Question E9F11:
What impedance does a 1/8-wavelength transmission line present to an RF generator when the line is open at the far end?  

Choices:  
A: The same as the characteristic impedance of the line  
B: An inductive reactance  
C: A capacitive reactance  
D: Infinite    

Correct Answer is  C: A capacitive reactance  

Correct answer: C — A capacitive reactance

Why this is correct:
- For a lossless line of length $l$ terminated in an open circuit, the input impedance is
  $$Z_\text{in} = -j Z_0 \cot(\beta l)$$
  where $\beta = \frac{2\pi}{\lambda}$.
- With $l = \frac{\lambda}{8}$, we have $\beta l = \frac{\pi}{4}$ and $\cot\left(\frac{\pi}{4}\right)=1$, so
  $$Z_\text{in} = -j Z_0$$
- The negative imaginary sign indicates a capacitive reactance. Therefore, an open-circuited 1/8-wave line looks like a capacitor to the generator.

Why the other choices are incorrect:
- A: “The same as the characteristic impedance of the line” — $Z_0$ is real (resistive), but $Z_\text{in} = -j Z_0$ is purely reactive. Although the magnitude equals $Z_0$, the impedance is not resistive, so this is wrong.
- B: “An inductive reactance” — Inductive would be positive imaginary ($+jX$). Here we have $-jZ_0$ (capacitive), so this is wrong.
- D: “Infinite” — An infinite input impedance occurs when $l = n\frac{\lambda}{2}$ for an open circuit. At $l=\frac{\lambda}{8}$, it is finite and reactive.

Useful formulas and rules of thumb:
- General input impedance:
  $$Z_\text{in} = Z_0 \frac{Z_L + jZ_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)}$$
- For special terminations:
  - Open circuit ($Z_L \to \infty$): $$Z_\text{in} = -j Z_0 \cot(\beta l)$$
  - Short circuit ($Z_L = 0$): $$Z_\text{in} = j Z_0 \tan(\beta l)$$
- Stubs:
  - Open-circuited stub: shorter than $\lambda/4$ behaves capacitive; between $\lambda/4$ and $\lambda/2$ behaves inductive.
  - Short-circuited stub: shorter than $\lambda/4$ behaves inductive; between $\lambda/4$ and $\lambda/2$ behaves capacitive.