I'm attempting to calculate the CRLB for a bandlimited time-delay system which has a triangular shaped signal spectrum, instead of the usual square one.
Currently I'm working on an active system.
Following the procedures done by Quazi et al. in "An overview on the time delay estimate in active and passive systems for target localization",
I can reach two different results. The first one, using his calculations for a High SNR, active system, don't really match my data, while the second one, computing for a Passive system and considering a $2$ factor (as he explains after equation 23) due to only one of the sources being contaminated by noise (which makes intuitive sense to me, as long as the SNR is high), actually fits my data very well.
As such, I am assuming I'm doing the first set of calculations wrong, and I was hoping some of you could help me spot my mistake.
So here goes, for both the calculations
PASSIVE
$$\frac{1}{\sigma^2} = 2T \int_{0}^{B}(2\pi f)^2 \frac{S(f)}{2N(f)}df$$
where my noise is square shaped, and my signal is triangle shaped, in baseband, so
$$\frac{1}{\sigma^2} = 2T \int_{0}^{B}(2\pi f)^2 \frac{S_0(1-f/B)}{2N_0}df$$
$$\frac{1}{\sigma^2} = 2T(2\pi)^2 \frac{S_0}{2N_0}\int_{0}^{B} f^2(1-f/B) df$$ $$\frac{1}{\sigma^2} = T \ 4\pi^2 \ SNR \ (B^3/3-B^3/4) $$ $$ \sigma^2= \frac{12}{4\pi^2T} \frac{1}{SNR} \frac{1}{B^3}$$
which, for an active system, dividing the variance by two since the source isn't noisy, I end up with $$ \sigma^2= \frac{12}{8\pi^2T} \frac{1}{SNR} \frac{1}{B^3}$$
This fits very well to my data, and the last argument is logical to me.
ACTIVE
So, he starts by defining the CRLB as $\sigma^2 = \frac{1}{d^2\beta^2}$ where
$$d^2 = \frac{2E}{N0}$$ where $E$ is the energy of the signal, so $E = ST$, where $T$ is the signal observation time
$$\beta^2 = \frac{\int_{-\infty}^{+\infty}\omega^2|F(\omega)|^2 d\omega}{\int_{-\infty}^{+\infty}|F(\omega)|^2 d\omega}$$
Now, the calculation of $\beta$ is similar to what I've done before, setting my $S(f) = S_0(1-\frac{f}{B})$
$$\beta^2 = \frac{\int_{0}^{B}(2\pi f)^2S_0(1-\frac{f}{B}) \ \ 2\pi df}{\int_{0}^{B} S_0(1-\frac{f}{B}) \ \ 2\pi df}$$
$$\beta^2 = \frac{(2\pi)^2\int_{0}^{B} f^2(1-\frac{f}{B}) \ \ df}{\int_{0}^{B} (1-\frac{f}{B}) \ \ df}$$
$$\beta^2 = \frac{(2\pi)^2\frac{B^3}{12}}{ \frac{B}{2}} = (2\pi)^2\frac{B^2}{6}$$
solving now for the cramer-rao
$$\frac{1}{\sigma^2}=\frac{2E}{N_0}(2\pi)^2\frac{B^2}{6}$$
multiply top and bottom by $B$ so $N = N_0B$, and $S/N = SNR$
$$\frac{1}{\sigma^2}=2T(2\pi)^2 \ SNR\ \frac{B^3}{6} = 8\pi^2T \ SNR \frac{B^3}{6}$$
$$ \sigma^2= \frac{6}{8\pi^2T} \frac{1}{SNR} \frac{1}{B^3}$$
Which does not fit my data as tightly.
I'd like to ask you where my intuition is going wrong with both solutions, or if I am making any dumb mistakes.
Thanks
