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标题: 惊天大骗局，44.1KHz的采样率真的够吗？ -------------- 后续！ [打印本页]
作者: zifzhu    时间: 2008-1-21 23:24
标题: 惊天大骗局，44.1KHz的采样率真的够吗？ -------------- 后续！
帖子又沉掉了，没办法，只能再开一个5 |" Q! a3 E6 u  Y" w* _' D2 z
http://we.pcinlife.com/thread-867632-1-1.html3 M5 n7 P; n$ _* x( O: H& Z
其实到最后也没得到什么结果，就这么不了了之。
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) E0 x4 k6 }! s; zHerculesVR老兄推荐我去问一下Dan Lavry。于是我去他论坛上发了个帖子，这些天几乎忘了这回事了，刚回去一看，似乎有些进展。先贴出来大家参考一下。后续有什么进展我会跟踪报道B)  期望我这个无聊的疑问会有一个好的结局。4 H% u4 s' s0 i
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zifzhu
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Posted: Wed Dec 26, 2007 7:37 am- _3 f4 S$ {' A' c6 y
   Post subject: 44.1Ksps sample rate enough for audio?
We are always be told 2x sample rate is enough. It seems 44.1Ksps for 20KHz audio is enough.
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* [( D. R$ Q3 E( @9 y$ y  o- }As this diagram, 2x sample rate regenrate the source signal perfectly. But the question is, it is in the best case. It samples at the peak luckly, but if it sample 90 degree before or after. Then it will simple samples all zero. 4 ]& l" ?) v( K& ^, Y# }: d: F

, s" I) B% u) z4 q; lIt seems 8x sample is the minimium reqirement for sample for hi-fi. Thus 192Ksps is the minimum spec to record.
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Am I wrong at this point?
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Brad Johnson
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Posted: Wed Dec 26, 2007 6:48 pm
% W+ Z7 f( C. r* I/ N   Post subject:
There is a "white paper" under the SUPPORT tab of the Lavry Engineering website that is a very complete discussion of this subject. . o' l: s5 D; ^6 V1 q4 k

6 j% s. p9 Q% c! d# U" M, f- DIt is entitled "Sampling Theory." % Q8 C! W- z- C4 \
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http://www.lavryengineering.com/documents/Sampling_Theory.pdf   s1 w% y2 x" {/ r# v

* Q$ b* I, P- d" nThere are many sources of inaccuracy in real-world conversion systems that can cause an apparent loss of high frequency information, and they are not the sample rate. There are also additional trade-offs which vary with the sample rate that occur in Digital Signal Processing; in both the processing that happens in contemporary converter technology and what commonly occurs in mixing or mastering after a signal is digitized . So in the end "higher" does not alway result in "better."
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$ v  P4 `, f! e/ C# d  a, KA recording made accurately at 44.1 kHz will always sound better (or more "analog") than an inaccurate recording made at 96 or 192 kHz. In many cases this would be mostly due to jitter in the AD and DA conversion, as versus the sample rate. But even in these cases, the degradation is almost certain to be caused by more than just this one source of inaccuracy. 6 Q) j5 H4 v, n4 `9 H$ n" ~
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Brad Johnson
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Posted: Fri Jan 04, 2008 7:35 pm% h* A6 g9 J% e9 U' g, f* X
   Post subject: Re: 44.1Ksps sample rate enough for audio?
zifzhu,
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Exactly half the fs is kind of a "special case". Check the "critical frequency" part here: $ |2 L" W4 m6 o  g
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem + M7 G. N+ a# y0 k, Z, |

) h; e* p- ~& SC.J., Finland
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"Science is a way for us to not fool ourselves." - Richard Feynman& e! J& j9 c9 J2 b
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作者: zifzhu    时间: 2008-1-21 23:25
Mike Derrick
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Posted: Sat Jan 05, 2008 8:43 pm
+ K! }. \$ @( r   Post subject:
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Brad Johnson wrote:
...There are also additional trade-offs which vary with the sample rate that occur in Digital Signal Processing; in both the processing that happens in contemporary converter technology and what commonly occurs in mixing or mastering after a signal is digitized . So in the end "higher" does not alway result in "better." / r* H3 P) j8 C1 O

+ a# J, \( z2 \! M" k9 qA recording made accurately at 44.1 kHz will always sound better (or more "analog") than an inaccurate recording made at 96 or 192 kHz. In many cases this would be mostly due to jitter in the AD and DA conversion, as versus the sample rate. But even in these cases, the degradation is almost certain to be caused by more than just this one source of inaccuracy. 9 E, {) z; W0 m! s7 H
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Brad Johnson
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Hi Brad, * H- ^2 o1 I7 R+ f- K

