Question 2 of 8: Parallel Open- and Short-Circuited Line Sections

Question 2: Parallel Open- and Short-Circuited Line Sections (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. Looking into the tap point, the line to the right is a 50 cm section ending in a short circuit, and the stub hanging off the tap is an identical 50 cm section ending in an open circuit; the two sit in parallel.

Find. The termination impedance at 300 MHz, the nearest other frequency at which it is again zero, and any frequency at which it becomes infinite.

Approach. Each section is a lossless stub whose input impedance is a pure reactance set by its electrical length; because they are in parallel, add admittances and read off where the total susceptance blows up (impedance zero) and where it vanishes (impedance infinite).

1. Write the two stub impedances. For lossless sections of length \(\ell\) and phase constant \(\beta=2\pi f/v_{p}\), $$Z_{sc}=jZ_{0}\tan\beta\ell,\qquad Z_{oc}=-jZ_{0}\cot\beta\ell$$ so that the parallel combination has admittance $$Y_{T}=\frac{1}{Z_{sc}}+\frac{1}{Z_{oc}}=\frac{j}{Z_{0}}\left(\tan\beta\ell-\cot\beta\ell\right)\equiv\frac{jb}{Z_{0}}$$ Everything that follows is read off the single function \(b(\,f\,)=\tan\beta\ell-\cot\beta\ell\), and note that \(Z_{0}\) cancels out of the frequencies entirely.
2. Evaluate the electrical length at 300 MHz. The wavelength is $$\lambda=\frac{v_{p}}{f}=\frac{3\times10^{8}}{300\times10^{6}}=1.00\ \text{m}$$ so each 50 cm section is exactly half a wavelength and $$\beta\ell=\frac{2\pi}{\lambda}\ell=\frac{2\pi}{1.00}(0.50)=\pi\ \text{rad}$$
3. Read the two sections at that length. A half-wavelength section repeats whatever terminates it. The shorted section gives \(Z_{sc}=jZ_{0}\tan\pi=0\) — the short is transported bodily to the tap — while the open stub gives \(Z_{oc}=-jZ_{0}\cot\pi\rightarrow\infty\), an open circuit at the tap. Putting a short in parallel with an open leaves the short: $$\boxed{Z_{T}(300\ \text{MHz})=0\ \Omega\ \ \text{(a short circuit)}}$$ This is the answer to part (i), and it is worth noticing that the open stub is doing nothing whatsoever at this particular frequency.
4. Locate every zero of the termination impedance. \(Z_{T}=Z_{0}/(jb)\) vanishes wherever \(b\) becomes infinite, which happens at the poles of \(\tan\beta\ell\) (the open stub becoming a quarter-wave short) and at the poles of \(\cot\beta\ell\) (the shorted section repeating its short). Together those are $$\beta\ell=n\frac{\pi}{2}\quad\Longleftrightarrow\quad \ell=n\frac{\lambda}{4}\quad\Longleftrightarrow\quad f_{n}=n\,\frac{v_{p}}{4\ell}=n\,\frac{3\times10^{8}}{4(0.50)}=n\,(150\ \text{MHz})$$ so the termination is a short at 150, 300, 450, 600 MHz and so on — every 150 MHz without exception.
5. Pick the nearest such frequency to 300 MHz. The neighbours of 300 MHz in that ladder are 150 MHz below and 450 MHz above, and they are equidistant: $$\boxed{f=150\ \text{MHz}\ \ \text{or}\ \ 450\ \text{MHz}\quad(\text{each }150\ \text{MHz from }300\ \text{MHz})}$$ Because the question says “a frequency” rather than “the frequency”, either is an acceptable answer; both should be quoted, together with the observation that the spacing is uniform. At 150 MHz the sections are a quarter wave, so the roles swap — the open stub is now the short and the shorted section is the open — but the parallel result is the same.
6. Locate every pole of the termination impedance. \(Z_{T}\) is infinite where the total susceptance vanishes: $$b=\tan\beta\ell-\cot\beta\ell=0\quad\Longrightarrow\quad\tan^{2}\beta\ell=1\quad\Longrightarrow\quad\beta\ell=\frac{\pi}{4}+n\frac{\pi}{2}$$ which in frequency reads $$f=(2n+1)\frac{v_{p}}{8\ell}=(2n+1)\frac{3\times10^{8}}{8(0.50)}=(2n+1)(75\ \text{MHz})$$ $$\boxed{f=75,\ 225,\ 375,\ 525\ \ldots\ \text{MHz}}$$
7. Confirm the mechanism at the lowest such frequency. At 75 MHz the wavelength is 4 m, each section is \(\lambda/8\) and \(\beta\ell=45^{\circ}\). The shorted section then presents \(+j50\ \Omega\) and the open stub \(-j50\ \Omega\): equal and opposite reactances in parallel, which is a textbook anti-resonance. Their parallel combination is \((j50)(-j50)/(j50-j50)\), an unbounded quantity, so the termination genuinely looks like an open circuit and the pair reflects everything back down the line. Note that this frequency depends only on \(\ell\) and \(v_{p}\); a different \(Z_{0}\) would change the individual reactances but not where they cancel.