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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

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  • $\begingroup$ For time delay estimation the CRLB is not a very good lower limit. You are rarely able to come close to it.You may want to look into Barakin bounds and the Ziv-Zakai bounds, which give better bounds and better represent what is practically achievable. $\endgroup$ – David Jan 19 '18 at 14:51
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Found the solution, I believe.

It's a dumb mistake. I'm calculating the SNR as $S_0 / N_0$ in one case and as $S/N$ in the other.

Since the noise is square shaped, and the signal is triangle shaped, the first is obviously wrong. Correcting to $\frac{S_0 B/2}{N_0 B}$ gives the factor that was missing from one to the other.

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