Title: **What bit-depth is required for low frequency sine**

Post by:**Serge Smirnoff** on **2018-09-18 00:47:12**

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If I understand correctly, for mathematically perfect quantization of 20Hz sine wave at 44.1k sampling rate at least 18 bits are required. If we assume that samples of the sine wave around its min/max (the slowest parts) should differ at least with one bit, then it's easy to find number of bits required for quantization of the whole sine period (for full scale signal):

*Bit-depth = log2(1/(1-cos(2*pi*F/Fs)))*, where

*Fs* - sampling freq.

*F* - freq. of sine wave

For F = 20 Hz:

Or I miss something?

(yes, I understand that this doesn't affect perceived sound quality )))

For F = 20 Hz:

- @44100 Bit-depth = 18 bit (17.91)
- @48000 Bit-depth = 19 bit (18.15)
- @96000 Bit-depth = 21 bit (20.15)
- @192000 Bit-depth = 23 bit (22.15)

Or I miss something?

(yes, I understand that this doesn't affect perceived sound quality )))

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**DVDdoug** on **2018-09-18 02:06:17**

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Quote

If I understand correctly, for mathematically perfect quantization of 20Hz sine wave at 44.1k sampling rate at least 18 bits are required.Quantization is never mathematically perfect.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**lvqcl** on **2018-09-18 04:42:16**

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If we assume that samples of the sine wave around its min/max (the slowest parts) should differ at least with one bit

That's rather strange assumption.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**saratoga** on **2018-09-18 05:18:17**

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Yeah if they have to differ by 1 bit then things like dsd or pcm with 1 bit dacs would be impossible. In reality you can have two samples in a row be identical for a sin wave and all that means is that you're oversampled, and all that does is give you a bit of quantization noise.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**fyrstormer** on **2018-09-18 06:49:59**

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I'm not sure why bit-depth would affect regeneration of a pure sinewave. D/A converters inherently use the mathematical properties of a pure sinewave as their model for interpolating missing data between samples. Even if the hardware designer *wanted* the D/A converter to produce a stepped waveform instead, the resistance and inductance of the wires carrying the output signal would round-off the edges of the stepped waveform to make it more sinusoidal. Hypothetically you could record a pure sinewave at a bit-depth of 1 and any D/A converter worth paying for should still be able to reconstruct a pure sinewave from the recording.

In fact, at the hardware level, this is precisely how PWM D/A converters work -- a bit value of 1 tells the D/A converter to increase the slope of the output waveform by a fixed rate, and a bit value of 0 tells the D/A converter to*decrease* the slope of the output waveform by a fixed rate. (alternating bits of 1 and 0 tell the D/A converter to not change the slope of the output waveform at all, by effectively cancelling each other out.)

In this way, a stream of 1's and 0's can describe a waveform of any shape -- the maximum detail of the output waveform is affected only the the sample rate, not the bit depth. Granted, digital audio is generally recorded and played using PCM instead of PWM, but the basic idea remains the same. So your pure sinewave should be able to be accurately encoded at any bit-depth, provided the sample rate is at least 2x the frequency of the sinewave.

In fact, at the hardware level, this is precisely how PWM D/A converters work -- a bit value of 1 tells the D/A converter to increase the slope of the output waveform by a fixed rate, and a bit value of 0 tells the D/A converter to

In this way, a stream of 1's and 0's can describe a waveform of any shape -- the maximum detail of the output waveform is affected only the the sample rate, not the bit depth. Granted, digital audio is generally recorded and played using PCM instead of PWM, but the basic idea remains the same. So your pure sinewave should be able to be accurately encoded at any bit-depth, provided the sample rate is at least 2x the frequency of the sinewave.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**Serge Smirnoff** on **2018-09-18 17:35:20**

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Thanks for the replies. The situation with sine wave is clear - it can be successfully quantized with two bits only. Agree.

