# [SOLVED]Variance of WSS white noise process

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#### hunter555persia

##### Member level 2
Hi, from the theory of stochastic processes we know that for WSS zero mean processes, variance would be the autocorrelation function at zero i.e. Rx(0).
But consider the white noise process: Rx(τ) = δ(τ)No/2 and Rx(0)=∞. But we always consider No/2 as noise variance. Why is that ??

Thanks for any help.

Hi Hunter

No/2 is not the noise variance (whose unities would be [Watts]) but the noise spectral density (whose unities would be [Watts/Hertz]).
Regards

Z

What is the variance of this white process then? Is Rx(0)=∞ the variance ?? ∫S(w) also yields ∞.

Right.
A "real life" process can not be "really" white.
"White" means "that has uniform power spectral density over all de bandwidth of interest".
Regards
Z

In performance analysis we are interested in a plot of BER Vs Eb/No. Eb/No is somehow a measure of SNR which means that No/2 was considered as noise power. And noise power equals its variance. Actual noise variance would be ∫S(w)dw over the receiver bandwidth. But No/2 is always considered as noise variance without considering the receiver bandwidth in the texts I've read e.g. Proakis.

Eb is energy. Units: [Joule].
No is power spectral density. Units: [Watt/Hz]=[Watt*second]=[Joule]
So, Eb/No is adimensional, as it must be.

No is not noise variance or power. It is power spectral density.
Sometimes it is said that "numerically, No equals the power of the noise in the bandwidth of 1 Hz". But this is rather confusing.

So why do we produce the noise with variance of No/2 in simulations ?

Is simulations you are in discrete time. Frequency does not cover -Inf to +Inf, but -pi to +pi [radians/sample], or -fs/2 to +fs/2 [Hz] (fs is sampling frequency).

The simulations are discrete time. But sampling frequency is not considered. The data and noise vector are sumed and then decoded. Noise is always produced with variance No/2.

You are using discrete time for simulate a continuous-time system, rigt?
I don't know what are you simulating and how, but since you mention that ∫S(w) yields ∞, the simulated system is continuous-time.
So, there is a sampling frequency.

I mean ∫S(w)dw=∞ theoretically. In simulation S(w) is not used. Not in my simulation, but in any BER simulation I have seen before, there is no concern about the sampling frequency. I should think about the sampling frequency though. It may be considered unity I guess. Thanks for pointing out this.

But the main question is the variance of a white process. It is infinity then, right ?

Right.
Take into account what I said in post #4.
Regards

Z

The white noise correlation is:

$R(\tau)=\frac{N_0}{2}\delta(\tau)$

So the power spectral density is:

$\mathcal{S}_{nn}(f)=\int_{-\infty}^{\infty}R(\tau)e^{-j2\pi f\tau}\,d\tau=\frac{N_0}{2}\int_{-\infty}^{\infty}\delta(\tau)e^{-j2\pi f\tau}\,d\tau=\frac{N_0}{2}$

and the autocorrelation is then:

$R(\tau)=\int_{-\infty}^{\infty}\mathcal{S}_{nn}(f)e^{j2\pi f\tau}\,df$

The average power is then:

$R(0)=\sigma_n^2=\int_{-\infty}^{\infty}\mathcal{S}_{nn}(f)\,df=\int_{-\infty}^{\infty}\frac{N_0}{2}\,df$

Now as zorro pointed out, the average power is taken over the bandwidth of interest. So:

$\sigma_n^2=\int_{-W}^{W}\frac{N_0}{2}\,df=N_0W$

Then the SNR will be:

$\text{SNR}=\frac{P}{\sigma_n^2}=\frac{P}{N_0W}=\frac{PT}{N_0}=\frac{E}{N_0}$

Hope this will help.

Regards

I guess this is what I was looking for: $\text{SNR}=\frac{P}{\sigma_n^2}=\frac{P}{N_0W}=\frac{PT}{N_0}=\frac{E}{N_0}$

Thanks guys for the help.

I made note concerning noise power in simulations. I hope it's clarifying.

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