Talk:Pulse compression

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[edit] Signal to noise or signal to clutter

Doesn't pulse compression improve the signal to clutter ratio and not the signal to noise ratio as stated on the first line? Greenshed 21:20, 18 May 2007 (UTC)

I believe there's some confusion here. Pulse compression does indeed improve the signal to noise ratio, as explained below in the main article. This is because pulse compression is only a particular application of the matched filtering theory. A matched filter is optimal (in the least squares sense) for enhancing the signal to noise ratio, when detecting a signal which shape is precisely known, and that is corrupted by additive stochastic noise. In the case of radar or sonar, this can be, for instance, thermal noise (one of the origins of the "static" hiss you hear when de-tuning a radio station, for instance). Clutter, on the other hand, is a word used in radar or sonar imaging to describe "unwanted" signal that is really the image of the environment surrounding the target. This can be clouds, the atmosphere for airspace surveillance radars, the ocean for marine surveillance radar, etc. For many reasons, clutter can generally be only described statistically; first because the environment is not known "exactly" because it is too complex, second because of specific phenomena occuring in radar or sonar imaging: speckle. But, the statistical properties of clutter are rarely those of an additive process (it is more often multiplicative). The increase of resolution brought by pulse compression may however help to better see the target in the clutter. Flambe 21:09, 3 July 2007 (UTC)


[edit] Chirping - Basic principles

In the formula, when t = T then the frequency equals f_0. Shouldn't it equal f_0 + Delta-f? —Preceding unsigned comment added by 201.150.79.138 (talk) 04:40, 28 January 2008 (UTC)

Answer: The chirped signal is:

s_c(t) = \left\{ \begin{array}{cl} A e^{i2\pi \left (f_0+\frac{\Delta f}{2T}t-\frac{\Delta f}{2}\right) t} & \mbox{ if } 0 \leq t < T \\ 0 & \mbox{else}\end{array}\right.

The phase of the chirped signal (that is, the argument of the complex exponential), is:

\phi(t) = 2\pi \left (f_0+\frac{\Delta f}{2T}t-\frac{\Delta f}{2}\right) t

The instantaneous frequency is:

f(t) = \frac{1}{2\pi}\left[\frac{d\phi}{dt}\right ]_t = f_0-\frac{\Delta f}{2}+\frac{\Delta f}{T}t

Thus we readily see that f(0) = f_0-\frac{\Delta f}{2}T and f(T) = f_0+\frac{\Delta f}{2}T, which is what was intended in the formula given in the wiki page. Flambe (talk) 19:54, 31 January 2008 (UTC)