+ Post New Thread

Results 1 to 16 of 16

- 7th January 2010, 17:59 #1

- Join Date
- Jan 2010
- Location
- Tennessee USA
- Posts
- 2
- Helped
- 0 / 0
- Points
- 530
- Level
- 4

## Problem with Friis's Equation for Free Space Path Loss

The well know Friis's Equation gives the free space loss from one isotropic antenna to another with variables being the frequency, wave length, and distance between antennas.

When the transmitting or receiving or both antennas are directional, then the gain of the antennas (in db) is subtracted from the Friis free space loss.

However, for high gain antennas, Friis's equation someti9mes gives impossible results.

For example, if frequency = 2 GHz, distance = 0.5 miles (about 800 meters),

and the gain of each antenna is 30 db, then the equation predicts a free space gain

instead of a free space loss. This is impossible, of course, since you can not get more power out of the receiving antenna than you put into the transmitting antenna.

The likely soure of the anomalous result is that Friis' equation is for the "far field" and does not take into account "near field" phenomena.

One wold think that half a mile separation would be far enough to be able to ignore the near field., but it apparently is not.

Does anybody know of a Free Space Attentuation formula that considers the "near field" in a simple enough way to aid in hand calculations. Or maybe I'm barking up the wrong tree.

Thanks for your help.

root-beer

- 7th January 2010, 19:59 #2

- Join Date
- Oct 2001
- Posts
- 4,237
- Helped
- 1187 / 1187
- Points
- 29,395
- Level
- 41

## Re: Problem with Friis's Equation for Free Space Path Loss

Friis equation for free space path loss doesn't have anything with antenna gain (you are confusing with link budget)

Free Space Path Loss(dB) = 27.6 – 20*LOG[Frequency(MHz)] – 20*LOG[Distance(m)]

2 members found this post helpful.

- 7th January 2010, 20:18 #3

- Join Date
- Jan 2008
- Location
- Bochum, Germany
- Posts
- 31,956
- Helped
- 9985 / 9985
- Points
- 191,299
- Level
- 100

## Problem with Friis's Equation for Free Space Path Loss

For the calculation of received power, it's correct to substract snedre and receiver antenna gain from free space loss.

But in the said example, it's still loss. You're right however that the equation isn't valid in near field.

- 7th January 2010, 20:18

- 9th January 2010, 02:27 #4

- Join Date
- Jan 2005
- Location
- South Korea
- Posts
- 14
- Helped
- 6 / 6
- Points
- 1,547
- Level
- 8

## Problem with Friis's Equation for Free Space Path Loss

hi root-beer,

please read any manual about antenna perfromance. it will help you to avoid any "impossible results".

- 11th January 2010, 16:07 #5

- Join Date
- Jan 2010
- Location
- Tennessee USA
- Posts
- 2
- Helped
- 0 / 0
- Points
- 530
- Level
- 4

## Re: Problem with Friis's Equation for Free Space Path Loss

Thanks so much for your comments, which I hope are pointing me in a more valid direction.

My confusion, was in part due to several "on-line link-budget" calculators found on the web. For example

1. Wikipedia also points to http://www.osischool.com/protocol/wireless/link-budget

which includes the Transmitting Antenna and the Receiving Antenna by

just subtracting the db gains for their respective gains.

2. Wikipedia also refers to:

http://www.random-science-tools.com/...nics/friis.htm

for Wireless Link Budget with the antenna gains taken into account by

the formula: Prx = Ptx + Gtx + Grx (in db)

3. Finally, the web site http://titanwireless.com has a tools page that offers

several link budget calculators with the same simple treatment of

transmitting and receiving antenna gains (but with an unspecified fudge factor).

It appear to me that there may be a widespread misconception of how Transmitting and Receiving Antenna gains effect the overall link budget.

It seems logical that that a Transmitting antenna with a 30 db gain will

really decrease the link loss budget by 30 db, since 1000 times more power is

being sent toward the receiver.

