Satellite internet · Session 2 of 8
Putting information on a wave
What sets the ceiling on how fast a wireless link can carry information?
Stand at opposite ends of a dark corridor and try to send the word “yes” by flashlight, having agreed nothing beforehand. One of you flashes. The other sees light, dark, light, and then a long dark — and cannot tell a pause from a finished message, or four zeros from an empty hallway. The light is working perfectly. What is missing is a clock that both of you trust.
By the end
- Use c = fλ to locate a channel on the electromagnetic spectrum and calculate its wavelength.
- Measure the bit rate of a hand-run link and express the cost of a repetition code in bits per second.
- Apply the Shannon–Hartley relation to explain why bandwidth and signal-to-noise, not transmitter power alone, set the ceiling on data rate.
The idea
Radio waves, microwaves, infrared and visible light are the same phenomenon at different scales: oscillating electric and magnetic fields that propagate through vacuum at c = 2.998 × 10⁸ m/s. They need no medium, which is exactly why they work in space and sound does not. Frequency and wavelength are locked together by c = fλ, so naming one names the other.
Your flashlight emits visible light around λ ≈ 500 nm, which is f = c / λ ≈ 6 × 10¹⁴ Hz — 600 terahertz. A Starlink user terminal works in the Ku band near 12 GHz, so λ = c / f = (3.00 × 10⁸) / (12 × 10⁹) = 2.5 cm. Those two channels are the same physics separated by a factor of 50,000 in wavelength, and that factor decides almost everything else: antenna size, how the beam spreads, what rain does to it.
Putting information on a wave is modulation. The flashlight uses the simplest kind, on-off keying: the carrier is present or absent, one bit per symbol. Real links squeeze more in — QPSK encodes 2 bits per symbol by choosing among four phases, 16-QAM encodes 4 by choosing among sixteen combinations of amplitude and phase. Bit rate = symbol rate × bits per symbol, which is why a link can get faster without the wave oscillating any faster.
There is a hard ceiling, and it is not about transmitter power. The Shannon–Hartley relation gives the maximum error-free rate of a channel as C = B · log₂(1 + S/N), where B is the bandwidth in hertz and S/N the signal-to-noise ratio. Because S/N sits inside a logarithm, raising transmitter power a hundredfold only about doubles the capacity — while doubling B doubles C outright, for far less energy. Spectrum, not wattage, is the scarce resource, which is why satellite operators fight over frequency allocations rather than over power supplies.
Higher frequencies carry more bandwidth but are more fragile. Ka band near 20 GHz (λ ≈ 1.5 cm) offers wider channels than Ku, and also suffers worse rain fade, because droplets a millimetre or two across interact strongly with centimetre waves. Between satellites there is no air at all, so operators use optical inter-satellite links at around 1550 nm infrared — enormous bandwidth, no rain, and no spectrum licence required in vacuum.
One more constraint shapes the hardware you can see from the ground. A beam from an aperture of diameter D spreads with an angle of roughly θ ≈ λ / D radians, so a shorter wavelength from the same dish gives a tighter beam and more of the power lands where you aimed it. A Starlink dish is a phased array: hundreds of small elements whose relative timing is adjusted electronically to steer the beam at a moving target, with nothing mechanical turning.
One frame, two readings
Gather
Flashlight, opaque card, paper, stopwatch, protractor, calculator and a partner. Use ordinary room lighting and never point the light towards anyone's eyes.
Build and investigate
- Agree a frame format before transmitting: a 3-flash start marker, then 8 data bits, each bit occupying a fixed 2-second slot, light on = 1 and light off = 0. Write the format down; it is your protocol.
- At a measured distance, send an agreed 8-bit message. The receiver records a value in every slot — including slots where nothing happens. Compare with the original and count correct bits out of 8.
- Repeat three times at the same distance, timing and message. Record correct bits and total elapsed time each run, and calculate the bit rate in bits per second.
- Change exactly one factor — distance, slot length, or a partly obscuring card — and run three more trials. Keep everything else fixed so the comparison means something.
- Now apply a triple-repetition code: each bit sent three times, majority decoding. Record accuracy and elapsed time, and calculate the new bit rate. State what you paid in bits per second for what you gained in accuracy.
- Calculate λ for your flashlight and for a 12 GHz downlink using c = fλ. Place both on a drawn spectrum line, log scale, and mark where rain, walls and vacuum each matter.
