Big fat paging bits

Posted by on Feb 15, 2013 in Blog, Simulcast, Tutorial | 0 comments

Back to our sponsor, paging, for a moment. In my very first post here, I took a comparative look at some distinctions between modern paging and digital cellular systems. I want to return to that idea, but with a slightly different point of view. This time, I want to consider the composition of the information-carrying components of these systems on a photon-by-photon basis.

First, let’s go back to the basic aspects of a bit. A base station (BS) transmitter radiates electromagnetic power in sending a communications symbol for a certain amount of time. In the case of a paging system, say FLEX™ or ReFLEX™, there will be about 300W sent for 1/6400 seconds for one symbol carrying two bits of information. It is worth noting though that in a serving area like, say, Dallas, there may be about 30 BSs sending the same symbol at the same time, synchronized by GPS clocks to within fractions of a microsecond, sending the same power level for the same symbol. This is simulcast. Because of this synchronization of efforts on the forward channel, information arriving at the mobile station (MS) does not compete against interference from other transmissions within the system; rather, it competes only against self-noise in the MS receiver.

In the case of a digital cellular system, and here I’m going to take GSM as a paradigmatic example, a BS will transmit at about 35W effective radiated power (ERP). This means that the actual power going to the antenna is less than 35 W but that the antenna shapes the delivery of this power so that the power sent out to the region occupied by users is equivalent to 35W. This is done by lowering the amount of power delivered straight up into the sky and straight down into the ground. Of course, many of you reading this already understand antenna patterns in great detail, so I’ll leave that topic alone. The bit rate for GSM is 270.833 kbit/s using Gaussian Minimum Shift Keying (GSMK). I’ve discussed in other posts here that modern digital paging systems also use this modulation method, but with 2 bits per symbol and different modulation indices than adopted in the GSM. This is one reason that I’m picking on the GSM as a comparison model, since one might argue that it is the digital cellular system that is closest to FLEX or ReFLEX.

With these facts in hand, one can work out the energy per bit, in Joules, for both a paging system and a GSM system. For paging we get,

$(300 \text{Watts} / \text{station}) (1/3200 \text{symbols} / \text{second}) (30 \text{stations} /\text{serving area}) / (2 \text{bit} / \text{symbol}) = 1.40625 \text{Joules} / \text{bit} / \text{serving area}$

For the GSM it is,

$(35 \text{Watts} / \text{cell}) (1/270000 \text{bit} / \text{second}) = 0.00012963 \text{Joules} / \text{bit} / \text{sector}$

This has a touch of comparing apples and oranges so far, but I’m going to address that in a moment. First, I want to consider the energy per photon for these communications systems. Since both operate at around 900 MHz, I’m going to choose that frequency in both cases. The energy per photon is then

$E = h f = 5.96346*10^{-25} \text{Joules}$

Clearly, there are jillions of photons per bit in either case. In paging, it would be

$2.35811*10^{24} \text{photons} / \text{bit} / \text{serving area}$

and in GSM, it’s

$2.17373*10^{20} \text{photons} / \text{bit} / \text{sector}$

The difference between these two expressions is the area covered by the two systems. As a rough approximation, the area served by the Dallas paging system’s nominal 30 BSs is around 1250 square kilometers. In contrast, an urban to suburban GSM cell may have a serving radius more like 2 to 10 km. The maximum radius for a GSM cell is around 35 km, without any extender technology; but this is very unlikely for capacity reasons in a dense urban market like Dallas. In fact, capacity considerations are the more likely design driver than pure coverage issues; and this would tend to drive cell radius down especially in the urban core. Also, to enable in-building coverage, it would be typical to deploy in-building cells with very small radius and much lower power; but again, we’d lose the apples to apples comparison if I were to include that technology in this simple-minded analysis. If I try to keep apples to apples, and assume that the number of BSs is the same in each system, I discover that the typical cell radius would be about 3.6km. That is, if I take the market area of around 1250 km and divide by 30 BSs and work out the radius, I get about 3.6km. This is not a bad average considering the market density. However, a GSM cell is divided into 3 sectors, and so the nominal area per BS would be lower than $\pi \text{3.6 km}^2$ by a factor of 3. Using this information, I can work out the density of photons per unit area as follows. For the paging system, it’s