' G! E& J3 I4 u! r$ z( SYou mentioned additional trade-offs in digital signal processing. Specifically could you elaborate on what those trade-offs might be within the context of the mixing? * g& l/ [; _0 _- S
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Can recording at 96k be as accurate as recording at 44.1k?
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& M% l. L+ @( G2 t  sMike' B+ c% u7 S/ Q+ U! F. A
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Brad Johnson
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Posted: Tue Jan 08, 2008 11:31 pm
- b7 p/ \- [" e4 V   Post subject:
The trade-offs that can occur in processing after a signal is digitized at higher sample rates are mainly in the amount of processing power needed for low audio frequency EQ or filtering. Since all real-world systems have a finite amount of processing power, this can lead to compromise in the accuracy of the processing, or taking much more time for non-real time processing. So this is probably not as much of a consideration as the compromise that happen in the actual conversion.
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4 O; H, j/ b; C& o. |In context of the original posting regarding the accuracy of the digitizing of high audio frequency signal- a 96kHz recording can be as accurate as a 44.1 kHz recording (if not more accurate). The point is that the sample rate is only one of many factors that can affect the accuracy of a digital recording. Present day converter technology can be very accurate at 88.2 or 96 kHz.
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In addition to possible trade-off's in noise-shaping that is typical of present-day over-sampling conversion, there are many possible "analog" sources of inaccuracy in converter circuitry. Effects such as stray capacitance in sample and hold circuitry or settling time of amplifiers become more significant as the sample rate is increased. This does not mean that 96 kHz conversion will alway be less accurate than 44.1, because it is the total effect of all of the factors involved. Careful design can minimize these effects, but if you consider that present-day converters typically are already sampling at many times the output sample rate , doubling or quadrupling these very high sample rates to achieve output sample rates higher than 96 kHz can make these "analog" problems that much more significant. If you don't raise the internal sample rate and output a higher sample rate, you lower the "bit accuracy" of the digital audio output, so you would compromise the quality of the signal across the audio spectrum.
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The question remains- what is the best over-all performance? Do we really want to make accurate recordings of supersonic frequencies at the expense of sound (music) we can hear?
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Brad Johnson. X9 E2 I7 f6 c3 C
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zifzhu
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Posted: Mon Jan 14, 2008 5:21 am
6 t8 j  a- r$ h5 `6 g* X   Post subject: Re: 44.1Ksps sample rate enough for audio?

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cj wrote:
zifzhu, + n4 y, g0 @# B  ~4 h# y1 M
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Exactly half the fs is kind of a "special case". Check the "critical frequency" part here: $ M; B! \* {6 J- J
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem % S  R4 N# ?# Q1 }* z! @

3 U. [3 p  |2 u2 B7 E" j# R! ?- XC.J., Finland

7 E& l: q5 }0 N+ T! @9 }8 F2 JUnable to open that link. Will try it later. 2 {: S, ^2 r$ c9 q6 @
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If exactly half the fs is not able to be sampled correctly. The next question is 'How much?'. 0.000000001% more? Or 100% more? There must be a value.
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作者: zifzhu    时间: 2008-1-21 23:26
lavrye# S! t, ?. a( E
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Posted: Tue Jan 15, 2008 7:58 pm% h4 Q4 c- l1 ~8 ^
   Post subject: Re: 44.1Ksps sample rate enough for audio?

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zifzhu wrote:
We are always be told 2x sample rate is enough. It seems 44.1Ksps for 20KHz audio is enough. # c; }1 z/ R+ q2 i% B- m

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; B6 y  C# t7 w& t( A2 EAs this diagram, 2x sample rate regenrate the source signal perfectly. But the question is, it is in the best case. It samples at the peak luckly, but if it sample 90 degree before or after. Then it will simple samples all zero. ! q3 S9 A5 u+ w. z! `( v4 u& f6 Z, y7 ^

. _- U: R" L) z( JIt seems 8x sample is the minimium reqirement for sample for hi-fi. Thus 192Ksps is the minimum spec to record.
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: C0 G# c0 W6 X- @) O6 b) dAm I wrong at this point?
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The sampling theorem states that you have to sample FASTER then twice the highest harmonic. 3 \, ?$ i) Q& a2 i) d6 c

7 M$ b- D$ c/ H! ?Many people look at a single cycle near Nyquist (say 20KHz at 44.1KHz), and wonder how it is possible to reconstruct the wave from 2 "dots". That is a wrong view of sampling. 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. % \# }; `& K0 V" W  u+ M0 x7 E

