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boB
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Posted: Mon Jan 21, 2008 8:45 pm Post subject: Re: 44.1Ksps sample rate enough for audio?
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lavrye wrote: / O, d5 x% H1 V
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You can not tell the wave shape of a cycle from 2 dots. To understand sampling, you need to take into account very many cycles prior to that "cycle with "2 dots". You will soon realize that the samples (dots) hit different values on each of the previous cycles. With data from many prior cycles, we can reconstruct the original sampled wave. ; v3 x0 j! ^0 R" d1 @1 y: E
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1 `, O0 v7 @! X- M& ]$ j/ DMaybe it's exactly the same thing, Dan, but I look at the re-make of the sine wave at close to FS/2 from two "dots" as reconstructing the sine wave because of the bandwidth limitation at FS/2 itself, rather than the "history" of the sampled waveform ... 7 G, I! J3 n, X+ e* |5 e7 b6 @
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But I suppose the history IS because of the same bandwidth constraint ???
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8 V+ Y# H1 h; I3 r" |" }Is this a valid thought, or another way of saying the same thing ?
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9 u }: g( t/ V! N) pPosted: Mon Jan 21, 2008 11:53 pm Post subject: Re: 44.1Ksps sample rate enough for audio?
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4 o& S6 p# h- {& _/ M/ V. p. nIf I'm not wrong, the 'beat cycle' is another kind of aliasing that Nyquist sampling theorem haven't told us.
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When the sample rate is close to 2x at the frequency. It will generate 'beat cycle'(or aliasing) no matter it's over or less than 2x. I will call it 'forward aliasing' and 'backward aliasing'. 9 q0 D6 T# C7 b/ K
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Use 44.1KHz sample rate to sample 20KHz freqency will generate a 2050 Hz aliasing. No matter if it is 'forward aliasing' or 'backward aliasing', that 2050Hz will be reconstructed. ) M) ~; N7 n% [
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Maybe the solution is sampling the frequency over 4X sample rate, then the 'beat cycle' will be over 20Khz, that will be easy to use a low pass filter.[/quote] % p- ]- T" C+ p* k" I
$ c2 F0 p; ?3 n- o) @0 |% m; D1 S+ _The term "beat cycle" is one I "generated" to help me explain a concept. $ ? ^" b1 t- Z& `1 g& ?2 ]
& J- \9 A- n% x) A, m$ [Nyquist did not miss a thing. I think you did not understand what I said.
" j0 K0 j7 c0 j- L0 W: R& _I think you missed my explanation.
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( |: N8 w% |1 vThe waves that you are looking at are BEFORE the reconstruction (before the filtering of energy that does not belong to the final outcome). ' `! F, _ |# j
% R; ~' R" x# H- L, X, }What you see as aliasing is not aliasing. The "modulating envelope" of the high frequency is not a signal, and it does not carry energy. & ~# l( Z2 P3 d0 ^
# G2 M$ L- I. K. W; AThat "modulating envelope" will disappear when you filter the signal with an anti imaging filter. The problem is that an anti imaging filter gets to be impractical when you are too close to Nyquist. That is why, as I already stated, we used to cut the bandwidth a little, to increase the margins and make filtering easier. , R2 K+ j0 Z0 [
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You said: "Maybe the solution is sampling the frequency over 4X sample rate, then the 'beat cycle' will be over 20Khz, that will be easy to use a low pass filter."
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If you read what I said again, you will note that I stated that these days (and for the last 15 or more years) we do use up sampling, to make filtering easier.
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You suggested to go X4 up sampling, and these days, it is common place to go X64, X128, X256 and even X1024! The comment about X4 is confusing and may be misleading, because it suggests 192KHZ, whias nothing to do with the question here.
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^. u+ z( o% c6 |* m$ m9 mThe fact is: you DO NEED to include many samples of the past history.
5 s2 [5 p+ g# u! \/ E% FSay you give me 8 points, instead of 2 sample points per cycle. I can draw INFINITE different shapes that go through those 8 points.
* i( S) z$ K; P0 f$ I. T+ HOne can not figure the original wave shape of a cycle from 8 points. The 8 points only define that the wave goes through those 8 points. What happens between the 8 points? You need a lot more information, thus you need to relay on a lot of previous history. 6 Z' ]3 d0 x, S5 s( Y
_0 _/ D7 B' T; X2 y( h- XAnd more importantly, music is not made out of repetitive identical cycles. With real music, the wave shape is changing all the time, making the signal itself much more complex, yet by filtering away all the energy above Nyquist (both at the AD and DA), you can reconstruct the signal perfectly.
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The concept was discovered by Dr. Nyquist, a man with a Phd. in Physics and a lot of solid math background. He was working at Bell Labs, with other greats such as Shannon (the father of information theory). The concept is not a theory it is a theorem. There is a PROOF that if you filter bellow the sample rate at both ends (AD and DA), you will end up with a perfect replica of the original waveform. In addition, Nyquist theorem stood the test of time (over 80 years in many areas of technology).
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. F, c' a- ?9 U j) iIt is difficult to construct a filter that will keep say 24049Hz intact but reject 22050Hz (1 cycle away). But when the PRACTICAL margin (from a filter standpoint) is met, there is no "beat cycle" and no "another kind of aliasing".
