I am not sure why one of the advantages of Digital vs Analog communication is higher SNR?

For example, when I want to transfer a 4.5v signal vs a 5v signal, how does converting these values to some digital levels first (e.g. using a 16-bit ADC) will help the communication path to have higher SNR?

Also, I read an article which points out the followings:

  • SNR of a digital signal determined by the maximum discrete amplitude and the resolution of the A/D (volts/count).
  • SNR of analog signal determined by power supply rails and by noise floor.

Could someone please explain what that means also? and why?

  • \$\begingroup\$ You'll have to provide a link to the article in order for us to comment meaningfully on it. \$\endgroup\$
    – Dave Tweed
    Feb 5, 2014 at 16:32
  • 1
    \$\begingroup\$ The point is probably something like a trade-off between bandwidth and SNR. If you convert a signal to digital at CMOS levels you could have 1V of noise in the communication channel and still faithfully transmit a signal accurate to 20 bits (1ppm, or 5uV with a 5V analog signal). \$\endgroup\$ Feb 5, 2014 at 16:48

1 Answer 1


Spehro's comment is basically the answer. Imagine a link which subtracts a random, varying noise value between 0 and 0.5V from the input. You measure 4.5V at one end, but that could mean either a 4.5V input with no noise or 5V minus noise. You can't tell, and have lost your signal entirely.

With a digital signal varying between 5V and 0V over the same link, you can declare everything over 2.5V to be a '1', and lose no signal.


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