$1.87652*10^{15} \text{photons} / \text{bit} / \text{square meter}$

and for the GSM

$1.55682*10^{13} \text{photons} / \text{bit} / \text{square meter}$

The ratio is about 120:1 or about 21 dB, expressed in an engineering format. Of course, these photons aren’t hanging around in any given square meter of serving area; they’re tearing along at the speed of light. We can also think about the number that are intercepted by the effective area of a receiving antenna, but such an area would be at right angles to the surface area of the earth that I’ve just worked out. So the two concepts aren’t really the same. As well, these numbers are averages over the served area and would vary with proximity to any BS antenna. One should also consider the uncertainty principle in terms of being able to localize a photon in this fashion. To the extent that BS transmissions are frequency-accurate; that is, there is a certain defined accuracy in both the carrier and in the modulation, and this accuracy is on the order of a few parts per million, then using $\Delta E \Delta t \leq \hbar$ we would have

$\Delta t \leq \frac{\hbar }{h f / 10^6 }= \frac{1}{2 \pi 900}$

Multiplying by the speed of light gives a photon width of about 53 km. Any given photon cannot be localized to within a square meter as a consequence of this simple calculation; we know its energy and momentum too well. That accuracy is essential for the process of modulation and information transmission. However, as a side effect of this blurring in space, I think that the notion of areal density is even a more reasonable metric. I could probably split some hairs here and make a distinction between the photon momentum in the radial direction and its conjugate position versus the photon momentum in the inclination angle and its conjugate momentum; (these are quite different) but I think I’ve made my point. All these photons are quite large relative to a square meter of serving area.

Now let’s think about the MS receiver side. At the paging MS, the receiver’s performance is noise limited, while in general at the GSM MS, the receiver’s performance is interference limited. The total noise at the paging MS is quite small because the receiver bandwidth is small; for example, around 25kHz. This is a larger than necessary BW, but is employed to deliver the same sort of processing gain, of around 10dB, that is a feature of CDMA systems. By using a higher passband width than receiver baseband width, the paging system can reduce noise by nearly 10dB (more like 8.9dB, but I’m rounding a lot here). This helps to discount the limitations of the first RF amplifier’s noise figure. Assuming a 50 ohm antenna, and about $1 nV/\sqrt{Hz}$ of noise at that impedance, and using a NF of about 3dB, we get around -120dBm of noise power at the receiver. Dropping this by the 10dB FM gain means around 130dBm of noise and a SNR of 42dB. A typical design level for a paging system is to deliver -88dBm of received signal strength at the ground level outside any building. (I’ve written about this elsewhere.) At -88 dBm, we are around 113 dB below the power of a single paging BS. The excess SNR of 32dB is designed to allow for 20dB of building penetration loss and still allow for an SNR of 12dB in-building and a 1% message error rate (MER) in a 40 character text message, with error correction.

At -88 dBm, we are receiving only $2.4764*10^{-16} \text{Joules} / \text{bit}$. This is still $4.15261*10^{8}$ photons per bit effecting the receive antenna. Dropping this by 20dB to consider the in-building case still leaves $4.15261*10^{6}$ photons per bit. The reason we need to accumulate the effect of so many photons is the minimal amount of energy per photon which comes along with operating at 900MHz instead of, say, optical or higher frequencies. I should be very clear here and state that just because the RF field strength at the MS antenna is around -88dBm doesn’t mean that every one of the photons that comprise that power level is being detected. The quantum efficiency of the antenna is pretty poor. Perhaps I’ll come back to that idea another time. For now, think about a photon that the uncertainty principle states has to be a wave packet 53 km long interacting with a single electron in a MS’s antenna at a distance of, say, 3km from the BS antenna, as it were, before it has been completely emitted from that antenna. Of all of the electrons, why that particular one? This is another reason why there have to be so many photons emitted from an RF antenna, there are just so many electrons to deal with in the scope of any given photon. Think of this as one huge multi-slit interference experiment.

