Why AWGN is additive?

Hi,

How do we know that the noise in AWGN is additive not multiplicative? on the same tone, how do we know that the fading noise is multiplicative not additive?

Thanks

The simple answer is the behaviour of noise in systems or communications channels is more complex than being simply addative or multatiplitive.

Thus it is realy to do with how "wee model the complex or "random signals" of which noise is comprised.

What we do is to make simple but realistic models of the various aspects of noise signal behaviour in a channel and deal with each seperatly. Thus as the adative aspects tend to be dominant in most channels we deal with those first and then go onto other asspects that are multitative then to various power laws.

Thus AWGN is the "simple" usually "dominant" case treated before the other more complex cases.

Have a look at,

Additive white Gaussian noise - Wikipedia, the free encyclopedia

I still do not understand. I mean in practice, we measure at the receiver a noisy version. How they have concluded that this noise is additive, or multiplicative?

The way they determin the effects is by experiment.

If you pick a particular measurment type and then run a test with the noise or the signal at various measured levels you can then draw a graph of the inputs and the outputs.

The difference between the two lines on the graph will tell you what the result is in effect.

Having worked out what the effect is you work backwards to the cause, then device a more specific test in the forward direction to verify the hypotheses, such is the scientific method.

But as I said noise behaves in all sorts of ways depending on what effect you are looking at.

Noise is in essence a statistical process and many of the calculated effects are derived from statistical measures. Like the growth rate of plants like grass you cannot predict exactly how fast a particular plant will grow but you can accuratly predict the growth rate of a large quantity of grass plants in a lawn or meadow.

So, it is a complicated process, and for I as a communication engineer have to accept this without verifying it, right?

David83 said:

I still do not understand. I mean in practice, we measure at the receiver a noisy version. How they have concluded that this noise is additive, or multiplicative?

Click to expand...

Hi David83,
For dealing with additive for AGWN, there are some experiments that you can check.
For example, if X1 is Gaussian R.V. and X2 is also Gaussian R.V., you can derive the summation of two R.V. (Y = X1+X2) is also a Gaussian R.V.
Based on this concept, you may record the noisy signal and check the additive property is still hold or not.

iamaeag said:

Hi David83,
For dealing with additive for AGWN, there are some experiments that you can check.
For example, if X1 is Gaussian R.V. and X2 is also Gaussian R.V., you can derive the summation of two R.V. (Y = X1+X2) is also a Gaussian R.V.
Based on this concept, you may record the noisy signal and check the additive property is still hold or not.

Click to expand...

I thought of this: the transmitted signal is not a Gaussian random variable, in general, but the received signal is (in AWGN channel) with mean equal to the transmitted signal. Then, we can conclude that the received signal is a noisy version of the transmitted signal, where the noise is in the form of additive Gaussian noise, because the sum of a constant with a Gaussian random variable is Gaussian with mean equals the constant. Right?

Thanks

It is not additive and can not be predicted about being additive subtractive or any thing else. BEcause of the randomness.
Why we take it additive???
Answer is simple to make our life simple in modelling the channel

tabraiz said:

It is not additive and can not be predicted about being additive subtractive or any thing else. BEcause of the randomness.
Why we take it additive???
Answer is simple to make our life simple in modelling the channel

Click to expand...

But then it will not be practical. I mean we will be modeling something else. Right?

David83 said:

I thought of this: the transmitted signal is not a Gaussian random variable, in general, but the received signal is (in AWGN channel) with mean equal to the transmitted signal. Then, we can conclude that the received signal is a noisy version of the transmitted signal, where the noise is in the form of additive Gaussian noise, because the sum of a constant with a Gaussian random variable is Gaussian with mean equals the constant. Right?

Thanks

Click to expand...

Hi David83,
Indeed the transmitted is also a R.V. and may not forced to be Gaussian. The received signal can be view as summation of two R.V. But when we want to analyze the noise signal from the received signal, we need to know what type of the transmitted signal is, i.e. BPSK or FSK modulated signal. If not, we can't say that the noise is gaussian when we found the additive property holds. Because there may have other R.V. that meet this criterion.
Finally go back to the interesting question, "Why AWGN is additive?". Before answer this question there are some assumptions that need to be made.
Thank you.
ps: please feel free to correct my reply.

iamaeag said:

Hi David83,
Indeed the transmitted is also a R.V. and may not forced to be Gaussian. The received signal can be view as summation of two R.V. But when we want to analyze the noise signal from the received signal, we need to know what type of the transmitted signal is, i.e. BPSK or FSK modulated signal. If not, we can't say that the noise is gaussian when we found the additive property holds. Because there may have other R.V. that meet this criterion.
Finally go back to the interesting question, "Why AWGN is additive?". Before answer this question there are some assumptions that need to be made.
Thank you.
ps: please feel free to correct my reply.

Click to expand...

Ok, so we can say that when the transmitted signal is not Gaussian, and the received signal is, then the noise is Gaussian and additive. To affirm this, we have to answer the question: is the addition of a non Gaussian RV with a Gaussian RV is the only way of getting a Gaussian received signal?

Any answer to this?

Regards

Yes i have the same question what happens when a IID RV whith unknown pdf is added to a zero mean gaussian noise.

Have you heard about Cramer's Decomposition theorem? It states the following, "If X and Y are independent real random variables and X+Y has normal distribution, then both X and Y are normally distributed."

Hi Pals,
I made some math calculations to the summation of two R.V.s.
Please feel free to correct my answer.
Thank you.