17. Link Budget in SDR Systems¶
Goal¶
Estimate whether a transmitted signal can be successfully received.
1. Basic equation¶
Prx = Ptx + Gtx + Grx - Lpath - Lcables
Where (all in decibels):
Ptx: transmitter output power, dBm;Gtx,Grx: gains of the transmit and receive antennas, dBi (not the receiver's amplifier gain);Lpath: propagation loss between the antennas;Lcables: cable, connector and attenuator losses.
In free space the path loss is
FSPL[dB] = 20·log10(d[m]) + 20·log10(f[Hz]) − 147.55
The receiver's own gain (LNA, SDR gain setting) raises the signal and the noise together, so it does not appear in the link budget; it decides whether the ADC sees the signal at a sensible level, not how far it is above the noise.
2. Components of the link¶
Transmitter¶
- output power;
- RF frontend characteristics.
Channel¶
- cable loss;
- free-space path loss;
- attenuation;
- reflections.
Receiver¶
- antenna gain;
- noise figure (how much noise the receiver adds);
- SDR gain settings (for the level at the ADC, see above).
3. Worked example¶
A tone at 915 MHz, Ptx = −10 dBm, simple antennas of about 0 dBi, 3 m apart, 1 dB of cable at each end:
FSPL = 20·log10(3) + 20·log10(915e6) − 147.55 ≈ 41.2 dB
Prx = −10 + 0 + 0 − 41.2 − 2 ≈ −53.2 dBm
Is that enough? Compare it with the receiver's noise floor over the observed bandwidth. Thermal noise is −174 dBm in 1 Hz; over 2.4 MHz it is −174 + 10·log10(2.4e6) ≈ −110.2 dBm, and a receiver with a 6 dB noise figure raises it to about −104.2 dBm. The tone is therefore about 51 dB above the noise across the whole 2.4 MHz, and much more inside a single FFT bin, so it will be easy to see.
The same arithmetic warns in the other direction: by cable, without the 41 dB of free-space loss, the same −10 dBm would arrive at about −12 dBm, enough to overload a receiver set to high gain. That is why the cable setups of the course always include an attenuator.
4. Practical SDR considerations¶
- RTL-SDR has limited dynamic range;
- too much gain leads to clipping;
- too little gain leads to poor SNR;
- the free-space formula assumes the far field: at 915 MHz the wavelength is about 0.33 m, so at a few metres it holds roughly, while at a few centimetres the coupling between antennas is unpredictable;
- indoors, reflections from walls and furniture make the real loss vary by several dB when an antenna moves by a few centimetres.
5. Diagram¶
flowchart TB
classDef rf fill:#FFE4E6,color:#0F172A,stroke:#E11D48;
classDef metric fill:#F1F5F9,color:#0F172A,stroke:#64748B;
TX["Transmitter"]:::rf
CH["Channel / losses"]:::rf
RX["Receiver"]:::rf
P["Received power"]:::metric
TX --> CH --> RX --> P
6. Review questions¶
- Why does the receiver's gain setting not appear in the link budget?
- How much does the free-space loss grow when the distance doubles? When the frequency doubles?
- In the worked example, what transmitter power would bring the tone down to 10 dB above the noise in 2.4 MHz?
- Why can a cable connection be more dangerous for the receiver than an over-the-air one?
7. Engineering conclusion¶
Link budget allows predicting whether a signal will be visible, and whether it will overload the receiver, before performing the experiment.