, L; S+ k, n" l9 A0 fIt is correct to note that more recent cycles (nearer in time to the point of interest) contribute more to the reconstruction. It is correct to note that what happened many cycles ago has less influence on the outcome. But one can not ignore previous samples, and the more "history" you take into account, the better the reconstructed wave. Of course there are practical limits to how far back we look. * ^7 D4 M5 J% ~& w$ ~5 H
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At low audio frequencies, there are many samples in each audio cycle.
6 g9 M7 f( Q4 y; tAs you increase the audio frequency, we get to a point where there are only 2 dots per cycle, and things get visually confusing for those that view only one cycle. Further examination shows that the samples "walk" across the audio cycles. Lets take some cycle as a "starting point". Many cycles later, we walk far enough to get back to (or pass by) a picture that looks like the "starting point". I will call that a "beat cycle". That beat cycle is repetitive (for a sine wave or any periodic input). 7 d2 f; w2 O: S* z
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If we wanted to reconstruct say 22049Hz with 44.1KHz (Nyquist at 22050Hz), the amount of data we would need to take (to reach even a crude approximation) would be huge. That is because the "distance" between 22049 and Nyquist is only 1HZ. The 1Hz difference has a "beat cycle" time of 1 second. That means that at 44.1KHz, a complete "beat cycle" consists of 44100 samples. That is a lot of data to collect and filter... ! \, V7 H% F+ e7 X! s" b

) V. O; j* l6 f3 mBut say we want to cleanly reproduce 20KHz with the same Nyquist, then the frequency difference is 2050Hz whias a "beat cycle time" of only 487usec. It takes only 21 samples of data to go through one "beat cycle". ! {6 p$ {6 i7 Y7 f2 m' Q: L

4 B, M* f; q# w$ }How many "beat cycle times" does one need for good reconstruction? This needs to be quantified, but clearly reconstruction of 22049Hz with 44.1KHz sampling is impractical! On the other hand, reconstruction of say 20KHz to very high quality is very practical. ( d$ b. @: i) U4 X6 e+ C8 c
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So instead of trying to do conversion up against Nyquist, we back of slightly and pre filter the signals so that the converter would not need to try for what is virtually impossible. Of course, with newer converters (over sampling front end) the Nyquist is higher by huge factors (in the MHz range) and the issue AD converter anti aliasing filters belong to the past (it was an issue until the early 1990'). The same holds for DA reconstruction which is filtered after a lot of up sampling. 8 i! \. c7 Z8 R, E) r
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The issue does exist for down sampling (decimation) to say 44.1KHz and that is a digital filter FIR's or IIR's. To get the FIRs closer to Nyquist, one needs longer filters, and to decimate to say 22,049KHz we would need probably over a hundred thousands samples and a huge compute engine. It is not worth it. If you are willing to settle for say 1000Hz below Nyquist, the problem becomes vastly easier (a beat cycle time is 1000 times shorter). You can do practicable decimation with reasonable amount of data. Similarly doing IIR has its limitations as well. 5 C1 X7 H7 G; }

1 [) b4 ], ^1 z; @. o* y, qRegards , s! E% ^; c7 v; ]% ^! F: X
Dan Lavry
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作者: zifzhu    时间: 2008-1-21 23:27
zifzhu2 B" @2 k! P9 @% _

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Posted: Mon Jan 21, 2008 3:12 pm( ^5 a* p6 I0 ~7 x% v
   Post subject: Re: 44.1Ksps sample rate enough for audio?

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lavrye wrote:
zifzhu wrote:
We are always be told 2x sample rate is enough. It seems 44.1Ksps for 20KHz audio is enough. * P" J6 r* a. |* G
+ S1 A" i6 d) @, _) V3 }5 G

3 z3 E& z  u, X) XAs this diagram, 2x sample rate regenrate the source signal perfectly. But the question is, it is in the best case. It samples at the peak luckly, but if it sample 90 degree before or after. Then it will simple samples all zero.
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; u1 a! p4 g  `: ~: F+ }It seems 8x sample is the minimium reqirement for sample for hi-fi. Thus 192Ksps is the minimum spec to record. ( E2 E" x: a- J; H  E+ t% e

2 H2 ~; y; l3 Y! hAm I wrong at this point?
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* U: l' _; R% NThe sampling theorem states that you have to sample FASTER then twice the highest harmonic. * J2 W" D8 |$ N/ k7 T% j
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Many people look at a single cycle near Nyquist (say 20KHz at 44.1KHz), and wonder how it is possible to reconstruct the wave from 2 "dots". That is a wrong view of sampling. 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. + f7 _& ^" x) P* `( T: S1 g! ?( M  A* `
( y; a$ O' h& Z. p. N
It is correct to note that more recent cycles (nearer in time to the point of interest) contribute more to the reconstruction. It is correct to note that what happened many cycles ago has less influence on the outcome. But one can not ignore previous samples, and the more "history" you take into account, the better the reconstructed wave. Of course there are practical limits to how far back we look.
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2 L& r+ ?, N' ]% `  u- ]At low audio frequencies, there are many samples in each audio cycle. ; o) n3 x' \; O  q$ @
As you increase the audio frequency, we get to a point where there are only 2 dots per cycle, and things get visually confusing for those that view only one cycle. Further examination shows that the samples "walk" across the audio cycles. Lets take some cycle as a "starting point". Many cycles later, we walk far enough to get back to (or pass by) a picture that looks like the "starting point". I will call that a "beat cycle". That beat cycle is repetitive (for a sine wave or any periodic input). ) Q+ w0 \  Q' E4 w
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If we wanted to reconstruct say 22049Hz with 44.1KHz (Nyquist at 22050Hz), the amount of data we would need to take (to reach even a crude approximation) would be huge. That is because the "distance" between 22049 and Nyquist is only 1HZ. The 1Hz difference has a "beat cycle" time of 1 second. That means that at 44.1KHz, a complete "beat cycle" consists of 44100 samples. That is a lot of data to collect and filter... + P( [6 Y) `( S& C0 n- x% K3 H
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But say we want to cleanly reproduce 20KHz with the same Nyquist, then the frequency difference is 2050Hz whias a "beat cycle time" of only 487usec. It takes only 21 samples of data to go through one "beat cycle". 7 ~* Z4 L* F8 J3 S( L  }  C* M

