Question 2 of 8: The Two Lowest Frequencies for a Short-Circuited Stub Across a Matched Load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.
Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).
Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.
Question 2: The Two Lowest Frequencies for a Short-Circuited Stub Across a Matched Load (20 marks)
Given. A line whose load already equals its characteristic impedance, spoiled only by a shorted stub hung across the load terminals.
Given data
Quantity
Symbol
Value
Characteristic impedance of line and stub
$Z_0$
50 Ω
Resistive load
$R_L$
50 Ω
Propagation velocity
$v_p$
$2\times10^{8}$ m/s
Stub length
$l$
50 cm
Stub termination
—
short circuit
Find. The lowest frequency at which no power reaches the load, and the lowest frequency at which the load is matched to the line.
The stub and the load share one node, so the stub susceptance simply parallels the load: no length of main line has to be transformed.
Approach. Note first that $R_L = Z_0$, so the load on its own is already matched and the stub is the only thing that can change matters. Then ask what the shorted stub’s input impedance must be in each case — an open circuit to leave the match alone, a dead short to starve the load — and solve $jZ_0\tan\beta l$ for those two conditions.
Write the stub’s input impedance. A lossless line of length $l$ terminated in a short presents $$Z_{\text{stub}} = jZ_0\tan\beta l, \qquad \beta = \frac{2\pi f}{v_p} .$$ It is purely reactive at every frequency: a stub can store energy and send it back, but it can never absorb any.
Match condition. The load alone gives a perfect match, so the line stays matched if and only if the stub draws no current at all, i.e. it must look like an open circuit: $|\tan\beta l| \to \infty$, which needs $\beta l = (2n+1)\pi/2$, i.e. $l = (2n+1)\lambda/4$. The stub is an odd number of quarter wavelengths long.
Lowest matched frequency. Taking $n = 0$, $\lambda = 4l = 4(0.50) = 2.00$ m, so $$f_{\text{match}} = \frac{v_p}{\lambda} = \frac{2\times10^{8}}{2.00} \;\Rightarrow\; \boxed{f_{\text{match}} = 100\ \text{MHz}} .$$ A shorted quarter-wave line inverts its own short into an open, which is exactly the classical quarter-wave transformer result $Z_{in}Z_L = Z_0^2$ evaluated at $Z_L = 0$.
Starvation condition. For no power to reach the load the stub must instead short out the load terminals, taking every ampere the line delivers: $\tan\beta l = 0$, which needs $\beta l = n\pi$, i.e. $l = n\lambda/2$. A half wavelength of line repeats whatever terminates it, so the short at the far end reappears at the load plane.
Lowest starved frequency. Taking $n = 1$, $\lambda = 2l = 1.00$ m, so $$f_{\text{null}} = \frac{2\times10^{8}}{1.00} \;\Rightarrow\; \boxed{f_{\text{null}} = 200\ \text{MHz}} .$$ At that frequency the load sees zero volts across it, and the reflection coefficient looking into the load plane is $\Gamma = -1$.
Check where the power goes. The stub is purely reactive, so at 200 MHz the power is not consumed anywhere near the load — it is reflected straight back down the line to the generator. Saying “the stub absorbs it” is physically impossible and is a standard mark loss.
Electrical length of the 50 cm stub against frequency. Where the line crosses 90° the stub looks open and the match survives; where it crosses 180° the stub looks like a short and the load is starved.
Because $\beta l$ rises linearly with frequency, the two families simply interleave: the load is matched at 100, 300, 500 MHz … (odd multiples of 100 MHz) and starved at 200, 400, 600 MHz … (multiples of 200 MHz). Neither family contains $Z_0$ anywhere in its derivation, which is a consequence of $R_L = Z_0$: the load was already matched, so the answer depends only on lengths and velocities.
