quick physics question

[iPie]

Okay so whilst revising for my A level physics exam tomorrow, I was doing some practice questions on my college website.
I did this question:

A signal of strength 0.2 Volts has a noise variation of 2 microvolts. How many bits should be used to digitise it?

using the following method:

0.2/(2x10^-6)=100,000
16 bits would give 65,536 levels and 17 bits would give 131,072 levels.
=> 17 bits

This was my answer which I thought was right but the quiz said I was wrong and should have chosen 16 bits?? am I going mad or is my original answer right?

(It was a multiple choice question with answers: 16, 17, 100000, and 8 bits, and gave 16 as the right answer.)

[Jim Nazium]

It depends if they are solid state volts or tube volts.

[iPie]

Originally Posted by Jim Nazium

It depends if they are solid state volts or tube volts.

I'm not sure if this is a joke I'm either very unknowledgeable about voltage or very gullible

[i_got_a_mohawk]

Got any links to the course work? We didn't do ADA in my day and I work in the photonics side of things.

I've been trying to find what we'd use to calculate the required bit depth, but I'm not having any luck :P

[Pilgrim]

I respect those who hold the mystic knowledge required to answer this question.

[Jimbob Jones]

EDIT: sorry; misread the OP.

The highest resolution you can measure to without the noise affecting your readings is 16 bits.

[LSquared]

You want your A/D resolution to be less than the signal noise. I'd say 16 bits

[iPie]

thanks for the replies Seems I was wrong, cleared this up for me now if i get an A in my exam i shall be thanking you guys

[i_got_a_mohawk]

Originally Posted by LSquared

You want your A/D resolution to be less than the signal noise. I'd say 16 bits

Of course *facepalm*, I should have thunk of that.

[capnjim]

42

[rr5025]

Wow we never did this in my physics courses, however I do understand the reasoning behind the answer (though that formula was complete magic to me haha).

[i_got_a_mohawk]

Originally Posted by rr5025

Wow we never did this in my physics courses, however I do understand the reasoning behind the answer (though that formula was complete magic to me haha).

The signal noise is 1/100,000th of the signal.

To reproduce the signal as clearly as possible, you want to use as many points as possible, in a digital sense. But you don't want to reproduce the noise in the analogue signal.

So if the signal is being digitised with points that are larger than the given noise (in 16 bit, the signal is split into 65,536 different points).

So the digital parts of the signal would be said to be in 1/65,536 th's of the whole. As those are larger than the threshold of noise, you don't have bits which are specifically representing the noise level.

(I've written that terribly poorly, but, I hope it kinda explains what's going on, the equation was simply (Signal Strength/Noise, giving 100,000) )

[fdeck]

I'm skeptical. Here's my answer:

Log2(100000) = 16.6 bits, so choose a 17-bit converter for the best possible overall system signal-to-noise.

Of course this is without knowing the statistical distribution of the 2 uV noise level. Is it Gaussian and specified in RMS terms, or something else? And to make matters worse, the 16-bit converter probably has a dynamic range of worse than 16 bits.

All of this is why multiple-choice exams have no place in a physics class. Note that my day job is as a physicist for a company that makes electronic instrumentation.

[i_got_a_mohawk]

Originally Posted by fdeck

All of this is why multiple-choice exams have no place in any class

Fixed it.

I hate multiple choice, it's nothing more than lazyness on the part of the examiners/assessors.

[aborgman]

Originally Posted by fdeck

I'm skeptical. Here's my answer:

Log2(100000) = 16.6 bits, so choose a 17-bit converter for the best possible overall system signal-to-noise.

Of course this is without knowing the statistical distribution of the 2 uV noise level. Is it Gaussian and specified in RMS terms, or something else? And to make matters worse, the 16-bit converter probably has a dynamic range of worse than 16 bits.

All of this is why multiple-choice exams have no place in a physics class. Note that my day job is as a physicist for a company that makes electronic instrumentation.

Yeah, that.

[Jimbob Jones]

Forgive me if I'm being dense, but I thought you always, always round down when working in digital? So you'd go with 16bit, even if the equation returned 16.9999...

[i_got_a_mohawk]

Originally Posted by fdeck

I'm skeptical. Here's my answer:

Log2(100000) = 16.6 bits, so choose a 17-bit converter for the best possible overall system signal-to-noise.

Of course this is without knowing the statistical distribution of the 2 uV noise level. Is it Gaussian and specified in RMS terms, or something else? And to make matters worse, the 16-bit converter probably has a dynamic range of worse than 16 bits.

This is a high school exam, they'll be talking in terms of ideals and keep it as simple as possible. I think they'll also be looking for reproduction as clearly as possible, without reproducing the noise.

Speculating tho, we didn't do ADC when I was at high school and my physics has been nearly all photonics and materials.

[DwaynieAD]

did someone say physics

YouTube - ‪Antigravity Cat-Toast Device‬‏

[i_got_a_mohawk]

[Hoover]

Originally Posted by rr5025

Wow we never did this in my physics courses

^^^This.

That's what they're teaching in high school physics? Wow. I mean, I'm aware that Time Marches On and all that, but that seems more like an electronics or DSP question. I guess high school physics now includes electronics and/or DSP? Man, when I was in high school it was mostly...

oh, wait, I never took physics in high school.

Nevermind.