Shorted stub that passes 300 MHz and blocks 400 MHz: 22-Elec-A7 Electromagnetics, December 2015, Q2

Question 2: Shorted stub that passes 300 MHz and blocks 400 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 matched 50 Ω system carries two separate frequency components, and a short-circuited stub of the same line is hung in parallel with the load to pass one and stop the other.

Given data
Quantity	Symbol	Value
Generator internal resistance	$R_g$	$50\ \Omega$
Load resistance	$R_L$	$50\ \Omega$
Line and stub impedance	$Z_0$	$50\ \Omega$
Propagation velocity	$v_p$	$3\times10^{8}\ \text{m/s}$
Component to be delivered	$f_1$	$300\ \text{MHz}$
Component to be blocked	$f_2$	$400\ \text{MHz}$

Find. The length $l$ of the short-circuited stub that leaves the 300 MHz match undisturbed while preventing the 400 MHz component from reaching the load.

Figure Q2. Shunt short-circuited stub of the same line hung across the load plane. The single length must look like an open circuit at 300 MHz and like a short circuit at 400 MHz.

Approach. Translate each requirement into a condition on the stub's input impedance, convert each condition into a family of lengths measured in that frequency's own wavelength, and find the shortest length common to both families.

Write the stub's input impedance. A lossless line of length $l$ terminated in a short circuit presents $$Z_{\text{stub}} = jZ_0\tan\beta l, \qquad \beta = \frac{2\pi}{\lambda} = \frac{2\pi f}{v_p}$$ It is purely reactive, so it can only add susceptance in parallel with the load — it can never absorb power itself.
State what “maintain the match” demands. The load already equals $Z_0$, so its normalised admittance is already 1. Any extra shunt susceptance would spoil that, so at 300 MHz the stub must contribute nothing: it must look like an open circuit, $Z_{\text{stub}}\to\infty$, i.e. $\tan\beta l\to\infty$, i.e. $$\beta l = \frac{\pi}{2} + n\pi \;\Longrightarrow\; l = (2n+1)\frac{\lambda_1}{4}$$
State what “prevent the 400 MHz from reaching the load” demands. To starve the load the stub must short it out, not open it: with $Z_{\text{stub}} = 0$ across the load terminals, the load voltage is forced to zero and no power at that frequency enters $R_L$. That requires $\tan\beta l = 0$, i.e. $$\beta l = m\pi \;\Longrightarrow\; l = m\frac{\lambda_2}{2}$$
Put in the two wavelengths. With $v_p = 3\times10^{8}\ \text{m/s}$, $$\lambda_1 = \frac{v_p}{f_1} = \frac{3\times10^{8}}{3\times10^{8}} = 1.00\ \text{m}, \qquad \lambda_2 = \frac{v_p}{f_2} = \frac{3\times10^{8}}{4\times10^{8}} = 0.75\ \text{m}$$ so the two families are $l \in \{0.25,\ 0.75,\ 1.25,\ 1.75,\ \ldots\}\ \text{m}$ (match at 300 MHz) and $l \in \{0.375,\ 0.75,\ 1.125,\ 1.50,\ \ldots\}\ \text{m}$ (block at 400 MHz).
Intersect the two families. Setting $(2n+1)\lambda_1/4 = m\lambda_2/2$ and cancelling gives $2(2n+1) = 3m$, whose smallest solution is $n=1,\ m=2$: $$\boxed{l = 0.75\ \text{m}}$$ which is $3\lambda_1/4$ at 300 MHz and exactly one full wavelength $\lambda_2$ at 400 MHz — a full-wave shorted stub is itself a short circuit, and a three-quarter-wave shorted stub is an open circuit, so both requirements are met by the same piece of cable.
Give the general family. The two conditions repeat with periods $\lambda_1/2 = 0.50\ \text{m}$ and $\lambda_2/2 = 0.375\ \text{m}$, whose least common multiple is 1.50 m, so every acceptable length is $$l = 0.75 + 1.50k\ \text{m}, \qquad k = 0,1,2,\ldots$$ The shortest is the practical answer; longer members waste cable and narrow the usable bandwidth around each design frequency.

It is worth noticing what the stub does not do. It never dissipates the blocked component; it reflects it back toward the generator, where the matched 50 Ω source resistance absorbs it. A reactive stub can only redistribute power, never consume it.

Final results
Quantity	Value
Wavelength at 300 MHz, $\lambda_1$	$1.00\ \text{m}$
Wavelength at 400 MHz, $\lambda_2$	$0.75\ \text{m}$
Condition at 300 MHz	stub open, $l=(2n+1)\lambda_1/4$
Condition at 400 MHz	stub short, $l=m\lambda_2/2$
Shortest stub length	$l = 0.75\ \text{m}$ ($=3\lambda_1/4 = 1\lambda_2$)
General family	$l = 0.75 + 1.50k\ \text{m}$

Topic. Reactive shunt stubs as frequency-selective elements on a matched line.

Key relations. Shorted stub: $Z=jZ_0\tan\beta l$ (open at $l=(2n{+}1)\lambda/4$, short at $l=m\lambda/2$). Open stub: $Z=-jZ_0\cot\beta l$ (the two roles swap). $\beta=2\pi f/v_p$.

Why this works. A stub of fixed physical length is seen through a different electrical length at every frequency, and that is the entire mechanism of the filter. Here the load is already matched, so the design reduces to demanding an open circuit at the wanted frequency and a short circuit at the unwanted one. Both are extremes of the same tangent function, separated by a quarter wavelength, so a single length can satisfy both whenever the frequency ratio makes the two quarter-wave grids commensurate — which the 3:4 ratio here does.

Common pitfall. Reaching for a short circuit to “pass” and an open circuit to “block” — the reverse of the truth. A shunt element starves its load when it is a short (it steals all the current) and is invisible when it is an open. The other frequent slip is using one wavelength for both conditions; each condition must be written in its own frequency's $\lambda$ before the two are intersected.

Source. Pozar, Microwave Engineering, 4th ed., §2.3 and §5.1 (shunt-stub matching); Sadiku, Elements of Electromagnetics, 7th ed., §11.5.

Question 2 of 8: Shorted stub that passes 300 MHz and blocks 400 MHz

Question 2: Shorted stub that passes 300 MHz and blocks 400 MHz (20 marks)