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sampling theory, recovery and resolution

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shanmei

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If a signal m(t) has a maximum frequency of 1KHz, and a 10 bit ADC with 10K samples/second rate to sample the signal m(t).

According to the sampling theory, if the sample rate is larger than twice of the maximum frequency of the signal m(t), then the signal m(t) can be recovered. The "recovered" means that the signal m(t) is fully recovered or fully reconstrued.

However, the ADC resolution is 10 bit, and the reconstructed signal from the ADC to recover the original signal m(t) then has an only 10-bit resolution even with very high sampling rate( > 2*maxium frequency of m(t)) . The higher solution informtion 11 bit , 12bit... are lost.

This is conflict with the sample theory that m(t) can be recovered without losing any information.

I am confused. Can you please help? Thanks.
 
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Hi,

then the signal m(t) can be recovered
Yes.
But be sure that the signal is
* sinusoid
* or a mixture of sinusoidal signal where each signal frequency is lower than f_sampling/2.

In other words: you can´t reconstruct a signal where the overtones (frequency) are higher than f_sample/2.

The higher solution informtion 11 bit , 12bit... are lost.
Yes, somehow.

But this does not mean that the ouput amplitude p-p of a sine can only have 10 bits of resolution.
Due to many samples per full wave you gain resolution of the reconstructed signal.

One can optimize this with dithering for example.
But for sure there are limitations.

The "lost infomation" causes noise. Quantisation noise.

The higher the voltage resolution, the ess the "lost information" the lower the quantisation noise.

Klaus
 

If a signal m(t) has a maximum frequency of 1KHz, and a 10 bit ADC with 10K samples/second rate to sample the signal m(t).

According to the sampling theory, if the sample rate is larger than twice of the maximum frequency of the signal m(t), then the signal m(t) can be recovered. The "recovered" means that the signal m(t) is fully recovered or fully reconstrued.

However, the ADC resolution is 10 bit, and the reconstructed signal from the ADC to recover the original signal m(t) then has an only 10-bit resolution even with very high sampling rate( > 2*maxium frequency of m(t)) . The higher solution informtion 11 bit , 12bit... are lost.

This is conflict with the sample theory that m(t) can be recovered without losing any information.

I am confused. Can you please help? Thanks.

For sure they are lost. But the technique surfices due to the fact that you select an adc based on requirements. If such level of details is important, then a higher resolution is the way out, otherwise the straight line approximation between every two consecutive points is enough information recovered.

Add: Realize that the actual information lies in the fundamental frequencies of each information and not in the harmonics so if you can sample at twice the highest frequency component of the signal, information, waveform, whatever, then all fundamental info are intact at recovery.
 
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If the sample is dithered with a small amount of noise of around 1-bit, then you can improve the resolution by the square-root of the number of samples above the Nyquist limit.
Thus for a 10kHz sample rate and a 1kHz maximum signal, the oversample rate is 5.
The theoretical improvement in resolution is thus √5 = 2.2 or an additional 1 bit.
 

This is conflict with the sample theory that m(t) can be recovered without losing any information.
.

The Information is Lost While Sampling Itself-How can you expect to recover Signal while reconstructing it...
Sampling Theorem Requires the Sampled Signal to be a Precise Value While Sampling. What you sample is what you recover.

This article about Sampling Theorem - National Semiconductor Gives good Insight
 

For sure they are lost. But the technique surfices due to the fact that you select an adc based on requirements. If such level of details is important, then a higher resolution is the way out, otherwise the straight line approximation between every two consecutive points is enough information recovered.

Add: Realize that the actual information lies around the fundamental frequencies of each information and not around the highest-frequency-component harmonics so if you can sample at twice the highest frequency component of the signal, information, waveform, whatever, then all fundamental info are intact at recovery.

Basically, the higher frequency that you sample, the more accurate signals around the fundamental frequencies become. The highest frequency component many many multiple times the frequency of the fundamental.

Let's say the fundamental is 100Hz and the sample rate is 10kHz, then you'll have 100 sample points per cycle. That's going to be accurate enough at recovery. Several higher harmonics are going to be also accurate and as such the information recovery is deemed accurate.

Take for instance if you're sampling music at a little above 2*20kHz, then the music still retains all their properties at recovery. That's much of what the criteria entails.
 
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Hi,

Now the first time is see "music" in this thread.

Digital audio are optimized in this. They have steep antialiasing filters at the ADC and almost perfect reconstruction filters (using oversampling and so on).
I assume you don´t have that perfect filters, thus you should expect a lot of drawback, especially at higher frequencies.

An example:
If you use
* 10kHz sampling frequency
* a first order filter with (oftten used, but in my eyes not useful) fc = 3000Hz
* an input signal of 2000kHz with a 3´rd overtone with 5%

* then the filter will attenuate the 2000Hz by 1.6dB with a phase shift of 34°. (Not sure if this matters for you)
* the 3rd overtone is 6000Hz. It is attenuated by about 7dB.. but still causing a 2.2% amplitude.
But after conversion you won´t see 6000Hz but you will see 4000 Hz alias frequency (still 2.2% amplitude).

A good audio device will have a passband ripple of less than 0.1dB passband ripple up to 0.45 x fs while giving a stop band attenuation of more than 50dB at 0.6 x fs.

***
Thus it may be a good idea to use an audio device for AD and DA conversion for very good results. And they are cheap and good available, too.
All depends on the requirement.

Klaus
 
Thanks for all your relies.

You are right that the sampling process is not ideal. For example, at the sampling time, if the original signal is 2V, but the sampled voltage might be 1.999V. The sampled voltage lost information. The original signal can not be recovered from the sampled signal.

If the ADC is ideal, which means the sampling circuit is perfect and the resolution is infinite. Therefore, the sampled voltage is 2V and the original signal can be recovered perfectly.

The sampling theory is right, but in real life, we do not have an ideal ADC. Only with an ideal ADC, we can recover the signal.

In the real life, no signal can be truly recovered due to the non-ADC. We say it is "recovered" because we only care about the limited resolution.

Thank you all.
 

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