TrueTears question thread

[TrueTears]

Cool thanks kurrymuncher!

And yeah when it converts the light, does it just remove the carrier wave (light) so all we are left is just the original electrical signal? Hence the term "demodulation"

[kurrymuncher]

Yep. The carrier wave is removed and the photodiode changes the signal into a current. This photodiode will be connected to a circuit, therefore the current  from the signal will cause a variation in Vout, and somehow they can detect the original signal.

I dont think that makes sense

also,did you know that a machine that can modulate and demodulate a signal is called a Modem. lol

[TrueTears]

Yep. The carrier wave is removed and the photodiode changes the signal into a current. This photodiode will be connected to a circuit, therefore the current  from the signal will cause a variation in Vout, and somehow they can detect the original signal.

I dont think that makes sense

also,did you know that a machine that can modulate and demodulate a signal is called a Modem. lol

You are one mad dog








changes to TT female version <3

[Mao]

For the modulation process at the LED:
LED is a diode, once reached the activation voltage, power output varies with current. That is, the electric signal (current variation) loses some energy (voltage) to release photons at the metal surface on the LED. The intensity of the light emitted is proportional to the power output hence current, i.e. the peaks in electric signal becomes peaks in light intensity. the carrier wave is amplitude-modulated.

At the photodiode (reverse biased), photons provide energy for the electrons to cross the depletion region. Hence the current is proportional to the number of photons (i.e. intensity of light). Peaks in light intensity becomes peaks in current, thus the signal is demodulated.

I'm not keen on using phrases such as 'carrier wave is removed' and things of that like.

[TrueTears]

Ahhh, thanks Mao!

[TrueTears]

Hey guys just another questions, this time on the detailed study "Structures and Materials"

We had to do a Practical SAC to work out the Young's Modulus of a fishing string. I have attached my data and everything that is required is in it.

The question I want to know is how do I work out the percentage error for each variable? Such as Force, New length, change in L, stress, strain. The ruler I used to measure the original length of string is just your normal standard ruler.

Our teacher has never taught us how to do it but he says you must put the percentage error for each variable, how do you work that out?

Many thanks!

[Over9000]

I think like this

U get original length of string is 150mm say for example
And ruler manufacturer states accurate
so then you have
So that's your error for that.
Then you have the weights we used
Say manufacturer states accurate
Where each was 100
so you have
Then u add both to have 5.33% error margin for those 2 factors

[TrueTears]

ahhh okay thanks over9000, how do we know the uncertainty for the certain equipments? It was never stated on them or anything :S

[Over9000]

ahhh okay thanks over9000, how do we know the uncertainty for the certain equipments? It was never stated on them or anything :S

Ask your teacher

[TrueTears]

This a linear optoisolator. The current travelling through the circuit with the LED is denoted by and the graph is given for against time. The current travelling through the circuit with the photodiode is denoted by , the LED current is proportional to the PD current . The question wants me to sketch the graph of the against time, which i know how to do.

I'm just wondering what does the graph of mean from time 10 - 30 ns, why does the current suddenly drop down to 10 mA for 20 ns and then go back up? And what's causing it?

Many thanks!

[TrueTears]

^^^ Thanks Mao

A few new questions:

1. A rower on the Yarra is able to reach a velocity of 10 relative to the water. An otter hitching a ride on a log, floats along in the water (at rest relative to the water). The water is moving at a velocity of 2 relative to the river bank.
a) With what velocity is the rower moving relative to the otter when rowing upstream?
b) What is the velocity of the rower relative to the riverbank?
c) What is the velocity of the otter relative to the rower?
d) What is the velocity of the rower relative to the riverbank when rowing downstream?

2. When Newton published his laws of motion he identified some assumptions about space and time that he assumed to be true and that underpin all his laws. Which one or more of the following would best describe his assumptions?
A. Time passes uniformly for observers in any frame of reference
B. The measurement of space is always relative to the observer's frame of reference
C. Time is not the same for all observers at very high speeds
D. Space is absolute therefore 1 m will be 1 m for all observers despite being in different reference frames
E. Time can only be measured by a clock
F. Distance can only be measured by a ruler

Many thanks!

[kamil9876]

1.)

a.)The otter "floats" which implies that it is stationary relative to the water. Hence we will pick a reference frame where the water is stationary. Relative to the water the guy rows at 10m/s. Hence in our reference frame the otter has speed of 0, while the rower has 10m/s. Hence 10m/s.

b.) The answer could be 8m/s or 12m/s, depending on whether he is rowing in the direction to the flow of water or against it. I think we are to assume that he is going upstream since part a says so. Hence 8m/s. This is easily done with vector addition. In one second he would travel 10m in the river's frame however because the river would move 2m back in that one second the displacement in the frame of the riverbank is 10+(-2)=8m in one second.

c.) 10m/s. (refer to part a)

d.) 12m/s as explained in part c.

[kamil9876]

2.) Issac says:

A,B(depends on what u mean by "space"),D.

B - the measurement of distance isn't any different in different frames, however the measurement of space co-ordinates is (and I think Galilleo made this explicit before Issac, hence "gallilean relativity").
Imagine the riverbank scenario again, the guy may be on the boat and say his bag is at co-ordinates (1,0,0) in his frame. In the riverbanks frame say the co-ordinates of the bag are the same at t=0, (1,0,0). However because the bag is moving at 12m/s after one second the bag will be at (13,0,0) in the river bank's frame. However in the rower's frame it is still at (1,0,0)

[/0]

I think B is the opposite of D.

[kamil9876]

that again depends on what i said... whether you are refering to space co-ordinates or distance between co-ordinates. co-ordinates themselves are relative, distance between them is not. How else would we measure two different velocities of the same object from different reference frames if velocity comes from space and time only and time is the same in newton