Margin is how far the available performance exceeds what the service requires. Positive margin means the link closes; how much margin is enough is a design (and risk) decision.

Margin Against a Received-Power Threshold

  • Start from the aligned, polarization-matched received power [𝑃𝑟0] and subtract the pointing and polarization losses:
    • [𝑃𝑟]=[𝑃𝑟0]−[𝐿pt]−[𝐿pr]−[𝐿pol]
    • [𝐿pt], [𝐿pr] = transmit and receive pointing losses
  • Margin against a required received power:
    • Margin=[𝑃𝑟]−[𝑃req]

Pointing and Polarization in the Link Budget

Illustrative linkValue
[𝑃𝑟0]: aligned, polarization-matched power−120 dBW
Pointing loss: [𝐿pt] transmit / [𝐿pr] receive1 dB / 2 dB
[𝐿pol]: polarization mismatch loss3 dB
Actual received power−126 dBW
  • For a −125 dBW threshold, the margin is −1 dB
  • These losses change whether this link supports the selected service

Don't double count

  • The aligned power already includes the other link terms
  • Use peak gains with pointing losses, or actual-direction gains without subtracting pointing loss again

Available and Required Eb/N0

  • Received power alone does not say whether the bits are recovered; the real comparison is in energy per bit
  • Available 𝐸𝑏/𝑁0 comes from received data power, noise density, and information rate
  • Required 𝐸𝑏/𝑁0 comes from a performance curve or a tested receiver specification
    • Example: ideal coherent uncoded BPSK needs about 9.6 dB for BER =10−5 in AWGN (BER curve)
  • Real receiver losses (implementation loss) and design reserve must also fit in the budget

Eb/N0 Margin

  • Margin=[𝐸𝑏/𝑁0]avail−[𝐸𝑏/𝑁0]req−[𝐿impl]
  • The margin must then cover any required design reserve

Lunar Link Margin at 1.0 Mbit/s

  • Available (from the lunar link budget and energy per bit):
    • [𝐸𝑏/𝑁0]=[𝑆data/𝑁0]−[𝑅𝑏]=74.5−60.0=14.5dB
  • Assume BER ≤ 10−5 requires 9.6 dB for ideal uncoded BPSK
  • Assume a separate implementation loss of 1.0 dB
  • Margin=14.5−9.6−1.0=3.9dB

Rate and Margin

Information rateAvailable 𝐸𝑏/𝑁0Margin
1.0 Mbit/s14.5 dB3.9 dB
2.0 Mbit/s11.5 dB0.9 dB
4.0 Mbit/s8.5 dB−2.1 dB
  • Margin=74.5−10log10(𝑅𝑏)−9.6−1.0dB
  • Bandwidth and receiver lock must also be adequate — power margin isn’t the only requirement
  • We need 3.0 dB reserve, but the 4.0 Mbit/s case has −2.1 dB margin. Which change alone is enough?
    1. Switch BPSK to QPSK at the same information rate
    2. Reduce the information rate to 1.0 Mbit/s
    3. Add 20 dB of ideal gain after the receiver front end
  • Answer: reduce the information rate (option 2)
    • The 1.0 Mbit/s case has 3.9 dB margin, which exceeds the 3.0 dB reserve
    • QPSK reduces bandwidth but does not change ideal bit-energy performance (BPSK vs. QPSK)
    • Gain after the front end amplifies signal and existing noise together (improving sensitivity)

Maximum Rate With Reserve

Lunar Link: Maximum Rate With 3 dB Reserve (optional slide)

  • Require 3.0 dB design reserve beyond the BER requirement and implementation loss
  • [𝑅𝑏,max]=74.5−9.6−1.0−3.0=60.9dB(bit/s)
  • 𝑅𝑏,max=1060.9/10≈1.2Mbit/s
  • This is a power-limited result; spectrum, coding, hardware, and acquisition can lower it