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7 k' B# X; I+ h: \3 z- T- V, {Regards # N+ E% ~7 S) b( e
Dan Lavry
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lavrye
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Posted: Tue Jan 22, 2008 12:12 am Post subject: Re: 44.1Ksps sample rate enough for audio? 6 V: Q5 I, v8 C( l0 V. v
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boB wrote:
9 n' ?$ v2 m Q6 r/ i$ H; {$ ?lavrye wrote:
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You can not tell the wave shape of a cycle from 2 dots. To understand sampling, you need to take into account very many cycles prior to that "cycle with "2 dots". You will soon realize that the samples (dots) hit different values on each of the previous cycles. With data from many prior cycles, we can reconstruct the original sampled wave.
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" Q5 @9 ]5 F5 X+ _3 A~~~~MORE STUFF~~~~ ~7 i) V3 ~! e; m' C" V
6 x- a1 y5 `4 t( ARegards 5 g# j6 k5 [; E7 K
Dan Lavry 4 h" M6 I/ }2 q: W) G3 \
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) `- S3 p s% {( \4 y. W5 _Maybe it's exactly the same thing, Dan, but I look at the re-make of the sine wave at close to FS/2 from two "dots" as reconstructing the sine wave because of the bandwidth limitation at FS/2 itself, rather than the "history" of the sampled waveform ...
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But I suppose the history IS because of the same bandwidth constraint ??? + u4 u7 H8 o @9 R5 m2 S
1 t9 @) L5 X/ |: K/ X, d3 ~Is this a valid thought, or another way of saying the same thing ?
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Thanks, 9 y( R/ S L9 C- q( I
boB
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, x) R- v. S D3 U I3 qI am not sure I understood what you said. Can you elaborate? . j9 X7 g1 ?+ e8 U% C" {- T) L
6 c6 Z1 U0 G! \7 _0 YWhen you sample at exactly fs/2, you violate Nyquist, because the points at each cycle are at the same location relative to the cycle. 4 o K5 A. k, A$ _0 N, E' z: \
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But when you sample a wave that is slightly below fs/2, look at previous cycles, and you can see that the dots are at a different points of each of the previous cycles. That is where the information is to enable the reconstruction of the waveform.
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I do not like talking about sine waves, because it simplifies matters too much. But OK, lets stay simple. Say you sample a sine wave at 22049Hz. Say the first sample is on a positive going zero crossing. The next one is very slightly earlier then the first negative zero crossing. The next sample is slightly earlier then the second positive zero crossing... and so on. If you take around 44100 cycles, and "lay them on top of each other" (think of a scope triggering on a positive zero crossing with a time base set to show one cycle time), you end up with a "solid looking sine wave" made out of 44100 points. In other words, the history of many previous cycles is what "fills in the gaps" between a single cycle made of only 2 dots.
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Of course we do not keep 44100 cycles and "lay them on top of each other". But we do need the information from many previous cycles to "steer" the signal to fill in the values between the samples. Filtering is what does it for us. ' e( b0 T+ i% Y) \5 r# E6 V1 o
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* z; o& R2 O% a. y" uPosted: Tue Jan 22, 2008 1:26 am Post subject: Re: 44.1Ksps sample rate enough for audio? - t+ u8 ` W$ x& S
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lavrye wrote:
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2 H: O. O. G: V$ ^( x' T+ AI am not sure I understood what you said. Can you elaborate? $ S! F, u& P( b! t; h, n4 i2 _, T
p( x: _: n4 i. EWhen you sample at exactly fs/2, you violate Nyquist, because the points at each cycle are at the same location relative to the cycle. 3 Y) p$ F) y2 r) C: X
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p1 i% _9 l3 }9 C {+ M; iDan Lavry
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' E2 l0 q2 B- h! z1 dI did indeed mean somewhat below FS/2 and that the filtering was filling in-between the dots. And, Sinewave at this 20kHz frequency ( G3 Z' \2 ]! J4 f& m
area because that is all that can really be represented at this fraction of sample rate.. These 2 requirements make it much easier for me
: }. F& n' }! j# T G# ]: `. jto think about this stuff.
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. I4 [. I+ W$ S7 ?I've thought that the bandwidth limit (reconstruction) was all that was necessary to "fill in" the rest of the waveform
1 e2 _. {+ ^6 s6 ubecause all of the energy of that sinewave is below FS/2. 5 b: W c, _) y( K8 t
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Is the "history" coming from the filtering itself ? Does this have to do with the sinc function ?
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In the past, I have seen unfiltered digital samples suddenly turn into sin-waves when the horizontal time base was just right. Is this related ?
2 l2 S& g: K' F/ \( I! BI can't remember what kind of scope it was but I seem to remember that it was an older one.
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zifzhu
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; Y2 A' u0 W/ Z% d4 gPosted: Tue Jan 22, 2008 4:53 am Post subject: * r+ w u1 u% n" T j" u
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My understanding about 'history' is just increase the sample count and period, so that will be more accurate and not lost 'something'. 1 r; L$ {9 U4 S( _" B
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I've create a excel file that demo the 44.1Ksps sample 20KHz sine wave. The amplitude of the frequency is just increasing and decreasing at a 'beat cycle'. It's not a simple sine wave not, but a harmonic. Hard to believe this wave will be reconstructed after filtering. At least, the total amplitude is decreased. The harmonic contains the 'beat cycle'.
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3 I+ n$ U' z3 p8 ?* J! G( q( j[ 本帖最后由 zifzhu 于 2008-1-22 12:59 编辑 ] |
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