For the GSM MS, the reference sensitivity level is specified as -102dBm; but this is noise performance and not carrier to interference ratio. In practice, designers would attempt to keep the received signal strength above this noise threshold by as much as 15 dB in order to guarantee interference-limited performance with a CIR of about 9dB or better at cell boundaries. At better than a 15dB SNR, the performance of a GSM MS is pretty good, even in bad urban conditions. This implies that the physical process that is the source of confusion at the MS is just the detection of photons from nearby co-channel BSs. If I take a 10dB CIR as a nice round number, then there are about $1/10$th as many photons from interfering BSs as there are from the intended BS. If I further assume that the received signal strength at the cell boundary is about 15dB above the noise threshold of -102dBm, I get a signal strength of -87dBm, which is so close to the design value for a paging system that it isn’t even funny. However, the part of this power that contributes energy to any single bit is far less, because the GSM bit time is so much smaller. This corresponds to around 2pW of power in the received signal. Dividing this value out by the GSM bit time and the energy per photon at 900MHz gives $1.23919*10^7 \text{photons} / \text{bit}$. From an assumption of 10dB CIR, we immediately get $1.23919*10^6 \text{photons} / \text{bit}$ of aggregate interference arriving from co-channel BSs. [I probably shouldn’t quote so many places of precision in these numbers given the numerical approximations that I’m making, but since I’m just copying and pasting from a Mathematica notebook, I leave it to you, gentle reader, to make the mental adjustments necessary.]

Now, we have a sort of interesting contrast in performance between the paging receiver and the GSM receiver, out there in mobile land. In the paging case, there is no co-channel interference. So when the user takes it indoors, the photon density may go down by a factor of 100 or so (20dB); but these desired photons are not competing with interfering photons at the MS antenna. In the GSM case, the opposite is true, the intensity of both the intended and desired photons are reduced by a factor of 100, but now the intended photons are competing against the noise threshold, and they are failing. They are failing in the GSM device and not the paging device for two reasons: there is no FM processing gain and the noise bandwidth is greater by a factor of about 10 (300kHz versus 25kHz). These two factors net to give the paging receiver a roughly 20dB advantage. The result is that the paging device continues working while the GSM device fails without an in-building extender.

Of course, these days the concept of in-building system extensions are taken to be a matter of course; so much so that nobody even considers them as anything but an essential component of a cellular network. I just bring out this detail to show that they are not an essential design element if one considers the overall design of the network to begin with.

Back to the paging receiver, what are its detected photons working against. The answer is nothing. It is one step down the process chain that the interference arises. After the paging photon excites an conduction electron in the MS antenna and gets it rolling, that electron’s motion must compete with the random thermal motion of other electrons in the input resistor of the first stage RF amplifier. Let’s assume that that resistor is of 50 Ohms. I mentioned earlier that this implies a noise of about $1 nV \sqrt{Hz}$. Working with a BW of 3.2 kHz gives a noise power of about -116dBm.

What does this mean physically? Well, I’m working from a formula for Johnson-Nyquist noise at a given temperature, bandwidth, and resistance; viz.

$\langle v\rangle =\sqrt{4 k_B T R \Delta f}$

where $\langle v\rangle$ is a mean thermal noise voltage in Volts, $k_B$ is Boltzmann’s constant in Joules per Kelvin, $T$ is temperature in Kelvin, $R$ is the resistance in Ohms, and $\Delta f$ is the bandwidth in Hertz. This formula itself is based on the fluctuation-dissipation theorem which is itself an extension of the Einstein relation in kinetic theory. What is all this junk, I hear you asking? Well, back in 1905, good old Albert published a paper on Brownian motion. This had to do with the diffusion of particles in a thermal system; for example, bits of dust in a gas at some temperature. A variation of Einstein’s relation applies to our friends, the electrons, in this front end resistor in the MS first-stage RF amplifier.

