About the variance of the coefficient of narrowband noise when using signal-space representation Assume that n(t) is a white Gaussian noise process with zero-mean and power spectrum density $N_0/2$. By using the signal-space representation, it can be expressed as:

$$n(t) =\sum_{j=1}^N n_j \phi_j(t)+n_o(t)$$
where $n_j = \int_{-\infty}^{\infty}n(t)\phi_j(t)\,\mathrm dt$ , $\phi_j(t)$ is the basis function and $n_o(t)$ is a part which cannot be expressed in terms of this basis function.


If the noise process $n(t)$ is a narrowband noise of bandwidth $B$ centered on frequency $f$, can it still be expressed in this way?

My understanding is that only after the matched filter, the part $n_o(t)$ is removed.
So the narrowband noise also can be expressed in this way.

Is my understanding correct?

In addition, the coefficient $n_j$ must be a random variable, how to derive the variance of the $n_j$ in the case of narrowband noise ?
Since the basis functions are narrowband, the formula still works and the variance of each $n_j$ remains $N_0/2$..

In other words, assume the bandwidth of the basis functions is $B$. You can then think of them as filters that ignore any frequency outside of $B$. So, they "don't care" if there is noise outside of $B$ or not.

The expression for $n_0(t)$ will change, though.