Talk:Free-space path 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)[reply]

Waves and freespace

Attenuation is frequency dependent due to the nature of photons. It's obvious:

- One photon at 1 GHz has the probability to be found inside a spheric space region with 30 cm of diameter, because the wavelength is about 30 cm
- One photon at 10 GHz has the probability to be found inside a spheric space region with 3 cm of diameter, because the wavelength is about 3 cm

So a 10 GHz photon as more energy respect to 1 GHz photon. If it's radiated the same power, at 10 GHz the number of photons will be less than the first case.

Practical example:

A certain quantity is radiated from 2 antennas, the power is equal:

- 100 photons at 1 GHz (from first antenna)
- 10 photons at 10 GHz (from second antenna)

The receiver will see less photon at 10 GHz, because the number of photons transmitted is distributed over a spheric surface having as radius the distance between antenna and receiver. The spheric surface covered by both antenna is the same but the number of photons at 10 GHz is less.

So the receiver at 10 GHz receive high energy photons but they are less, more rare. Proportion can be this:

100 photons (1 GHz) transmitted -> 25 photons received
10 photons (10 GHz) transmitted -> 2 photons received

it's not possible receive 2.5 photons, particles are discrete.

--Artemis 10:22, 23 January 2013 (UTC)[reply]

The entire reason the FSPL has a wavelength dependence is that the Friis Transmission Formula it is derived from states, by definition, a dependence on receive antenna area. The poorly named derivative, Free-Space Path Loss, includes this behavior despite the confusing name. The article text has been adjusted appropriately and shrunk to address this point. crcwiki (talk) 15:37, 11 September 2017 (UTC)[reply]

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)[reply]

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)[reply]

If EM signal strength is $s_{0}$ at distance $r_{0}$ 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 $r_{1}$ to distance $r_{2}$ 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)[reply]

Path Loss is defined w.r.t. an antenna of fixed size in terms of lambda. This means that as lambda is increasing your antenna is actually shrinking which results in less energy captured and thus greater path loss. If the antenna was kept the same size the directivity of the antenna would :UnsignedIP -->

I have added some more explanation regarding frequency dependency (with references). Hope it is more clear now. Ev2geny (talk) 23:50, 17 March 2014 (UTC)[reply]

As the FSPL is a direct decendent of the Friis Transmission Formula, that incorporates receive antenna capture are by definition, I have removed the additional explanation section and woven its meanings directly in to the introduction and definition. This helps to make the point up front and keeps this article short and encyclopedic. crcwiki (talk) 15:30, 11 September 2017 (UTC)[reply]

The dB equation constants

For those who are confused (like I was), here is the derivation of where the +32.45 constant came from:

${\begin{aligned}{\mbox{FSPL(dB)}}=20\log _{10}(d)+20\log _{10}(f)-147.55\end{aligned}}$

For typical radio applications, it is common to find $\ f$ measured in units of MHz and $\ d$ in km...

OK, so if I want to count with km and MHz, I derive the new constant from the equation above this way:

{\begin{aligned}{\mbox{FSPL(dB)}}&=20\log _{10}(10^{3}d_{km})+20\log _{10}(10^{6}f_{MHz})-147.55\\&=20\log _{10}(d_{km})+20\log _{10}(10^{3})+20\log _{10}(f_{MHz})+20\log _{10}(10^{6})-147.55\\&=20\log _{10}(d_{km})+60+20\log _{10}(f_{MHz})+120-147.55\\&=20\log _{10}(d_{km})+20\log _{10}(f_{MHz})+32.45\end{aligned}}

Then for $\ d$ in meters and $\ f$ in MHz, the constant would be $\ -147.55+120=-27.55$ .

Dulik (talk) 14:45, 3 March 2011 (UTC)[reply]

Comments of using dezibel (dB)

We are incorrect when using log functions on real values. A log function always has to have the unit [1], you can't take the log function of e.g. log(1 GHz).

Strictly spoken one has to rewrite these formulas, or indicate that $d_{km}=d[km],f_{MHz}=f[MHz]$ , meaning d measured in unit km and f measured in unit MHz.

The results are the same for the constants, but the formula should be written as:

{\begin{aligned}{\mbox{FSPL(dB)}}&=10\log _{10}\left(\left({\frac {4\pi }{c}}df\right)^{2}\right)\\&=20\log _{10}\left({\frac {4\pi }{c}}df\right)\\&=20\log _{10}(d_{m})+20\log _{10}(f_{Hz})+20\log _{10}\left({\frac {4\pi }{c_{[m/s]}}}\right)\\&=20\log _{10}(d_{m})+20\log _{10}(f_{Hz})-147.55\end{aligned}}

with distance d in measured in m and frequency f measured in Hz: $d_{m}=d[m],f_{Hz}=f[Hz]$

JosefNoll (talk) 14:48, 21 October 2013 (UTC)[reply]

I would suggest not to use indices, but instead do what is actually done: Divide and multiply by the reference measure, i.e.

