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Additive white Gaussian noise (AWGN) is a basic noise model used ininformation theory to mimic the effect of many random processes that occur in nature. The modifiers denote specific characteristics:

• Additive because it is added to any noise that might be intrinsic to the information system.
• White refers to the idea that it has uniformpower spectral density across the frequency band for the information system. It is an analogy to thecolor white which may be realized by uniform emissions at all frequencies in thevisible spectrum.
• Gaussian because it has anormal distribution in the time domain with an average time domain value of zero (Gaussian process).

Wideband noise comes from many natural noise sources, such as the thermal vibrations of atoms in conductors (referred to as thermal noise orJohnson–Nyquist noise),shot noise,black-body radiation from the earth and other warm objects, and from celestial sources such as the Sun. Thecentral limit theorem ofprobability theory indicates that the summation of many random processes will tend to have distribution called Gaussian or Normal.

AWGN is often used as achannel model in which the only impairment to communication is a linear addition ofwideband orwhite noise with a constantspectral density (expressed aswatts perhertz ofbandwidth) and aGaussian distribution of amplitude. The model does not account forfading,frequency selectivity,interference,nonlinearity ordispersion. However, it produces simple and tractable mathematical models which are useful for gaining insight into the underlying behavior of a system before these other phenomena are considered.

The AWGN channel is a good model for manysatellite and deep space communication links. It is not a good model for most terrestrial links because of multipath, terrain blocking, interference, etc. However, for terrestrial path modeling, AWGN is commonly used to simulate background noise of the channel under study, in addition to multipath, terrain blocking, interference, ground clutter and self interference that modern radio systems encounter in terrestrial operation.

The AWGN channel is represented by a series of outputs${\displaystyle Y_i}$ at discrete-time event index${\displaystyle i}$.${\displaystyle Y_i}$ is the sum of the input${\displaystyle X_i}$ and noise,${\displaystyle Z_i}$, where${\displaystyle Z_i}$ isindependent and identically distributed and drawn from a zero-meannormal distribution withvariance${\displaystyle N}$ (the noise). The${\displaystyle Z_i}$ are further assumed to not be correlated with the${\displaystyle X_i}$.

${\displaystyle Z_i\sim \mathcal{N}(0,N)}$

${\displaystyle Y_i=X_i+Z_i.}$

The capacity of the channel is infinite unless the noise${\displaystyle N}$ is nonzero, and the${\displaystyle X_i}$ are sufficiently constrained. The most common constraint on the input is the so-called "power" constraint, requiring that for a codeword${\displaystyle (x_1,x_2,\dots ,x_k)}$ transmitted through the channel, we have:

${\displaystyle \frac{1}{k}\sum _{i=1}^k{x}_i^2\le P,}$

where${\displaystyle P}$ represents the maximum channel power.Therefore, thechannel capacity for the power-constrained channel is given by:^{[clarification needed]}

${\displaystyle C=max\left\{I(X;Y):f\text{ s.t. }E\left(X^2\right)\le P\right\}}$

where${\displaystyle f}$ is the distribution of${\displaystyle X}$. Expand${\displaystyle I(X;Y)}$, writing it in terms of thedifferential entropy:

But${\displaystyle X}$ and${\displaystyle Z}$ are independent, therefore:

${\displaystyle I(X;Y)=h(Y)-h(Z)}$

Evaluating thedifferential entropy of a Gaussian gives:

${\displaystyle h(Z)=\frac{1}{2}\log (2\pi eN)}$

Because${\displaystyle X}$ and${\displaystyle Z}$ are independent and their sum gives${\displaystyle Y}$:

${\displaystyle E(Y^2)=E((X+Z{)}^2)=E(X^2)+2E(X)E(Z)+E(Z^2)\le P+N}$

From this bound, we infer from a property of the differential entropy that

${\displaystyle h(Y)\le \frac{1}{2}\log (2\pi e(P+N))}$

Therefore, the channel capacity is given by the highest achievable bound on themutual information:

${\displaystyle I(X;Y)\le \frac{1}{2}\log (2\pi e(P+N))-\frac{1}{2}\log (2\pi eN)}$

Where${\displaystyle I(X;Y)}$ is maximized when:

${\displaystyle X\sim \mathcal{N}(0,P)}$

Thus the channel capacity${\displaystyle C}$ for the AWGN channel is given by:

