Question 2 of 8: Placing the protective stubs of a two-band diplexer
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2018, 16-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions, all of equal value; the rubric says any five constitute a complete paper, so a candidate answers five — but all eight are worked here, because the set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $e = 1.6\times10^{-19}$ C.
Reference texts. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.
Constants used throughout. Question 6 hands us $\omega = 2\pi\times10^{10}$ rad/s together with $k = (2\pi/3)\ \text{cm}^{-1}$, so the setter’s phase velocity is $\omega/k = 3.000\times10^{8}$ m/s exactly; Question 3’s cut-off arithmetic only closes on round numbers under the same choice. Accordingly $c = 3\times10^{8}$ m/s and $\eta_0 = 120\pi = 376.99\ \Omega$ are used here. The paper’s own aid list would give $c = 2.99863\times10^{8}$ m/s and $\eta_0 = 376.83\ \Omega$ — a 0.05 % shift that changes no answer except at one deliberate boundary case, flagged in Question 3.
Question 2: Placing the protective stubs of a two-band diplexer (20 marks)
Given. A single feeder carrying both bands splits at a junction into two matched branches, each fitted with one reactive stub whose job is to block the other band.
Given data
Quantity
Symbol
Value
Characteristic impedance, every section
$Z_0$
50 Ω
Propagation velocity, every section
$v_p$
$2\times10^{8}$ m/s
Signal A (to amplifier A)
$f_A$
200 MHz
Signal B (to amplifier B)
$f_B$
400 MHz
Amplifier input impedances
—
50 Ω each (matched)
Stub across line A
—
1/4 m, short-circuited
Stub across line B
—
1/2 m, open-circuited (see note)
Find. How far from the junction each stub must be tapped so that both signals reach their own amplifier with the common feeder matched.
The diplexer. Each stub is a dead short at the band it must block; standing a quarter wavelength back from the junction turns that short into an open, so the unwanted branch disconnects itself.
Approach. Convert each stub length into electrical degrees at both frequencies to confirm which band it kills, then choose the standoff distance that transforms the resulting short into an open at the junction.
Wavelengths in the two bands. $$\lambda_A = \frac{v_p}{f_A} = \frac{2\times10^{8}}{2\times10^{8}} = 1.00\ \text{m}, \qquad \lambda_B = \frac{v_p}{f_B} = \frac{2\times10^{8}}{4\times10^{8}} = 0.50\ \text{m}.$$ Everything that follows is a ratio of lengths to these two numbers; no answer in this question depends on $Z_0$ at all.
Check the stub on line A. Its input impedance is $Z_{sc} = jZ_0\tan\beta\ell$. At 200 MHz, $\ell/\lambda_A = 0.25/1.00 = \tfrac14$, a quarter wave, so the short is transformed into an OPEN and amplifier A never notices it. At 400 MHz, $\ell/\lambda_B = 0.25/0.50 = \tfrac12$, a half wave, which repeats the short unchanged. The quarter-metre shorted stub is therefore exactly right: invisible at 200 MHz, a dead short at 400 MHz.
Check the stub on line B. An open stub has $Z_{oc} = -jZ_0\cot\beta\ell$. At 200 MHz the stated half-metre gives $\ell/\lambda_A = \tfrac12$, and at 400 MHz it gives $\ell/\lambda_B = 1$ — a half wave and a full wave respectively, and an open stub of either length simply repeats its own open circuit. As printed it is transparent in both bands and protects nothing. For the stated function the open stub must be an ODD quarter wave at the band it kills and a whole half wave at the band it passes, i.e. $\ell = \lambda_A/4 = 0.25$ m (or $3\lambda_A/4 = 0.75$ m). The design below uses that value; see the note under this question.
The design rule for the standoff. Each stub creates a dead short at its own tap point in the band it blocks. A short seen through a quarter wave of line appears as an open, because $$Z_{in} = Z_0\,\frac{Z_L + jZ_0\tan\beta\ell}{Z_0 + jZ_L\tan\beta\ell} \quad\text{becomes}\quad \infty \quad\text{when } Z_L = 0 \text{ and } \beta\ell = \pi/2 .$$ Placing each stub an odd quarter wave from the junction therefore makes the dead branch disconnect itself instead of loading the source.
Locate stub A. It blocks the 400 MHz signal, so the standoff is counted in $\lambda_B$: $$d_A = \frac{\lambda_B}{4} = \frac{0.50}{4} = \boxed{0.125\ \text{m} = 12.5\ \text{cm}}$$ from the junction, and equally any odd multiple — 12.5, 37.5, 62.5 cm and so on, the family repeating every $\lambda_B/2 = 25$ cm.
Locate stub B. It blocks the 200 MHz signal, so its standoff is counted in $\lambda_A$: $$d_B = \frac{\lambda_A}{4} = \frac{1.00}{4} = \boxed{0.250\ \text{m} = 25\ \text{cm}}$$ from the junction, repeating every $\lambda_A/2 = 50$ cm: 25, 75, 125 cm.
Confirm the junction is matched in both bands. At 200 MHz branch B presents an open circuit while branch A is an unloaded 50 $\Omega$ line into a 50 $\Omega$ amplifier, so the junction admittance is that of one matched branch and the feeder sees $Z_{in} = 50\ \Omega$, SWR 1. At 400 MHz the roles swap and the result is identical. Each amplifier receives its whole band; the rejected band is reflected back into the other branch, never absorbed, since a lossless reactive stub cannot dissipate power.
Check: the half-metre open stub as printed cannot work. At 200 MHz it is a half wavelength and at 400 MHz a full wavelength, and an open stub of either length repeats its own open circuit, so it is invisible in both bands and amplifier B would receive the 200 MHz signal in full. The complementary length that performs the stated function is a quarter wavelength at 200 MHz, i.e. 1/4 m (or 3/4 m) — which also makes the two stubs the same physical length, one shorted and one open, the conventional form of this diplexer. The locations quoted above are the answer to the question as intended and are unaffected by the stub length itself, since they depend only on the quarter-wave standoff at the blocked frequency.
