Question E2A03:
How is an upload signal processed by an inverting linear transponder?  

Choices:  
A: The signal is detected and remodulated on the reverse sideband  
B: The signal is passed through a nonlinear filter  
C: The signal is reduced to I and Q components, and the Q component is filtered out  
D: The signal is mixed with a local oscillator signal and the difference product is transmitted    

Correct Answer is  D: The signal is mixed with a local oscillator signal and the difference product is transmitted  

Correct answer: D

Why D is correct
- An inverting linear transponder is a “bent-pipe” that linearly frequency-translates and amplifies the entire uplink passband without demodulating it.
- The inversion occurs when the mixer uses a local oscillator (LO) above the uplink band and the transponder selects the difference product:
  $$f_{out} = f_{LO} - f_{in}$$
- This mapping makes the output decrease as the input increases, flipping sidebands. If the uplink has components at $f_c \pm \Delta$, then:
  $$f_{out} = f_{LO} - (f_c \pm \Delta) = (f_{LO} - f_c) \mp \Delta$$
  So an uplink USB becomes a downlink LSB (and vice versa).

Why the other choices are incorrect
- A: “Detected and remodulated” implies demodulation/remodulation (a regenerative transponder). A linear transponder is transparent; it does not detect the signal.
- B: “Nonlinear filter” contradicts “linear transponder” and filtering alone does not invert the spectrum.
- C: “Reduce to I/Q and filter out Q” describes a phasing method for generating SSB, not how a linear transponder works. Dropping Q would severely distort signals.

Useful formulas and rules of thumb
- Mixer products:
  $$f_{out} = |f_{LO} \pm f_{in}|$$
- Inversion occurs when:
  $$f_{out} = f_{LO} - f_{in} \quad \Rightarrow \quad \frac{df_{out}}{df_{in}} < 0$$
  Sidebands flip: USB uplink → LSB downlink.
- Non-inverting translation (sidebands preserved) when:
  $$f_{out} = f_{LO} + f_{in} \quad \text{or} \quad f_{out} = f_{in} - f_{LO}$$