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My SNR is better with DC Free transmission. Why?

Example:

1.) Datasignal with DC:

Datastream consisting of 0V and 1V.

2.) Datasignal without DC:

Datastream consisting of -0.5V and 0.5V.

The variance is the same at both examples.

So I don't get, why the SNR is better without DC.

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  • \$\begingroup\$ It's not as a rule, unless you are sensitive to 1/f noise, show your assumptions or real issue \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 19 at 23:08
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Noise performance depends on the difference between levels. Since both your signals have a difference of 1 volt, they have the same resistance to noise.

However, they have different energy. Let's say that your data signal has a period of 1 second. Then, the pulses you transmit have the following energy:

  • The 0V pulse has energy 0 J.
  • The 1V pulse has energy 1 J.

Then, on average your signal requires 0.5 J per pulse.

The second signal uses pulses of the following energy:

  • The 0.5 V pulse has energy 0.25 J.
  • The -0.5 V pulse has energy 0.25 J.

So, on average the energy in this signal is 0.25 J per pulse.

So, removing DC results in a signal with the same error performance but less energy. This is equivalent to saying that you have a larger SNR.

(Note that I have made an assumption that is common in communications, that the load resistance is \$R_L = 1\,\Omega\$. In any case, when comparing the SNR of the two signals, \$R_L\$ cancels out, so its actual value is not important).

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