Length of a short-circuited stub that matches at 300 MHz and isolates the load at 600 MHz: 22-Elec-A7 Electromagnetics, December 2017, Q2

Question 2: Length of a short-circuited stub that matches at 300 MHz and isolates the load at 600 MHz

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 line already matched by its own load carries one shunt short-circuited stub of the same characteristic impedance.

Given data
Quantity	Symbol	Value
Characteristic impedance (line and stub)	$Z_0$	$50\ \Omega$
Propagation velocity	$v_p$	$3 \times 10^{8}\ \text{m/s}$
Load resistor	$R_L$	$50\ \Omega$
Frequency at which the match must survive	$f_1$	$300\ \text{MHz}$
Frequency at which the load must be isolated	$f_2$	$600\ \text{MHz}$
Stub termination	—	short circuit

Find. The stub length $\ell$ that leaves the 300 MHz match untouched while cutting the load off completely at 600 MHz.

Figure 2.1 — the matched line with one shunt short-circuited stub. Its position along the line is not constrained by either requirement.

Approach. Because the load already equals $Z_0$, the line is matched with no stub at all; so the stub must be invisible (an open circuit) at 300 MHz and a dead short at 600 MHz. Turn each requirement into a condition on $\tan\beta\ell$ and intersect the two families.

Write the two wavelengths. $\lambda_1 = v_p/f_1 = (3 \times 10^{8})/(300 \times 10^{6}) = 1.00\ \text{m}$ and $\lambda_2 = v_p/f_2 = 0.50\ \text{m}$.
State the stub's input impedance. A lossless line of length $\ell$ terminated in a short presents$$Z_{stub} = jZ_0 \tan\beta\ell , \qquad \beta = \frac{2\pi}{\lambda}$$Note that $Z_0$ multiplies the whole expression, so neither requirement below will depend on it — which is exactly why the paper can specify a stub "identical with the driving line" without that mattering.
Impose the match at 300 MHz. The load already terminates the line in $Z_0$, so the match is preserved only if the stub draws no current at all, i.e. it must look like an open circuit. That needs $\tan\beta\ell \to \infty$:$$\beta_1\ell = \frac{\pi}{2} + n\pi \quad\Longrightarrow\quad \ell = (2n+1)\,\frac{\lambda_1}{4} = (2n+1) \times 0.25\ \text{m}$$
Impose isolation at 600 MHz. To stop power reaching the load the stub must short the line out, clamping the voltage there to zero. That needs $\tan\beta\ell = 0$:$$\beta_2\ell = m\pi \quad\Longrightarrow\quad \ell = m\,\frac{\lambda_2}{2} = m \times 0.25\ \text{m}$$
Intersect the families. Because $f_2 = 2f_1$ we have $\lambda_1/4 = \lambda_2/2 = 0.25\ \text{m}$, so the odd multiples of $0.25\ \text{m}$ satisfy both conditions at once. The shortest such stub is$$\boxed{\ell = 0.25\ \text{m} = 25\ \text{cm}}$$with the further admissible lengths $0.75\ \text{m}$, $1.25\ \text{m}, \ldots$, i.e. $\ell = (2n+1) \times 0.25\ \text{m}$.
Verify both bands. At 300 MHz the electrical length is $\beta_1\ell = 2\pi(0.25)/1.00 = 90^\circ$: a quarter-wave shorted line, which indeed appears open. At 600 MHz it is $\beta_2\ell = 2\pi(0.25)/0.50 = 180^\circ$: a half-wave shorted line, which repeats its own short. So the junction reflection coefficient goes from $0$ at 300 MHz to $-1$ at 600 MHz.
Say where the blocked power goes. A reactive stub cannot dissipate anything, so the 600 MHz component is reflected back toward the generator rather than absorbed. Neither condition involves the stub's position, so it may be placed anywhere between load and generator; the position only sets what reactance the generator sees at 600 MHz, through $Z_{in} = jZ_0\tan\beta_2 s$ for a spacing $s$.

Figure 2.2 — electrical length of the 25 cm stub against frequency. It crosses 90° (looks open) at 300 MHz and 180° (looks short) at 600 MHz.

Final results
Quantity asked	Result
Shortest stub length	$\ell = 0.25\ \text{m} = 25\ \text{cm}$
Complete family of solutions	$\ell = (2n+1) \times 0.25\ \text{m}$
Spacing between admissible lengths	$0.50\ \text{m}$
Electrical length at 300 MHz	$90^\circ$ — stub looks open
Electrical length at 600 MHz	$180^\circ$ — stub looks short
Junction reflection coefficient at 600 MHz	$\Gamma = -1$ (total reflection)
Dependence on $Z_0$	none — the design is pure length ratios

Topic. A single shunt stub used as a frequency-selective switch on an already matched line.

Key relations. Shorted stub $Z_{stub} = jZ_0\tan\beta\ell$ (open at odd multiples of $\lambda/4$, short at multiples of $\lambda/2$); open stub $Z_{stub} = -jZ_0\cot\beta\ell$ (the two families swap); $\beta = 2\pi f/v_p$; a zero impedance across the line gives $\Gamma = -1$.

Why this works. Once the load equals $Z_0$ there is nothing left to match, so the design problem inverts: instead of choosing a susceptance to cancel a mismatch, the stub must contribute zero susceptance in one band and infinite susceptance in the other. Both conditions are conditions on electrical length alone, and because electrical length is proportional to frequency, one physical length can satisfy both whenever the two frequency ratios line up. The harmonic relation $f_2 = 2f_1$ is what makes the quarter-wave family at $f_1$ coincide with the half-wave family at $f_2$.

Common pitfall. The commonest slip is to reason that "isolation needs a quarter-wave stub" and put $\ell = \lambda_2/4$: a quarter-wave shorted stub looks open, which is the opposite of what isolation requires. A second slip is to treat the reflected 600 MHz power as dissipated in the stub — a purely reactive element cannot absorb power, so it all returns to the source.

Source. Pozar, Microwave Engineering 4th ed., §2.5 (the Smith chart and stub tuning); Sadiku, Elements of Electromagnetics 7th ed., ch. 11.

Question 2 of 8: Length of a short-circuited stub that matches at 300 MHz and isolates the load at 600 MHz

Question 2: Length of a short-circuited stub that matches at 300 MHz and isolates the load at 600 MHz