terry9, No worries on being confused about this. I find that many audio writers, many people in the audio industry period, and certain many (most) on audio forums do not get this concept. When you do the math (no literally go through the math), which I have not done in years, it becomes quite obvious how it works (after the 3rd of 4th reading).
Let me do a more real world signal. We have a 24 bit audio system, so it captures with a resolution of about 1/16.7 million, though practically will be closer to 1/1-2 million. Let’s say the system is sampling at 100Khz, and the system is bandwidth limited to 20KHz. Now let’s say we have 10KHz signal.
One key concept in a bandwidth limited system is that you cannot have just a pulse 1 waveform long, i.e. you can’t have a 1Khz waveform that last exactly 1 cycle. That would violate the bandwidth of the system because in a bandwidth limited system you cannot start and stop instantly. You can’t start and stop instantly in the real world either.
Here is where it gets harder. So these two signals, both 1KHz tones, separated by 1 microsecond arrived at these two ADCs. Let’s assume that Signal B arrives at Channel 2, 1 microsecond before Signal A arrives at Channel 1. To make the math easy for me, let’s assume that Signal A arrives at exactly 0 phase. Here are the digital outputs for the first 10 samples at 1KH and 20KHz. This is a DC offset AC signal, so the numbers go from 1 to 2^24.
You can easily tell these numbers do not represent the same signal, there is definitely something different about them. Your next question may be about accuracy / resolution. Jitter will obviously impact the inter-channel timing accuracy. I have not looked at the math in a while, but as you approach the SNR, I remember there is an increase in the inter-channel timing uncertainty.
We can do it at 20Khz as well
Let me do a more real world signal. We have a 24 bit audio system, so it captures with a resolution of about 1/16.7 million, though practically will be closer to 1/1-2 million. Let’s say the system is sampling at 100Khz, and the system is bandwidth limited to 20KHz. Now let’s say we have 10KHz signal.
One key concept in a bandwidth limited system is that you cannot have just a pulse 1 waveform long, i.e. you can’t have a 1Khz waveform that last exactly 1 cycle. That would violate the bandwidth of the system because in a bandwidth limited system you cannot start and stop instantly. You can’t start and stop instantly in the real world either.
Here is where it gets harder. So these two signals, both 1KHz tones, separated by 1 microsecond arrived at these two ADCs. Let’s assume that Signal B arrives at Channel 2, 1 microsecond before Signal A arrives at Channel 1. To make the math easy for me, let’s assume that Signal A arrives at exactly 0 phase. Here are the digital outputs for the first 10 samples at 1KH and 20KHz. This is a DC offset AC signal, so the numbers go from 1 to 2^24.
You can easily tell these numbers do not represent the same signal, there is definitely something different about them. Your next question may be about accuracy / resolution. Jitter will obviously impact the inter-channel timing accuracy. I have not looked at the math in a while, but as you approach the SNR, I remember there is an increase in the inter-channel timing uncertainty.
- 1KHZ
- Ch1 / Ch2
- 8,388,608 / 8,441,314
- 8,915,333 / 8,967,925
- 9,439,979 / 9,492,249
- 9,960,476 / 10,012,218
- 10,474,769 / 10,525,779
- 10,980,830 / 11,030,906
- 11,476,660 / 11,525,605
- 11,960,303 / 12,007,923
- 12,429,850 / 12,475,958
- 12,883,448 / 12,927,862
We can do it at 20Khz as well
- Ch1 / Ch2
- 8,388,608 / 9,439,979
- 16,366,648 / 16,628,630
- 13,319,308 / 12,429,850
- 3,457,907 / 2,646,210
- 410,567 / 798,368
- 8,388,607 / 9,439,979
- 16,366,648 / 16,628,630
- 13,319,308 / 12,429,850
- 3,457,907 / 2,646,210
- 410,567 / 798,368
- Ch1 / Ch2
- 265 / 298
- 281 / 314
- 298 / 331
- 314 / 347
- 331 / 362
- 347 / 378
- 362 / 393
- 378 / 407
- 393 / 421
- 407 / 434
terry91,067 posts11-15-2019 12:29amThe nature of the signals is irrelevant. It is the relative timing of the encoding that matters. If the sampling rate is not high enough, or the jitter rate not low enough, then two signals differing by 1 microsecond will be encoded as identical.
Perhaps an example will help you to understand my confusion. It seems to me that if sampling is done at a frequency of 1Hz, and two signals differing by 1 us are detected, they will be encoded in the same pulse about 999,999 times out of 1,000,000. Which logically implies that sampling rate is intrinsic to the issue.
Perhaps you could point out the source of my confusion.