a few photonics questions

[Shark 774]

Just these questions confused me a bit. For the first one (attached) the answer says that it can't be negative because the diode won't let a current go backwards, however it's a photodiode so won't it let a current through both ways? For the second one (also attached) I had absolutely no idea what each wave would look like... Any tips or hints? And for those who have checkpoints I got incorrect answers for Q415, 416, 435 and 452/455 (didn't think they should be clipped?) and I am not sure why I got them wrong; some help with these would be greatly appreciated as my teacher is a bit dodge!

[xZero]

If the voltage is negative, its saying that the photo diode sends a signal back to the laser diode, which doesnt make sense coz photo diode only receives and laser diode only transmits signal. 

ROFL i did the second question for my sac Q is the carrier signal, which is the 1 with the highest frequency and not modulated, so its A. The signal at P should be resembling a human voice wave, which will be B. R is the modulated wave since its in the transmission line, so its C. S should be B since its passed the demodulator.

[Vincezor]

For the first one, is the answer D?

A and C are wrong as a laser diode emits light, not voltage.

B is wrong because brightness cannot be below zero.

For the second picture:

P - B
Q - A
R - C
S - B

We know that C is definitely the modulated signal due to it being the combination of the A and B, so that will be at R.

The carrier wave's frequency must be higher than the input signal and this comes from the RF Source.

P is the input signal (the dude singing into the mic or whatever) and S is the output signal (The voice signal sent to the speakers)

[Shark 774]

For the first one, is the answer D?

A and C are wrong as a laser diode emits light, not voltage.

B is wrong because brightness cannot be below zero.

For the second picture:

P - B
Q - A
R - C
S - B

We know that C is definitely the modulated signal due to it being the combination of the A and B, so that will be at R.

The carrier wave's frequency must be higher than the input signal and this comes from the RF Source.

P is the input signal (the dude singing into the mic or whatever) and S is the output signal (The voice signal sent to the speakers)

C is the correct answer, not D because point Y is a point in the circuit, hence the change is in voltage not brightness (the voltage changes because of a change in brightness)

[Vincezor]

But isn't the question asking for the graph at point X?

[cranberry]

alth

But isn't the question asking for the graph at point X?

yeah but isn't it in optical/light form as a signal? therefore "carries" the input signal's voltage..

[Shark 774]

But isn't the question asking for the graph at point X?

Ah shit yeah... oops. Well then yeah I agree with you in saying D, however the correct answer was C. Not sure why though......

[Vincezor]

Just checked the 2007 assessment report. Answer says D.

I'm guessing you encountered that question in checkpoints? I remember seeing that there when I did it a while ago...

@Cranberry

Isn't the voltage proportional to the light intensity? so the light pulses from the laser diode would eventually be converted back to the voltage when it is demodulated...

[Shark 774]

Just checked the 2007 assessment report. Answer says D.

I'm guessing you encountered that question in checkpoints? I remember seeing that there when I did it a while ago...

@Cranberry

Isn't the voltage proportional to the light intensity? so the light pulses from the laser diode would eventually be converted back to the voltage when it is demodulated...

Yeah I got it from checkpoints. They are so f***ing dodgy!!!