Talk:Free-space loss

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There is an inconsistancy between this article and the article on attenuation (http://en.wikipedia.org/wiki/Attenuation) and also the article on free space loss (http://en.wikipedia.org/wiki/Free_space_loss) - i.e., this article states that free space loss is attenuation but the article on attenuation states that attenuation includes all forms of decreases in intensity of a signal, etc. "not including the reduction due to geometric spreading". The article on free space loss states that 'free space loss assumes the transmitter and receiver are both located in free space and does not consider other sources of loss such as reflections, cable, connectors etc.', which I believe is correct and, therefore, I am changing this article to what I believe is the correct definition and also the definition that maintains consistency between the relevant articles. Note: I did notice the source for this definition and have sent an email to ATIS suggesting that they revise their definition.

Free-space loss is attenuation and attenuation is not (necessarily) free-space loss. There is no contradiction there. A cow is an animal but an animal is not necessarily a cow. --MarSch 11:27, 21 March 2007 (UTC)

[edit] Habicht - Why is the FSL frequency dependent?

I don't think that free space loss depends on the frequency because in free space you have no material that could attenuate the wave. Therefore the loss or attenuation arises from the geometric spreading of the wave which results in an decrease of the field intesity. As a wave normally propagates spherically the free space loss should be:
free\ space\ loss = ( \frac{1}{4\pi R^2})
or expressed in dB
free\ space\ loss = 10 * \log{ ( \frac{1}{4\pi R^2})}
Note that this formual is also state on the german wikipedia article about freiraumdämpfung. Habicht 16:35, 7 February 2007 (UTC)

It seems very strange to me also. It seems very strange to me that there should be any terms which do not depend on the distance. Also the loss over a distance of 0 should be 0dB. Instead the current formulas have a pole of minus infinity at zero distance, which would mean that the signal would be amplified infinitely. --MarSch 11:35, 21 March 2007 (UTC)

If EM signal strength is s0 at distance r0 from the source, than it should be s_0 \left(\frac{r_0}{r}\right)^2 at distance r assuming the source beams equally in all directions because of conservation of energy. The loss from distance r1 to distance r2 from the source would be s_0 \left(\frac{r_0}{r_1}\right)^2 - s_0 \left(\frac{r_0}{r_2}\right)^2. The factor by which the signal decreased is new signal strength divided by old: \frac{ s_0 \left(\frac{r_0}{r_2}\right)^2 }{ s_0 \left(\frac{r_0}{r_1}\right)^2 } = \left(\frac{r_1}{r_2}\right)^2. To make decibels from this we need to take the logarithm. Signal loss in decibels is \log(\left(\frac{r_1}{r_2}\right)^2) = 2 (\log r_1 - \log r_2) which is zero when r_1 and r_2 coincide as it should. --MarSch 12:06, 21 March 2007 (UTC)

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