Question 2 of 8: Shorted-Stub Length for Match and Isolation

Question 2: Shorted-Stub Length for Match and Isolation (equal value)

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. Main line and stub both have $Z_0=50\ \Omega$, $v=3\times10^8$ m/s; the main line is terminated in $Z_0$ (matched, no reflection of its own); a shorted stub of length $l$ is connected in shunt across the line; frequencies of interest $f_1=300$ MHz (want the tap "matched", i.e. the stub invisible) and $f_2=600$ MHz (want the tap to "isolate" the load, i.e. the stub shorts the line).

Find. The shortest stub length $l$ that presents an open circuit (invisible, load stays matched) at $f_1$ and a short circuit (isolates everything downstream) at $f_2$.

Approach. A shorted stub of length $l$ has input reactance $Z_{in}=jZ_0\tan(\beta l)$; being "invisible" in shunt requires $Z_{in}\to\infty$ (open), being "isolating" requires $Z_{in}=0$ (short). Solve both conditions and find the smallest common $l$.

1. Wavelengths at the two frequencies. $$\lambda_1=\frac{v}{f_1}=\frac{3\times10^8}{300\times10^6}=1\ \text{m},\qquad \lambda_2=\frac{v}{f_2}=\frac{3\times10^8}{600\times10^6}=0.5\ \text{m}.$$
2. Open-circuit (matched, invisible) condition at $f_1$. $\tan(\beta_1 l)\to\infty$ requires $\beta_1 l=\pi/2+n\pi$, i.e. $$l=(2n+1)\frac{\lambda_1}{4},\qquad n=0,1,2,\dots$$
3. Short-circuit (isolating) condition at $f_2$. $\tan(\beta_2 l)=0$ requires $\beta_2 l=m\pi$, i.e. $$l=m\,\frac{\lambda_2}{2},\qquad m=0,1,2,\dots$$
4. Smallest length satisfying both. With $\lambda_1=1$ m and $\lambda_2=0.5$ m, both families reduce to multiples of 0.25 m: the first condition gives $l=0.25,0.75,1.25,\dots$ (odd multiples of $\lambda_1/4$) and the second gives $l=0,0.25,0.5,0.75,\dots$ (any multiple of $\lambda_2/2$). The smallest length common to both (smallest odd multiple of 0.25 m) is $$\boxed{l=\frac{\lambda_1}{4}=0.25\ \text{m}=25\ \text{cm}}.$$ Check: at $f_1$, $\beta_1 l=(2\pi/1)(0.25)=\pi/2\Rightarrow Z_{in}\to\infty$ (open, matched); at $f_2$, $\beta_2 l=(2\pi/0.5)(0.25)=\pi\Rightarrow Z_{in}=0$ (short, isolates). Both requirements are met.