${\displaystyle C=\frac{1}{2}\log \left(1+\frac{P}{N}\right)}$

Suppose that we are sending messages through the channel with index ranging from${\displaystyle 1}$ to${\displaystyle M}$, the number of distinct possible messages. If we encode the${\displaystyle M}$ messages to${\displaystyle n}$ bits, then we define the rate${\displaystyle R}$ as:

${\displaystyle R=\frac{\log M}{n}}$

A rate is said to be achievable if there is a sequence of codes so that the maximum probability of error tends to zero as${\displaystyle n}$ approaches infinity. The capacity${\displaystyle C}$ is the highest achievable rate.

Consider a codeword of length${\displaystyle n}$ sent through the AWGN channel with noise level${\displaystyle N}$. When received, the codeword vector variance is now${\displaystyle N}$, and its mean is the codeword sent. The vector is very likely to be contained in a sphere of radius$\sqrt{n(N+\epsilon )}$ around the codeword sent. If we decode by mapping every message received onto the codeword at the center of this sphere, then an error occurs only when the received vector is outside of this sphere, which is very unlikely.

Each codeword vector has an associated sphere of received codeword vectors which are decoded to it and each such sphere must map uniquely onto a codeword. Because these spheres therefore must not intersect, we are faced with the problem ofsphere packing. How many distinct codewords can we pack into our${\displaystyle n}$-bit codeword vector? The received vectors have a maximum energy of${\displaystyle n(P+N)}$ and therefore must occupy a sphere of radius$\sqrt{n(P+N)}$. Each codeword sphere has radius${\displaystyle \sqrt{nN}}$. The volume of ann-dimensional sphere is directly proportional to${\displaystyle r^n}$, so the maximum number of uniquely decodeable spheres that can be packed into our sphere with transmission powerP is:

${\displaystyle \frac{(n(P+N){)}^{n/2}}{(nN{)}^{n/2}}=2^{(n/2)\log \left(1+P/N\right)}}$

By this argument, the rateR can be no more than${\displaystyle \frac{1}{2}\log \left(1+\frac{P}{N}\right)}$.

In this section, we show achievability of the upper bound on the rate from the last section.

A codebook, known to both encoder and decoder, is generated by selecting codewords of lengthn, i.i.d. Gaussian with variance${\displaystyle P-\epsilon }$ and mean zero. For large n, the empirical variance of the codebook will be very close to the variance of its distribution, thereby avoiding violation of the power constraint probabilistically.

Received messages are decoded to a message in the codebook which is uniquely jointly typical. If there is no such message or if the power constraint is violated, a decoding error is declared.

Let${\displaystyle X^n(i)}$ denote the codeword for message${\displaystyle i}$, while${\displaystyle Y^n}$ is, as before the received vector. Define the following three events:

1. Event${\displaystyle U}$:the power of the received message is larger than${\displaystyle P}$.
2. Event${\displaystyle V}$: the transmitted and received codewords are not jointly typical.
3. Event${\displaystyle E_j}$:${\displaystyle (X^n(j),Y^n)}$ is in${\displaystyle A_{\epsilon }^{(n)}}$, thetypical set where${\displaystyle i\ne j}$, which is to say that the incorrect codeword is jointly typical with the received vector.

An error therefore occurs if${\displaystyle U}$,${\displaystyle V}$ or any of the${\displaystyle E_i}$ occur. By the law of large numbers,${\displaystyle P(U)}$ goes to zero as n approaches infinity, and by the jointAsymptotic Equipartition Property the same applies to${\displaystyle P(V)}$. Therefore, for a sufficiently large${\displaystyle n}$, both${\displaystyle P(U)}$ and${\displaystyle P(V)}$ are each less than${\displaystyle \epsilon }$. Since${\displaystyle X^n(i)}$ and${\displaystyle X^n(j)}$ are independent for${\displaystyle i\ne j}$, we have that${\displaystyle X^n(i)}$ and${\displaystyle Y^n}$ are also independent. Therefore, by the joint AEP,${\displaystyle P(E_j)=2^{-n(I(X;Y)-3\epsilon )}}$. This allows us to calculate${\displaystyle P_e^{(n)}}$, the probability of error as follows:

Therefore, asn approaches infinity,${\displaystyle P_e^{(n)}}$ goes to zero and. Therefore, there is a code of rate R arbitrarily close to the capacity derived earlier.

