Question 2 of 8: Which branch receives which frequency — a stub diplexer

Question 2: Which branch receives which frequency — a stub diplexer (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. Two branches, each a 50 Ω line ending in a matched 50 Ω resistor, hang in parallel from one junction; each branch carries a single 50 cm stub, and the two stubs differ in their terminations and in their distance from the junction.

Find. Which resistor, A or B, absorbs the 300 MHz signal and which absorbs the 150 MHz signal.

Approach. Work out the two wavelengths, decide at which frequency each stub becomes a short circuit and at which it disappears, then transform that short along the tap distance to see what the junction sees.

1. Wavelengths. With $v_p = 3\times 10^{8}$ m/s, $$\lambda_{300} = \frac{3\times 10^{8}}{300\times 10^{6}} = 1.00\ \text{m}, \qquad \lambda_{150} = \frac{3\times 10^{8}}{150\times 10^{6}} = 2.00\ \text{m}.$$ The 50 cm stubs are therefore a half wave long at 300 MHz and a quarter wave long at 150 MHz — the whole question turns on that single observation.
2. What the open stub does. An open-circuited stub presents $Z = -jZ_0\cot\beta\ell$, i.e. the shunt susceptance $Y = jY_0\tan\beta\ell$. At 300 MHz $\beta\ell = \pi$ so $\tan\beta\ell = 0$ and $Y = 0$: the stub is invisible. At 150 MHz $\beta\ell = \pi/2$ so $Y \to \infty$: the stub is a dead short across branch A.
3. What the shorted stub does. A short-circuited stub presents $Z = jZ_0\tan\beta\ell$, i.e. $Y = -jY_0\cot\beta\ell$. The roles simply exchange: at 300 MHz ($\beta\ell = \pi$) the half-wave stub repeats its own short, $Y \to \infty$, and at 150 MHz ($\beta\ell = \pi/2$) $\cot\beta\ell = 0$ and the stub vanishes. Each branch is thus short circuited at exactly one of the two frequencies: $$\text{branch A shorted at }150\ \text{MHz}, \qquad \text{branch B shorted at }300\ \text{MHz}.$$
4. Transform each short back to the junction. A short seen through a quarter-wave line looks like an open circuit, and both tap distances are exactly a quarter wave at the frequency their own stub kills: $$\frac{d_A}{\lambda_{150}} = \frac{0.50}{2.00} = \tfrac{1}{4}, \qquad \frac{d_B}{\lambda_{300}} = \frac{0.25}{1.00} = \tfrac{1}{4}.$$ Hence $Z_{\text{in},A}(150\ \text{MHz}) \to \infty$ and $Z_{\text{in},B}(300\ \text{MHz}) \to \infty$: the dead branch disconnects itself instead of loading the junction.
5. What the junction sees at each frequency. At the frequency it passes, a branch is just matched line into $50\ \Omega$, so its input impedance is $50\ \Omega$: $$Z_{\text{junction}} = 50\ \|\ \infty = \boxed{50\ \Omega \text{ at both } 300 \text{ and } 150\ \text{MHz}} .$$ The driving line therefore runs at $\text{SWR} = 1$ at both frequencies, and each signal is delivered in full to the one branch that is transparent to it.
6. Answer. Branch A is transparent at 300 MHz and short circuited at 150 MHz; branch B is the reverse. So $$\boxed{R_A \text{ receives the } 300\ \text{MHz signal}, \qquad R_B \text{ receives the } 150\ \text{MHz signal}.}$$
7. Where the rejected power goes. A lossless reactive stub cannot dissipate anything. The 150 MHz wave that reaches branch A is totally reflected at the stub and returns to the junction, where it finds branch B transparent and is delivered there instead — which is exactly why the junction still looks like a matched $50\ \Omega$ and no power is wasted. The network is a two-way diplexer, not a filter that burns the unwanted band.