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.

$\endgroup$

1 Answer 1

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.