* f" Y  C+ Z! O, F' f& RHow many "beat cycle times" does one need for good reconstruction? This needs to be quantified, but clearly reconstruction of 22049Hz with 44.1KHz sampling is impractical! On the other hand, reconstruction of say 20KHz to very high quality is very practical.
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So instead of trying to do conversion up against Nyquist, we back of slightly and pre filter the signals so that the converter would not need to try for what is virtually impossible. Of course, with newer converters (over sampling front end) the Nyquist is higher by huge factors (in the MHz range) and the issue AD converter anti aliasing filters belong to the past (it was an issue until the early 1990'). The same holds for DA reconstruction which is filtered after a lot of up sampling.
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( @( _& X2 c; `7 {# y' nThe issue does exist for down sampling (decimation) to say 44.1KHz and that is a digital filter FIR's or IIR's. To get the FIRs closer to Nyquist, one needs longer filters, and to decimate to say 22,049KHz we would need probably over a hundred thousands samples and a huge compute engine. It is not worth it. If you are willing to settle for say 1000Hz below Nyquist, the problem becomes vastly easier (a beat cycle time is 1000 times shorter). You can do practicable decimation with reasonable amount of data. Similarly doing IIR has its limitations as well.
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Regards
0 P+ H9 K8 |3 {) v6 G7 FDan Lavry

$ `9 L4 Z. s2 @0 H( wIf I'm not wrong, the 'beat cycle' is another kind of aliasing that Nyquist sampling theorem haven't told us. : N: f. i  P+ d5 o7 U7 a$ [# e6 l
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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'.
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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. % _8 o  S2 c' m. g, s
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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.
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作者: q38466621    时间: 2008-1-22 08:33
:funk: 又来
作者: bill_max    时间: 2008-1-22 08:38
翻译一下吧～～～
作者: supermouse    时间: 2008-1-22 09:00
楼主其实很无聊...
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! A+ e2 k! U9 [别人写了那么多东西, 楼主就来个标题"惊天大骗局", 说的是自己的这个帖子?
作者: adinx    时间: 2008-1-22 09:17
:sweatingbullets:  太长没耐心看完。。。。
作者: poposun    时间: 2008-1-22 09:49
楼主还是没明白，不是说得很清楚了吗0 ^: T! z9 p! h' ~8 `! g
从你上个帖子里发的图看，你似乎错误的认为采样后重建信号是简单的利用单周期采样结果，但实际上，有限频域信号在使用Nyquist采样所得进行重建时，纯理论的说法是频域有限时域无限，因此可以利用无限多的采样点重建原始输入，当采样频率->2*Fs，采样点数目->无穷，工程上则是在能够接受的范围（数据处理能力、存储器能力、成本等）内，利用有限采样点进行重建（但绝少会出现单周期）) X9 V7 @) W/ C4 S! O
PS：以上采样实际上均为采样后的量化数据
作者: zifzhu    时间: 2008-1-22 12:56
boB
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8 s/ B# N4 o1 F0 BPosted: Mon Jan 21, 2008 8:45 pm    Post subject: Re: 44.1Ksps sample rate enough for audio?   
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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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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 ... 0 Z% p' k1 }# h' h! ~4 a
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But I suppose the history IS because of the same bandwidth constraint ???
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! u% K' C2 a- A/ `  [, S% Q& ]Is this a valid thought, or another way of saying the same thing ?
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Posted: Mon Jan 21, 2008 11:53 pm    Post subject: Re: 44.1Ksps sample rate enough for audio?   
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If I'm not wrong, the 'beat cycle' is another kind of aliasing that Nyquist sampling theorem haven't told us. 8 I: q# l* |* m% F

- r& l( L  e$ I/ I6 w- mWhen 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'.
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. H, {1 C! _* EUse 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. * ?; j, J9 y( {, D& P2 \
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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]
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The term "beat cycle" is one I "generated" to help me explain a concept. + z2 o9 f) A9 V' O  y# o  M
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Nyquist did not miss a thing. I think you did not understand what I said. ! d8 l% i9 a" c  u- \7 j5 R
I think you missed my explanation. & g$ M5 i3 {% y7 M: ?

