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Given the noise spectral density ${N_0\over 2} = a$, what can be deduced about the variance?

In the case of a two-symbol canal(/cable), and when $T_0 = 1$.

Is it that, $\sigma_X^2 = {N_0 \over 2} *T_0$?

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Given a low-pass signal $x$ that is a realization of random variable $X$ with constant spectral density $a=N_0/2$ in $f=-B\ldots B$ (and zero otherwise) we can calculate the mean power of that signal by $\bar P = 2aB=N_0B$, where $B$ is the signal bandwidth. The mean power of a random variable is defined as the expectation of $X^2$, i.e. $\bar P = \mathrm E[X^2]$. The correspondance between mean power and variance $\sigma_X^2$ of a random variable $X$ is $$ \sigma_X^2=\mathrm E[X^2]-m_X^2 $$ where $m_X$ is the mean value of $X$. That is, for zero mean (which is often true for noise) we get $$ \sigma_X^2=\mathrm E[X^2] = N_0B $$

Concerning your second question, I assume that $T_0$ is the symbol duration. Thus the symbol rate $R_\mathrm{S} = 1/T_0$. The bandwidth of a signal with symbol rate $R_\mathrm{S}$ is often approximated by $B=R_\mathrm{S}/2$ and thus $$ \sigma_X^2 = N_0B = N_0R_\mathrm{S}/2 = N_0/(2T_0) $$ (You can also check the units in your equation: $N_0$ is power per Hz and $T_0$ is in sec. But the result should be a power)

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  • $\begingroup$ Most people would say that a signal whose Fourier transform has support $[-B/2,+B/2]$ Hz has a bandwidth of $B/2$ Hz, not $B$ Hz. Some people might even add the words "low-pass" as an adjective describing the signal. Indeed, the very reason that $N_0/2$ (watts/Hz) is used for the parameter of interest is to get the simple answer that white noise when passed through an ideal lowpass filter of bandwidth $B$ has noise power $N_0B$. Thus, your $\bar{P} = aB$ should be parsed as $aB = (N_0/2)B = N_0(B/2)$. $\endgroup$ – Dilip Sarwate Feb 11 '14 at 21:19
  • $\begingroup$ @DilipSarwate You're absolutely right. I've edited my answer accordingly. $\endgroup$ – Deve Feb 12 '14 at 15:44
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There are, in my opinion, several ambiguities in the question asked, and also in Deve's answer which the OP has accepted.

The phrase "$\ldots$ noise has spectral density $\frac{N_0}{2}\ldots$" is usually interpreted to mean that the noise is a continuous-time white noise process which is a mathematical abstraction that is useful and convenient in many analyses. This noise process is often assumed to be Gaussian as well (white Gaussian noise) which leads to the acronym AWGN or WGN. Any resemblance to the signals broadcast by a certain Chicago radio and TV station are purely coincidental... In any case, the random variables constituting this mythical process have zero mean, because if not, then the power spectral density would have a impulse at $f=0$ in contradiction to the assumption that the spectral density has fixed value $\frac{N_0}{2}$ for all $f$. The random variables also have infinite variance (cf. the answer to this question) -- that is, the process does not exist in nature and cannot be observed directly. As pointed out in the answer to the cited question, what can observed is the effect that this mythical process produces at the output of a filter in which case the result is a wide-sense-stationary random process in which all the random variables have variance $\sigma^2$ given by $$\sigma^2 = \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,\mathrm df$$ where $H(f)$ is the transfer function of the filter. Since the input process has zero mean, so does the output process have zero mean, that is, all the random variables constituting the process have zero mean. For the case of WGN, the filter output is a strictly stationary Gaussian process, meaning that all the random variables are Gaussian random variables. As a special case of all this, if the filter is an ideal bandpass filter of bandwidth $B$ Hz with unit gain in the passband, then the noise variance is $N_0B$. Such a filter has nonzero gain for $|f| \leq B$ (lowpass filter) or for $|f| \in [f_c-B/2,f_c+B/2]$ (bandpass filter with center frequency $f_c$ Hz), and of course for any filter, the noise variance is $N_0B$ where $B$ is the equivalent noise bandwidth of the filter.

Turning to the question asked by the OP, it is not clear what is meant by _"two-symbol canal (cable)" and what is meant by $T_0$. One possible interpretation is that we are transitioning from a continuous-time process to a discrete-time process by sampling the white noise at intervals of $T$ seconds. Of course, the white noise is observable only through the sampler which has finite bandwidth and so what is actually being sampled is a finite-variance process as described in the previous paragraph. More practically, one is usually concerned with observing some finite-bandwidth lowpass signal $x(t)$, and so the signal (plus the white noise) is first passed through a lowpass filter of bandwidth $B$ where $B$ denotes the signal bandwidth, and now it is this bandwidth, (and not the bandwidth of the sampler which can be orders of magnitude larger) that determines the noise variance. But what about the sampling interval $T$? That has no effect on the variance of the noise samples. What $T$ affects is the covariance between different samples. The correlation between two random variables separated by $t$ seconds in time from bandlimited noise of bandwidth $B$ Hz is $\operatorname{sinc}(Bt) = \frac{\sin(\pi Bt)}{\pi Bt}$, that is, samples separated by multiples of $B^{-1}$ seconds are uncorrelated (and in the case of Gaussian noise, independent) and so if the sampling interval $T$ is a multiple of $B^{-1}$ seconds, the random variables are uncorrelated/independent which is a model used very often. But if the samples are taken at other sampling rates, then the noise samples are correlated. See this question for more discussion of this point.

Finally, if "two-symbol" means that something like a BPSK system with symbol interval $T$ and a correlation receiver is under consideration, then the noise at the correlator output has variance $$\sigma^2 = \frac{N_0T}{2}$$ (cf. my comment in response to a query from moderator PeterK. in the question Variance of white Gaussian noise cited earlier.) Note the difference between this and the answer $N_0/(2T_0)$ given by @Deve.

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