You can think of these electrons in the conduction band of the resistor as gas particles in a container. Just like any old gas at thermal equilibrium, they have some average kinetic energy due to equipartition; namely, $3/2 k_B T$. Just like a collection of simple monatomic gas molecules, these electrons will diffuse within the volume of the resistor with this mean kinetic energy, which is equivalent to some average velocity. At the conditions we’re talking about here, this thermal velocity is pretty much Gaussian distributed about the mean.

And meanwhile, our photons are imparting their energy, $E = h f$, to the electrons in the antenna. When an electron in the conduction band of the antenna acquires this energy, it goes sailing off with a kinetic energy that imparts to it the corresponding velocity. When these electrons enter the first stage resistor, we now have the interesting problem of being able to distinguish an electron drifting through the resistor with a kinetic energy of $h f$ versus a thermal electron with an energy $3/2 k_B T$. Well, recall that $h f = 5.96346*10^{-25} \text{Joules}$, but $3/2 k_B T = 6.21292*10^{-21}$. The energy of a single RF photon is not enough to be distinguished from the thermal energy of any single electron, not by a long shot. Hence, we need to accumulate enough photons to exceed this thermal energy in the resistor. If we had about 10,500 photons, we could just equal the thermal energy in the resistor, per electron. Luckily, we’re managing to get $4.15261*10^{8}$ photons per bit. This allows for an accumulated energy in drift electrons to be distinguishable from that of thermal electrons. Even though each photon delivers its energy only to a single electron, by the time these electrons begin moving at the antenna, their net energy is gradually coupled to other electrons through the conductors that communicate their flow (aka current) to the first stage RF amplifier’s front-end resistor.

For any given resistor, there will be a certain number of conduction band electrons based on the details of its composition. Say, one valence electron per atom, and so many atoms of a given element based on its density and volume. The resistance, in Ohms, will depend upon the exact features of this composition and the physical dimensions of the device. In operation in the RF receiver we’re thinking of here, these electrons will be moving in part due to their thermal kinetic energy and in part due to energy received from capture of RF photons at the antenna, which is coupled to the resistor by conductors.

While it would take me too far down this rabbit hole right now, one could even work out the thermodynamic entropy of the electron gas in the resistor and compare that with the Shannon entropy of the signal that is driving drift electrons through the resistor. When you think of where the “rubber hits the road”, as it were, in communications, it becomes increasingly apparent that Shannon was spot on in terms of his introduction of entropy in the analysis. The problem would be exactly equivalent to the issue of detecting drift molecules from diffusion molecules in a gas in which the drift would be due to a superimposed electric field and in which some of the gas molecules had been ionized. This form of analysis is also used in semiconductor physics; for example, in the Ebers-Moll model of a bipolar junction transistor. Not surprisingly, in most RF front ends, right after this now-famous first resistor will be a transistor that amplifies the signal using these self-same physical principles.

Anyway, enough of this faulty bagnose. I mainly wanted to contrast the design of a modern digital paging system with that of a cellular system. I picked on FLEX/ReFLEX versus the GSM since they operate with about as similar a set of conditions as one could find, right down to modulation method. I hope you found it interesting. There are several loose threads here; and I’ll try to tie them up in future posts.

PS. I’ve made a cute little circuit diagram for this front-end amp below. It is extremely simplified and no one would really do it quite this simplistic way, but it gets the basic idea across. The effective input resistance, R, is just the parallel combination of the two bias resistors, R1 and R2.

Antenna and amplifier

PPS. Using the mass-energy equivalence $E = mc^2, the energy of a paging bit of$latex 1.40625 Joules\$ is equivalent to a rest mass of $15.625 picograms$. Just saying.