Run the numbers
Everything here follows from two relations: one connects frequency to wavelength, the other sets the ceiling on rate.
| Relation | Expression | What it means |
|---|---|---|
| Wave relation | c = fλ | c = 2.998 × 10⁸ m/s in vacuum |
| Bit rate | R = symbol rate × bits per symbol | OOK = 1, QPSK = 2, 16-QAM = 4 bits/symbol |
| Channel capacity | C = B · log₂(1 + S/N) | B in Hz; S/N as a ratio, not decibels |
| Decibels | dB = 10 · log₁₀(S/N) | 20 dB means S/N = 100 |
| Beam spread | θ ≈ λ / D radians | shorter λ or wider aperture → tighter beam |
Core · Grade 7–8
- Calculate the wavelength of a 12 GHz downlink and of a 20 GHz Ka-band link. Which has the shorter wavelength, and which would you expect rain to disturb more?
- Your link runs at one bit per 2-second slot. What is its rate in bits per second? How long would your link take to deliver one 6.2 MB video frame from Lesson 1? Express the answer in years.
- Your triple-repetition run sent 24 slots for 8 bits. Calculate the effective bit rate and compare it with the uncoded rate as a percentage.
Deeper · Grade 9–12
- A channel has B = 250 MHz and S/N = 100 (20 dB). Calculate C from the Shannon–Hartley relation. Then recalculate with the power raised a hundredfold, S/N = 10,000, and state the factor by which C improved. Why is doubling the bandwidth a better investment?
- Compare your hand-run flashlight link with that 250 MHz channel. Express the ratio as a power of ten, and identify which term in C = B·log₂(1 + S/N) accounts for nearly all of the difference.
- A phased array 0.5 m across transmits at 12 GHz. Estimate the beam spread θ ≈ λ / D in radians and in degrees, then calculate the beam's diameter where it meets a satellite 550 km away. Comment on whether the beam is aimed at one satellite or illuminating much of the sky.
Measure and record
Record trial, distance in metres, slot length in seconds, coding scheme, correct bits out of 8, elapsed time in seconds and the calculated bit rate. Repeated copies carry no extra message information — count data bits, not slots.
Worked example
At one bit per 2-second slot, 8 bits occupy 16 s of data slots, giving a rate of 8 ÷ 16 = 0.5 bit/s, excluding the start marker. Triple repetition needs 24 slots — 48 s — so the rate falls to 8 ÷ 48 = 0.167 bit/s, a third of the original, in exchange for surviving one error per triple. Now put that beside a real channel: B = 250 MHz with S/N = 100 gives C = 250 × 10⁶ × log₂(101) = 250 × 10⁶ × 6.66 ≈ 1.66 × 10⁹ bit/s. That is 3.3 × 10⁹ times your corridor link. Almost all of the gap is B: your hand supplies a bandwidth of well under a hertz, and a brighter flashlight would not close it, because S/N enters only through a logarithm — raise it a hundredfold and C merely doubles, from 1.66 to 3.32 Gbit/s.
Pause and explain
A book blocks your flashlight completely. Your partner concludes the same book would block a 12 GHz Starlink signal. Evaluate that conclusion using a number from this lesson.
Check your reasoning after your attempt
The conclusion does not follow. The two waves differ in wavelength by a factor of about 50,000 — 500 nm against 2.5 cm — and absorption, reflection and diffraction all depend strongly on how the wavelength compares with the structures in the material. Longer waves diffract around and penetrate obstacles that stop visible light; that is why radio reaches into a building and a torch does not. The experiment is valid evidence about visible light in those conditions and says nothing about 12 GHz. Note also what is transferable: line of sight matters more as wavelength shortens, which is why optical inter-satellite links need an unobstructed path and a very precise aim.
Save in your mission notebook
Your written frame format; the full trial table with calculated rates; λ for both channels with working; the Shannon–Hartley calculation; one sentence on which term dominates the gap.
Parent / teacher notes
Insist on the written frame format before the first flash — learners who skip it rediscover the ambiguity the hard way, which is the point, but only if you let the failure stand before fixing it. Use written cards instead of flashes for anyone sensitive to flashing light; the protocol reasoning is unchanged. Grade 7–8 can stop after c = fλ and the rate arithmetic. Grade 9–12 should do the Shannon–Hartley comparison, which needs log₂ — on a calculator, log₂ x = ln x ÷ ln 2.
Suggested timing: 10 minutes predict, 15 read and explain, 30 build and investigate, 25 measure and calculate, 10 independent explanation. Adapt the activities rather than rush unfinished reasoning.
Take it further
Design a 3-bit code for the letters A–H and send a two-letter word. Then design a variable-length code that gives your most common letter a shorter codeword, and measure whether your message got shorter. You have just rediscovered the principle behind Huffman coding.