Here we show that rates above the capacity${\displaystyle C=\frac{1}{2}\log \left(1+\frac{P}{N}\right)}$ are not achievable.

Suppose that the power constraint is satisfied for a codebook, and further suppose that the messages follow a uniform distribution. Let${\displaystyle W}$ be the input messages and${\displaystyle \hat{W}}$ the output messages. Thus the information flows as:

${\displaystyle W⟶X^{(n)}(W)⟶Y^{(n)}⟶\hat{W}}$

Making use ofFano's inequality gives:

${\displaystyle H(W\mid \hat{W})\le 1+nR{P}_e^{(n)}=n{\epsilon }_n}$ where${\displaystyle {\epsilon }_n\to 0}$ as${\displaystyle P_e^{(n)}\to 0}$

Let${\displaystyle X_i}$ be the encoded message of codeword indexi. Then:

Let${\displaystyle P_i}$ be the average power of the codeword of index i:

${\displaystyle P_i=\frac{1}{2^{nR}}\sum _w{x}_i^2(w)}$

where the sum is over all input messages${\displaystyle w}$.${\displaystyle X_i}$ and${\displaystyle Z_i}$ are independent, thus the expectation of the power of${\displaystyle Y_i}$ is, for noise level${\displaystyle N}$:

${\displaystyle E(Y_i^2)=P_i+N}$

And, if${\displaystyle Y_i}$ is normally distributed, we have that

${\displaystyle h(Y_i)\le \frac{1}{2}\log 2\pi e(P_i+N)}$

Therefore,

We may apply Jensen's equality to${\displaystyle \log (1+x)}$, a concave (downward) function ofx, to get:

${\displaystyle \frac{1}{n}\sum _{i=1}^n\frac{1}{2}\log \left(1+\frac{P_i}{N}\right)\le \frac{1}{2}\log \left(1+\frac{1}{n}\sum _{i=1}^n\frac{P_i}{N}\right)}$

Because each codeword individually satisfies the power constraint, the average also satisfies the power constraint. Therefore,

${\displaystyle \frac{1}{n}\sum _{i=1}^n\frac{P_i}{N},}$

which we may apply to simplify the inequality above and get:

${\displaystyle \frac{1}{2}\log \left(1+\frac{1}{n}\sum _{i=1}^n\frac{P_i}{N}\right)\le \frac{1}{2}\log \left(1+\frac{P}{N}\right).}$

Therefore, it must be that${\displaystyle R\le \frac{1}{2}\log \left(1+\frac{P}{N}\right)+{\epsilon }_n}$. Therefore,R must be less than a value arbitrarily close to the capacity derived earlier, as${\displaystyle {\epsilon }_n\to 0}$.

In serial data communications, the AWGN mathematical model is used to model the timing error caused by randomjitter (RJ).

The graph to the right shows an example of timing errors associated with AWGN. The variable Δt represents the uncertainty in the zero crossing. As the amplitude of the AWGN is increased, thesignal-to-noise ratio decreases. This results in increased uncertainty Δt.^[1]

When affected by AWGN, the average number of either positive-going or negative-going zero crossings per second at the output of a narrow bandpass filter when the input is a sine wave is

where

ƒ₀ = the center frequency of the filter,

B = the filter bandwidth,

SNR = the signal-to-noise power ratio in linear terms.

In modern communication systems, bandlimited AWGN cannot be ignored. When modeling bandlimited AWGN in thephasor domain, statistical analysis reveals that the amplitudes of the real and imaginary contributions are independent variables which follow theGaussian distribution model. When combined, the resultant phasor's magnitude is aRayleigh-distributed random variable, while the phase is uniformly distributed from 0 to 2π.

The graph to the right shows an example of how bandlimited AWGN can affect a coherent carrier signal. The instantaneous response of the noise vector cannot be precisely predicted, however, its time-averaged response can be statistically predicted. As shown in the graph, we confidently predict that the noise phasor will reside about 38% of the time inside the 1σ circle, about 86% of the time inside the 2σ circle, and about 98% of the time inside the 3σ circle.^[1]

• Ground bounce
• Noisy-channel coding theorem
• Gaussian process

• Noise (electronics)
• Time series models

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