8 l, j) |6 Q% s6 g. s% O9 HThe waves that you are looking at are BEFORE the reconstruction (before the filtering of energy that does not belong to the final outcome). $ q; @. I/ _2 r, ~7 b9 L, L
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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.
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That "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.
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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." + m8 j  V3 a8 e% f& ?

  K# Y8 ~; Q. C% a# w, n  CIf 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. , R9 E# N; ]5 L7 q, W

- S  f: e- I# Y5 E5 hYou 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. 6 u* [, L) E2 q
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The fact is: you DO NEED to include many samples of the past history. 2 F7 s  K% A" {% u
Say you give me 8 points, instead of 2 sample points per cycle. I can draw INFINITE different shapes that go through those 8 points. . o7 w9 }! a0 N1 ~' X: G
One 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. # p6 n7 H$ N9 |6 W
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And 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. 4 d7 P0 @! J/ R7 S. \( G

* {' j) m0 m* C1 y) S$ J4 AThe 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). 7 J! M0 u7 k( U) c) S% r
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It 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". $ n& \8 S' x  p" t( r2 m& _

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4 B% @* J& M* [. ?7 EPosted: Tue Jan 22, 2008 12:12 am    Post subject: Re: 44.1Ksps sample rate enough for audio?   $ F  l/ q( _4 b. m7 [9 F
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boB wrote: 8 A/ ?$ S; n6 d. b! l
lavrye wrote:
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* u) w8 N' D. g( XYou 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. , y& K. Y9 P- _3 o

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Dan Lavry
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- \4 B' J  J9 j0 Z% {0 `/ [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 ??? 5 `" ~/ z2 B8 d" K
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Is this a valid thought, or another way of saying the same thing ?
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Thanks, ( O3 s' U# k# m0 P
boB
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# Q) Z- t, e" y  a4 z; ^I am not sure I understood what you said. Can you elaborate? 0 f5 L* @! \7 _) D# ^  c% a, Y
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When you sample at exactly fs/2, you violate Nyquist, because the points at each cycle are at the same location relative to the cycle. $ X" Q( `7 s/ S
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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. 2 H* p2 G+ v) Q7 I* I
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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. + a$ R- b% I1 n( q1 w# ^+ H" V
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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.
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Posted: Tue Jan 22, 2008 1:26 am    Post subject: Re: 44.1Ksps sample rate enough for audio?   
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5 [) G9 [* E* k9 Vlavrye wrote:
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7 l& x: F/ W( t2 m+ g; qI am not sure I understood what you said. Can you elaborate?
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/ O+ T( F2 N; `& w4 [3 CWhen you sample at exactly fs/2, you violate Nyquist, because the points at each cycle are at the same location relative to the cycle.
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; ?) A# T9 c! y4 wRegards # u# O* l8 `: A' u: G, y4 z& w
Dan Lavry
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5 }. f% ~" Q  z1 YI did indeed mean somewhat below FS/2 and that the filtering was filling in-between the dots. And, Sinewave at this 20kHz frequency
5 B0 t% c- L, ~( S/ Y; f+ l8 ~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
1 G, K/ z- _$ D. T  W8 q9 y0 bto think about this stuff.
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2 D9 `+ e9 ]6 S& m  v# }I've thought that the bandwidth limit (reconstruction) was all that was necessary to "fill in" the rest of the waveform   N/ V3 X% b& F+ P
because all of the energy of that sinewave is below FS/2.
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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 ? ' v  u* m0 t* B6 Y5 ]
I can't remember what kind of scope it was but I seem to remember that it was an older one.
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* F6 i8 I$ u  W0 CPosted: Tue Jan 22, 2008 4:53 am    Post subject:     
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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'. , z8 _6 r5 K9 U2 b7 k1 x# [
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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'. 1 ?! H5 ^+ U' K* h: R

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! Y5 k, \. [8 d! u. m& V[ 本帖最后由 zifzhu 于 2008-1-22 12:59 编辑 ]
作者: zifzhu    时间: 2008-1-22 13:00
原帖由 poposun 于 2008-1-22 09:49 发表
# H* s( k( d/ ?8 h5 V# w楼主还是没明白，不是说得很清楚了吗
: E  h3 W. S& ]从你上个帖子里发的图看，你似乎错误的认为采样后重建信号是简单的利用单周期采样结果，但实际上，有限频域信号在使用Nyquist采样所得进行重建时，纯理论的说法是频域有限时域无 ...

