In what country? The FCC has be screwing around with the 5.7GHz power and spectra rules until nobody seems to know what's legal and what's not. This is a bit old but covers the basics and the politics:
the section starting with "Power Levels". Basically, for point to point, 1 watt tx maximum into any directional antenna up to 23dBi gain. After that, a reduction in tx power is required for larger antennas.
Point-to-point UNII-3 systems can employ the transmitting antennas with a directional gain up to 23 dBi without any corresponding reduction in peak transmission power. For antennas with a gain greater than 23 dBi, a dB reduction in output power is required for every corresponding dB increase in excess of 23 dBi.
EIRP = Peak transmission power (30 dBm or 1 W) + 23 dBi = 53 dB (200 W)
Note that Cisco has not obtained regulatory approval for Saudi Arabia, assuming such approval is required.
Wrong. You must SUBTRACT your losses and add your gains. EIRP is: +24dBm + 28dBi -1.3dB = 50.7dBm EIRP I'm not sure where you got your numbers, but the -1.3dB cable and connector losses are unrealisticly small. If you describe the RF components a bit better, we might be able to conjur a more realistic number.
Yep, it really is about 200 watts EIRP (with the numbers you supplied). That's one nice thing about microwave frequencies. You can get literally huge antenna gains, which result in sky high EIRP levels. However, the much higher free space loss means that there's no free lunch and that you can't win.
Higher EIRP is allowed if the antennas are directional in nature.
Systems operating in a point-to-point operation may employ transmitting antennas with directional gain greater than 6 dBi provided the maximum output power of the transmitter is reduced by 1 dB for every 3 dB that the directional gain of the antenna that exceeds 6 dBi. Maximum transmitter power versus largest antenna table for PTP:
Transmitter RF power Antenna Gain EIRP in watts 30dBm 1W 6dBi
The higher the frequency, the smaller the antenna. Translation - a 5 dB gain 42 MHz Yagi will get torn off the headache rack on my pickup when driving through a Burger King, but a 5.8 GHz antenna is smaller than a Barbie Doll accessory.
There really isn't greater attenuation, rather the antenna is "less sensitive" at higher frequencies.
Yep. For a given antenna aperature size, the gain increases with frequency. For example, the gain of a parabolic dish is: dBi = 20 * Log(N * Pi) where N = diameter in wavelengths. You can see that as the frequency increases, the number of wavelengths in the dish diameter also increases, thus increasing the gain. Double the number of wavelengths in the diameter, and the dish gain increases by 6dB.
Correct. As the frequency goes up, so does the free space losses.
Double the frequency, and the free space loss goes up 6dB, which cancels the the 6dB gain increase in the antenna gain.
Try again. The free space loss calculation does not in any way involve the antenna. It is the loss between two isotropic antennas at a given frequency and a given distance only. If there were anything involving the antenna (capture area, aperature size, gain, beamwidth, etc), then it would appear in the free space loss formulas and calculations. They don't. There's no antenna gain in any of the formulas on this page:
What do you mean by "total path loss"? Are you including the transmitter, antennas, receiver, coax cables, and propagation imparements? If so, you're talking about system gain/loss, which is a different animal, and certainly includes the antenna gains. If you want to measure your path loss between the antenna ports, that's fine, but it's NOT the free space loss.
Quiz: You have a broadband dish antenna at both ends of a fixed
2.4GHz link. Shove 0dBm into one antenna and you get -60dBm at the other end of the link. Now, double the frequency to 4.8GHz. What's the signal level at the other end?
I'll rephrase it this way.... The effective free space loss does indeed increase (as shown by the on-line calculator above), but not due to attenuation traveling through free space; rather less signal is captured by the antenna.
Ok, slow down...I don't think were on the same page yet.
Its not obvious in the formula for the on-line calculator, but if you examine how that formula was derived, you can see how antenna capture area was used.
read down to the Physical explanation section and you'll see "The second effect is that of the receiving antenna's aperture, which describes how well an antenna can pick up power from an incoming electromagnetic wave...Note that this is entirely dependent on wavelength, which is how the frequency-dependent behaviour arises."
