Question E9F10:
What impedance does a 1/8-wavelength transmission line present to an RF generator when the line is shorted at the far end?  

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

Correct Answer is  C: An inductive reactance  

Correct answer: C — An inductive reactance

Why C is correct:
- For a lossless line, the input impedance is:
  $$Z_\text{in} = Z_0 \frac{Z_L + j Z_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)}$$
- With a short at the far end, $Z_L = 0$, so:
  $$Z_\text{in} = j Z_0 \tan(\beta l)$$
- For $l = \lambda/8$, $\beta l = 2\pi/\lambda \cdot \lambda/8 = \pi/4$, and $\tan(\pi/4) = 1$, so:
  $$Z_\text{in} = j Z_0$$
- This is a purely inductive reactance (positive imaginary), with magnitude equal to $Z_0$.

Why the others are incorrect:
- A: A capacitive reactance — Not for a shorted stub of length less than $\lambda/4$. A shorted stub is inductive for $0 < l < \lambda/4$.
- B: The same as the characteristic impedance of the line — The magnitude equals $Z_0$, but it is not resistive; it is purely reactive ($j Z_0$), so it’s not “the same as $Z_0$” in the usual resistive sense.
- D: Zero — A shorted line looks like a short (zero impedance) only at lengths $l = n\lambda/2$. At $l=\lambda/8$ it is not zero.

Useful formulas and rules of thumb:
- Shorted stub: $$Z_\text{in} = j Z_0 \tan(\beta l)$$
  - Inductive for $0 < l < \lambda/4$
  - Infinite (open) at $l = \lambda/4$
  - Capacitive for $\lambda/4 < l < \lambda/2$
- Open stub: $$Z_\text{in} = -j Z_0 \cot(\beta l)$$
  - Capacitive for $0 < l < \lambda/4$
  - Zero (short) at $l = \lambda/4$