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看我上面的回复o:)
作者: 287381906    时间: 2008-1-22 14:56
用个PLL不就结了？:huh:
作者: zifzhu    时间: 2008-1-23 14:00
boB
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Posted: Tue Jan 22, 2008 6:44 am
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Hello zifzhu... I hope all is well in Suzhou, China !
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$ \+ a; Y9 F. I* B! L4 |) }* E( c( |I have seen this very same Excel example before. One of the problems is that the beat and triangle waves you see are not real because the bandwidth
1 _) R( N* h: h4 E' dis not limited to FS/2... I think that if you were to filter the sample points, you will end up with a sinewave again because a triangle wave has harmonics
- j$ i5 g! J. |* W: E  z$ tabove the frequency it has re-constructed. This cannot happen with a filtered reconstruction of a sampled signal close to FS/2. The reconstruction can 0 i5 e6 [' E/ K6 v! K
ONLY be a sinewave, however it could take on a high or low amplitude.
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Does this make sense ?? There is a very good web site that talks about this very same example and if I find it again I will definately post the URL here. 0 T% ]$ [7 {. u' @% T
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Posted: Tue Jan 22, 2008 6:05 pm
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zifzhu wrote:
Hard to believe this wave will be reconstructed after filtering. At least, the total amplitude is decreased.
/ ]; @( f* u/ vThe harmonic contains the 'beat cycle'. * s, w% ]+ c9 G; h

" q7 l' l2 I2 G9 @# R7 ?Hard to believe...? OK, here you have it right before your eyes.
& [( a3 E' B8 \# uI've used this same graph many times over the years. It shows the sample points of a 10kHz sine wave.
, ?3 n/ ^; |1 t- }- \: SIn the upper wave lines are simply drawn between the sample points (Sound Forge).
. L8 {2 \7 U' H8 t% v/ cIn the lower wave, Adobe Audition has correctly reconstructed the samples back to a sine wave. " o$ e% J# k8 `5 T' m; }+ R' p1 Z
The sample values are exactly the same in both waves.
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作者: zifzhu    时间: 2008-1-23 14:00
lavrye  [" w0 [. y  E  @# s' e
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Posted: Wed Jan 23, 2008 3:20 am8 d9 \, U6 D% i- i5 x7 V0 L) Q
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cj wrote:
zifzhu wrote:
Hard to believe this wave will be reconstructed after filtering. At least, the total amplitude is decreased.
- v$ V/ s! G8 ?The harmonic contains the 'beat cycle'. 0 H) e0 |% `5 S' Z9 R

+ t( P/ Q7 p: {7 c" L: U  f1 OHard to believe...? OK, here you have it right before your eyes. - ]0 D" l) ^9 x/ a/ C
I've used this same graph many times over the years. It shows the sample points of a 10kHz sine wave. : g9 D+ R7 f  Q& h/ v
In the upper wave lines are simply drawn between the sample points (Sound Forge).
% ~* H9 y4 M# t. ]7 WIn the lower wave, Adobe Audition has correctly reconstructed the samples back to a sine wave.
+ d) O! X! u6 I+ K; s4 _( lThe sample values are exactly the same in both waves.
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The Excel file shows the wave BEFORE FILTERING. I keep saying that we need to filter. It is very difficult to make a filter that will pass 22049Hz, and reject 22050Hz, but if you could do it, and if your file was much bigger then 44100 points (you would need a lot of "history"), your filtered wave would be a sine wave.
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  E$ a" v5 e  m0 b& a5 mNo one seems to have a problem with say 1KHz sine wave. Why? Because it looks intuitivly (visualy) correct. The dots are "tracking" a sine wave shape so the process seems OK. At 1KHz, no one is questioning what happens to the signal between the sample points, and we all accept that the filter will "connect" the dots correctly. At 1KHz, look at one cycle and you see 44 points clearly on a sine wave....We did not need Nyquist to tell us that...
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) w6 b& ^* h& aSo lets increase the frequency to say 5KHz. Now we have only about 9 points per cycle, and if you connect them with stright lines, it still looks "sort of" like a sine wave, but with some "sharp corners". You filter it and it becomes a sine wave - the "sharp corners" are smoothed out....
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CJ shows you the case of 10Kz, with about 4.5 points per cycle. Connect the dots with stright lines, and things are starting to look rather "non sinusoidal". At 15KHz it looks worse, that is when you connect with stright line. But filtering "brings back" the sine wave... 0 s4 M; L/ j/ p' h

! k# l" ]2 `$ H7 i( t3 G! y8 x  ZSay you start at 1Hz signal. The filtering is very easy - pass signals to 1Hz, reject signals at 22050Hz. ! `1 e  d( z- q  P, p
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Raise the signal 1Hz at a time and look at the required filter. At 1KHz, the filter needs to pass signals to 1KHz but block at 22050Hz. That is an easy filter, with a pass band of 1KHz and transition band of around 21Khz. At 10Khz, the filter gets tougher, the slope is about twice as steep, but still "doable". At 20KHz, you need a filter that passes 20KHz but blocks 22050Hz, that is a very steep filter - transition band to pass band ratio of 10:1. At 22049Hz the filter is so steep it would take a super computer to simulate... The transition to pass band ratio is 22049:1
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There is no "sudden" break at some "magic frequency". Why is 1Khz OK and 22049Hz not OK? Or is there a "sudden break" at say 19KHz? Why not 12KHz? The whole thing is gradual - as you increase the gradualy the tone frequency, you need to make the filter steeper... ; x, I# \( `5 Z: h4 q: O. q# E1 D