See above. Since just about all of my equipment is similar, I don't need to input any variables such as different antennas, I'm just so used to saying "total path loss" to describe the free space attenuation. Sorry for the confusion.
Unwittingly, that could be a trick question.
If you had a unity gain quarter-wave antennas at the receiving end and doubled the frequency, the answer is -66 dBm as the 4.8 GHz antenna has one fourth the capture area.
If you had increased the gain of the receiving antenna by making it four times longer (as in a four element co-linear array), you increase the capture area by four times, so the answer would be -60 dBm.
Since you didn't specify that you used a smaller antenna when you doubled the frequency, rather retained the same square area and thus increased the gain by four times, the answer is -60 dBm.
Look at the equations again. See any term that involves an antenna? The free space loss is dependent on frequency and distance, but nothing else (including aperture size, capture area, size of the antennas, position of the moon, or anything that does NOT involve frequency and distance). Now, you can include the antenna gain and aperture if you want into the free space loss calculations, but then you'll need to remove them from the antenna calculations, which already consider both gain and aperture.
No, I can't see an antenna gain, or aperture size in any of the equations on the Wikipedia page, or the derivations on:
Where are you seeing this reference to path loss?
I think you might mean this quote: "The second effect is that of the receiving antenna's aperture, which describes how well an antenna can pick up power from an incoming electromagnetic wave." Quoting myself two messages ago: "Double the frequency, and the free space loss goes up 6dB, which cancels the 6dB gain increase in the antenna gain." Translation: For a given antenna aperture size, the gain of the antenna increases with frequency in exactly the same amount that the path loss decreases. For a given antenna aperture, an increase in frequency results in no net change in received signal strength.
Total path loss is usually used in reference to all the real world impediments to transmission, such a Fresnel Zone obstructions, knife edging, atmospheric attenuation, water vapor attenuation, foliage attenuation, dust, smog, and so on. These are the sum total of everything involved in the actual path, but do *NOT* include anything involving the antennas, coax cables, and other losses that are NOT path dependent.
It's quite clear. The only trick is the term "broadband dish antenna" which implies that the dish diameter remains constant, and that the feed is efficient and functional over at least an octave of bandwidth. That's not all that easy to do in practice, but not impossible.
If you ignore my stipulation that the dish diameter remain constant, then you can contrive all manner of effects caused by the antenna. For example, if you were to use a pair of log periodic antennas, which have a constant gain at all frequencies within it's operating range, then doubling the frequency would NOT increase the two antenna gains, thus resulting in a net loss of -6dB due solely to the path loss.
Yep. That's the answer I was looking for. The received signal level does NOT change when the frequency is changed, as long as the dish feed is sufficiently broadband. However, I was hoping that Mr Kaon would catch the point as he seems a bit lost.
OK...lets look and see how the formula was DERIVED and you'll see where antenna aperture is figured in.
1) Determine flux density: Pfd = Pt / (4 Pi R squared)
2) Determine the power received by multiplying the antenna aperture: Pr = A [Pt / (4 Pi R squared)]
3) Given the aperture of an antenna is related to it's gain: A = G (wavelength squared / 4 Pi)
4) Doing a little substitution and rearranging the formula: Pr = G (wavelength squared / 4 Pi) x [Pt / (4 Pi R squared)]
Pr = G x Pt x [wavelength / (4 Pi R)] squared
5) Since we'll use isotropic antennas at both ends with a gain of 1: G = 1 then Pr = Pt x [wavelength / (4 Pi R)] squared
6) Since free space loss = Pr / Pt, we can say: FSL = [wavelength / (4 Pi R)] squared or FSL = [velocity of light / (4 Pi R)] squared
7) Now we'll covert it to logarithms and add a correction factor to convert to miles and MHz: FSL = 32.4 + 20log (frequency) + 20log (distance)
If I missed something, its because its after midnight and I'm on my
3rd Jack & Coke after getting thrown off my ATV when a 2,000 bull blindsided me this afternoon out in the pasture.