But a 30 db gain in the Receiving antenna apparently does not have the same factor of 1000 increase in the reciving signal power. In the case of a parabolic receiving antenna, the effect of the reflector is to give a larger aperture for picking us the waves from the transmitting antenna. For large transmission distances, the transmitted beam (at half power points) would be considerably broader than the parabolic dish.

Another confusing issue is the Antenna Reciprocity Theorem which states that

"reciprocity implies that antennas work equally well as transmitters or receivers, and specifically that an antenna's radiation and receiving patterns are identical"

see: http://en.wikipedia.org/wiki/Reciprocity_(electromagnetism)

I think that "reciprocity" is also widely misunderstood. Given the radiation patterns are the same for tranmitting as receiving, but signal gains are not equal.

It seems to be common practice to specify the power gain of a antenna in db) no matter whether it is going to used for transmitting or receiving.

Even if the receiving antenna would have a 30 db gain rateing if it were used for Transmitting, the actual gain when receiving would be 30 db, but only for the portion of the transmitting beam that actually "hits" the receiving antenna aperture, viewing the signal from the Transmitting antenna as a number straight rays that diverge from the antenna.

If one has a choice of selecting a parapolic disk receiving antenna or a wave-guide receiving antenna (e.g. a cantenna), both with the same rated gain for transmitting, it seems logical (to me) that the effective gain for both antenna (used as a receiver) would be reduced by the same amount), even though the physical "aperture" of antenna might be quite different. one buys two identical antennas with, say, a 30 db gain (measured as a transmitting antenna),

It seems that a simple way of relating the actual gain of a receiving antenna to the "rated" gain as a Transmitting antenna) would be to take the ratio of the area of the transmitted beam (at its half power points) measured at the location of the receiver, and the effective area of the aperature of the receiving antenna.

My guess is that the gain as a parabolic Transmitting antenna and the effective aperture of a transmitting antenna related by the aperature area. Perhaps this relation holds for all types of antennas (not just parbolic antennas).

I've looked in several antenna handbooks and text books, and have not found a simple formula for dealing with the receiving antenna. If anyone knows of a reliable reference for a similar formula for the Path budget that correctly takes into account the Receiving antenna, please let me know.

Thanks ...

... Ross

P.S. I think that one lesson learned is not to believe literally everything that is on Wikipedia.

- 11th January 2010, 18:09 #6

- Join Date
- Aug 2001
- Posts
- 414
- Helped
- 24 / 24
- Points
- 3,555
- Level
- 14

## Re: Problem with Friis's Equation for Free Space Path Loss

"This is impossible, of course, since you can not get more power out of the receiving antenna than you put into the transmitting antenna."

Root-Beer: you should read carefully on the definition of "antenna gain"--the "gain" is relative to isotropic antenna, not absolute power gain.

- 12th January 2010, 05:49 #7

- Join Date
- Jul 2009
- Location
- Sydney, Australia
- Posts
- 766
- Helped
- 176 / 176
- Points
- 5,187
- Level
- 17

## Re: Problem with Friis's Equation for Free Space Path Loss

Originally Posted by**loucy**

a 30dB gain dish has that gain for TX and RX that can be relatively tested

in the field with an antenna setup and a properly calibrated receiver.

As Loucy inferred antenna gain has to be related to something and is usually

dBi ( gain over a isotropic radiator) or dBd ( gain over a dipole radiator)

and that gain figure is constant for TX and RX

To say an antenna has a gain of 25dB is meaningless ... relative to what ???

a wet piece of string ?? :D

The main pathloss program I use for my amateur radio microwave work is

ukWTools.

cheers

Dave

VK2TDN

- 12th January 2010, 05:49

- 15th January 2010, 01:11 #8

- Join Date
- Feb 2009
- Location
- Durham, USA
- Posts
- 73
- Helped
- 4 / 4
- Points
- 1,184
- Level
- 7

## Re: Problem with Friis's Equation for Free Space Path Loss

Originally Posted by**root-beer**

If we're assuming no losses in the antenna/no mismatch/etc, so all the power gets radiated by the antenna, then two antennas with identical radiation patterns have identical gain by definition.