! k1 A/ o  h* [, AI said that one needs to look at a lot of history to steer the filter outcome to track the signal appropriatly. Your graph competly violated that, because connecting adjacent samples with a stright line completly discounts all the history except for a single previous sample. That is not how one filters a signal. You do need to filter the signal.
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CJ's example is very good. Take a 10KHz sine signal and it looks "terrible", but when you filter it it looks like a perfect sine wave. Those that do not like math can use a scope to show it, and it will match the plot that CJ posted - the "strange" wave, when filtered, will become a clean sine wave. 8 D2 S- A5 w& k+ k- }

. a6 p" `: f! e7 B7 nBetter yet, take a DA converter, a good one (with a filter). Run a 10KHz tone through it, the look at the output with a scope. It does not look like 4-5 dot per cycle connected by a stright line. It looks like a clean sine wave! 5 H1 z: h* W2 ~

3 g$ f9 O, ?+ |  w3 Y6 b; |! JAnd before you point out that the DA is up sampled, I will remind you that up sampling makes a lot more sample points (samples) out of the few original sample point by a process no other FILTERING. - T+ y, ?* C7 Q
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Regards
8 J5 J0 D% B1 h; g( |Dan Lavry$ T( ^1 |& N3 v0 J$ r8 I! d% O' t
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Dan Lavry
作者: zifzhu    时间: 2008-1-23 14:01
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Joined: 26 Dec 2007! V- J& n2 V% z  ?$ J$ `, T9 y
Posts: 6
: D/ w$ _7 Z6 r, T/ N' Z* w3 A$ v3 RLocation: Suzhou, China  z; v* B; v5 t# X; G% d
Posted: Wed Jan 23, 2008 5:28 am
* @1 ~! |9 e: ]/ H/ n# _   Post subject:
Hello boB and Cj. I was also confusion about filtering that triangle wave. The low pass filter should like just smooth the wave. The better idea is to use fourier transform to analyze it's frequency energy.
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1 Y8 o9 t( k% L4 j4 rI've found a excel FFT analyze tool from maxim. http://www.maxim-ic.com.cn/appnotes.cfm/an_pk/3292
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After input 128 samples of 2.2X sampling wave. The FFT shows it does contain other frequency(not a straight bar in 20Khz). It contains 18Khz at -30db, 11KHz at -40db, 0-1.7KHz at -43.4db. # v1 W4 U0 f6 a; T2 B

- k! B% n% j8 e" A1 {# ]# NAgain, I input 4.4X sampling wave. The FFT shows a better result. 0-1KHz is only -51db. About 7.6 db better than 2.2X sampling. ' w9 \7 u0 j" y& p
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So. Can I put a conclusion '2X sample rate can not do a perfect sample'. To do a better sample. A higher sample rate is prefered.
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0 E" v" U3 }& f6 V[ 本帖最后由 zifzhu 于 2008-1-23 14:09 编辑 ]
作者: zifzhu    时间: 2008-1-23 14:02
zifzhu9 s) V: `6 u/ y3 W7 y

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1 n1 L  ]" l( g7 ^Joined: 26 Dec 2007. G8 j, {2 Y: ?% `+ E
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Location: Suzhou, China1 F' }- n* j$ q0 A" O! u
Posted: Wed Jan 23, 2008 5:54 am
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lavrye wrote:
cj wrote:
zifzhu wrote:
Hard to believe this wave will be reconstructed after filtering. At least, the total amplitude is decreased. $ }9 b+ R, Y1 s+ ]
The harmonic contains the 'beat cycle'.
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% n& Z0 w; s6 H4 \Hard to believe...? OK, here you have it right before your eyes. ' X7 b, g# U; ?' v3 s2 b
I've used this same graph many times over the years. It shows the sample points of a 10kHz sine wave.
+ F& w+ Q6 f( ^In the upper wave lines are simply drawn between the sample points (Sound Forge).
9 v3 [- ?9 T8 |0 i7 r7 m* kIn the lower wave, Adobe Audition has correctly reconstructed the samples back to a sine wave. 1 C/ s( U2 v5 _" n% {% S6 ^
The sample values are exactly the same in both waves. * i1 q% |% o+ }+ {8 P- X3 f
C-J
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The Excel file shows the wave BEFORE FILTERING. I keep saying that we need to filter. It is very difficult to make a filter that will pass 22049Hz, and reject 22050Hz, but if you could do it, and if your file was much bigger then 44100 points (you would need a lot of "history"), your filtered wave would be a sine wave.   R7 A. B( w( T9 U5 I0 {1 ^