{\begin{aligned}{\mbox{FSPL(dB)}}&=10\log _{10}\left(\left({\frac {4\pi }{c}}df\right)^{2}\right)\\&=10\log _{10}\left(\left({\frac {4\pi }{c}}{\frac {d}{1{\text{m}}}}\cdot 1{\text{m}}\cdot {\frac {f}{1{\text{MHz}}}}\cdot 1{\text{MHz}}\right)^{2}\right)\end{aligned}}

you may even insert $d_{0}$ and $f_{0}$ as reference measures. This clarifies the process and makes the conversion "constant" easily adaptable.

Cweickhmann (talk) 09:33, 29 October 2014 (UTC)[reply]

Unreferenced text copying

Parts of the text are very similar to page 49 of "Sensors and Low Power Signal Processing" by AvSyed Kamrul Islam and Mohammad Rafiqul Haider.

http://books.google.com/books?id=0sIZoDVh7uMC&pg=PA49&dq=free+space+path+loss&hl=sv&ei=9mSVTo2DE-f34QTw-IHkBw&sa=X&oi=book_result&ct=book-preview-link&resnum=3&ved=0CD0QuwUwAg#v=onepage&q=free%20space%20path%20loss&f=false

This book should be referenced, alternativly the common source should be cited. — Preceding unsigned comment added by 130.243.26.101 (talk) 10:11, 12 October 2011 (UTC)[reply]

An error in dB koefficient estimation?

From the equation

20\log _{10}(d)+20\log _{10}(f)+20\log _{10}\left({\frac {4\pi }{c}}\right)=20\log _{10}(d)+20\log _{10}(f)-147.55

follows, that $20\log _{10}\left({\frac {4\pi }{c}}\right)=-147$ , but careful estimation leads us to a different solution (matlab/octave code)

20 * log10((4*pi)/ (2.99792458*10e8)) == -167.552

Why the reference?

Why is there a reference to someone's PhD thesis paper that's mostly about channel codes and only peripherally about propagation? — Preceding unsigned comment added by 64.142.50.80 (talk) 16:24, 17 March 2012 (UTC)[reply]

Well, because it was free on the Web. --Wtshymanski (talk) 17:09, 17 March 2012 (UTC)[reply]

There is a problem defining Pt as the EIRP (Equivalent Isotropic Radiate Power). In fact, in the thefinition of the pathloss formula, we find Pt*Gt*Gr/Pr, which means that Pt*Gt is actually the EIRP and not Pt. Pt is infact the true transmitted power arriving to the antenna which undergo an amplification Gt in some direction. — Preceding unsigned comment added by 41.227.63.8 (talk) 16:49, 7 February 2015 (UTC)[reply]

Graph units

I find the units employed on the graphs confusing (e.g. "d/m"). I think what it means is distance expressed in meter, but the "d/m" is nonstandard and therefore open to interpretation. I propose to replace it by "d (m)", and similarly elsewhere. JanCeuleers (talk) 07:13, 19 September 2016 (UTC)[reply]

I cannot do this myself because it's embedded in graphs someone else contributed. JanCeuleers (talk) 09:51, 22 September 2016 (UTC)[reply]

The graphs don't seem to have much purpose as a visual aid in understanding FSPL and have been removed. crcwiki (talk) 15:32, 11 September 2017 (UTC)[reply]

What's in a name "Free Space Path Loss"

I see an anonymous user removed my wording "so called" in front of the phrase "Free-space path loss" claiming the name is its "scientific name." Okay fair enough that this is the "well known" name of this phenomena, but I challenge anyone to find a good text book reference that specifies this is anyway scientific. The Friis formula is what is actually used in professional circles. The term "free space" in the name is quite inaccurate and almost hopelessly misleading by not inferring the effects of antenna aperture. crcwiki (talk) 02:45, 23 September 2017 (UTC)[reply]

@Crcwiki Rappaport, Theodore S. (2010). Wireless communications: principles and practice (Second edition, twentieth impression 2019, Indian subcontinent adaption ed.). Noida: Pearson India Education Services. p. 107. ISBN 978-81-317-3186-4 . Check the section 'Free Space Propagation Model' he has defined path loss as dB difference between transmitted power and received signal. It is consistent with the article. Salekin.sami36 (talk) 13:16, 10 January 2024 (UTC)[reply]