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22-Elec-A7 Electromagnetics · May 2017

Question 3 of 8: Distributed parameters of a two-ribbon strip line

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

Notes on this paper

Paper format. National Exams, May 2017 — 16-Elec-A7, Electromagnetics. Three hours, closed book; one of two approved calculators (Casio or Sharp) permitted. Eight questions of equal value (20 marks each). The paper states that any five questions constitute a complete paper and that only the first five appearing in the answer book are marked — all eight are solved here, because this set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times 10^{-12}$ F/m and $\mu_0 = 4\pi\times 10^{-7}$ H/m, which together fix $c = 2.9986\times 10^{8}$ m/s and $\eta_0 = 376.8\ \Omega$. Where a question states its own propagation velocity (Questions 1 and 2 both give $3\times 10^{8}$ m/s), that stated value is used verbatim.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, stubs, waveguides); F. T. Ulaby, Fundamentals of Applied Electromagnetics (transients, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (statics, Faraday's law, guided waves); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, waveguides); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element radiation).

Question 3: Distributed parameters of a two-ribbon strip 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 parallel-ribbon (parallel-plate) line whose conductors are much wider than their separation, so the field between them may be taken as uniform.

Given data
QuantitySymbolValue
Ribbon width$w$$2$ cm $= 0.02$ m
Dielectric thickness (separation)$d$$0.5$ mm $= 5\times 10^{-4}$ m
Relative permittivity$\varepsilon_r$$2.25$
Relative permeability$\mu_r$$1$ (non-magnetic dielectric)
Aspect ratio$w/d$$40$ (fringing negligible)

Find. The per-metre capacitance and inductance, and from them the characteristic impedance and phase velocity of the line.

E dielectric, ε(r) = 2.25 ribbon ribbon w = 2 cm d = 0.5 mm cross-section (not to scale: w/d = 40, so fringing is negligible) field is confined between the ribbons and is uniform
Figure 3.1 — cross-section. With w/d = 40 the field is essentially confined and uniform, which is exactly the licence to ignore fringing.

Approach. Treat the cross-section as an ideal parallel-plate capacitor for $C'$ and as a wide flat solenoid of one turn for $L'$, then combine them in the two standard TEM line formulas.

  1. Distributed capacitance. For a uniform field between plates of width $w$ separated by $d$, the charge per unit length gives $$C' = \frac{\varepsilon_0\varepsilon_r w}{d} = \frac{(8.85\times 10^{-12})(2.25)(0.02)}{5\times 10^{-4}} = \boxed{796.5\ \text{pF/m}} .$$
  2. Distributed inductance. The magnetic field between the ribbons is $H = I/w$ and fills the cross-section $d \times 1$ m, so the flux linked per metre is $\mu_0 (I/w) d$ and $$L' = \frac{\mu_0 d}{w} = \frac{(4\pi\times 10^{-7})(5\times 10^{-4})}{0.02} = \boxed{31.42\ \text{nH/m}} .$$ The dielectric is non-magnetic, so $\varepsilon_r$ does not appear here — a useful check on the algebra.
  3. Characteristic impedance. For a lossless TEM line, $$Z_0 = \sqrt{\frac{L'}{C'}} = \sqrt{\frac{31.42\times 10^{-9}}{796.5\times 10^{-12}}} = \sqrt{39.44} = \boxed{6.28\ \Omega} .$$ The same number follows from the intrinsic impedance, $Z_0 = \eta_0 d/(w\sqrt{\varepsilon_r}) = 376.8 \times 5\times 10^{-4}/(0.02 \times 1.5) = 6.28\ \Omega$, which is why such a wide, thin line has an impedance of only a few ohms.
  4. Phase velocity. Multiplying instead of dividing, $$v_p = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(31.42\times 10^{-9})(796.5\times 10^{-12})}} = \boxed{2.00\times 10^{8}\ \text{m/s}} .$$
  5. Cross-check. $L'C' = \mu_0\varepsilon_0\varepsilon_r$ identically for this geometry (the $d/w$ factors cancel), so $v_p$ must equal $c/\sqrt{\varepsilon_r} = 2.9986\times 10^{8}/1.5 = 1.999\times 10^{8}$ m/s regardless of the ribbon dimensions — it agrees. Only $Z_0$ carries the geometry.
Final results
Quantity askedResult
Distributed capacitance$C' = 796.5$ pF/m ($0.797$ nF/m)
Distributed inductance$L' = 31.42$ nH/m
Characteristic impedance$Z_0 = 6.28\ \Omega$
Phase velocity$v_p = 2.00\times 10^{8}$ m/s $= c/1.5$
Consistency check$L'C' = \mu_0\varepsilon_0\varepsilon_r$, so $v_p = c/\sqrt{\varepsilon_r}$