0
$\begingroup$

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 ?

Improve this question
$\endgroup$
4
  • $\begingroup$ the formula you've posted makes no sense! The sum includes $n_2$, which can by, the integral formula below, be expressed as coefficient to the basis function $\phi_2$. There's something wrong in the formula. $\endgroup$ – Marcus Müller Feb 24 '19 at 9:11
  • $\begingroup$ Sorry for my confused text expression. What I mean by that formula is the noise process cannot be completely expanded in terms of the basis functions. The first part of the noise process that can be expanded in terms of basis functions. The other part which cannot expressed in terms of this basis function is denoted by n2(t). $\endgroup$ – Berman Song Feb 24 '19 at 9:49
  • $\begingroup$ So, can we just rename $n_2$ to $n_o$, with $o$ like "orthogonal"? You can't just use $n_j$ and $n_2$ in a formula and say $n_j, j=2$ is not the same as $n_2$... $\endgroup$ – Marcus Müller Feb 24 '19 at 10:57
  • $\begingroup$ by the way, your first question doesn't make too much sense: you literally ask whether anything can be represented as a sum of something plus something else – yes, that is possible for everything. You will probably want to define how your $n_j$ and $\phi_j$ relate to your matched filter. $\endgroup$ – Marcus Müller Feb 24 '19 at 18:08
0
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I see your point. But, I still don't understand that if the noise $n(t)$ becomes a narrowband noise, and we calculate the covariance of $n_i$ and $n_j$ as $COV[n_in_j]=E[\int_{0}^{T}n(t)\phi_i(t)dt \int_{0}^{T}n(s)\phi_j(s)ds]=\int_{0}^{T} \int_{0}^{T} E[n(t)n(s)] \phi_i(t) \phi_j(s) dt ds$. Doesn't its autocorrelation function becomes $N_0Bsinc(\pi B\tau)cos(2 \pi f\tau)$ ? And by letting $\tau=0$, we can get the variance by solving the integral. Is that right? $\endgroup$ – Berman Song Mar 5 '19 at 14:24
  • $\begingroup$ I haven't looked super closely at your math but it looks correct. In many applications, such as digital communications, it is enough that the noise samples become uncorrelated at specific instants, which is the case in your example. $\endgroup$ – MBaz Mar 5 '19 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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