Comment by ohazi | Hacker News Reader

Comment by ohazi | Hacker News Reader

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Please. This is actually a common misconception about the Friis equation. It assumes an isotropic antenna with an effective area that is dependent on lambda. But physical antennas have an effective area that varies with lambda squared (obvious on something like a horn or a dish with a real 2-D aperture, but even simpler antennas behave this way). It turns out that this perfectly cancels out the frequency dependent term in the equation when the transmitter and receiver use the same antenna, leaving you with only the reduction in power density as you move away from the transmitter.

I had a professor give an epic lecture about this several years ago, but this is the best I could find on short notice [1].

As it turns out, there is a real source of frequency dependence in your path loss, which is the atmosphere. Your link budget calculation is really just an annoying geometry problem. Conservation of energy still holds, so you should be immediately suspicious when the naively applied Friis equation (which does not take atmospheric effects into account) makes received energy vanish just because you cranked up the frequency. Just as you should be suspicious when you receive more energy than you transmit when you're in the antenna's near field. Always know what your approximations imply.

The receivers are actually a bit more sensitive than that (the TI chip is ~-120 dBm at the target frequency; the whole package could be close to -114 dBm if designed well).

The claim I made was that you can't get half a km on 16 milliwatts. I cited 16 milliwatts because it's the max transmit power of the TI chip alone. You cannot make this go half a km with the antennas shown on the kickstarter page. You'd either need more power, or one of the antennas would need to be a dish. Even in an unreasonably quiet environment, you'd run out of energy at the receiver way before running into frequency or bitrate dependent effects.

But it turns out that 16 milliwatts is not the output power used. TI makes an amplifier that increases the output power to half a watt. In this case you'll easily get the half-km range.

[1]: http://www.dslreports.com/forum/r24210307-Reconciling-the-Fr...

Can you explain more what you mean? I think you're getting at something I agree with, but you're also saying some things I don't think are quite right...

Firstly I don't know what you mean by the Friis equation "assumes an isotropic antenna". It explicitly accounts for the gain of each antenna, in the direction of the link, relative to an isotropic radiator. The fact that it's relative to an isotropic radiator is just how all gains are measured. In fact, the antenna pattern doesn't matter at all to the link, only the gain in the direction of the link (Friis assumes no multipath). Whether the receive and transmit antenna are "the same" or "different" really doesn't make a difference.

This is the distinction I think you're trying to make, and which I agree with:

If gain is held constant, link margin improves as frequency decreases because of reduced path loss at lower frequencies. If instead antenna aperture is held constant as frequency is lowered, antenna gain will decrease at the same rate as path loss improves and there is no net effect to the link.

Both answers are technically correct, in a fixed gain scenario where your antenna can grow as large as necessary to hold gain constant, (or the scenario doesn't allow for high gain, narrow beamwidth antennas) lower frequencies will make longer links. In systems where aperture is constant (which is the case for many practical systems) antenna gain will improve as quickly as path loss degrades when you go to higher frequencies, and there is no net advantage at any frequency.

Sure, sorry.

So you're right that the two antennas being the same doesn't really matter. What I meant was that if the two antennas are the same, the entire part of the Friis equation that deals with frequency dependence goes away (the lambda / (4piR) part). If the antennas are different, the frequency dependence still goes away, but there's some new scale factor.

There are two interesting things that you want to know about your antenna. The gain, which measures directivity, and the effective area, which roughly corresponds to the cross section of sky that the antenna can listen to. Going from the transmitter to the receiver, you have some transmit power going into the antenna. You now want to figure out what the power density is in the vicinity of the receiver. You get this by spreading the power over a sphere and multiplying by the antenna gain of the transmitter.

Now you need to know how much of that power density is seen by the receiver. This is slightly more complicated than the transmit case, because you now have to take into account the antenna gain of the receiver (i.e. where it's pointing), which the Friis equation considers, as well as how big a chunk of sky it's listening to (i.e. effective area), which the Friis equation does not consider.

It turns out that the effective area is a function of lambda^2 (an antenna of some size and ideal frequency will have an easier time collecting higher frequency signals, and a harder time collecting lower frequency signals). So the lambda^2 from the effective area cancels the 1/(lambda^2) from the Friis equation.

You're right we're speaking a bit tangentially. Hadn't heard of the Friis argument that frequency doesn't matter as long as antennas on RX and TX are the same. I'm not sure I understand it yet.

Nevertheless, let's get some numbers on the page to see if I'm misunderstanding the subject.

Here's a rough link budget with some estimates

TX/RX antenna gain is around 0dBi (seems reasonable [1]) TX power = 16mW = 12dBm [2] RX sensitivity = -90dBm (personally, better than -100dBm seems aggressive)

Roughly, our link budget formula can be... Path loss = TX power + TX gain + RX gain - Rx power [3] Path loss = 102dB

Ideal free space loss over 1km at 900MHz is around 91dB [4]. Therefore, we have about 10dB margin in our link budget to transmit 1km. We also have a few dB margin in the antenna and receive sensitivity estimate.

[1] http://store.rfdesign.com.au/antenna-900mhz-2dbi-monopole-rp... [2] http://www.cpcstech.com/dbm-to-watt-conversion-information.h... [3] http://skypilot.trilliantinc.com/pdf/link_budget.pdf [4] http://en.wikipedia.org/wiki/Path_loss

Take this as the kind of half-cooked thought it is at this point (long day, about to go to bed and don't want to crack open the textbook), but the fact that your path loss leaves 10dBm of power above your Rx sensitivity doesn't mean you have 10dBm of link margin. I'm not completely sure, but I'd guess it means you have 10dB of SNR (theres a units from dBm to dB thing I haven't completely resolved in my head). Whether that makes your link work or not depends on the bitrate, modulation you choose and BER you can accept (and etc). If your modulation scheme needs 9dB of SNR (proportional to Eb/N0) to operate at an acceptable BER, your system may only have 1dB of actual link margin.

We're circling around some good stuff. Night though, thanks for the information to think about.

They cancel perfectly when the antennas are the same. There's a scale factor when they're different, but the frequency dependence still goes away.

So I'm adding about 10dB for noise, and getting a path loss that's about 15 dB higher, giving me an expected range closer to 100 m. What are you using for the antenna's effective area?

Do you agree with lpmay's explanation of Friis? Path loss increases as frequency increases. However, if aperture is constant while frequency decreases, which increases path loss, gain increases, which offsets the increase in path loss, and there is no change in link budget.

My receive sensitivity has noise included. It's why I think -100dBm seems for a 900MHz TI chip (could be wrong? didn't design this system...). Factor in noise/packet loss/PER or however you want to quantify, and I'd guess you'll get to a measured RX sensitivity of ~-90dBm.

Mentally, I equated RX sensitivity at data rate with RX power, which shouldn't be done for explanation purposes, but my original formula should still hold.

What are you using for path loss [1]?

[1]: http://en.wikipedia.org/wiki/Path_loss

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