Question E9F12:
What impedance does a 1/4-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: The same as the input impedance to the generator  
C: Very high impedance  
D: Very low impedance    

Correct Answer is  D: Very low impedance  

To analyze a 1/4-wave line, use the lossless transmission-line input impedance formula:
$$Z_{\text{in}} = Z_0 \frac{Z_L + j Z_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)}$$
For a quarter wavelength, $l = \lambda/4 \Rightarrow \beta l = \pi/2$, so $\tan(\beta l) \to \infty$. This reduces to the quarter-wave transformer relation:
$$Z_{\text{in}} \approx \frac{Z_0^2}{Z_L}$$

1) Why the correct answer (D: Very low impedance) is correct
- The line is open at the far end, so $Z_L \to \infty$.
- Using $Z_{\text{in}} = Z_0^2 / Z_L$, as $Z_L \to \infty$, $Z_{\text{in}} \to 0$.
- A “0 Ω” input is a short, i.e., a very low impedance. A 1/4-wave line inverts open ↔ short.

2) Why the other choices are incorrect
- A: Same as the characteristic impedance of the line: Only true when $Z_L = Z_0$ (matched). With an open load, the 1/4-wave line transforms it to a short, not $Z_0$.
- B: Same as the input impedance to the generator: The generator’s internal impedance is unrelated. The line presents its own transformed impedance to the generator; with an open load at λ/4, that is near 0 Ω.
- C: Very high impedance: That would be true for a shorted load at the far end of a 1/4-wave line ($Z_L = 0 \Rightarrow Z_{\text{in}} \to \infty$). Here the load is open, so the result is very low.

3) Useful formulas and rules of thumb
- General input impedance (lossless line):
  $$Z_{\text{in}} = Z_0 \frac{Z_L + j Z_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)}$$
- Quarter-wave transformer:
  $$Z_{\text{in}} = \frac{Z_0^2}{Z_L}$$
- Impedance inversion at λ/4: high ↔ low, open ↔ short.
- Half-wave line repeats the load: at $l = \lambda/2$, $Z_{\text{in}} = Z_L$.