per unit length (fringing neglected, $w/d=40\gg1$ so this is a good approximation), then combine via $Z_0=\sqrt{L'/C'}$ and $v_p=1/\sqrt{L'C'}$.

  1. Capacitance per unit length. $$C'=\frac{\varepsilon_0\varepsilon_r w}{d}=\frac{(8.85\times10^{-12})(2.25)(0.02)}{0.0005}=\boxed{796.5\ \text{pF/m}}.$$
  2. Inductance per unit length. For a non-magnetic parallel-plate line, $L'=\mu_0 d/w$: $$L'=\frac{\mu_0 d}{w}=\frac{(4\pi\times10^{-7})(0.0005)}{0.02}=\boxed{31.42\ \text{nH/m}}.$$
  3. Characteristic impedance. $$Z_0=\sqrt{\frac{L'}{C'}}=\sqrt{\frac{31.42\times10^{-9}}{796.5\times10^{-12}}}=\boxed{6.28\ \Omega}.$$ (Equivalently, $Z_0=\eta_0\,d/(w\sqrt{\varepsilon_r})$ for a TEM parallel-plate line, which reproduces the same value.)
  4. Propagation velocity. Since $L'C'=\mu_0\varepsilon_0\varepsilon_r$ (the $w,d$ geometry cancels), $$v_p=\frac{1}{\sqrt{L'C'}}=\frac{c}{\sqrt{\varepsilon_r}}=\frac{3\times10^8}{\sqrt{2.25}}=\boxed{2.00\times10^{8}\ \text{m/s}}.$$
εr = 2.25 d=0.5mm w = 2 cm Parallel-strip line (fringing neglected) C′ = 796.5 pF/m, L′ = 31.42 nH/m, Z0 = 6.28 Ω, vp = c/√εr
Parallel-strip line cross-section (fringing neglected): dielectric fills the full gap between the two 2 cm ribbons, 0.5 mm apart.
Final results
QuantityValue
Capacitance per metre $C'$796.5 pF/m
Inductance per metre $L'$31.42 nH/m
Characteristic impedance $Z_0$6.28 Ω
Propagation velocity $v_p$$2.00\times10^8$ m/s