ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Shark 774 on May 18, 2011, 07:25:19 pm
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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!
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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 :P 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.
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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)
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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)
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But isn't the question asking for the graph at point X?
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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..
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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......
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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...
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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!!!