Bobthelost
Diamond Member
- Dec 1, 2005
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Originally posted by: smack Down
Besides the sampling at zero problem sampling a frequency at 2x its frequency will result in a sampled singal that has twice the amplitude as the orginial.
It will?
Originally posted by: smack Down
Besides the sampling at zero problem sampling a frequency at 2x its frequency will result in a sampled singal that has twice the amplitude as the orginial.
Originally posted by: Bobthelost
The higher frequency sine waves get attenuated more than the lower frequency sine waves. Look at the diagram provided at one of the sites you linked to: http://members.aol.com/ajaynejr/nyquist.htm. Look at the difference between the left hand diagram and the one on the right, attenuation.
Originally posted by: Bobthelost
Edit: Don't delete it! Crossposting happens and some/all the questions were good ones!
I hate it when people ask for proofs, but such is life...
Ok, example 1 you have a sine wave, it is a nice sine wave, all the curvy bits in the right places and it crosses the x axis exactly twice per cycle. As i said, it's a good sine wave. Now you decide to sample aforementioned wave, so you take it's frequency (Y hz) and then double it to furfill Nyquist's theorem (2Y hz). However there's a problem, you're sampling at 180 degrees and 360 degrees, or in other words at the zero crossing points.
Your signal has just made like elvis and left the building.
Of course this is slightly silly, you'd have to sample the signal at exactly the right phase to zero it all out, (on the flip side unless you're sampling at the exact right place you'll never get the max amplitude of the signal either. So your signal will be attenuated)
Remember our lovely little sine wave? Well now move the sampling points a tiny bit earlier, it's now sampling just before the wave crosses the axis (say 179 degrees and 359). The signal is back, but attenuated to near death. If you're just listening for a pure sine wave then that's ok, but if that sine wave is part of your latest britney collection then the higher frequency notes are going to be too quiet. So they will get lost compared to the lower freq stuff.
Ok, so at nyquist's freq we can lose the signal completely, or it can be attenuated so badly as to ruin the music, (remembering my example this may be a goal in itself). What happens if we increase the sampling frequency a tiny bit? Now the first sample is taken on a crossing point, the second one just next to a crossing point, the third slightly further away and so on. In other words you'll get the right frequency output, but the amplitude will be increasing as the sample points get closer to the peaks of the waves, and then smaller as they move further away. So britney is warbling a nice little solo bit, but the higher tones are even more unstable than a drunk on a tightrope.
Or what i've refered to as the AM bit.
Now if you want mathematical proof for all of this then i'm going to sit in the corner and sulk.![]()
Nice try, but let's count the amount of samples you are taking. Say we have a sine wave at 1hz. And as you said, say we sample at 180degrees and 360 degrees. We get all zero points and no way of figuring out what the amplitude of the 1hz sine wave is. But we only took 2 samples over a period of 1 second; we only sampled at 2 hz, which was not greater than the nyquist rate.
Try sampling at 2.2 Hz
Originally posted by: Mday
um, the MINIMUM nyquist frequency is 2x the max freq in a signal. you will need a PERFECTLY square filter to do any type of filtering with the sample. The reason why more than 2x is necessary is because a perfect square filter does not exist. And attenuation is a bad thing (which is the best any real square filter can do).
Originally posted by: Tick
Infinite sampling does not equal analogue. Same way a reimann sum is not an integral, their just equivelent.
Originally posted by: Peter
You just can't argue those things with people who don't grasp the concept of infinity. See also: 0.99999... = 1.
Originally posted by: Tick
Infinite sampling does not equal analogue. Same way a reimann sum is not an integral, their just equivelent.
Originally posted by: Navid
the samples will never be close enough together
Originally posted by: Peter
Originally posted by: Peter
You just can't argue those things with people who don't grasp the concept of infinity. See also: 0.99999... = 1.
Originally posted by: Tick
Infinite sampling does not equal analogue. Same way a reimann sum is not an integral, their just equivelent.
Originally posted by: Navid
the samples will never be close enough together
See what I mean? :roll:
Originally posted by: Navid
Originally posted by: Tick
Infinite sampling does not equal analogue. Same way a reimann sum is not an integral, their just equivelent.
They are not "equal" but they are just "equivalent"!
Could you explain what you mean exactly?
Anyway, my point was that there is no point in sampling if you are going to sample at every point in time. You might as well have an analog system. There will be no advantage in sampling with respect to the original non-sampled system if you sample at every possible point in time.
Given an analog waveform, and an infinite number of sampling points, ...
The problem is that "infinity" is just a silly word people use to refer to really big numbers.
Originally posted by: Mday
analogue signals attain any value. sampling may introduce rounding or digitization errors.
Originally posted by: BigT383
I'm having an argument with this guy at work...
This is one of those "never possible in reality" things...
Given an analog waveform, and an infinite number of sampling points, can't you choose your sampling points so that there are no gaps between them- IE, produce a sampled function that, given any real number, returns sampled data without any interpolation?
My conjecture is that you can, and at that point your perfectly sampled function is the same as the original waveform. He believes that there would still be gaps between your sample points. I said that yes, you could have an infinite number of sample points and choose to still have gaps (only sample for rational numbers, for instance), but you didn't need to.
This all started from an argument we were having about whether it was possible for digital representations to ever be as good as analog. He said it was not possible, I said it was.
Correct me if I'm wrong about this: but in the real world there comes a point when, no matter how good your detector of the original waveform is there is still a margin of error (uncertainty principle), and that as soon as the distance between every two adjacent sample points becomes less than that margin of error, the digital representation becomes just as good as the analog.
