# Sprite Signals: Challenges

Sprites are tiny; there’s a reason they are called chip satellites. They are powered by a pair of TASC (triangular advanced solar cells) that are less than 5 centimeters square in surface area, and their antennas are even smaller. Is it even possible for one of these devices to send a signal 500 km back to Earth?

To answer this question, I’ll start with the most simplistic model of signal power reduction as it travels to Earth: free space path loss. This model describes the power of a signal along a path ray pointing outward from the source as it spreads through vacuum, and it gives the best case scenario for a Sprite signal arriving at a ground station on Earth. Free space path loss takes two phenomena into account.

The first phenomenon this model takes into account is the inverse square law. Mathematically,

$S(d) = P_t \frac{1}{4 \pi d^2}$,

where $S(d)$ is the power per unit area (spatial power density) as a function of distance $d$ from the power source and $P_t$ is the total power transmitted from the power source in watts. For the Sprites, the distance $d \approx 500 \text{ km}$ and the total power transmitted $P_t \approx 10 \text{ milliwatts}$. Thus,

$S(500 \text{ km}) \approx 10 \text{ milliwatts} \cdot \frac{1}{4 \pi (500 \text{ km})^2} = 10 \times 10^{-3} \text{ watts} \cdot \frac{1}{4 \pi (500 \times 10^3 \text{ m})^2} \approx 3.18 \times 10^{-15} \frac{\text{watts}}{\text{m}^2}$.

The second phenomenon this model takes into account has to do with how well the receiving antenna absorbs power from an incoming signal. Mathematically, the power received

$P_r (d, \lambda) = S(d) \frac{\lambda^2}{4 \pi}$

for an isotropic (ideal) receiver, where $\lambda$ is the wavelength of the incoming signal. Sprites transmit at a frequency of $437 \text{ MHz}$, so

$\lambda = \frac{c}{\nu} \approx \frac{3 \times 10^8 \frac{\text{m}}{\text{s}}}{437 \times 10^6 \text{ s}^{-1}} \approx 0.686 \text{ m}$.

Using a wavelength of $0.686 \text{ m}$, then

$P_r ( 500 \text{ km}, 0.686 \text{ m} ) \approx S(500 \text{ km}) \frac{(0.686 \text{ m})^2}{4 \pi} \approx 3.18 \times 10^{-15} \frac{\text{ watts}}{\text{ m}^2} \cdot 0.0374 \text{ m}^2 \approx 1.19 \times 10^{-16} \text{ watts}$.

So, in a best case scenario, Sprite signals will reach a ground station antenna with a power of only 119 attowatts. That’s incredibly small! Surely there is no hope of picking up such a weak signal?

Before abandoning the project, let’s examine the signal strength of a different group of satellites: GPS. Satellites that are part of the GPS constellation orbit the Earth at an altitude of approximately $20200 \text{ km}$ and transmit $1227.60 \text{ MHz}$ signals at a power of $50 \text{ watts}$. A $1227.60 \text{ MHz}$ signal has a wavelength

$\lambda = \frac{c}{\nu} \approx \frac{3 \times 10^8 \frac{\text{m}}{\text{s}}}{1227.60 \times 10^6 \text{ Hz}} \approx 0.244 \text{ m}$.

$P_r( 20200 \text{ km}, 0.244 \text{ m} ) \approx 50 \text{ watts} \cdot \frac{1}{4 \pi (20200 \times 10^3 \text{ m})^2} \cdot \frac{(0.244 \text{ m})^2}{4\pi} \approx 4.63 \times 10^{-17} \text{ watts}$.

In other words, GPS signals received by smartphones have a power no greater than 46 attowatts!

It seems as if these incredibly small values are not so unusual. In fact, they are common enough that a different unit of measurement, the dBm, is often used instead of watts so as to avoid calculations with overly small numbers. By defining one dBm to equal the power ratio in decibels of some measured power $P$ referenced against one milliwatt, the range of values tightens considerably. Mathematically, for some power $P$ in watts the equivalent value in dBm is given by the expression

$10\log_{10} P + 30$.

So the Sprites’ 119 attowatt signals can be alternately expressed as

$10\log_{10} (119 \times 10^{-18}) + 30 \approx -129.2 \text{ dBm}$.

If a smartphone can receive 46 attowatt signals from GPS satellites, then a ground station can receive 119 attowatt signals from a Sprite. However, the extremely low power of these signals means that they can be easily corrupted by background noise in the electromagnetic spectrum. Signals of any power can be received and amplified without a problem. But unless some sort of filtering is applied for weak signals, noise will drown out their message. In order to ensure that received signals are readable, both the sender and receiver must make use of some clever tricks. I will cover some of these tricks – such as forward error correction and CDMA (code division multiple access) – in follow-up posts.