  ]3 K* Q: u" r3 }- l, D# \No one seems to have a problem with say 1KHz sine wave. Why? Because it looks intuitivly (visualy) correct. The dots are "tracking" a sine wave shape so the process seems OK. At 1KHz, no one is questioning what happens to the signal between the sample points, and we all accept that the filter will "connect" the dots correctly. At 1KHz, look at one cycle and you see 44 points clearly on a sine wave....We did not need Nyquist to tell us that... # {3 R# K& p7 f: z/ q# J+ c) F  S2 E
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So lets increase the frequency to say 5KHz. Now we have only about 9 points per cycle, and if you connect them with stright lines, it still looks "sort of" like a sine wave, but with some "sharp corners". You filter it and it becomes a sine wave - the "sharp corners" are smoothed out....
, n2 n) h" b7 z/ y4 S; E9 N. |8 w# a) v+ M/ f& M
CJ shows you the case of 10Kz, with about 4.5 points per cycle. Connect the dots with stright lines, and things are starting to look rather "non sinusoidal". At 15KHz it looks worse, that is when you connect with stright line. But filtering "brings back" the sine wave... 2 l5 K! m" f% E) G

; n& [$ {* x6 w8 A$ ~' m7 `4 jSay you start at 1Hz signal. The filtering is very easy - pass signals to 1Hz, reject signals at 22050Hz. 9 S4 z$ U1 o" z% [% [

/ J" R, t, A/ v5 V2 @) SRaise the signal 1Hz at a time and look at the required filter. At 1KHz, the filter needs to pass signals to 1KHz but block at 22050Hz. That is an easy filter, with a pass band of 1KHz and transition band of around 21Khz. At 10Khz, the filter gets tougher, the slope is about twice as steep, but still "doable". At 20KHz, you need a filter that passes 20KHz but blocks 22050Hz, that is a very steep filter - transition band to pass band ratio of 10:1. At 22049Hz the filter is so steep it would take a super computer to simulate... The transition to pass band ratio is 22049:1 5 C) ~, F/ m* O3 f! M. F: E

( o) {4 R; f/ F9 B# rThere is no "sudden" break at some "magic frequency". Why is 1Khz OK and 22049Hz not OK? Or is there a "sudden break" at say 19KHz? Why not 12KHz? The whole thing is gradual - as you increase the gradualy the tone frequency, you need to make the filter steeper...
8 x. h1 _* r5 S1 n' F1 u9 L& V/ D9 |+ o; B1 S! |" C
I said that one needs to look at a lot of history to steer the filter outcome to track the signal appropriatly. Your graph competly violated that, because connecting adjacent samples with a stright line completly discounts all the history except for a single previous sample. That is not how one filters a signal. You do need to filter the signal. 1 ^. `1 l5 D( ^* X
# ]9 @% ^% A1 u* Y
CJ's example is very good. Take a 10KHz sine signal and it looks "terrible", but when you filter it it looks like a perfect sine wave. Those that do not like math can use a scope to show it, and it will match the plot that CJ posted - the "strange" wave, when filtered, will become a clean sine wave.
4 I% v7 b. J" b( b" M! n
$ k( S# _8 e0 a9 EBetter yet, take a DA converter, a good one (with a filter). Run a 10KHz tone through it, the look at the output with a scope. It does not look like 4-5 dot per cycle connected by a stright line. It looks like a clean sine wave!
) p4 \2 F: [2 v. Y2 ~8 E# D* ?) H# k' O: U! P, L
And before you point out that the DA is up sampled, I will remind you that up sampling makes a lot more sample points (samples) out of the few original sample point by a process no other FILTERING. 1 Y0 E# K( m* M9 y) a  b+ l! x/ ]

$ L& c% `+ K- |. Q4 f3 @Regards
4 s. a: P" Z0 X" i+ Q1 X/ @  CDan Lavry

8 ^. h: V! f" f  z! M9 F7 N, XHi Dan,
4 l+ \, o- d- o! ~. i, T& iI think the problem is the harmonic contains not only higher frequency but also lower frequency, that is not able to be filted by a low pass filter. As my previous post, the higher sample rate can get a cleaner lower frequency.
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There is no PERFECT sampling. The higher sample rate loss less information. Nyquist frequency is just a point that avoid aliasing that means samples the freqency correctly, not perfectly.* J, @9 q6 S0 x: F, y% W4 T! o
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作者: zifzhu    时间: 2008-1-23 14:03
又有新的进展了:a)
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估计很快就会有结论了:p
作者: 辛心枫    时间: 2008-1-23 22:50
楼主很执着:lol:
作者: kingcole    时间: 2008-1-24 11:38
44.1足够了，不是技术决定的，是市场决定的



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