1
$\begingroup$

I'm interested in the case where I have received a signal as a composition of three echoes of the transmitted signal $x[n]$ (each echo has its amplitude and time delay). I know that in order to compare the signals I use cross-correlation. My question is if I use $r_{xx}[m] = \delta[m]$, would that make it easier to calculate the parameters of echoes? And, if so, can you explain me why?

$\endgroup$
  • $\begingroup$ Found similar notation elsewhere, $r_{xx}$ is autocorrelation of $x$. $\endgroup$ – Olli Niemitalo Oct 31 '17 at 22:16
1
$\begingroup$

The mathematical justification would be that the Cramer Rao Bound on the estimate of time delay is inversely proportional to the time bandwidth product of the pulse, so in cases where there are many distinct echoes, the shorter the pulse is better to resolve each arrival, so one would increase the bandwidth to minimize the bound on time delay.

In the case where the signal is long, as illustrated by the matlab code below for a single delay:

clear all
x=randn(1,16*32768);
x1=[x zeros(1,.5*128)];
x2=[zeros(1,.5*128) x];
figure(1)
pwelch(x1+x2)

enter image description here

The psd has periodic nulls, and you can get an estimate from the null spacing, and the wider the band (white noise), the more nulls that can be resolved.

$$ H(\omega)=1+\alpha e^{\jmath \omega \tau} $$

For 3 delays it will be a bit more of a bother, but doable. Actually if you don't know the waveform, like a passive SONAR problem, this is an essential approach.

$\endgroup$
1
$\begingroup$

If the echoes are far apart and interfere minimally, and you know the peak shape to be sinc, you can calculate the peak time more simply and using less data points. See my answer to "How to calculate a delay (correlation peak) between two signals with a precision smaller than the sampling period?".

If the signal covers the whole spectrum up to half the sampling frequency, then there is no bandwidth headroom left where could be only noise. Alternatively with a lower bandlimit the noise could be filtered without affecting the wanted signal.

$\endgroup$

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.