Question 2 of 8: Parallel Short-Circuited and Open-Circuited Stub Termination
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A3, Electromagnetics — National Exam, December 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.) — transmission-line theory and stub matching; Balanis, Antenna Theory (4th ed.) — short-dipole far field.
Question 2: Parallel Short-Circuited and Open-Circuited Stub Termination (equal value)
Given. Two 50 cm sections of $Z_0=50\ \Omega$ line, $v=3\times10^8$ m/s: one continues to a short circuit, the other taps off in parallel (shunt) to an open circuit — the "termination" seen looking into the tap point is these two 50 cm stub input impedances in parallel.
Find. (i) the termination impedance at $f=300$ MHz; (ii) the nearest other frequency at which it is again zero; (iii) a frequency at which it is infinite.
Looking into the tap point: a 50 cm short-circuited section in parallel with a 50 cm open-circuited stub.
Approach. Write each stub's input reactance from the standard shorted/open lossless-line formulas, combine them as reactances in parallel, and simplify to a single closed-form function of frequency whose zeros and poles can be read off directly.
Individual stub reactances. With $\beta=2\pi f/v$ and stub length $L=0.5$ m,
$$X_{sc}=Z_0\tan(\beta L),\qquad X_{oc}=-Z_0\cot(\beta L).$$
Combine in parallel and simplify. Writing $s=\sin(\beta L)$, $c=\cos(\beta L)$,
$$X_{tot}=\frac{X_{sc}X_{oc}}{X_{sc}+X_{oc}}=\frac{-Z_0^2}{Z_0(s^2-c^2)/(sc)}=\frac{Z_0}{2}\tan(2\beta L).$$
So the whole termination behaves like a single line of length $2L=1$ m — its zeros and poles repeat every time $2\beta L$ advances by $\pi$.
Part (i): evaluate at 300 MHz. At $f=300$ MHz, $\lambda=v/f=1$ m, so $L=0.5$ m $=\lambda/2$ and $2\beta L=2\pi$, giving $\tan(2\pi)=0$:
$$X_{tot}(300\text{ MHz})=\boxed{0\ \Omega\ \text{(short circuit)}}.$$
(The shorted stub alone is a half-wave line, so it repeats its own $0\ \Omega$ load; a short in parallel with anything is a short.)
Part (ii): the next zero. $X_{tot}=0$ whenever $2\beta L=n\pi$, i.e. at $f=n\cdot v/(4L)=n\times150$ MHz. The zeros are therefore uniformly spaced every $150$ MHz: $\dots,150,300,450,\dots$ By this spacing, both neighbours of $300$ MHz are equally close:
$$\boxed{f=150\ \text{MHz and } f=450\ \text{MHz, each } 150\ \text{MHz away}}.$$
Part (iii): a frequency of infinite impedance. $X_{tot}\to\infty$ where $\tan(2\beta L)$ has a pole, i.e. $2\beta L=\tfrac{\pi}{2}+n\pi$, giving $f=\tfrac{v}{8L}(1+2n)=75(2n+1)$ MHz:
$$\boxed{f=75\ \text{MHz}}\ \ (\text{also }225,\,375,\,525\ \text{MHz},\dots).$$
Physically this is where the shorted stub's inductive reactance exactly cancels the open stub's capacitive reactance — a parallel-resonance condition.
