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22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2013

Question 1 of 8: Stub-Loaded Line — Frequencies of Short Circuit and of a Matched 50 Ω Termination

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

Notes on this paper

Paper format. National Exams, May 2013 — 07-Elec-B9, Electromagnetic Field, Transmission Lines, Antennas, and Radiation. Three hours, closed book; an approved Casio or Sharp calculator is permitted. Eight questions are printed and five constitute a complete paper, each of equal value (20 marks); the first five appearing in the answer book are the ones marked. The cover page supplies only two aids, $\varepsilon_0 = 8.85\times 10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, plus a Smith chart on page 4. Candidates are urged to state any interpretation assumptions with their answers. All eight questions are solved below, because this set is a study resource rather than a three-hour sitting.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, the Smith chart, waveguides and cavities); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. and W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, oblique incidence, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, arrays, apertures); F. T. Ulaby & U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. (radar equation, ground-plane imaging). SI units throughout; the paper's own $c = 3\times 10^{8}\ \text{m/s}$ and $Z_0 = 377\ \Omega$ are used so that the printed aids reproduce exactly.

Check: reading of the printed data. Some symbols are printed loosely. Question 3 prints “10GH” and “1KWs total energy”; these are read as a 10 GHz carrier and a total pulse energy of $1\ \text{kW}\cdot\text{s} = 1000\ \text{J}$. Question 6 prints “15 MH” (15 MHz) and the aid $R_{\text{rad}} = (2\pi/3)Z_0 (l/\lambda)^2$. Question 7 prints “10GH” (10 GHz) and the aid $\theta_0 = 1.2\lambda/d$. Question 8 prints “an 600 angle” ($60^\circ$).

Question 1: Stub-Loaded Line — Frequencies of Short Circuit and of a Matched 50 Ω Termination (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 uniform lossless line of characteristic impedance $Z_0 = 50\ \Omega$ and phase velocity $v_p = 3\times 10^{8}\ \text{m/s}$ is terminated by three elements in parallel: an open-circuited stub, a short-circuited stub, and a resistor.

Given data
QuantitySymbolValue
Characteristic impedance$Z_0$50 Ω
Phase velocity$v_p$$3\times 10^{8}\ \text{m/s}$
Open-circuited stub length$l$0.20 m
Short-circuited stub length$l$0.20 m
Shunt resistor$R$50 Ω

Find. The lowest frequencies at which the load presents a short circuit to the driving line, and the lowest frequency at which it presents a pure 50 Ω resistance.

fromgenerator50 Ω linev = 3 x 10^8 m/sload planeopencircuit20 cm stubshortcircuit20 cm stub50 Ωthree branches in parallel
Figure 1.1 — The load plane. Both stubs are cut from the same 50 Ω line and are the same length, so their susceptances track one another as the frequency is swept.

Approach. Add the three branch admittances at the load plane, collapse the two stub terms with the identity $\tan\theta - \cot\theta = -2\cot 2\theta$, and then read off the frequencies at which the total admittance is infinite (a short) and at which it equals $Y_0$ (a 50 Ω termination).

  1. Express the electrical length of a stub. Each stub is $l = 0.20\ \text{m}$ of the same line, so its electrical length is $$\theta = \beta l = \frac{2\pi f l}{v_p}.$$ Because both stubs have the same length, one variable $\theta$ describes both.
  2. Write the branch admittances. For a lossless stub of characteristic admittance $Y_0 = 1/Z_0$, $$Y_{\text{oc}} = j Y_0 \tan\theta, \qquad Y_{\text{sc}} = -j Y_0 \cot\theta, \qquad Y_R = \frac{1}{50\ \Omega} = Y_0 .$$ The resistor happens to equal $Z_0$, so it contributes exactly one unit of normalised conductance.
  3. Combine the three branches. Adding the admittances and using $\tan\theta - \cot\theta = -2\cot 2\theta$, $$\boxed{\ \frac{Y_L}{Y_0} = 1 + j\left(\tan\theta - \cot\theta\right) = 1 - 2j\cot 2\theta\ }$$ The load is therefore always one unit of conductance in parallel with a susceptance that sweeps through every value as the frequency changes.
  4. Locate the short circuits. $Y_L \to \infty$ requires $\cot 2\theta \to \infty$, i.e. $2\theta = n\pi$, so $\theta = n\pi/2$ and $$f_{\text{sc}} = \frac{n\, v_p}{4l} = n\,\frac{3\times 10^{8}}{4(0.20)} = n \times 375\ \text{MHz}, \qquad n = 1, 2, 3,\dots$$ Physically, at $l = \lambda/4$ the open stub inverts into a short, and at $l = \lambda/2$ the shorted stub repeats its own short — either one crowbars the whole load plane.
  5. Locate the 50 Ω points. $Y_L = Y_0$ requires $\cot 2\theta = 0$, i.e. $2\theta = (2n-1)\pi/2$, so $\theta = (2n-1)\pi/4$ and $$f_{50} = \frac{(2n-1)\, v_p}{8l} = (2n-1)\times 187.5\ \text{MHz}, \qquad n = 1, 2, 3,\dots$$ At $l = \lambda/8$ the open stub is $+jY_0$ and the shorted stub is $-jY_0$: the two susceptances cancel exactly and only the resistor is left.
  6. Order the critical frequencies and answer. The two families interleave, and the lowest four members are 187.5 MHz (50 Ω), 375 MHz (short), 562.5 MHz (50 Ω) and 750 MHz (short). Answering the question as posed — the two lowest short-circuit frequencies and the lowest 50 Ω frequency — $$\text{(i)}\quad f = 375\ \text{MHz\ and\ } 750\ \text{MHz}; \qquad \text{(ii)}\quad f = 187.5\ \text{MHz}.$$

The pattern repeats indefinitely: every 187.5 MHz the load alternates between a pure 50 Ω resistance and a dead short, because the pair of equal stubs behaves as a series-resonant/parallel-resonant pair whose combined susceptance $-2Y_0\cot 2\theta$ passes through zero and through infinity twice per half-wavelength of stub. Note also that at every 50 Ω point the line is genuinely matched, so a generator driving it sees no reflection at all.

Final results — Question 1
FrequencyStub electrical lengthLoad presented
187.5 MHz$\theta = \pi/4$  ($l = \lambda/8$) $Z_L = 50\ \Omega$ — answer to (ii)
375 MHz$\theta = \pi/2$  ($l = \lambda/4$) short circuit — answer to (i)
562.5 MHz$\theta = 3\pi/4$  ($l = 3\lambda/8$) $Z_L = 50\ \Omega$ (next matched point)
750 MHz$\theta = \pi$  ($l = \lambda/2$) short circuit — answer to (i)

Check: the question's own grouping. Taken literally, the three lowest critical frequencies are 187.5 MHz (50 Ω), 375 MHz (short) and 562.5 MHz (50 Ω) — that is two matched points and one short, the reverse of the (i)/(ii) split the question implies. The split the examiner describes is recovered by taking the two lowest short-circuit frequencies (375 MHz, 750 MHz) together with the lowest 50 Ω frequency (187.5 MHz), which is what is reported above. The intervening 562.5 MHz matched point is listed so that nothing is hidden; an exam answer should state the general families and then name the three frequencies asked for.

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