I expressed my idea not clearly. I wanted to talk about different aspect of quantization. So, I will ask another way.

If we take any band-limited signal sampled at 44100 and normalized to 1.0, what will be the minimum distance between two neighboring samples. I noticed that there is a clear bottom limit for that distance even for 64bit signals. And my thought was that this min distance is determined by the lowest frequency component of the signal.

I expressed my idea not clearly. I wanted to talk about different aspect of quantization. So, I will ask another way.

If we take any band-limited signal sampled at 44100 and normalized to 1.0, what will be the minimum distance between two neighboring samples. I noticed that there is a clear bottom limit for that distance even for 64bit signals. And my thought was that this min distance is determined by the lowest frequency component of the signal.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**saratoga** on **2018-09-18 17:42:54**

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It depends on the frequency of the signal. If it is at the Nyquist limit, the distance between samples has to be equal to the amplitude. If your signal is well below the Nyquist limit, you will have smaller and smaller differences, culminating in a DC signal for which the difference is always zero (in the absence of dither and noise shaping at least).

You can easily calculate the minimum step size for a given frequency and sampling rate by the way. It's just the change in amplitude between two sequential samples on either size of the maximum or minimum of a sin wave. When that step size gets less than the quantization level, and if you are not dithering or using noise shaping, the step size will be zero.

You can easily calculate the minimum step size for a given frequency and sampling rate by the way. It's just the change in amplitude between two sequential samples on either size of the maximum or minimum of a sin wave. When that step size gets less than the quantization level, and if you are not dithering or using noise shaping, the step size will be zero.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**Serge Smirnoff** on **2018-09-18 18:03:53**

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Yes, that's what I did in my first post. Do we have similar relationship for non-periodic signals (but band-limited at low and high freq.)?

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**saratoga** on **2018-09-18 18:16:08**

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Yes, that's what I did in my first post. Do we have similar relationship for non-periodic signals (but band-limited at low and high freq.)?

Not really. For an arbitrary signal anything is possible. Image an isolated square pulse for instance, which is bandlimited but could have both a full scale change between two samples (the rising/falling edges), and then no change for several samples (the top of the pulse).

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**Serge Smirnoff** on **2018-09-18 18:26:24**

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Yes. Right you are (I checked)). The question is closed. Thank you.

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**KMD** on **2018-09-27 14:25:13**

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To prevent the sample of the sine wave dwelling at one quantization level the sine wave is dithered between adjacent quantisation levels, then the D/A filter can do its job.

ref, Resolution Below the Least Significant Bit in Digital Systems with Dither John Vanderkooy and Stanley P. Lipshitz

ref, Resolution Below the Least Significant Bit in Digital Systems with Dither John Vanderkooy and Stanley P. Lipshitz

Title: **Re: What bit-depth is required for low frequency sine**

Post by:**jsdyson** on **2018-09-28 07:31:16**

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I know that the answer is answered -- but I want to help make it clear that the dithering really works... Cell systems often have receivers with perhaps 12bit/14bit (more or less) A/D converters. Even with such coarse grained A/D converters a very wide dynamic range of radio signals can be very accurately received. A SW receiver that is sensitive to between 1uV (uselessly sensitive for SW) all the way to perhaps 1V can be digitized by an A/D of 12bits. It all depends upon dithering, over sampling and a bit of DSP. Intuitively, one might htink that there might be troubles lower than 250uvolts, but that isnt' true. Digital signals can easily transmit information that is less than one bit in signal level. I ihave heard of studies that people can detect tones with their hearing at less that one bit in level.

Once you read about what is going on, and a little bit of re-enforcing experience (the experience REALLY helps), then the whole matter of 'stair stepping' becomes unimportant (as long as the signal is dithered and filtered/processed correctly.)

Once you read about what is going on, and a little bit of re-enforcing experience (the experience REALLY helps), then the whole matter of 'stair stepping' becomes unimportant (as long as the signal is dithered and filtered/processed correctly.)