I think you should use any good book that treats the Friis formula or link budget and go through the equation's derivation from first principles. That should help you. Your original confusion is that the other side of the link budget equation is not loss or gain but received power. Pr=Pt + Gt + Gr -Lp. Friis formula just gives you Lp, the path loss. The equation I gave is all in dB of course. In non-dB, Pr=Pt*Gt*Gr/Lp. In your own words, the path loss from Friis assumes isotropic radiators, so you add the gains to account for the different antennas.

- 4th February 2010, 02:06 #9

- Join Date
- May 2008
- Posts
- 1
- Helped
- 1 / 1
- Points
- 821
- Level
- 6

## Re: Problem with Friis's Equation for Free Space Path Loss

Root beer:

1) Friis assumes you are in the far zone of the antenna...

2) In addition to the r >> lambda far zone requirement we also need to consider r>>D and r>2D^2/lambda requirements.

3) As far as I can tell (playing with Friis in MATLAB) the only time you realize positive gain is when you are either VERY close.. or you have HUGE gain (which ultimately translates to HUGE D)... probably something could be said about the r/D ratio and break down of Friis...

4) Also, I think you should re-calculate your received power.. for 2 30dBi antennas operating at 2.4GHz at a distance of 1mile (1.6km) I get: Pr/Pt = 3.9e-5... good enough for a pretty good com link!!

:]

1 members found this post helpful.

- 3rd May 2010, 10:47 #10

- Join Date
- May 2010
- Posts
- 1
- Helped
- 0 / 0
- Points
- 466
- Level
- 4

## Re: Problem with Friis's Equation for Free Space Path Loss

I don’t see a problem as long as you use the correct form of the Friis equation. For f in GHz and d in miles that would be:

LdB = 96.6 + 20Log f + 20Log d

For a half-mile path at 2 GHz:

LdB = 96.6 + 20Log 2 + 20Log 0.5

LdB = 96.6 + 6 – 6

LdB = 96.6

The path-length and the frequency cancel each other. With 30 dB antennas at each end of the hop, the section loss would be 36.6 dB transmit antenna in to receive antenna out – not including line-loss.

The Friis equation is based on spherical area. For a separation of one wave length:

LdB = 20Log (4pi)

LdB = 20Log (12.566)

LdB = 22.0

For two 2-GHz dipoles separated by 5.9 inches – one wavelength at 2-GHz – the loss would be 22 dB. You can use this to estimate coupling loss between two antennas on the same tower. Measure the separation, convert to wavelengths, find the 20Log, and add 22.

But does this work on a long path? Let’s see, there are 5,368 2-GHz wavelengths in a half-mile path, so:

LdB = 22 + 20Log 5,368

LdB = 22 + 74.6

LdB = 96.6

I think Harald T. Friis knew precisely what he was talking about.

My apologies if this is a duplicate reply....waw...

- 31st January 2011, 19:10 #11

- Join Date
- May 2010
- Location
- Santa Barbara
- Posts
- 2,637
- Helped
- 772 / 772
- Points
- 14,895
- Level
- 29

## Re: Problem with Friis's Equation for Free Space Path Loss

To your concerns about that applying the Friis formula for a radio link while using too high-gain antennas, so you seem to expect the antenna would give a higher that transmitted power...

Please refer to the basic physics of energy conservation: you cannot "generate" energy, you can change its forms. If you use a lens to focus the heat from the Sun to a small spot, you can set afire a piece of paper. But if you can MEASURE the spot temperature, it CANNOT reach more than the temperature of the Sun, some 6000K. This is again the principle of energy conservation which holds.

- 31st January 2011, 21:03 #12

- Join Date
- Jan 2008
- Location
- Bochum, Germany
- Posts
- 31,956
- Helped
- 9985 / 9985
- Points
- 191,299
- Level
- 100

## Re: Problem with Friis's Equation for Free Space Path Loss

it CANNOT reach more than the temperature of the Sun, some 6000K. This is again the principle of energy conservation which holds.**can**generate a spot above 6000K, or convert solar light to electricity and you don't have a principle temperature limit. I guess, you have been thinking about 2nd fundamental law of thermodynamics, but it doesn't show this way.

