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What is the SNR in this ADC ?

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silver_sparrow

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Finding SNR

Hello,

This equation is for an ADC:

SNR = 6.02n + 1.76 dB , where n is the number of bits

However, this equation has a few assumptions and is invalid
if your input clips and saturates the ADC.
Assuming my input signal does clip and has a guassian distribution,
what is the SNR then?

can someone point me here to a source/literature?

thanks!
sparrow
 

Re: Finding SNR

This is the maximum SNR that you can reach by using this ADC. And keep in mind that this implies your signal has a peak to average power ratio of 0dB. For a Gaussian distributed signal, you have to have a big headroom for the signal not to clip.
 

Re: Finding SNR

It is farely easy.

Given the average input power, with Gaussian distribution, you can easily calculate the distortion power due to clipping. In the case of large number of bits in ADC, this will be the main contribution of the noise. The next, if you are still concern about the noise caused by rounding. The power is delta/12, where delta represents the power of the MSB.

Usually you can ignore the rounding noise when you have a lot of clipping.

By the way why do you want the signal to be clipped?
 

Re: Finding SNR

The equation also is for sampling at the Nyquist rate and not filtering the output to just include the signal frequencies.

I have seen graphs of what you are seeking, but not equations. The SNR with signal level graph has a slope of 1 on the low signal side and [edited part: 3] on the high signal side of the maximum. It is rounded on the peak.

To be more specific, the graphs are really of Noise Power Ratio which includes the effects of the quantization noise and the IMD from clipping added together.

Since writing this I have found a document on the topic. It is in .ps format and requires acrobat distiller to convert to .pdf. Since you may not have this I have done it and uploaded it here.
 

Re: Finding SNR

A signal with guassian distribution will have -9dB less than the maximun of SNR = 6.02n + 1.76 dB.

Bastos
 

Re: Finding SNR

One final thought. Both of the above are for ideal converters. Many years ago I measured real hardware and the peak was more rounded than shown in the graphs. I suspect that it was due to slew rate limiting starting and reducing the gain of the feedback loops in the converter. Also, the LSB usually only contributed 3 dB to the NPR so we ended up using a converter with more bits than we needed and did not connect the lower ones.
 

Re: Finding SNR

tohji: sometimes you want to clip your signal if you have a very high
peak-to-average power ratio. This way you tradeoff clipping for
better usage of the ADC range for most part of your signal.

flatulent: thnx for the troubles. the paper has what i'm looking for.
i owe you a beer ;)

bastos4321: -9dB less...? what does this assume about the variance of the
signal? hmmm... bastos you joined elektroda since
'Joined: 01 Jan 1970'? heh!
 

Re: Finding SNR

I have seen that number (-9dB) some time ago in a paper. The variance will be equal to the signal power. Assuming a zero mean gaussian distribution.

Bastos.

Yeh Im very old ..... :lol:
 

Re: Finding SNR

In the WCDMA design. I suggested to use 7 bits A/D given a analog filter with 40dB adjacent channel rejection ratio. The headroom suggested is 12dB. However, some other people suggested to use 20dB of headroom. Do you think it is necessary?
 

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