2
$\begingroup$

I am sorry for such a noobie question.

Continuous time (Clear part)

Suppose that we have continuous time "AWGN" noise $n(t)$ with bandwidth $W$ and power spectral density $N_0/2$, i.e. the Fourier transform $\widehat{R_n}(f)$ of the autocorrelation function is:

(1) $\widehat{R_{n}}(f)$ is $N_0/2$ for all $f \in [-W,W]$, 0 otherwise.

Now since the power of continuous noise power is defined as $E(\frac{1}{2T}lim_{T\rightarrow \infty}\int_{-T}^{T}n^2(t)dt)$, it is actually $E(n^2(0))$. In turn is the variance $\sigma^2$ of continuous time Gaussian variable n(0). Moreover, by calculating the inverse Fourier transform of Fourier transform of autocorrelation (see (1)) we have $\sigma^2=N_0W$: $\int_{-\infty}^{\infty}\widehat{R_n}(f)e^{2\pi ift}df=N_0W$.

Thus we obtain that continuous time noise power as

(2) $E(n(0)^2)=N_0W=\sigma^2$ .

Suppose we sample with Nyquist rate $1/2W$, i.e. at times $\frac{k}{2W}$, $k\in Z$:

Question 1 How we define the sampled noise power for sampled model then?

Question 2 What happens if not the Nyquist sampling is used?

I can naturally define the discrete noise power as the expected value of $2W\sum_{k\in{0..2W-1}}n^2(k/2W)$, since the sum is the sum of random variables. Due to linearity of expectation, it is a equal to continuous noise power $E(n(0)n(0)^*)=\sigma^2=N_0W$, which is very different for the text book value $N_0/2$ for discrete time AWGN variance.

p.s. Proakis defines the discrete time AWGN power as the integral of spectral noise density $N_0/2$ from $-1/2$ to $1/2$, so $N_0/2$ total, but the intuition behind is unclear for me.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

No need to apologize; this is a very good question.

What you are looking at is a band-limited (lowpass in this case instead bandpass) white Gaussian noise process $\{N(t)\colon -\infty < t < \infty\}$ which means is that each random variable $N(t)$ is a zero-mean Gaussian random variable with variance $\sigma^2 = N_0W$ where $W$ is the bandwidth (in Hertz). The power spectral density of this lowpass white noise process is $$S_N(f) = \begin{cases}\frac{N_0}2, &-W \leq f \leq W,\\ 0, & |f| > W\end{cases} = \frac{N_0}2 \operatorname{rect}\left(\frac{f}{2W}\right)$$ while its autocorrelation function is $$R_N(\tau) = E[N(t)N(t+\tau) = \frac{N_0}2\cdot 2W \operatorname{sinc}(2W\tau) = N_0W\operatorname{sinc}(2W\tau).$$ Since $\operatorname{sinc}(2W\tau) = 0$ whenever $\tau$ is such that $2W\tau$ is a nonzero integer, we have that the zero-mean random variables $N(t)$ and $N(t+\frac{1}{2W})$ spaced $\frac{1}{2W}$ seconds apart are uncorrelated random variables, and since they are jointly Gaussian, we can also claim that they are independent random variables; ditto for spacing $\frac{n}{2W}$ seconds apart. However, all other pairs of random variables are indeed correlated (and thus not independent) though this correlation decreases as $|n|$ increases.

Turning to sampling this process to create a discrete-time process, we see that sampling at the Nyquist rate ($2W$ samples per second) means the samples are exactly $\frac{1}{2W}$ seconds apart and thus are independent, as described above}. All sampling rates other than those at integer multiples of $2W$ samples per second will give rise to correlated samples. An alternative viewpoint is that the samples are of the (non-bandlimited) white Gaussian process as observed through a filter/sampler of very large bandwidth $\gg W$, and so the samples may be treated as independent zero-mean Gaussian random variables all having the same constant variance $\sigma^2$, with the connection between $N_0$ and $\sigma^2$ remaining unspecified, or set to $N_0W$ if we so prefer.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.