What is additive noise? What actually additive means?

I have tried to search in the internet but the only answer that I get is the noises gets added therefore it is additive, which make me think that noises are not destructive in nature am I right?


noises gets added therefore it is additive


which make me think that noises are not destructive

incorrect :(

simple thought experiment: You flip a fair coin $X$ (Head = -1 / Tail = 1) and tell me the result. The entropy here is 1 bit, i.e. the (expected) information ($I(X=\xi) = -\log_2 \left[P(X=\xi)\right]$) of each outcome is 1 bit.

Then there's additive noise $N$ that takes one of the values $\{-2,0,+2\}$ with equal probability.

When you receive a -1, you can't know whether the coin was Head and there was 0 noise, or the coin was Tail and there was -2 noise. Both are equally likely!¹

So, your additive noise is absolutely able to destroy information and hence is very destructive to your signal.

If you're more coming from a wireless communications background: Your $X\in\{-1,+1\}$ can be interpreted as BPSK. Now you see how even benign Gaussian noise destroys your reception when its sign is the opposite of your transmit symbol!

¹ we can even formalize that. Since $X$ (2 options) and $N$ (3 options) are independent, and each of them equidistributed, there's six possible combinations, each of them equally likely

 X | N  | Y = X+N
-1 | -2 | -3
-1 |  0 | -1
-1 | +2 | +1
+1 | -2 | -1 
+1 |  0 | +1
+1 | +2 | +3

Thus, we have four possible outcomes for the sum of signal and additive noise, -3, -1, +1 and +3.

  • If we see +3 or -3, we get 1 bit of the 1 bit info in the coin toss. (It must have been +1, otherwise we couldn't get +3, or -1 for -3, respectively.) That happens in 2 out of 6 times, so with probability 1/3.
  • If we see -1, we don't know whether it was +1-2 or -1+2, so we have zero bit of the 1 bit of the coin toss. Same for +1. That happens 4 out of 6 times, so with probability of 2/3.

Thus, the expected information to get out of this channel is 1/3·1 + 2/3·0 bit = 1/3 bit, where put in full 1 bit! That's a very destructive additive noise channel.

  • $\begingroup$ The written explanation does not conform to the math. How can one "receive a $0$" when the input can be $\pm 1$ and noise is in $\{-2,0,+2\}$? As Marcus's own table shows, the only possible received values are $\{\pm 1, \pm 3\}$ and so it is impossible to "receive a $0$" as in the explanation. $\endgroup$ – Dilip Sarwate Oct 2 '20 at 14:42
  • $\begingroup$ ah shoot, an error introduced when simplifying things, @DilipSarwate. I'll fix this in 30s. $\endgroup$ – Marcus Müller Oct 2 '20 at 14:44
  • $\begingroup$ Thank you very much... I Never expected such a beautiful and simple explanation from both of you!!! $\endgroup$ – Suraj Kumar Oct 2 '20 at 16:45

To complement Marcus' answer:

Say you have a resistor (nothing is connected to it) at a certain temperature above absolute zero. The heat causes electrons to move around at random, creating a random current. This current through the resistor creates a random voltage.

If you connect a sensitive-enough voltmeter to the resistor, you can detect this voltage -- but, in practice, you have to be careful not to measure the random currents inside the voltmeter itself!

Now, imagine you connect a signal source to one end of the resistor, and you ground the other end. The source may be, for example, an antenna. The signal source will create a voltage across the resistor.

Now this is the key part: the voltage created by the source will add up with the random voltage caused by heat. This happens because a resistor is linear, in the sense that all currents applied to it are added. That's just the way a resistor works (I don't know if there is a fundamental explanation for this).

In brief, if the random noise is called $n(t)$, and the signal source is called $v(t)$, then the voltage across the resistor is $v(t)+n(t)$ -- and that is why $n(t)$ is called additive noise.


  • This example is about thermal noise -- there are other kinds of noise, most of them additive.

  • Since the noise $n(t)$ is the cumulative effect of billions of electrons moving at random, the central limit theorem applies and the probability density function of the noise will be Gaussian.

  • 1
    $\begingroup$ nice, thank you! $\endgroup$ – Marcus Müller Oct 2 '20 at 14:15
  • 1
    $\begingroup$ Cleared my doubt thanks a lot!! $\endgroup$ – Suraj Kumar Oct 2 '20 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.