Pracitce Exam Problem Thread

[Shark 774]

Hi guys, I have some questions regarding practice exam questions and I'm sure I'll have many more to come so rather than clogging up the system, as I've been doing, I'll put them here!

Ex 1: I said LED and Photodiode, but the given answer was laser diode instead of LED. I understand that an LD has a faster response time, but for this circumstance I thought and LED would do be sufficient because a photodiode wouldn't even be able to keep up with the response time of an LD if it was modulated as fast as possible.

Ex 2: I had absolutely no idea whatsoever.
 
Ex 3 pt1: I picked the points (4.8V, 0.003V) and (4.0V, 0.00325V) and found the gradient to be (4.8 - 4.0)/(0.003-0.00325) = (0./(-0.00025) = - 3200. However they took the values at the points just before saturation/cut-off and got a value of 2457.14. What's gone wrong, because the gradient is constant so you should be able to pick any two points and get the same gain value (within the linear region)?

Ex 3 pt2: Since the amplifier is not biased and there is no mention of anything to do with biasing it, I thought the graph of the output would be the correct shape for input values of 0.8mv and above (highest input value is 1.5mv) but would be clipped below that. However in the answer it is the exact same shape, just inverted and peaks at -3.68V and +3.68V. No idea how they managed that???

I'll continue to post questions on here leading up to the exams, and feel free to do the same. If anyone can be bothered sharing their knowledge, as much detail as possible would be VERY GREATLY appreciated because my teacher is slack and makes me more confused when I ask him qusetions!!

Thanks in advance for your help guys. Much appreciated.

[Shark 774]

Ex 4: I had absolutely no idea how to answer this one. Any ideas? When you combine two waves do you use addition of ordinates or what??

Cheers -

[onur369]

Pretty sure its not in the study design. Just let someone confirm it.

[cranberry]

for EX 1, laser diodes are exactly the same as LEDs (only difference is the emitting range..somthing like 1-5nm for LD, ~30-50 for LED?..)

for EX 2, the demodulation splits the modulated signal into a carrier and information signal (attached - bottom middle of page)

[cranberry]

for Ex 3, always remember gain = the change in Vout/the change in Vin (max clip to min clip)
part 2 - negative gradient of linear region= inverting

[Shark 774]

for EX 1, laser diodes are exactly the same as LEDs (only difference is the emitting range..somthing like 1-5nm for LD, ~30-50 for LED?..)

for EX 2, the demodulation splits the modulated signal into a carrier and information signal (attached - bottom middle of page)

Thanks: Ex1, I thought laser diodes also had faster reaction times, hence would be more suitable in some situations (although you would then need an opto-electric transducer that has a faster response time than a photodiode in order to keep up with the LD, right?)

Ex2: I still don't understand how you figure out what the information signal will look like?

[Shark 774]

for Ex 3, always remember gain = the change in Vout/the change in Vin (max clip to min clip)
part 2 - negative gradient of linear region= inverting

Part 1: Ok that makes sense
Part 2: I get that it's inverted but by looking at the in/out graph there should definitely be clipping because the input is exceeding the linear region.

[cranberry]

Ex2: I still don't understand how you figure out what the information signal will look like?

i forgot the actual terms, but wherever the given signal is narrow, its a minimum turning point on the demodulated signal. wherever the signal is broadest, its a maximum on the sinusoidal signal...its in that image i attached.

laser diodes are identical to LEDs except for their range...

and easiest way to treat amplifier graphing q's when they are not zero-biased (through the origin), is to plot points.

[Shark 774]

Ex2: I still don't understand how you figure out what the information signal will look like?

i forgot the actual terms, but wherever the given signal is narrow, its a minimum turning point on the demodulated signal. wherever the signal is broadest, its a maximum on the sinusoidal signal...its in that image i attached.

laser diodes are identical to LEDs except for their range...

and easiest way to treat amplifier graphing q's when they are not zero-biased (through the origin), is to plot points.

Right ok that makes sense. That's what I did do for the graphing of the output and my answer was totally different to theirs. The answer they gave is what you would expect it to look like if it WAS biased, which is wasn't. Maybe they just stuffed up??

[Vincezor]

Right ok that makes sense. That's what I did do for the graphing of the output and my answer was totally different to theirs. The answer they gave is what you would expect it to look like if it WAS biased, which is wasn't. Maybe they just stuffed up??

I'm guessing you use the characteristics from the graph in question 5, right? If so, then wouldn't the output voltage be off the graph?  like it'll be clipped at 10V for a lot of the time? But yes, I've seen incorrect answers for this whereby the answer is literally just the input voltage * gain, not taking into account any clipping

[Shark 774]

Right ok that makes sense. That's what I did do for the graphing of the output and my answer was totally different to theirs. The answer they gave is what you would expect it to look like if it WAS biased, which is wasn't. Maybe they just stuffed up??

I'm guessing you use the characteristics from the graph in question 5, right? If so, then wouldn't the output voltage be off the graph?  like it'll be clipped at 10V for a lot of the time? But yes, I've seen incorrect answers for this whereby the answer is literally just the input voltage * gain, not taking into account any clipping

Yes I agree. I think that's what they did; they just used the gain value and didn't consider the fact that the amp is not biased.

[pi]

Ex 4: I had absolutely no idea how to answer this one. Any ideas? When you combine two waves do you use addition of ordinates or what??

Cheers -

I remember this problem! Best way is to remember two things:
1. the period of the modulated wave is the same as the 2nd wave (the one with a fixed period and huge amplitude)
2. use the first dodgy wave as endpoints (upper endpoints or lower endpoints, just pick one) for the other wave when sketching


Weird question though...

[Shark 774]

Ex 4: I had absolutely no idea how to answer this one. Any ideas? When you combine two waves do you use addition of ordinates or what??

Cheers -

I remember this problem! Best way is to remember two things:
1. the period of the modulated wave is the same as the 2nd wave (the one with a fixed period and huge amplitude)
2. use the first dodgy wave as endpoints (upper endpoints or lower endpoints, just pick one) for the other wave when sketching


Weird question though...

Thanks, but what do you mean by upper endpoints or lower endpoints? In the answer they put the peak of the dodgy wave on top of the first maximum of the second wave. Don't know why at all...