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?

correct!

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.

• 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. – Dilip Sarwate Oct 2 '20 at 14:42
• ah shoot, an error introduced when simplifying things, @DilipSarwate. I'll fix this in 30s. – Marcus Müller Oct 2 '20 at 14:44
• Thank you very much... I Never expected such a beautiful and simple explanation from both of you!!! – Suraj Kumar Oct 2 '20 at 16:45

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.

Notes:

• 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.

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