22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2016

Question 6 of 8: Two Shorted Stubs that Block 300 MHz and Pass 200 MHz

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

Question 6: Two Shorted Stubs that Block 300 MHz and Pass 200 MHz (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 50 Ω line with $v = 3\times10^{8}\ \text{m/s}$ feeds a 50 Ω load and carries 300 MHz and 200 MHz simultaneously. Two short-circuited stubs of the same line are connected in parallel across the load terminals; their lengths $l_1$ and $l_2$ are the unknowns.

Find. Lengths that short the load out completely at 300 MHz while leaving it entirely undisturbed at 200 MHz.

Approach. Translate the two requirements into conditions on the stubs' input admittances — infinite at 300 MHz, zero in total at 200 MHz — show that no single stub can meet both, and then solve the pair.

1. Write the two wavelengths. $$\lambda_{300} = \frac{3\times10^{8}}{300\times10^{6}} = 1.00\ \text{m}, \qquad \lambda_{200} = \frac{3\times10^{8}}{200\times10^{6}} = 1.50\ \text{m}.$$
2. Write the admittance of a shorted stub. A short-circuited length $l$ has input impedance $Z = jZ_0\tan\beta l$, hence $$Y(l) = \frac{1}{jZ_0\tan\beta l} = -\,\frac{j\cot\beta l}{Z_0}.$$ It is purely susceptive at all lengths, as it must be: a lossless stub can store energy but cannot absorb any.
3. State the two design conditions. "Prevented from reaching the load" means the stub bank must be a dead short across the load at 300 MHz, i.e. the total admittance is infinite. "Unobstructed" means it must be an open circuit at 200 MHz, i.e. the total admittance is zero: $$\text{at }300\ \text{MHz}: \ Y_1 + Y_2 \to \infty, \qquad \text{at }200\ \text{MHz}: \ Y_1 + Y_2 = 0.$$
4. Show why one stub cannot do the job — this is the point of the question. A single stub is a short at 300 MHz only if $\tan\beta_{300}l = 0$, i.e. $l = n\lambda_{300}/2 = 0.5n\ \text{m}$. It is an open at 200 MHz only if $\beta_{200}l$ is an odd multiple of $90^\circ$, i.e. $l = (2m-1)\lambda_{200}/4 = 0.375(2m-1)\ \text{m}$. Equating gives $0.5n = 0.375(2m-1)$, so $n = 0.75(2m-1)$ — and since $2m-1$ is always odd, $n$ can never be an integer. No single length works at all, which is precisely why the paper supplies two sections.
5. Recognise how two stubs escape the deadlock. The total admittance goes infinite if either stub alone is a short, so the 300 MHz condition can be loaded onto one stub; the 200 MHz condition, being a cancellation, then has the second stub free to satisfy it. Take $$l_1 = \frac{\lambda_{300}}{2} = 0.500\ \text{m},$$ a half wavelength at 300 MHz, which repeats its own short-circuit termination at its input and shorts the load out regardless of what the other stub is doing.
6. Choose the second length to cancel the first at 200 MHz. At 200 MHz, $l_1$ is $\beta_{200}l_1 = (360^\circ)(0.500/1.50) = 120^\circ$ long, so it presents $Z_1 = j50\tan120^\circ = -j86.60\ \Omega$: capacitive. The second stub must present the equal and opposite susceptance, which requires $$\cot\beta_{200}l_2 = -\cot(120^\circ) \;\Longrightarrow\; \beta_{200}l_2 = 60^\circ \;\Longrightarrow\; l_2 = \frac{60}{360}(1.50)$$ $$\boxed{l_1 = 0.500\ \text{m}, \qquad l_2 = 0.250\ \text{m}}$$ Then $Z_2 = j50\tan60^\circ = +j86.60\ \Omega$, inductive, and the two in parallel form a parallel resonance at 200 MHz: $Y_1 + Y_2 = +j0.011547 - j0.011547 = 0$, an open circuit.
7. Check the pair at 300 MHz. There, $l_1$ is a half wavelength and is a short as designed; $l_2 = 0.250\ \text{m}$ is a quarter wavelength and presents an open circuit, contributing zero admittance. The combination is a short, so the 300 MHz signal is entirely reflected back up the feed and none reaches the load. Both conditions hold exactly, not approximately.
8. Note the general family. Either stub can be lengthened by $\text{lcm}(\lambda_{300}/2,\ \lambda_{200}/2) = \text{lcm}(0.50,\ 0.75) = 1.50\ \text{m}$ without changing its behaviour at either frequency. More generally, any $l_1 = 0.50n\ \text{m}$ ($n$ not a multiple of 3, so that stub 1 is not also a short at 200 MHz) paired with $l_2 = 0.75k - l_1 > 0$ works; the next-shortest pair is $l_1 = 0.50\ \text{m}$ with $l_2 = 1.00\ \text{m}$ (both stubs short at 300 MHz, while their $120^\circ$ and $240^\circ$ susceptances cancel at 200 MHz). The shortest pair is the practical choice: longer stubs are bulkier, lossier, and far more sensitive to frequency drift.