Talk:Pulse compression

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)[reply]

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)[reply]

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)[reply]

Answer: The chirped signal is:

s_{c}(t)=\left\{{\begin{array}{cl}Ae^{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)[reply]

Signal description

"The simplest signal a pulse radar..."

How does the article define "simple?" Is the "simplest" finite-duration signal a sinusoid multiplied by a square window? Or, is it something smoother with fewer harmonics? The complexity (or lack thereof, simplicity) could be measured by the information content, by one method or another, e.g. time-bandwidth. —Preceding unsigned comment added by 24.126.222.79 (talk) 01:49, 22 December 2010 (UTC)[reply]

Less-mathematical explanation

It would be nice to have a section, preferably at the beginning of the article, that discusses the principle in a less-mathematical, more real-world, practical manner. Jim, K7JEB (talk) 16:43, 18 January 2011 (UTC)[reply]

Graphs: Filter response time offset?

In the example "Before matched filtering" the response times are $t_{r}=\{3,\ 5\}$ According to the formula for the cross-relation $<s,r>(t)$ should have its maximum (echo) for $t_{e}=t_{r}$ .

But in the graph "After matched filtering" the echo-times are $t_{e}=\{3.5,\ 5.5\}$ .

What's the reason for the offset of $0.5$ ? — Preceding unsigned comment added by 86.33.223.67 (talk) 09:55, 15 July 2011 (UTC)[reply]

The correlation peak is located where the reference signal

s

and the received signal

r

"overlap most"; this is not at the beginning of the signal but at its middle (this is "said with the hands" -- it's not a precise mathematical description but it works well for both the truncated sine wave and the chirp signal). In the examples provided in the main page the signal begins to be received at

t_{r}=\{3,\ 5\}

but since it lasts for one unit, the correlation peaks are only reached at

t_{e}=\{3.5,\ 5.5\}

. To position the correlation peak at the beginning of the signal, for this example, you'd need to artificially delay the reference signal of 0.5 units OR delay the result of the correlation of 0.5 units. Actually someone did a better job than me at explaining this using an animation with a square pulse (taken from Convolution article):

Correlation of a square pulse (as input signal) with another square pulse. The integral of their product is the area of the yellow region.

You may wonder why the animation is taken from the convolution article, but the idea behind both convolution and correlation is the exactly the same (when doing a correlation, you just don't reflect one of the signals as is the case for convolution), and in the specific case shown in the animation, the correlation and the convolution are strictly identical. See that the "received signal" in blue is beginning at t=-0.5, yet the peak of the correlation is only reached at 0. Hope this helps.

Flambe (talk) 22:42, 21 August 2011 (UTC)[reply]