- 30th August 2011, 04:11 #13

- Join Date
- Oct 2009
- Posts
- 13
- Helped
- 0 / 0
- Points
- 628
- Level
- 5

## Re: Problem with Friis's Equation for Free Space Path Loss

This is correct.

There is an implicit assumption hidden in Friis formula, that the EM wave arrives at receiver antenna as plane wave, therefore the power intensity/EM field intensity doesn't change within the antenna. In other word, effective aperture only makes sense under plane wave. If the antenna gain is too high relative with communication distance, the EM wave cannot spread enough before reaching receiver antenna, Friis formula breaks.

I guess the exact result of received power involves some kind of integration over receiver antenna with power intensity.

- 30th August 2011, 07:27 #14

- Join Date
- Apr 2007
- Posts
- 40
- Helped
- 4 / 4
- Points
- 1,685
- Level
- 9

## Re: Problem with Friis's Equation for Free Space Path Loss

Friis Formula, link budget and reciprocity are right, but it must be considered that antenna gain is a far field concept thus you must be far enough (more than 2*D^2/Lmabda) to use formula. In the other hand the more antenna gain the more its aperture and teh more aperture results in greater 2*D^2/Lmabda and more spacing and more loss.

The Formula is:

Pt/Pr=(Gt*Gr/((4*pi*R/Lambda)^2)

and the problem is that it shouldn't be greater than 1. Considering G=4*pi*A/(Lambda^2) and manipulating above formula gives:

Pt/Pr=At*Ar/((R*Lambda)^2)

and assuming Transmit and receive antenna the same (Dt=Dr=D)

At<=D^2

Ar<=D^2

then

Pt/Pr<=(D^4)/(((R*Lambda)^2)) , R>(2*(D^2)/Lambda)

then

Pt/Pr<=(D^4)/(((2*(D^2)/Lambda*Lambda)^2)) = (D^4)/((4*D^4)) <1

thus

Pt/Pr<1Last edited by AmirMohammad; 30th August 2011 at 09:05.

- 30th August 2011, 08:18 #15

- Join Date
- Feb 2006
- Posts
- 99
- Helped
- 7 / 7
- Points
- 2,013
- Level
- 10

## Re: Problem with Friis's Equation for Free Space Path Loss

Hey Root Beer

your initial calculation seems to be wrong:

f=2GHz; d=800m=0.8km

let's calculate the free space loss according to Friis: L=32.5+20Log(f[MHZ]*d[km])=32.5+20log(2000*0.8)=32.5+64.08=96.6dB

if your antennas have a gain of 30dBi then the attenuation between the two will be 96.6-30*2=16.6dB, so not a gain but loss..

- 30th August 2011, 17:47 #16

- Join Date
- May 2010
- Location
- Santa Barbara
- Posts
- 2,637
- Helped
- 772 / 772
- Points
- 14,895
- Level
- 29

## Re: Problem with Friis's Equation for Free Space Path Loss

All above arguments are correct. The Friis' formulation of a radio link is based upon a distance between the end antennas where there is a "fee-space" planar-wave electromagnetic field. Many antennas, mostly high-gain ones, form their main beams quite far from the aperture. Neglecting this is wrong. Many real-world conditions do not equal "free-space": closeness of the ground cause reflections and power cancellation at a receiver antenna. Indoor propagation causes multipath, again quite difficult condition.

There were recently many experimental and theoretical works devoted to solving similar difficult conditions. Some use different propagation-loss definition; instead of reverse-square, up to reverse sixth order is used, for the multipath conditions, rather delay spread is important than power loss; for mobile networks, Doppler shift and delay spread combined are important.

Use the basic equations in "standard" situations where they are valid. Be cautious when you face different situations. Experiments are always decisive in such situations. Theory is based on limiting conditions, often unknown or not stated by authors.

+ Post New Thread

Please login